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ETI E ....
Untran S. Inan Ko~
University Stanford University
Aziz S. Inan University of Portland
Ryan K. Said Vaisala Inc.
Second Edition
Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
Credits and acknowledgments borrowed from other sources and reproduced, with permission, in this textbook appear on appropriate page within text. Copyright © 2015, 2006 by Pearson Education, Inc., publishing as Prentice Hall, 1 Lake Street, Upper Saddle River, NJ 07458. All rights reserved. Manufactured in the United States of America. This publication is protected by Copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. To obtain permission(s) to use material from this work, please submit a written request to Pearson Education, Inc., Permissions Department, imprint permissions address. Many of the designations by manufacturers and seller to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps. The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Library of Congress CataloginginPublication Data Inan, Umran S. Engineering electromagnetics and waves I Umran S. Inan, Aziz S. Inan, Ryan K. Said. pages cm Includes bibliographical references and index. ISBN 9780132662741 (alk. paper) 1. Electromagnetic theory. I. Inan, Aziz S. II. Said, Ryan K. III. Title. QC670.152 2014 530.14 11dc23 2014002226 10 9 8 7 6 5 4 3 2 1
0132662744 ISBN 10: ISBN 13: 9780132662741
To our families
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~ontents
•
Preface
XI
•
Biography 1
2
XXI
Introduction
1
1.1
Lumped versus Distributed Electrical Circuits
5
1.2
Electromagnetic Components
1.3
Maxwell's Equations and Electromagnetic Waves
1.4
Summary
14 15
17
Transient Response of Transmission Lines
23
2.1
Heuristic Discussion of Transmission Line Behavior and Circuit Models 25
2.2
Transmission Line Equations and Wave Solutions
2.3
Reflection at Discontinuities
2.4
Transient Response of Transmission Lines with Resistive Terminations 47
2.5
Transient Response of Transmission Lines with Reactive Terminations 60
29
36
v
•
Contents
VI
3
4
2.6
TimeDomain Reflectometry
2.7
Transmission Line Parameters
2.8
Summary 78
70 75
SteadyState Waves on Transmission Lines
99
3.1
Wave Solutions Using Phasors
101
3.2
Voltage and Current on Lines with Short or OpenCircuit Terminations 105
3.3
Lines Terminated in an Arbitrary Impedance
3.4
Power Flow on a Transmission Line
3.5
Impedance Matching
3.6
The Smith Chart
3.7
Sinusoidal SteadyState Behavior of Lossy Lines
3.8
Summary
117
138
147
164 176
193 211
The Static Electric Field 4.1
Electric Charge
213
4.2
Coulomb's Law
218
4.3
The Electric Field 226
4.4
The Electric Potential 239
4.5
Electric Flux and Gauss's Law
4.6
Divergence: Differential Form of Gauss's Law
4. 7
Metallic Conductors
4.8
Poisson's and Laplace's Equations
4.9
Capacitance
4.10
Dielectric Materials
4.11
Electrostatic Boundary Conditions
4.12
Electrostatic Energy
328
4.13
Electrostatic Forces
337
4.14
Microelectromechanical Systems (MEMS)
4.15
Summary
257
276 291
297
354
305 321
343
268
••
Contents
5
6
7
VII
Steady Electric Currents
367
5.1
Current Density and the Microscopic View of Conduction
5.2
Current Flow, Ohm's Law, and Resistance
5.3
Electromotive Force and Kirchhoff's Voltage Law
5.4
The Continuity Equation and Kirchhoff's Current Law
5.5
Redistribution of Free Charge
5.6
Boundary Conditions for Steady Current Flow
5.7
Duality of J and D: The ResistanceCapacitance Analogy
5.8
Joule's Law
5.9
Surface and Line Currents
5.10
Summary
368
374 381 385
387 389 395
400 402
404
The Static Magnetic Field
415
6.1
Ampere's Law of Force
6.2
The BiotSavart Law and Its Applications
6.3
Ampere's Circuital Law
6.4
Curl of the Magnetic Field: Differential Form of Ampere's Law
6.5
Vector Magnetic Potential
6.6
The Magnetic Dipole
6.7
Divergence of B, Magnetic Flux, and Inductance
6.8
Magnetic Fields in Material Media 491
6.9
Boundary Conditions for Magnetostatic Fields
6.10
Magnetic Forces and Torques
6.11
Summary
417 424
438
446
459
467 473
504
508
517
TimeVarying Fields and Maxwell's Equations
7.1
Faraday's Law
7.2
Induction Due to Motion 546
7.3
Energy in a Magnetic Field 556
7.4
Displacement Current and Maxwell's Equations
531
534
568
•••
Contents
VIII
8
9
10
7.5
Review of Maxwell's Equations
7.6
Summary
579
584
595
Waves in an Unbounded Medium 8.1
Plane Waves in a Simple, SourceFree, and Lossless Medium
596
8.2
TimeHar1nonic Unifor1n Plane Waves in a Lossless Medium
604
8.3
Plane Waves in Lossy Media
8.4
Electromagnetic Energy Flow and the Poynting Vector
8.5
Polarization of Electromagnetic Waves
8.6
Arbitrarily Directed Unifor1n Plane Waves
8. 7
Nonplanar Electromagnetic Waves
8.8
Summary
615 635
653 667
673
674
Reflection, Transmission, and Refraction of Waves at Planar Interfaces 9.1
Normal Incidence on a Perfect Conductor
690
9.2
Normal Incidence on a Lossless Dielectric
700
9.3
Multiple Dielectric Interfaces
9.4
Normal Incidence on a Lossy Medium
9.5
Oblique Incidence upon a Perfect Conductor
9.6
Oblique Incidence at a Dielectric Boundary
9.7
Total Internal Reflection
9.8
Oblique Incidence on a Lossy Medium
9.9
Summary
689
708 721 734 747
765 777
787
ParallelPlate and Dielectric Slab Waveguides 10.1
Waves between Parallel Metal Plates
10.2
Dielectric Waveguides
10.3
Wave Velocities and Waveguide Dispersion
10.4
Summary
876
811
814
844 864
•
Contents
11
A
IX
885
FieldMatter Interactions and Metamaterials
11.1
Wave Propagation in Ionized Gases (Plasmas)
887
11.2
Frequency Response of Dielectrics and Conductors
11.3
Metamaterials
11.4
Summary
899
906
924
Vector Analysis
929
A.1
Vector Components, Unit Vectors, and Vector Addition
A.2
Vector Multiplication
A.3
Cylindrical and Spherical Coordinate Systems
A.4
Vector Identities
930
932 935
943
B
Uniqueness Theorem
947
C
Derivation of Ampere's Circuital Law from the BiotSavart Law
951
Symbols and Units for Basic Quantities
955
General Bibliography
961
Answers to OddNumbered Problems
963
Index
975
This book provides engineering students with a solid grasp of electromagnetic fundamentals and electromagnetic waves by emphasizing physical understanding and practical applications. The topical organization of the text starts with an initial exposure to transmission lines and transients on highspeed distributed circuits, naturally bridging electrical circuits and electromagnetics. Engineering Electromagnetics and Waves is designed for upperdivision (3rd and 4th year) college and university engineering students, for those who wish to learn the subject through selfstudy, and for practicing engineers who need an uptodate reference text. The student using this text is assumed to have completed typical lowerdivision courses in physics and mathematics as well as a first course on electrical engineering circuits.
Key Features The key features of this textbook are • Modern chapter organization, covering transmission lines before the development of fundamental laws • Emphasis on physical understanding • Detailed examples, selected application examples, and abundant illustrations • Numerous endofchapter problems, emphasizing selected practical applications • Historical notes on the great scientific pioneers • Emphasis on clarity without sacrificing rigor and completeness • Hundreds of footnotes providing physical insight, leads for further reading, and discussion of subtle and interesting concepts and applications •
XI
••
XII
Preface
Modern Chapter Organization
We use a physical and intuitive approach so that this engineering textbook can be read by students with enthusiasm and interest. We provide continuity with electric circuit theory by first covering transmission lines an appropriate step, in view of the importance of transmission line concepts, not only in microwave and millimeterwave applications but also in highspeed digital electronics, microelectronics, integrated circuits, packaging, and interconnect applications. We then cover the fundamental subject material in a logical order, fallowing the historical development of human understanding of electromagnetic phenomena. We base the fundamental laws on experimental observations and on physical grounds, including brief discussions of the precision of the fundamental experiments, so that the physical laws are easily understood and accepted. Once the complete set of fundamental laws is established, we then discuss their most profound implications: the propagation of electromagnetic waves. We begin this discussion with the propagation of waves through empty space or unbounded simple media. We then discuss the reflection and refraction of electromagnetic waves from simple planar boundaries, followed by the guiding of electromagnetic waves within planar metallic or dielectric structures. We conclude with an introduction to fieldmatter interactions and electromagnetic wave propagation in metamaterials. Emphasis on Physical Understanding
Future engineers and scientists need a clear understanding and a fir1n grasp of the basic principles so that they can interpret, for1nulate, and analyze the results of complex practical problems. Engineers and scientists nowadays do not and should not spend time obtaining numerical results by hand. Most of the number crunching and formula manipulations are left to computers and packaged application and design programs, so a solid grasp of fundamentals is now more essential than ever before. In this text we maintain a constant link with established as well as new and emerging applications (so that the reader's interest remains perked up), while at the same time emphasizing fundamental physical insight and solid understanding of basic principles. We strive to empower the reader with more than just a working knowledge of a dry set of vector relations and formulas stated axiomatically. We supplement rigorous analyses with extensive discussions of the experimental bases of the laws, of the microscopic versus macroscopic concepts of electromagnetic fields and their behavior in material media, and of the physical nature of the electromagnetic fields and waves, often from alternative points of view. Description of the electrical and magnetic properties of material media at a sufficiently simple, yet accurate manner at the introductory electromagnetics level has always been a challenge, yet a solid understanding of this subject is now more essential than ever, especially in view of many applications that exploit these properties of materials. To this end we attempt to distill the essentials of physicallybased treatments available in physics texts, providing quantitative physical insight into microscopic behavior of materials and the representation of this behavior in terms of macroscopic parameters. Difficult threedimensional vector differential and integral concepts are discussed when they are encountered again, with the emphasis being on physical insight.
Preface
•••
XIII
Detailed Examples and Abundant Illustrations We present the material in a clear and simple yet precise and accurate manner, with interesting examples illustrating each new concept. Many examples emphasize selected applications of electromagnetics. Over 190 illustrative examples are detailed over eleven chapters, with five of the chapters having at least 20 examples each. Each example is presented with an abbreviated topical title, a clear problem statement, and a detailed solution. In recognition of the importance of visualization in the reader's understanding, especially in view of the threedimensional nature of electromagnetic fields, over 500 diagrams, graphs, and illustrations appear throughout the book.
Numerous EndofChapter Problems Each chapter is concluded with a variety of homework problems to allow the students to test their understanding of the material covered in the chapter, with a total of over 400 exercise problems spread over eleven chapters. The topical content of each problem is clearly identified in an abbreviated title (e.g., ''Digital IC interconnects'' or ''Inductance of a toroid''). Many problems explore interesting applications, and most chapters include several practical ''reallife'' problems to motivate students.
Historical Notes The history of the development of electromagnetics is laden with outstanding examples of pioneering scientists and development of scientific thought. Throughout our text, we maintain a constant link with the pioneering giants and their work, to bring about a better appreciation of the complex physical concepts as well as to keep the reader interested.
Emphasis on Clarity without Sacrificing Rigor and Completeness This textbook presents the material at a simple enough level to be readable by undergraduate students, but it is also rigorous in providing references and footnotes for indepth analyses of selected concepts and applications. We provide the students with a taste of rigor and completeness at the level of classical reference texts combined with a level of physical insight that was so well exemplified in some very old texts while still maintaining the necessary level of organization and presentation clarity required for a modern textbook. We also provide not just a superficial but a rigorous and indepth exposure to a diverse range of applications of electromagnetics, in the body of the text, in examples, and in endofchapter problems.
Hundreds of Footnotes In view of its fundamental physical nature and its broad generality, electromagnetics lends itself particularly well to alternative ways of thinking about physical and xengineering problems and also is particularly rich in ter1ns of available scientific literature and many
•
XIV
Preface
outstanding textbooks. Almost every new concept encountered can be thought of in different ways, and the interested reader can explore its implications further. We encourage such scholarly pursuit of enhanced knowledge and understanding by providing many footnotes in each chapter that provide further comments, qualifications of statements made in the text, and references for indepth analyses of selected concepts and applications. Over 550 footnotes are spread over eleven chapters. These footnotes do not interrupt the flow of ideas and the development of the main topics, but they provide an unusual degree of completeness for a textbook at this level, with interesting and sometimes thoughtprovoking content to make the subject more appealing and satisfying.
Electromagnetics and Waves in Engineering The particular organization of this textbook, as well as its experimentally and physically based philosophy, are motivated by our view of the current status of electromagnetics in engineering curricula. Understanding electromagnetics and appreciating its applications require a generally higher level of abstraction than most other topics encountered by electrical engineering students. Beginning electrical engineers learn to deal with voltages and currents, which appear across or flow through circuit elements or paths. The relationships between these voltages and currents are determined by the characteristics of the circuit elements and by Kirchhoff s current and voltage laws. Voltages and currents in lumped electrical circuits are scalar quantities that vary only as a function of time, and are readily measurable, and the students can relate to them via their previous experiences. The relationships between these quantities (i.e., Kirchhoff s laws) are relatively simple algebraic or ordinary differential equations. On the contrary, electric and magnetic fields are threedimensional and vector quantities that in general vary in both space and time and are related to one another through relatively complicated vector partial differential or vector integral equations. Even if the physical nature of electric and magnetic fields were understood, visualization of the fields and their effects on one another and on matter requires a generally high level of abstract thinking. Most students are exposed to electromagnetics first at the freshman physics level, where electricity and magnetism are discussed in ter1ns of their experimental bases by citing physical laws (e.g., Coulomb's law) and applying them to relatively simple and symmetrical configurations where the field quantities behave as scalars, and the governing equations are reduced to either algebraic equations of firstorder integral or differential relationships. Freshman physics provides the students with their first experiences with fields and waves as well as some of their measurable manifestations, such as electric and magnetic forces, electromagnetic induction (Faraday's law), and refraction of light by prisms. The first course in electromagnetics, which most students take after having had vector calculus, aims at the development and understanding of Maxwell's equations, requiring the utilization of the full threedimensional vector for1n of the fields and their relationships. It is this very step that makes the subject of electromagnetics appear insurmountable to many students and turns off their interest, especially when coupled with a lack of presentation and discussion of important applications and the physical (and
Preface
xv
experimental) bases of the fundamental laws of physics. Many authors and teachers have attempted to overcome this difficulty by a variety of topical organizations, ranging from those that start with Maxwell's equations as axioms to those that first develop them from their experimental basis. Since electromagnetics is a mature basic science, and the topics covered in introductory texts are well established, the various texts primarily differ in their organization as well as range and depth of coverage. Teaching electromagnetics was the subject of a special issue of IEEE Transactions on Education [vol. 33, February, 1990]. Many of the challenges and opportunities that lie ahead in this connection were summarized well in an invited article by J. R. Whinnery. 1 Challenges include (1) the need to return to fundamentals (rather than relying on derived concepts), especially in view of the many emerging new applications that exploit unusual properties of materials and that rely on unconventional device concepts, 2 submillimeter transmission lines, 3 and optoelectronic waveguides,4 and (2) the need to maintain student interest in spite of the decreasing popularity of the subject of electromagnetics and its reputation as a difficult and abstract subject. 5 Opportunities are abundant, especially as engineers working in the electronics industry discover that as devices get smaller and faster, circuit theory is insufficient in describing system performance or facilitating design. Transmission line concepts are not only important in microwave and millimeterwave applications; due to modern GHz clock rates and nanoscale fabrication technology, they are also necessary in highspeed digital electronics, microelectronics, integrated circuits, interconnects, 6 and packaging applications. 7 In addition, issues of electromagnetic interference (EMI) and electromagnetic compatibility (EMC) limit the performance of system, board, and chiplevel designs, and electrostatic discharge phenomena have significant impacts on the design and performance of integrated circuits. 8 The need for a basic understanding of electromagnetic waves and their guided propagation is underscored by the explosive expansion of the use of optical fibers, which enables extremely high data rates, ranging to 100 Gbits/s 9 . Fundamental bandwidth and power constraints in traditional copperbased transmission lines is also driving the development of optical interconnects with perchannel bandwidths
1J.
R. Whinnery, The teaching of electromagnetics, IEEE Trans. on Education, 33(1), pp. 37, February
1990. 2 D. GoldhaberGordon, M. S. Montemerlo, J. C. Love, G. J. Opiteck, and J. C. Ellenbogen, Overview of nanoelectronic devices, Proc. IEEE, 85(4), pp. 521540, April 1997. 3 L. P. B. Katehi, Novel transmission lines for the submillimeter region, Proc. IEEE, 80(11), pp. 17711787, November 1992. 4 R. A. Soref, Siliconbased optoelectronics, Proc. IEEE, 81(12), December, 1993. 5 M. N. 0. Sadiku, Problems faced by undergraduates studying electromagnetics, IEEE Trans. Education, 29(1), pp. 3132, February, 1986. 6 S. H. Hall and L. H. Heck, Advanced signal integrity for highspeed digital designs, John Wiley & Sons, 2011. 7 H. B. Bakoglu, Circuits, Interconnections, and Packaging for VLSI, Addison Wesley, 1990. 8 J. E. Vinson and J. J. Liou, Electrostatic discharge in semiconductor devices: an overview, Proc. IEEE, 86(2), pp. 399418, February 1998. 9 N. Cvijetic, D. Qian, and J. Hu, 100 Gb/s optical access based on optical orthogonal frequencydivision multiplexing, Communications Magazine, IEEE 48(7), pp. 7077, 2010.
•
Preface
XVI
10 applications.
in excess of 10 Gbits/s for highperformance computing Other important applications that require better understanding of electromagnetic fields are emerging in biology 11 and medicine. 12 In organizing the material for our text, we benefited greatly from a review of the electromagnetic curriculum at Stanford University that one of us conducted during the spring quarter of 1990. A detailed analysis was made of both undergraduate and graduate offerings, both at Stanford and selected other schools. Inquiries were also made with selected industry, especially in the aerospace sector. Based on the responses we received from many of our colleagues, and based on our experience with the teaching of the twoquarter sequence at Stanford, it was decided that an emphasis on fundamentals and physical insight and a traditional order of topics would be most appropriate. It was also determined that transmission line theory and applications can naturally be studied before fields and waves, so as to provide a smooth transition from the previous circuits and systems experiences of the typical electrical engineering students and also to emphasize the importance of these concepts in highspeed electronics and computer applications.
New to this Edition This book represents an effort to merge the most important concepts from our two pre13 14 vious textbooks: Engineering Electromagnetics and Electromagnetic Waves . Some of the advanced topics from these two books, such as using transmission lines as resonant circuits and cylindrical waveguides, were moved to a web addendum (see Online Addendum section below). By moving some of these sections to the web, we are better able to focus the reader on the core concepts central to transmission lines, electromagnetics, and electromagnetic waves. We also introduce two new sections on increasingly relevant modem topics: Microelectromechanical Systems (MEMS) and Metamaterials. While these are relatively advanced topics, some of the fundamental physics underpinning these two areas of active research and development connect directly to the core ideas presented in this book, and so they give concrete examples of how a solid foundation in electromagnetics and waves is still very relevant to modem technology. The list below summarizes the changes and additions we introduced in the second edition: • We merged topics from the two first edition textbooks into a single volume covering both engineering electromagnetics and electromagnetic waves. • We added two new sections: Microelectromechanical Systems (Section 4.14) and Metamaterials (Section 11.3). 10 L.
Chrostowski and K. Iniewski (Eds.), Highspeed Photonics Interconnects (Vol. 13), CRC Press., 2013. 11 R. H., Funlc, T. Monsees, and N. Ozkucur, Electromagnetic effectsFrom cell biology to medicine. Progress in histochemistry and cytochemistry, 43(4), 177264, 2009. 12 E. J. Bond, et al., Microwave imaging via spacetime beamforming for early detection of breast cancer, Antennas and Propagation, IEEE Transactions on, 51(8), pp. 16901705, 2003. 13 U.S. Inan and A. S. Inan, Engineering Electromagnetics, Addison Wesley Longman, 1999. 14 U.S. Inan and A. S. Inan, Electromagnetic Waves, Prentice Hall, 2000.
••
Preface
XVII
• We added an appendix with a proof of the uniqueness theorem for Poisson's equation (Appendix B). • We moved several advanced topics in the first two editions to standalone addendum chapters that are available on the web. • We introduced numerous edits throughout the text to add clarity and improve the presentation of some of the more challenging topics. • We updated, modified, or added over 100 new endofchapter problems. • We corrected errata that have been reported since the publication of the first two editions.
Recommended Course Content The wider breadth of topics covered by this single volume allows the instructor to tailor the content based on the duration of the course. Tables 1 and 2 list the suggested course content for a single course and two course sequence, respectively. Each table details suggested content based on the quarter system (32 contact hours per course per quarter) and semester system (42 contact hours per course per semester). The sections marked under ''Cover'' are recommended for complete coverage, including illustrative examples, whereas those marked ''Skim'' are recommended to be covered lightly, although the material provided is more complete in case individual students want to have more indepth coverage. The onecourse sequences provide the students with (1) a working knowledge of transmission lines, (2) a solid, physically based background and a fir1n understanding of Maxwell's equations and their experimental bases, and (3) an introduction to electromagnetic waves. In addition to a more indepth coverage of the transmission lines chapters and the development of Maxwell's equations, the twocourse sequences give the student a working knowledge of electromagnetic wave phenomena and their applications. TABLE 1
SUGGESTED COURSE CONTENT: SINGLE QUARTER OR SEMESTER
Quarter Course (32 Hours) Chapter 1 2 3 4 5 6 7 8 9 10 11
Cover All 2.12.4 3.13.3 4.14.9 5.15.5 6.16.7 7.1, 7.2, 7.4 8.1, 8.2
Skim 2.7 4.10 6.8 7.3, 7.5
Semester Course (42 Hours) Cover All 2.1  2.5 3.13.6 4.14.10 5.15.5 6.16.8 7 .1, 7 .2, 7 .4 8.18.4
Skim 2.7 4.12 5.7 6.10 7.3, 7.5
•••
Preface
XVIII
TABLE 2
SUGGESTED COURSE CONTENT: TWO QUARTERS OR SEMESTERS
Tuo Quarter Course (64 Hours Total) Chapter 1 2 3 4 5 6
Cover
Skim
All All 3.13.6 4.14.12 5.15.7 6.16.9
Tuo Semester Courses (84 Hours Total) Cover
All All All 4.14.13 5.8
All
Semester Break
All
All All All All
11.1, 11.2
11.1, 11.2,
8.18.6 9.19.3, 9.59.7
4.14
All All
Quarter Break 7 8 9 10 11
Skim
11.3
Instructor's Manual
We firmly believe that practice is the key to learning and that homework and exams are all instruments of teaching although they may not be regarded as such by the students at the time. In our own courses, we take pride in providing the students with detailed solutions of homework and exam problems, rather than cryptic and abbreviated answers. To aid the instructors who choose to use this text, we have thus taken it upon ourselves to prepare a welllaidout solutions manual, describing the solution of every endofchapter problem, in the same stepbystep detailed manner as our illustrative examples within the chapters. This instructor's manual is available to instructors upon request at www .pearsonhighered.com. Supplemental information about the book and errata will be available at www.pearsonhighered.com/inan. As authors of this book, we are looking forward to interacting with its users, both students and instructors, to collect and respond to their comments, questions, and corrections. We can most easily be reached by electronic mail at [email protected] (http://vlf.stanford.edu/), [email protected] (http://www.ku.edu.tr/en/aboutku/president), [email protected] (faculty .up.edu/ainanl), and [email protected] Online Addendum
The topics in this book were carefully selected to give the student a solid foundation of transmission lines, Maxwell's equations, and the propagation and guiding of electromagnetic waves. The goal of this book is to develop an intuitive understanding of these fundamental concepts, so that the student is well equipped to apply these principles to
•
Preface
XIX
TABLE 3
Addendum Addendum Addendum Addendum Addendum Addendum
ADDENDUM: ADVANCED TOPICS (AVAILABLE ONLINE)
A B C D E F
Transmission Lines: Advanced Topics Miscellaneous Wave Topics Cylindrical Waveguides Cavity Resonators FieldMatter Interactions: Advanced Topics Electromagnetic Radiation and Elementary Antennas
new challenges and to expand his/her study to more advanced topics. Due to the breadth of discussion given to each topic, in order to maintain a manageable page count, a few of the more advanced topics from the first edition are moved to an online addendum. This addendum, which is available for free at www.pearsonhighered.com/inan, contains the chapters listed in Table 3. The Addendum chapters cover advanced topics that expand on the core material of this text. Addendum A includes selected advanced transmission line topics, including transients on lossy transmission lines and the use of transmission lines as resonant circuits. The application examples of Addendum A present further reading that can be covered after Chapters 2 and 3. The remaining addendum chapters cover advanced material related to the subject matter from the final four chapters of the book. Addendum B expands on two advanced topics introduced in Chapters 8 and 9: examples of nonplanar waves and obliqueincidence reflection from a good conductor. Addendum C extends the treatment of planar waveguides from Chapter 10 to those that are bounded in two dimensions. Addendum D introduces cavity resonators, whose treatment follows naturally from the twodimensional waveguides covered in Addendum C. Addendum E extends the fieldmatter interaction topics encountered in Sections 11.1 and 11.2. Addendum F introduces electromagnetic radiation and elementary antennas, which connect the electromagnetic fields studied in Chapters 811 to their sources.
Acknowledgments We gratefully acknowledge those who have made significant contributions to the successful completion of this text. We thank Professor J. W. Goodman of Stanford, for his generous support of textbook writing by faculty throughout his term as department chair. We thank Professor Gordon Kino and Dr. Timothy F. Bell of Stanford, for coursetesting a preliminary version of the manuscript for our first edition. We thank numerous colleagues and for1ner students who have identified errors and suggested clarifications. We thank Mrs. JunHua Wang for typing parts of the first edition manuscript and drawing some of the illustrations. We thank Dr. Robert Marshall of Stanford for preparing the two first edition manuscripts for use in a single unified book, and Giine~ Aydindogan for typesetting large portions of the solution manual. We owe special thanks to our reviewers on the first edition for their valuable comments and suggestions, including J. Bredow of University of Texas Arlington; S. Castillo of New Mexico State University; R. J. Coleman of University of North Carolina Charlotte; A. Dienes of University of California Davis; J. Dunn
xx
Preface
of University of Colorado; D. S. Elliott of Purdue University; R. A. Kinney of Louisiana State University; L. Rosenthal of Fairleigh Dickinson University; E. Schamiloglu of University of New Mexico; T. Shumpert of Auburn University; D. Stephenson of Iowa State University; E. Thomson of University of Florida; J. Volakis of University of Michigan; and A. Weisshaar of Oregon State University. We greatly appreciate the efforts of our managing editor at Pearson, Scott Disanno, and his staff, including Michelle Bayman, William Opaluch, Joanne Manning, and Julie Bai. We also thank Haseen Khan and her staff at Laserwords for their dedication and attention to detail in the layout and production of the book. We dedicate this book to our parents, Mustafa and Hayriye lnan, for their dedication to our education; to our wives, Elif and Belgin, for their persistent support and understanding as this project expanded well beyond our initial expectations and consumed most of our available time for too many years; and to our children, Ayse, Ali, Baris, and Cem, and grandchildren, Ayla and Nisa, for the joy they bring to our lives.
Umran S. !nan Aziz S. !nan I would like to thank Professors Umran lnan and Aziz lnan for giving me the opportunity to contribute to this manuscript. When I was a new graduate student at Stanford University, I had the pleasure of learning electromagnetics from the two first edition textbooks that form the basis for this second edition. It is an honor to help create a unified book that spans the content of these courses that I so enjoyed taking over a decade ago. I dedicate this book to my parents, James and Connie, for cultivating my passion for science, and to my sister, Amirah, for her unwavering moral support.
Ryan K. Said
Umran S. Inan is President of
Ko~
University in Istanbul, Turkey where he is Professor of Electrical Engineering and Physics. He is also Professor (Emeritus) of Electrical Engineering at Stanford University where he supervised 60 PhD dissertations. lnan is a Fellow of IEEE, American Geophysical Union and American Physical Society. He is the recipient of Appleton Prize of URSI and Royal Society, Allan Cox Medal of Stanford, Special Science Award of the Scientific & Technological Research Council of Turkey, Stanford Tau Beta Pi Award for Excellence in Teaching, and numerous NASA and ESA Group Achievement Awards.
Aziz S. Inan is Professor of Electrical Engineering at the University of Portland, where he has also served as Department Chairman. A winner of the University's faculty teaching award, he conducts research in electromagnetic wave propagation in conducting and inhomogeneous media. He is a member of Tau Beta Pi and IEEE.
Ryan K. Said received his Ph.D. in Electrical Engineering from Stanford University. As a research scientist and systems engineer at Vaisala Inc., his work focuses on the development of lightning detection technologies and their applications. He is a recipient of the HMEI Award for Young Engineers and a member of the American Geophysical Union.
•
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This book is an introduction to the fundamental principles and applications of electromagnetics. The subject of electromagnetics encompasses electricity, magnetism, and electrodynamics, including all electric and magnetic phenomena and their practical applications. A branch of electromagnetics, dealing with electric charges at rest (static electricity) named electrostatics, provides a framework within which we can understand the simple fact that a piece of amber, when rubbed, attracts itself to other small objects. 1 Another branch dealing with static magnetism, namely magnetostatics, is based on the 2 facts that some mineral ores (e.g., lodestone) attract iron and that currentcarrying wires produce magnetic fields. 3 The branch of electromagnetics known as electrodynamics deals with the time variations of electricity and magnetism and is based on the fact that 4 magnetic fields that change with time produce electric fields. Electromagnetic phenomena are governed by a compact set of principles known as Maxwell's equations, 5 the most fundamental consequence of which is that electromagnetic energy can propagate, or travel from one point to another, as waves. The propagation of electromagnetic waves results in the phenomenon of delayed action at a distance; in other words, electromagnetic fields can exert forces, and hence can do work, at distances far away from the places where they are generated and at later times. Electromagnetic radiation is thus a means of transporting energy and momentum from one set of electric charges and currents (at the source end) to another (those at the receiving end).
1First
discovered by the Greek mathematician, astronomer, and philosopher Thales of Miletus
[640548 B.C.]. 2First noted by the Roman poet and philosopher Lucretius [99?55? B.c.], in his philosophical and scientific poem titled De re rum natura (On the Nature of Things). 3 First noted by Danish physicist H. C. Oersted in 1819. 4 First noted by British scientist M. Faraday in 1831. 5 J. C. Maxwell, A Treatise in Electricity and Magnetism, Clarendon Press, Oxford, 1892, Vol. 2, pp. 247262.
1
2
Introduction
Chap. 1
Since whatever can carry energy can also convey information, electromagnetic waves thus provide the means of transmitting energy and information at a distance. The latter half of this book provides an introduction to electromagnetic waves, their propagation in empty space or material media, their reflection from boundaries, and their guiding within planar boundaries. 6 The concept of waves is one of the great unifying ideas of physics and engineering. Our physical environment is full of waves of all kinds: seismic waves, waves on oceans and ponds, sound waves, heat waves, and even traffic waves. The idea of delayed action as manifested in wave phenomena is familiar to us when we hear a sound and its echo 7 or when we create a disturbance in a pool of water and observe that waves reach the edge of the pool after a noticeable time. We also appreciate that it might take minutes or hours for heat to penetrate into objects; that the thunderclap is delayed with respect to the lightning flash by many seconds; and that when we are lined up in front of a traffic light, it often takes a long time for us to be able to move after the light turns green. Light, or electromagnetic waves, travel so fast that their delayed action is not perceptible to our senses in our everyday experiences. On the other hand, in astronomy and astrophysics we deal with vast distances; light waves from a supernova explosion may arrive at earth millions of years after the brightness that created them has been extinguished. Example 1.1: Time delay between Mars and Earth. The distance between Earth and Mars varies from 54.6 x 106 to 401 x 106 km. How long does it take for a message sent from Earth to reach NASA's Curiosity Rover on Mars? Solution:
The travel time t is determined by the distance l and speed of propagation v: l
t = 
v
All electromagnetic waves, including radio waves used in broadcast telecommunications, travel in free space at the speed of c ~ 3 x 108 ms 1 . Using v = c, the travel time from Earth to Mars ranges from 54.6 x 106 km
tmin =       ·
3 x 10s
kms 1
1 minute 60 s
~
3.03 minutes
~
. 22 .3 ffilnutes
to tmax
6
401 x 106 km = 3 x 10s kms 1
·
1 minute 60 s
For an excellent qualitative discussion, see J. R. Pierce, Almost All About Waves, MIT Press, Cambridge, Massachusetts, 1974. For more extensive treatment of waves of all kinds, see K. U. Ingard, Fundamentals of Waves and Oscillations, Cambridge University Press, Cambridge, England, 1990. 7 On the scale of a pond, we can simply think of dropping a stone; on a larger scale, earthquakes in oceans produce giant tsunami waves. A 9meterhigh tsunami produced by the 1964 Alaskan earthquake hit the Hawaiian islands (at a distance of 2000 km) about 5 hours later, causing more than 25 million dollars of damage.
Chap. 1
3
Introduction f(z  vt)
/(() p
tz > t1
t = t1
p
p t = t2
v,~, I I
(i
Z1
(a)
= (i + Vt1
I I I I
'
\ \
' .....
Z2 = (i + vtz
z
(b)
Figure 1.1 Example of a wave. (a) An arbitrary function f
(~).
(b) The function f (z  vt), where v is a positive constant, plotted versus z at t = t 1 and t = t2 . The wave nature is evident as the pattern in space at time t1 is shifted to other values of z at a later time t2.
At a qualitative level, we recognize a wave as some pattern in space that appears to move in time. Wave motion does not necessarily involve repetitive undulations of a physical quantity (e.g., the height of the water surface for water waves in a lake). If a disturbance that occurs at a particular point in space at a particular time is related in a definite manner to what occurs at distant points in later times, then there is said to be wave motion. To express this mathematically, let z be distance, t be time, and v be a fixed positive parameter. Consider any arbitrary functionf(~) of the argument~= (z  vt). Figure l. la shows a sketch off(~), identifying a point P on the curve that corresponds to f (~ 1 ). Note that this peak value of the function (i.e., point P) is obtained whenever its argument equals ~1. Shown in Figure 1.1 b is a sketch off (z  vt1) with respect to z at a fixed time t1, where the point Pis at z1 =~I+ vt1. A plot of the function at t = t2 >ti (i.e., f (z  vt2)) is also shown in Figure 1.1 b, where we see that the point P has now moved to the right, to the new location z2 =~I + vt2. It is clear that the entire curvef (~), which comprises the function f (z  vt), moves in the z direction as time elapses. The velocity of this motion can be determined by observing a fixed point on the curve for example, point P. Since this point is defined by the argument of the function being equal to ~1, we can set (z  vt) = ~1, which upon differentiation yields dz/ dt = v, since ~1 is a constant. It thus appears that the speed with which point P moves to the right is v, which is identified as the velocity of the wave motion. Note that the function f (·) could represent any physically observable entity; it may be a scalar, 8 such as voltage, or it may be a vector, such as the velocity of an object in motion. If f (·) is a vector, each of its components must be a function of (z  vt) for it to be a propagating wave. Quantities varying as functions of (z  vt) constitute natural solutions of the fundamental equations of electromagnetics and distributed electrical circuits. Chapters 2, 3, and 811 of this textbook are devoted to the study of voltage, current, and electromagnetic waves that vary in space and time as functions of (z  vt). 8A
scalar is a quantity that is completely specified by its value, such as the number of coins in your pocket, the number of people or the density of air in a room, pressure, or temperature. Other physical quantities have direction; for example, velocity, momentum, force, or displacement. Specification of a vector quantity requires both a magnitude and direction. A brief review of basic principles of vector analysis is provided in Appendix A.
4
Introduction
Chap. 1
Most waves travel through substances, whether they be earth, water, air, steel, or quartz, without actually carrying the substance bodily with them. 9 Like moving objects, traveling waves carry energy, albeit by different amounts depending on the nature of the waves and the medium they propagate in. Electromagnetic waves have the special property that they can also propagate in vacuum, without any matter present. However, the propagation of electromagnetic waves is nevertheless affected by the presence of matter, and this property often allows us to confine or guide waves and in doing so utilize them more efficiently. Electromagnetic engineering problems generally involve the design and use of materials that can generate, transmit, guide, store, distribute, scatter, absorb, and detect electromagnetic waves and energy. The 20th century has witnessed rapid advances in electrical engineering, which have largely come about by our ability to predict the perfor1nance of sophisticated electrical circuits accurately. Central to this tremendous progress is our ability to utilize the simple but powerful tool called electric circuit theory. Classical circuit theory considers a voltage or current source applied to an electrical circuit consisting of series and/or parallel connection of simple lumped (see Section 1.1) circuit elements, such as resistances, capacitances, inductances and dependent sources, which may be idealized models of more complex physical devices. The behavior of circuits is described by ordinary differential equations that are derived on the basis of Kirchhoff s voltage and current laws. Circuit theory is a simplified approximation to the more exact electromagnetic theory. 10 The classical theory of electricity and magnetism relies on a set of physical laws known as Maxwell's equations, which are based on experimental facts and which govern all electromagnetic phenomena. Electromagnetic theory is inherently more complicated than circuit theory, primarily because of the larger number of variables involved. In general electromagnetic problems, most of the physical quantities that we deal with are vectors, whose values may depend on all three coordinates of space (i.e., x, y, and z in rectangular coordinates) and time (t). In classical circuit theory, on the other hand, voltages and currents are scalar quantities and are typically functions of only one variable, namely time. The theory of distributed circuits (see Section 1.1), or transmission lines, represents an intermediate level of complexity where, in many cases, we can continue to deal with scalar quantities, such as voltages and currents, that are now functions of two variables, namely a single spatial dimension and time. In this regime, we can continue to benefit from the relative simplicity of circuit theory, while treating problems for which the lumped circuit theory is not applicable. In this text, and in view of the preceding discussion, we choose to study distributed circuits or transmission lines using a natural extension of circuit theory before we formally introduce the physical laws of electricity and magnetism. This approach presents the general fundamental concepts of waves and oscillations at the outset, which 9 Leonardo
da Vinci [14521519] wrote of waves, ''The impetus is much quicker than the water, for it often happens that the wave flees the place of its creation, while the water does not; like the waves made in a field of grain by the wind, where we see the waves running across the field while the grains remain in place'' [J. R. Pierce, Almost All About Waves, MIT Press, Cambridge, Massachusetts, 1974]. 10 Kirchhoff's voltage and current laws, which provide the basis for classical circuit theory, can be derived from the more general electromagnetics equations; see Chapter 4 of S. Ramo, J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics, 3rd ed., John Wiley & Sons, Inc., New York, 1994.
Sec. 1.1
Lumped versus Distributed Electrical Circuits
5
are expanded upon later when we study propagation of electromagnetic waves. In this way, the reader is provided an unhindered initial exposure to properties of waves such as frequency, phase velocity, wavelength, and characteristic impedance; to energy relations in oscillating systems and waves; and to concepts such as reflection and bandwidth, as well as to fundamental mathematical techniques necessary to describe waves and oscillations. All of these concepts subsequently extend to more complicated problems and applications where a full electromagnetic treatment becomes necessary. An initial exposure to transmission line analysis also enables us to address a wide range of increasingly important engineering applications that require the use of wave techniques, but for which a full vector electromagnetic analysis may not be necessary. 11 In our coverage of transmission lines or distributed circuits, we assume that the reader is familiar with the elementary physics of electricity and magnetism (at the level of freshman physics) and with electrical circuits at the level of understanding Kirchhoff s voltage and current laws and terminal behavior (i.e., voltagecurrent relationships) of circuit elements such as inductors, capacitors, and resistors. A more complete discussion of the concepts of inductance, capacitance, and resistance is provided in later chapters using the concepts and principles of electromagnetic fields as we introduce the fundamental laws of electromagnetics.
1.1 LUMPED VERSUS DISTRIBUTED ELECTRICAL CIRCUITS A typical electrical engineering student is familiar with circuits, which are described as lumped, linear, and timeinvariant systems and which can be modeled by ordinary, linear, and constantcoefficient differential equations. The concepts of linearity and time invariance refer to the relationships between the inputs and outputs of the system. The concept of a lumped circuit refers to the assumption that the entire circuit (or system) is at a single point (or in one ''lump''), so that the dimensions of the system components (e.g., individual resistors or capacitors) are negligible. In other words, current and voltage do not vary with space across or between circuit elements, so that when a voltage or current is applied at one point in the circuit, currents and voltages of all other points in the circuit react instantaneously. Lumped circuits consist of interconnections of lumped elements. A circuit element is said to be lumped if the instantaneous current entering one of its terminals is equal to the instantaneous current throughout the element and leaving the other ter1ninal. Typical lumped circuit elements are resistors, capacitors, and inductors. In a lumped circuit, the individual lumped circuit elements are connected to each other and to sources and loads within or outside the circuit by conducting paths of negligible electrical length. 12 Figure l .2a illustrates a lumped electrical circuit to which an input voltage of Vin = Vo is applied at t = 0. Since the entire circuit is considered in one lump, the effect of the input excitation is instantaneously felt at all points in the 11 Examples
are onchip and chiptochip interconnections in digital integrated circuits and many other computer engineering applications. See A. Deutsch, et al., When are transmissionline effects important for onchip interconnections, IEEE Trans. Mic. Th. MTT, 45(10), pp. 18361846, October 1997. 12The electrical length equals the physical length of the circuit element divided by the wavelength. As we discuss in Section 1.1.3, this ratio determines whether a given circuit element should be treated as a lumped or distributed element.
6
Introduction
Chap. 1
Lumped circuit
/
Rs
/ /
I •
's vo
..,.,,,,,......... I ""' .... ....
•
I
I
f1 (t)
I
"N
''
J\10
""
'
I Lt () \
\
+ V1(t) .l
I
\
' I
I
'
\ \
I I
J2(t)
\ \
•
.
J\10~I
' ' .... .... _ ....
I
t
I
"TI'

_... ""'
.
•
I
/ /
~
(a)
•
I
A
&
•
A
•
•
s 
I
I
/in(t) Vo
I
•
J1(t)
•
J2(t)
... 1. . 1. . 1. .
Distributed circuit
I
~
~

(b) Figure 1.2 Lumped versus distributed electrical circuits. (a) When a step voltage Vin (t) is applied to a lumped circuit, we assume that all currents and voltages start to change at t = 0, implicitly assuming that it takes zero travel time for the effect of the input to move from any point to any other point in the circuit. (b) In a distributed circuit, the nonzero travel time of the signal from one point to another cannot be neglected. For example, when a step voltage Vin (t) is applied at one end of the circuit at t = 0, the load current at the other end does not start to change until t = l /v.
circuit, and all currents and voltages (such as 11, 12, V1, and the load current IL) either attain new values or respond by starting to change at t = 0, in accordance with the natural response of the circuit to a step excitation as determined by the solution of its corresponding differential equation. Many powerful techniques of analysis, design, and computeraided optimization of lumped circuits are available and widely used. The behavior of lumped circuits is analogous to rigidbody dynamics. In mechanics, a rigid body is postulated to have a definite shape and mass, and it is assumed that the distance between any two points on the body does not change, so that its shape is not deformed by applied forces. Thus, an external force applied to a rigid body is assumed to be felt by all parts of the body simultaneously, without accounting for the finite time it would take for the effect of the force to travel elastically from one end of the body to another.
Sec. 1.1
Lumped versus Distributed Electrical Circuits
7
With the ''lumped'' assumption, one does not have to consider the travel time of the signal from one point to another. In reality, however, disturbances or signals caused by any applied energy travel from one point to another in a nonzero time. For electromagnetic signals, this travel time is determined by the speed of light, 13 c ~ 3 x 108 ms 1 = 30 cm(ns) 1. In practical transmission systems, the speed of signal propagation is determined by the electrical and magnetic properties of the surrounding media and the geometrical configuration of the conductors and may in general be different from c, but it is nevertheless of the same order of magnitude as c. Circuits for which this nonzero travel time cannot be neglected are known as distributed circuits. An example of a distributed circuit is a long wire, as shown in Figure l .2b. When an input voltage Vo is applied at the input terminals of such a distributed circuit (i.e., between the input end of the wire and the electrical ground) at t = 0, the voltages and currents at all points of the wire cannot respond simultaneously to the applied excitation because the energy corresponding to the applied voltage propagates down the wire with a finite velocity v. Thus, while the input current Iin (t) may change from zero to Io at t = 0, the current 11 (t) does not flow until after t = 11/v, I2(t) does not flow until after t = 12/v, and no load current IL(t) can flow until after t = l /v. Similarly, when a har1nonically (i.e., sinusoidally) timevarying voltage is connected to such a line, the successive rises and falls of the source voltage propagate along the line with a finite velocity so that the currents and voltages at other points on the line do not reach their maxima and minima at the same time as the input voltage. In view of the fundamentally different behavior of lumped and distributed circuits as illustrated in Figure 1.2, it is important, in practice, to determine correctly whether a lumped treatment is sufficiently accurate or whether the circuit in hand has to be treated as a distributed circuit. In the following, we quantify on a heuristic basis the circumstances under which the travel time and/or the physical size of the circuit components or the length of the interconnects between them can be neglected. Note that in problems related to heat, diffusion, sound waves, water waves, traffic waves, and so on, the travel time is readily observable and almost always has to be accounted for. In the context of electromagnetics, on the other hand, we find a wide range of applications where lumped analysis is sufficiently accurate, is substantially simpler, and provides entirely satisfactory results. However, in an equally wide range of other applications we find that the lumped treatment is not sufficiently accurate and that one has to resort to field and wave techniques, which are generally more involved, both mathematically and conceptually. It is thus important, particularly in the context of electromagnetic applications, that we develop criteria by which we can determine the applicability of lumped circuit formulations. We provide below a heuristic discussion from three different but related points of view.
1.1.1 Rise Time versus Travel Time It is apparent from the preceding discussion that we can consider the delay time (travel time) over the signal path as long or short, important or negligible only relative to 13 The
more accurate empirical value of the speed of light is c = 299,792,458 ms 1 [CRC Handbook of Chemistry and Physics, 76th ed., CRC Press, Inc., Boca Raton, Florida, 1995].
8
Introduction
Vo
Chap. 1

0.9 V 0
Meter f\,..;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;p?J......J~ .._
  [  
0.1 V 0


t,
I .. I

t
Figure 1.3 Rise time versus oneway travel time.
or in comparison with some other quantity. In ter1ns of positive step changes in an applied signal, we note that the input signal typically exhibits a nonzero rise time (usually measured as the time required for the signal to change from 10% to 90% of its final value), which may be denoted as tr (Figure 1.3). We can then compare tr to the oneway propagation time delay through the signal path (also called the oneway transit time or time of flight), td = l /v, where v is the velocity of propagation and l is the length of the signal path. For example, in practical design of interconnects between integrated circuit chips, one rule of thumb is that the signal path can be treated as a lumped element if (tr/td) > 6, whereas lumped analysis is not appropriate for (tr/td) < 2.5. Whether lumped analysis is appropriate for the inbetween range of 2.5 < (tr /td) < 6 depends on the application in hand and the required accuracy. To summarize: (tr /td) > 6 (tr/td) < 2.5
(lumped) (distributed)
(1.1)
In ter1ns of the particular application of highspeed integrated circuits, the onchip rise times range from 0.52 ns for CMOS to 0.020.1 ns for GaAs technologies. The speed of signal propagation within the chips depends on the material properties; for example, it is rv0.5 lc for Si02 • Thus, for onchip interconnections (typical l ~ 1 cm), lumped circuit analysis breaks down (i.e., tr /td < 2.5) for rise times less than rv0.165 ns. For printed circuit boards made of a commonly used glass epoxy material, the speed of propagation is rv0.47c, so that for a rvlO cm interconnect, lumped analysis is not appropriate for rise times less than rvl.8 ns. As clock speeds increase and rise times become accordingly shorter, distributed analyses will be required in a wider range of digital integrated circuit applications. The importance of considering rise time versus travel time is underscored by advances in the generation of picosecond pulses. 14 For such extremely short pulse durations, with subpicosecond rise times, distributed circuit treatment becomes necessary for oneway propagation time delay of td > 10 13 seconds. Assuming propagation at the 14 See
D. W. Van der Weide, Allelectronic generation of 0.88 picosecond, 3.5 V shockwaves and their application to a 3 Terahertz freespace signal generation system, Appl. Phys. Lett., 62(1), pp. 2224, January 1993.
Sec. 1.1
9
Lumped versus Distributed Electrical Circuits
speed of light, the corresponding distances are> 0.03 mm! In other words, for picosecond rise times, lumped analysis is not appropriate for circuits with physical dimensions longer than a few tens of microns ( 1 micron = 1 µm). Example 1.2: Lumped or distributedcircuit element. Consider a 10cm long microstrip transmission line having a propagation speed of 15 cmns 1 . For digital signal transmission with a 100 ps rise time, should this microstrip line be treated as a lumped or distributedcircuit element? Solution:
The oneway time delay of the microstrip line can be found as td = l /v = 10 cm/15 cmns
1
:::'.
0.67 ns
Based on ruleofthumb criteria (1.1), since t7 /td :::'. 0.1 ns/0.67 ns = 0.15 < 2.5, the microstrip line should be modeled as a distributedcircuit element.
1.1.2 Period versus Travel Time For sinusoidal steadystate applications, the suitability of a lumped treatment can be assessed by comparing the oneway propagation delay td with the period T of the propagating sine wave involved. As an illustration, consider the telephone line shown in Figure 1.4. If a sinusoidal voltage at frequency f is applied at the input of the line so that VAA'(t) = Vocos(2nft), the voltage at a distance of l from the input is delayed by the travel time, td = l /v. In other words, td VBB'(t) = Vo cos [2nf (t  td)] = Vo cos 2nft  2nT
(1.2)
where T = 1/f is the period of the sinusoidal signal. It is apparent from (1.2) that if td O.OlA.
(lumped) (distributed)
(1.6)
Example 1.4: Component size versus wavelength. A wireless unit operating at 4 GHz is connected to an antenna via a 2.5cm long microstrip transmission line. Determine whether this microstrip transmission line can be approximated as a lumpedcircuit element. Assume a propagation speed of 15 cmns 1 for the microstrip line. Solution:
The wavelength in the microstrip transmission line can be calculated as v
).. 

 f 
(15 cmns 1) 4 x 109 Hz
109 ns 1s
= 3.75 cm
The ratio between the length L of the microstrip transmission line and the wavelength is
L/A. = 2.5 cm/3.75 cm::'.'. 0.67 Using the ruleofthumb criteria (1.6), a lumpedcircuit model is not appropriate to use in this case.
1.2 ELECTROMAGNETIC COMPONENTS The electromagnetic circuit components used at high frequencies can differ conspicuously in appearance from the often more familiar lumpedelement circuits used at low frequencies. The connecting wires of conventional circuits provide conductors for the electric currents to flow, and the resistors, capacitors, and inductors possess simple relationships between their terminal currents and voltages. Often overlooked is the fact that the wires and circuit components merely provide a framework over which charges move and disperse. These charges set up electric and magnetic fields that per1neate the circuit, often having almost indescribably complicated configurations. It would, in principle, be possible to treat the behavior of circuits entirely in ter1ns of these electromagnetic fields instead of the usual practice of working in terms of circuit voltages and currents. It can be argued, however, that much of the progress in modern electrical and electronic applications would not have come about if it were not for the simple but powerful circuit theory. As the operating frequency of circuits increases, however, the effects of stray capacitances and inductances alter the effective circuit behavior radically compared with its lowfrequency characteristics. Radiation from the circuit also increases rapidly with frequency, which can cause significant power loss. 17 This radiative power loss may be prevented by confining the fields to the interior of metallic enclosures. For example, 17 In
antenna applications, we take advantage of such radiation ''losses'' when we design antennas to maximize the radiated power in selected directions.
Sec. 1.3
Maxwell's Equations and Electromagnetic Waves
15
Coaxial cables and other transmission lines with lengths comparable to wavelength •
. ...
••
•• •• ••
••
.. . .. . ... ,
>>.A
...
R
•• •• ••
••
•• •• ••
... ·····,,,, ••
_ ___._Iw1Lumped elements with dimensions much smaller than wavelength .A Optical fibers with cross sections comparable to or much larger than wavelength
Waveguides with cross sections comparable to wavelength
Figure 1.8 Different categories of electromagnetic circuit components.
in hollow metallic tubes (referred to as waveguides) the charges move exclusively on the interior surfaces of conductors, and because of the simple geometry of the enclosures, the electromagnetic fields can have simple analytical forms. Since it is often not possible to define voltages and currents uniquely within waveguides, analysis of waveguides is usually carried out on the basis of the full electromagnetic theory. At even higher frequencies, metallic enclosures become too lossy and impractical. Efficient guiding of electromagnetic energy at optical frequencies occurs in optical fibers, consisting of hairthin glass strands. The light wave in an optical fiber is guided along the fiber by means of multiple reflections from its walls. The principles underlying the guiding of electromagnetic energy in metallic enclosures is discussed in Chapter 10. In summary, component sizes in electromagnetic applications can be categorized as shown in Figure 1.8: • Component sizes much smaller than A (lumped elements). Examples: Resistors, capacitors, inductors, ICs, transistors, and interconnects used for < rv30 MHz. • Component sizes comparable to A. Examples: Hollow waveguides, long coaxial cables, some optical fibers, and some antennas. • Component sizes much greater than A. Examples: Graded index optical fibers and some antennas.
1.3 MAXWELL'S EQUATIONS AND ELECTROMAGNETIC WAVES
The principles of guiding and propagation of electromagnetic energy in these widely different regimes differ in detail, but are all governed by one set of equations, known as Maxwell's equations, 18 which are based on experimental observations and which provide 18 J.
1892.
C. Maxwell, A Treatise in Electricity and Magnetism, Clarendon Press, Oxford, Vol. 2, pp. 247262,
16
Introduction
Chap. 1
the foundations of all electromagnetic phenomena and their applications. The sequence of events in the late 19th century that led to the development of these fundamental physical laws is quite interesting in its own right. Many of the underlying concepts were developed by earlier scientists, especially Michael Faraday, who was a visual and physical thinker but not enough of a mathematician to express his ideas in a complete and selfconsistent for1n to provide a theoretical framework. James Clerk Maxwell translated Faraday's ideas into strict mathematical form and thus established a theory that predicted the existence of electromagnetic waves. The experimental proof that electromagnetic waves do actually propagate through empty space was supplied by the experiments of Heinrich Hertz, carried out many years after Maxwell's brilliant theoretical work. 19 Maxwell's equations are based on experimentally established facts, namely Coulomb's law,20 Ampere's law,21 Faraday's law, 22 and the principle of conservation of electric charge. When most of classical physics was fundamentally revised as a result of Einstein's introduction23 of the special theory of relativity, Maxwell's equations 24 remained intact. To this day, they stand as the most general mathematical statements of the fundamental natural laws that govern all of classical electrodynamics. The basic justification and validity of Maxwell's equations lies in their consistency with physical experiments over the entire range of the experimentally observed electromagnetic 22 spectrum (see Table 8.1), extending from cosmic rays at frequencies greater than 10 Hz to the socalled micropulsations at frequencies of 103 Hz. The associated practical applications cover an equally wide range, from the use of gamma rays (10 181022 Hz) for cancer therapy to the use of waves at frequencies of a few hertz and below for geophysical prospecting. Electromagnetic wave theory as embodied in Maxwell's equations has provided the underpinning for the development of many vital practical 19 H.
Hertz, On the finite velocity of propagation of electromagnetic actions, Sitzb. d. Berl. Akad. d. Wiss., Feb. 2, 1888; for a collected English translation of this and other papers by H. Hertz, see H. Hertz, Electric Waves, MacMillan, London, 1893. 2°Coulomb's law states that electric charges attract or repel one another in a manner inversely proportional to square of the distance between them; C. A. de Coulomb, Premiere memoire sur l 'electricite et magnetisme (First Memoir on Electricity and Magnetism), Histoire de l'Academie Royale des Sciences, p. 569, 1785. 21 Ampere's law states that currentcarrying wires create magnetic fields and exert forces on one another, with the amplitude of the magnetic field (and thus force) depending on the inverse square of the distance; A. M. Ampere, Recueil d'observations electrodynamiques, Crochard, Paris, 18201833. 22 Faraday's law states that magnetic fields that change with time induce electromotive force or electric field; M. Faraday, Experimental Researches in Electricity, Taylor, London, Vol. I, pp. 1109, 1839. 23 Einstein, A., Annalen der Physik, 1905. The English translation of this paper is remarkably readable and is available in a collection of original papers titled The Principle of Relativity, Dover Publications, New York. 24 Maxwell's formulation was in fact one of the major motivating factors that led to the development of the theory of special relativity. The fact that Galilean relativity was consistent with classical mechanics but inconsistent with electromagnetic theory suggested that either Maxwell' s equations were incorrect or that the laws of mechanics needed to be modified. For discussions of the relationship between electromagnetism and the special theory of relativity, see Section 15 of D. M. Cook, The Theory of the Electromagnetic Field, PrenticeHall, Inc., New Jersey, 1975; Chapter 10 of D. J. Griffiths, Introduction to Electrodynamics, 2nd ed., PrenticeHall, Inc., New Jersey, 1989; Chapter 2 of R. S. Elliott, Electromagnetics, IEEE Press, Piscataway, New Jersey, 1993; Chapter 11 of J. D. Jackson, Classical Electrodynamics, 2nd ed., Wiley, New York, 1975.
Sec. 1.4
Summary
17
tools of our technological society, including broadcast radio, radar, television, cellular phones, optical communications, the Global Positioning Systems (GPS), microwave heating and processing, and Xray imaging. Maxwell's equations embody all of the essential features of electromagnetics, including the ideas that light is electromagnetic in nature, that electric fields that change in time create magnetic fields in the same way as timevarying voltages induce electric currents in wires, and that the source of electric and magnetic energy resides not only on the body that is electrified or magnetized but also, and to a far greater extent, in the surrounding medium. However, arguably the most important and farreaching implication of Maxwell's equations is the idea that electric and magnetic effects can be transmitted from one point to another through the intervening space whether that be empty or filled with matter. To appreciate the concept of propagation of electromagnetic waves in empty space, it is useful to think of other wave phenomena that we may observe in nature. When a pebble is dropped into a body of water, the water particles in the vicinity of the pebble are immediately displaced from their equilibrium positions. The motion of these particles disturbs adjacent particles, causing them to move, and the process continues, creating a wave. Because of the finite velocity of the wave, a finite time elapses before the disturbance causes the water particles at distant points to move. Thus the initial disturbance produces, at distant points, effects that are retarded in time. The water wave consists of ripples that move along the surface away from the initial disturbance. Although the motion of any particular water particle is essentially a small upanddown movement, the cumulative effects of all the particles produce a wave that moves radially outward from the point at which the pebble is dropped. Another excellent example of wave propagation is the motion of sound through a medium. In air, this motion occurs through the toandfro movement of the air molecules, but these molecules do not actually move along with the wave. Electromagnetic waves consist of timevarying electric and magnetic fields. Suppose an electrical disturbance, such as a change in the current through a conductor, occurs at a point in a region. The timechanging electric field resulting from the disturbance generates a timechanging magnetic field. The timechanging magnetic field, in turn, produces an electric field. These timevarying fields continue to generate one another in an everexpanding region, and the resulting wave propagates away from the location of the initial disturbance. When electromagnetic waves propagate in vacuum, there is no vibration of physical particles as in the case of water and sound waves. Nevertheless, the velocity of wave propagation is limited by the speed of light, so that the fields produced at distant points are retarded in time with respect to those near the source.
1.4 SUMMARY There are two kinds of electrical circuits: lumped and distributed. In classical lumped circuit theory, the entire circuit is assumed to be at a single point such that all parts of the circuit respond to an excitation at the same time. Distributed circuits or transmission lines can be treated using a natural extension of classical circuit theory by taking account of the nonzero travel time of signals from one point to another in a distributed circuit,
18
Introduction
Chap. 1
the speed of travel being bounded by the speed of light. In any given application, a number of criteria can be used to determine the applicability of a lumped analysis. A given system must be treated as a distributed one if: • The rise time tr of the applied signal is less than 2.5 times the oneway travel time td across the circuit, that is, tr < 2.5td. • The oneway travel time td across the circuit is greater than onehundredth of the period T of the applied sinusoidal signal, that is, td > O.OlT. • The physical dimensions of the circuit are a significant fraction of the wavelength at the frequency of operation. In Chapter 2, we will assume that all step sources are ideal and have zero rise time. We will also assume that all time delays are due to transmission lines, and consider lumped elements such as resistors, capacitors, and inductors to be ideal with zero time delay. Electromagnetic components can be categorized in three basic groups with sizes much smaller than, comparable to, or much larger than the wavelength. Electromagnetic applications involving any one of these three classes of components are governed by a set of physical laws known as Maxwell's equations, which are based on experimental facts and describe the propagation of electromagnetic waves through empty space or material media and, in general, all other electrical and magnetic phenomena. In the next two chapters, we will study a special class of electromagnetic waves called transverse electromagnetic (TEM) waves, which propagate on electrical transmission lines consisting of two metallic conductors. In the case of TEM waves, both the electric and magnetic fields are everywhere perpendicular to each other and to the direction of propagation. The behavior of TEM waves on twoconductor lines can be understood in the context of a voltagecurrent description by modeling the transmission line as a distributed circuit and utilizing many of our electrical circuits concepts. The fact that the voltage and current waves that propagate on such twoconductor transmission lines are indeed transverse electromagnetic waves will become clear in Chapter 10. After we study TEM waves in the context of electrical transmission lines in Chapters 2 and 3, we will introduce and discuss the fundamental components of Maxwell's equations in Chapters 4 through 7. In Chapter 4 we begin by studying the branch of electromagnetics that deals with electric charges at rest, called electrostatics. In this chapter, we investigate the consequences of Coulomb's law, an experimentally derived statement on the forces between charged particles. In Chapter 5 we consider the steady motion of charged particles through a material, which is sustained by electrostatic forces. We then note another experimental fact: wires containing steadystate current exert a force on each other. This observation, which is captured by Ampere's law, underpins the branch of electromagnetics called magnetostatics, the topic of Chapter 6. Finally, Chapter 7 introduces a third experimental fact that involves currents that change with time: Faraday's law of electromagnetic induction. This final observation, coupled with the fundamental fact of charge conservation, completes the formulation of Maxwell's equations, and ultimately leads to the phenomenon of electromagnetic waves.
Chap. 1
19
Problems
In the remainder of the book, we take a deeper look at the propagation of electromagnetic waves. In Chapter 8 we restrict our attention to electromagnetic waves in an unbounded, simple, and sourcefree medium. In Chapter 9 we remove the first condition and consider the reflection and transmission of electromagnetic waves at planar boundaries between different material media. Many practical problems encountered in electromagnetics involve reflection of waves from interfaces between dielectrics and perfect conductors (e.g., air and copper) or between two different dielectrics (e.g., glass and air), and the treatment of such problems requires that we take into account the complicating effects of the boundary surfaces. In Chapter 10 we consider electromagnetic waves guided by planar metallic and dielectric boundaries. Finally, in Chapter 11, we return to considering an unbounded medium but now allow it not to be simple, but rather have material properties which significantly affect the nature of electromagnetic waves. PROBLEMS
1.1
Travel time around the Earth. Calculate the approximate time it takes for an electromagnetic signal to travel around the circumference of the Earth. Assume the Earth to be a perfect sphere with an average diameter of 1.274 x 104 km.
1.2 1.3
1.4
1.5 1.6
Travel time between the Earth and the Moon. The average distance from Earth to the Moon is 384,400 km. How long will it take an electromagnetic wave to travel this distance? EarthMoon communication. During the Moon landing of one of the Apollo spaceflights, a spoken message from Earth is heard returning to Earth approximately 2.7 seconds later. The message signal traveled from Earth to the Moon, from the earphone to the microphone attached to the astronaut's space helmet, and back to Earth. 25 Use this time delay to estimate a rough value for the speed of light. Time delay of a radar signal. Radars use radio waves to determine the distance to a target. Radar systems consist of a transmitting antenna that emits a radio pulse, and a receiving antenna that measures the (typically small) reflected component of the emitted waves. If the transmitter and receiver are colocated, and if the receiver detects the pulses reflected from a target 40 f.LS after the radar pulses leave the transmitter, what is the distance of the target from the radar? Echo from a cliff. A man shouts in the direction of a cliff, which causes an echo that he hears 3 seconds later. If sound travels in air at 340 ms 1, how far away is the cliff? Sonar. Sonar stands for SOund NAvigation and Ranging and is a device that uses reflected sound waves to measure underwater depths. A ship sends a sonar signal downward into the ocean and receives a return signal 6 seconds later. Assuming the speed of sound waves to be the same at all depths given by 1.5 x 103 ms 1 , how deep is the ocean at that location?
1.7
Sonar. It takes a sonar pulse 3.7 seconds to return to its ship after being transmitted vertically
1.8
towards the ocean floor. Assuming a constant speed of 1.5 kms 1, what is the depth of the ocean at this point? Autofocus camera. An autofocus camera can focus on objects by use of an ultrasonic sound wave. Just like radar, the camera sends out sound waves which reflect off the distant objects and return back to the camera. A sensor detects the time it takes for the waves to 25 David
Keeports, ''Estimating the speed of light from EarthMoon communication," The Physics Teacher, Vol. 44, pp. 414415, October 2006.
20
Introduction
Chap. 1
return and then determines the distance of the object from the camera. If an ultrasonic sound wave returns to the camera 0.1 second after leaving the camera, how far is the object from the camera? Assume speed of the ultrasonic sound wave to be 340 ms 1 . (Polaroid developed an ultrasonic autofocus system that used 50 kHz waves, well outside the human hearing range of 20 Hz20 kHz. Modem autofocusing cameras typically use reflected infrared light to triangulate objects in the field of view.)
1.9
Lightning and thunder. If a person heard the sound of a thunder approximately 5 seconds after observing the lightning flash at a distance, how far away from this person did the lightning strike? Assume the speed of sound to be 340 ms 1 .
1.10 A lightyear. A lightyear (ly) is a unit of distance and is defined as the distance that light can travel in 1 year. Find the equivalent of 1 lightyear in kilometers.
1.11 A lightnanosecond. What is 1 lightnanosecond in meters? 1.12 1 Astronomical Unit. The average distance between the Earth and the Sun is defined as 1 Astronomical Unit (AU) is about 1.5 x 108 km. What is 1 AU in lightseconds?
1.13 Distance between Proxima Centauri and Earth. The distance between Proxima Centauri, the nearest known star to Earth (besides our Sun), is about 4 x 10 13 km. Find this distance in light years. 1.14 Seismic waves. A geological disturbance in California is detected at a seismograph station in Washington approximately 900 km away from the epicenter of the earthquake. If the seismic wave traveled at an average speed of 5 kms 1, determine the time delay between the epicenter and seismograph station. 1.15 Tsunami waves. Tsunami is a Japanese word. It means harbor (''tsu'') waves (''nami''). Tsunamis are fairly common in Japan and caused many thousands of deaths in recent centuries. A massive earthquake hit the Pacific Ocean near Honshu, Japan in 2011 producing a tsunami wave which reached Crescent City, California about 10 hours later. If the distance between the epicenter of the earthquake and Crescent City is about 8020 km, determine the average speed of the tsunami wave. 1.16 The Indian Ocean tsunami. The Indian Ocean tsunami triggered by a massive underwater earthquake that occurred in 2004 devastated the shores of Indonesia, Sri Lanka, India, Thailand, and other countries with waves up to 15 m high, even reaching as far as the east coast of Africa. The epicenter of the earthquake was about 160 km west of Sumatra, Indonesia and 4500 km from Africa. Assuming the average speed of the tsunami to be 800 km per hour, approximately how long did it take for the tsunami to travel from the epicenter to Sumatra? How about to Africa?
1.17 Microbats. Microbats use a form of radar called echolocation to navigate and find their prey such as flying insects. They locate the surrounding objects by bouncing sound wave pulses off these objects and detecting the time delay between the emitted pulses and the reflected pulses. Determine the time delay between the pulse emitted by the microbat and the detected pulse reflected from an insect located 10 m away from the microbat. Assume the approximate speed of sound waves to be 340 ms 1 .
1.18 Overhead power lines. The alternating current (ac) highvoltage overhead power lines in
1.19
most countries in the world operate at a standard frequency of 50 Hz. Assuming propagation at the speed of light, determine the maximum line length for lumped analysis. Maximum path length. Consider an electrical signal path in a highspeed digital circuit application to carry signals with rise times as low as 250 ps. If the speed of propagation is v = c /3, determine the maximum path length for lumpedcircuit analysis.
Chap. 1
Problems
21
1.20 Maximum coax length. Consider a commercial coaxial cable with v = 2 x 108 ms 1 . If this cable is to be used at frequencies around 900 MHz, what is the maximum cable length for lumpedcircuit analysis? 1.21 Travel time of a microstrip transmission line. What is the oneway travel time of a 9 cm long microstrip transmission line with v = 1.7 x 108 ms 1 ? 1.22 Maximum cable length. A ham radio operator connects his 30 MHz transmitter to an antenna using a coaxial cable. Assuming the speed of propagation along the cable to be 2 x 108 ms 1 , determine the maximum coaxial cable length beyond which it can't be modeled as a lumpedcircuit element. 1.23 Microstrip transmission line. A 6 cm long microstrip signal trace with speed of propagation v = 2c /3 is to be used to carry 1 ns rise time signals. Is it appropriate to neglect transmission line effects and use lumped circuit analysis? 1.24 Stripline transmission line. A 12 cm long stripline transmission line having a speed of propagation v = c /2 is used to carry digital signals with 0.5 ns rise time. Is it appropriate to consider transmission line effects in this case? 1.25 Onchip GaAs interconnect. Consider an onchip GaAs interconnect with speed of propagation v = 8 x 107 ms 1 to be used to carry a digital signal with rise time tr = 50 ps. What is the maximum interconnect length for lumpedcircuit analysis? 1.26 A coaxial cablelumped or distributed analysis? A coaxial cable of length 10 m having a speed of propagation v = 0.75c is used to connect the antenna and the receiver of a microwave communication system. Assuming sinusoidal steady state, what is the highest frequency for lumpedcircuit analysis?
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o Transmission Lines
Wave motion is said to occur when a disturbance of a physical quantity at a particular point in space at a particular time is related in a definite manner to what occurs at distant points in later times. Transient waves occur in response to sudden, usually brief disturbances at a source point, leading to temporary disturbances at distant points at later times. They are thus different from steadystate waves, which are sustained by disturbances involving periodic oscillations at the source point. Transient waves are of importance in many different contexts. Consider, for example, a line of cars waiting at a red traffic light. When the light turns green, the cars do not all start moving at the same time; instead, the first car starts to move first, fallowed by the car behind it, and so on, as the act of starting to move travels backwards through the line. This wave travels at a speed that depends on the response properties of the cars and the reaction times of the drivers. As another example, when the end of a stretched rope is suddenly moved sideways, the action of moving sideways travels along the rope as a wave whose speed depends on the tension in the rope and its mass. If the rope is infinitely long, the disturbance simply continues to propagate away from its source. If, on the other hand, the distant end of the rope is held fixed, the wave can be reflected back toward the source. Other examples of transient waves include the thunderclap, the sound wave emitted from an explosion, and seismic waves launched by an earthquake. Transient waves are often used as tools to study the disturbances that create them. The sound wave from a blast can be used for detecting the source of the blast from a long distance; a seismograph measures the strength of a distant earthquake based on tiny transient motions of the earth. The purpose of this chapter is to study voltage and current transients on electrical transmission lines. A transmission line may consist of two parallel wires (as it will often be illustrated in this book), of coaxial conductors, or of any two conductors separated by an insulating material or vacuum. Some types of twoconductor transmission lines are shown in Figure 2.1.
23
24
Transient Response of Transmission Lines
Chap.2
,, /
,,
/ /
/
/
/
/
/
/
/
/
/ / /
(b)
(a)
(d)
(c)
(e)
Figure 2.1 Different types of uniform transmission lines: (a) parallel twowire; (b) coaxial; (c) parallelplate; (d) stripline; (e) microstrip.
Many important electrical engineering applications involve transients: temporary variations of current and voltage that propagate down a transmission line. Transients are produced by steplike changes (e.g., sudden on or off) in input voltage or current. Digital signals consist of a sequence of pulses, which represent superposition of successive steplike changes; accordingly, the transient response of transmission lines is of interest in most digital integrated circuit and computer communication applications. Transient transmission line problems arise in many other contexts. The transient response of lines can be used to generate rectangular pulses; the earliest applications of transmission lines involved the use of rectangular pulses for telegraphy. When lightning strikes a power transmission line, a large surge voltage is locally induced and propagates to other parts of the line as a transient. This chapter is unique; the following chapters are mostly concerned with applications that either involve static quantities, which do not vary in time or involve steady signals that are sinusoids or modified sinusoids. However, with the rapid advent of digital integrated circuits, digital communications, and computer communication applications, transient responses of transmission lines are becoming increasingly more important. Increasing clock speeds make signal integrity 1 analysis a must for the design of highspeed and highperfor1nance boards. Managing signal integrity in today's highspeed printed circuit boards and multichip modules involves features such as interconnect lengths, vias, bends, terminations, and stubs and often requires close attention to transmission line or distributed
1The
term signal integrity refers to the issues of timing and quality of the signal. The timing analysis is performed to ensure that the signal arrives at its destination within a specified time interval and that the signal causes correct logic switching without false switching or transient oscillations, which can result in excessive time delays. See R. Kollipara and V. Tripathi, Printed wiring boards, Chapter 71 in J. C. Whitaker (Ed.), The Electronics Handbook, CRC Press, Boca Raton, Florida, pp. 10691083, 1996.
Sec. 2.1
Heuristic Discussion of Transmission Line Behavior and Circuit Models
25
2 effects.
circuit It is thus fitting that we start our discussion of engineering electromagnetics by studying the transient response of transmission lines. Also, analysis of transients on transmission lines requires relatively little mathematical complexity and brings about an intuitive understanding of concepts such as wave propagation and reflection, which facilitates a better understanding of the following chapters.
2.1 HEURISTIC DISCUSSION OF TRANSMISSION LINE BEHAVIOR AND CIRCUIT MODELS
Typically, we explain the electrical behavior of a twoconductor transmission line in terms of an equal and opposite current flowing in the two conductors, as measured at any given transverse plane. The flow of this current is accompanied by magnetic fields set up around the conductors (Ampere's law), and when these fields change with time, a voltage (electromotive force) is induced in the conductors (Faraday's law), which affects the current flow. 3 This behavior can be represented by a small inductance associated with each shortlength segment of the conductors. Also, any two conductors separated by a distance (such as the short sections of two conductors facing one another) have nonzero capacitance, so that when equal and opposite charges appear on them, there exists a potential drop across them. Hence, each short section of a twoconductor line exhibits some series inductance and parallel capacitance. 4 The values of the inductance and capacitance depend on the physical configuration and material properties of the twoconductor line, including the surface areas, crosssectional shape, spacing,5 and layout of the two conductors, as well as the electrical and magnetic properties 6 of the substance filling the space between and around the conductors. 2.1.1 Heuristic Discussion of Transmission Line Behavior
We can qualitatively understand the behavior of a twoconductor transmission line by considering a lossless circuit model of the line, consisting of a large number of series inductors and parallel capacitors connected together, representing the short sections ~z of the line, as illustrated in Figure 2.2. 2 See
R. Goyal, Managing signal integrity, IEEE Spectrum, pp. 5458, March 1994. 3 Detailed discussion of these experimentally based physical laws will be undertaken later; here we simply rely on their qualitative manifestations, drawing on the reader's exposure to these concepts at the freshman physics level. 4 Neglecting losses for now. 5 This can be seen at a qualitative level, from the reader's understanding of capacitance and inductance at the freshman physics level. For example, the closer the conductors are to each other, the larger the capacitance is. On the other hand, the inductance of a twoconductor line is smaller if the conductors are closer together, since the magnetic field produced by the current flow is linked by a smaller area, thus inducing less voltage. 6 At the simplest level, the magnetic properties of a material represent the ability of a material medium to store magnetic energy. Similarly, by electrical properties we refer to the ability of a material to store electric energy or its response to an applied electric field. The microscopic behavior of the materials that determines these properties will be discussed in Chapters 4 and 6.
26
Transient Response of Transmission Lines 1 t=O
Rs1
Vo=
2 Rs2
Lia
Lza
I I
I I
: C1
: C2
I I I I I
I I I I I
L1b
Lka

Chap.2
L(k+ l)a
ck+ 1
ck

L2b

Lkb
L(k+ l)b
Liz Figure 2.2 Circuit model of a twoconductor lossless line.
To illustrate the behavior of a lossless transmission line, we now consider the simplest possible transient response: the step response, which occurs upon the sudden application of a constant voltage. At t = 0, a battery of voltage Vo and source resistance Rsi is connected to the infinitely long twoconductor transmission line represented by the lossless circuit shown in Figure 2.2, where each pair of inductances corresponds to the inductance of a short section of the line of length ~z, and each capacitor corresponds to the capacitance of the same section of length ~z. Initially, the transmission line is completely discharged, so all the capacitances have zero charge (and thus zero voltage) and the inductances have zero current flowing through them. The switch in Figure 2.2 is moved to position 1 at t = 0, so that, starting at t = o+, the source voltage Vo appears across the source resistance Rsi and the ter1ninals of the capacitance Ci, which takes time to charge;7 and until it charges, there is no voltage across it to drive currents through Lia and Lib· As 8 the voltage across Ci builds up, the currents in Lia and Lib also take time to increase. When these currents increase enough to cause appreciable flow through C2, this capacitance now takes some time to charge. As it charges, current starts to flow in inductors L2a and L2b, but this takes time. This same process continues all the way down the line, with the capacitor Ck not starting to charge until the preceding capacitors are charged, just as if it did not know yet that the voltage step had been applied at the input. In this way, the information about the change in the position of the switch travels down the line. When the switch moves back to position 2 at t =ti, the reverse happens: Ci has to discharge through R82, which pulls current (not suddenly) from Lia and Lib, which in turn allows C2 to discharge, and so on. All of this takes time, so Ck is not affected by the removal of the input signal until the preceding capacitors are discharged first. The rate of charging and discharging depends only on the circuit element values, so the charging and discharging disturbances both continue down the line at the same speed, since Lia = Ljb for all i, j and all Ci values are equal, assuming a uniform transmission line. Note from the above discussion that if the inductance of the line segments is negligible, the line can be approximated as a lumped capacitor (equal to the parallel combination of all of its distributed shunt capacitances); all the points on the line are then at the same potential, and travelingwave effects are not important. The line inductance 7 Voltage
across a capacitance cannot change instantaneously. 8 Current through an inductor cannot change instantaneously.
Sec. 2.1
Heuristic Discussion of Transmission Line Behavior and Circuit Models
27
becomes important if the line is relatively long or if the rise time of the applied signal (as defined in Figure 1.3) is so fast that the current through the inductor increases very rapidly, producing appreciable voltage drop (°V = L d!P / dt) across the inductor even if the value of L is small. By the same token, it is intuitively clear that, even if the line is long (and thus L is large), transmission line effects will be negligible for slow enough rise times, as was discussed in Chapter 1.
2.1.2 Circuit Models of Transmission Lines It is often more useful to describe transmission line behavior in ter1ns of inductance and capacitance per unit length rather than viewing the line as an infinite number of discrete inductances and capacitances, as implied in Figure 2.2. We must also note that, in general, the conductors of a transmission line exhibit both inductance and resistance and that there can be leakage losses through the insulating material surrounding the conductors. The inductance per unit length (L) of the line, in units of henrys per meter, depends on the physical configuration of the conductors (e.g., the separation between conductors and their crosssectional shape and dimensions) and on the magnetic properties of the material surrounding the conductors. The series resistance per unit length R, in units of ohms per meter, depends on the crosssectional shape, dimensions, and electrical conductivity of the conductors 9 and the frequency of operation. Between the conductors there is a capacitance ( C), expressed as farads per meter; there is also a leakage conductance ( G) of the insulating material surrounding the conductors, in units of siemens per meter. The capacitance depends on the shape, surface area, and proximity of the conductors as well as the electrical properties of the insulating material surrounding the conductors. The conductance depends on the shape and dimensions of the outside surface of the conductors 10 and on the degree to which the insulating material is lossy. A uniform transmission line consists of two conductors of unifor1n crosssection and spacing throughout their length, surrounded by a material that is also uniform throughout the length of the line. An equivalent circuit of a unifor1n transmission line can be drawn in terms of the perunitlength parameters, which are the same throughout the line. One such circuit is shown in Figure 2.3, where each short section of length ~z of the line is modeled as a lumped circuit whose element values are given in ter1ns of the perunit parameters of the line. The electrical behavior of a unifor1n transmission line can be studied in ter1ns of such a circuit model if the length of the line (~z) represented by a single LRCG section is very small compared with, for example, the wavelength of electromagnetic waves in the surrounding material at the frequency of operation. Four different circuit models are shown in Figure 2.4. Expressions for L, R, C, and G for some of the commonly used uniform transmission lines shown in Figure 2.1 are provided in Section 2. 7. The values of these quantities depend on the geometric shapes and the crosssectional dimensions of the 9 The
resistance simply represents the ohmic losses due to the current flowing through the conductors; hence, it depends on the crosssectional area and the conductivity (see Chapter 5) of the conductors. 10The leakage current flows from one conductor to the other, through the surrounding material, and in the direction transverse to the main current flowing along the conductors of the line; hence, it depends on the outer surface area of the conductors.
28
Transient Response of Transmission Lines
Chap.2
1 Lt).z Rt).z 12 2 I I I I I I I I I I I I
~
i
Lt).z Rt).z
~
2
12
I
I I•
.. 1
Figure 2.3 Distributed circuit of a uniform transmission line. r~
I
I
.....;.~
I I I I 1
I
Rt).z 2
Lt).z 2
Rt).z
Lt).z 2
2
r~
I

I
rr 1
Lt).z
Rt).z
I I I I
I I
I IL _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ j
IL _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ j
t).z
t).z
(a)
(b)
__ Rt).z
r ~
I  "
,,..
Lt).z Lt).z I Rt).z I 2 2 2 2 I I N N I I I I ________________ j L
~
C3
r~
I
I I I
Gt).z 1
2
I
Lt).z Rt).z Ct).z
I I
Ct).z
I 2 2 I I I IL _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
t).z
t).z
(c)
(d)
1
I I I I
Gt).z
2
j
Figure 2.4 Lumpedcircuit models for a short segment of a uniform transmission line.
line, the electrical conductivity of the metallic conductors used, the electrical and magnetic properties of the insulating surrounding medium, and the frequency of operation. The expressions for L, R, C, and G for the various transmission lines can be obtained by means of electromagnetic field analysis of the particular geometries involved. For some cases (such as the parallel twowire, coaxial line, and parallel plate lines shown in Figure 2.1), compact analytical expressions for R, L, C, and G can be found. For other, more complicated structures (e.g., the stripline and microstrip lines in Figure 2.1),
Sec. 2.2
Transmission Line Equations and Wave Solutions
29
calculation of the R, L, C, and G parameters usually requires numerical computation. Methods for the determination of transmission line parameters are discussed in Chapters 4 through 6 as we introduce the governing electromagnetic equations, so that we can formally derive the expressions for the transmission line parameters. For our distributed circuit analyses of transmission lines in this and the following chapter, it suffices to know that the values of L, R, C, and G are directly calculable for any unifor1n transmission line configuration. We can thus proceed by using their values as specified by the expressions in Section 2.7, as given in handbooks, or as measured in specific cases.
2.2 TRANSMISSION LINE EQUATIONS AND WAVE SOLUTIONS In this section we develop the fundamental equations that govern wave propagation along general twoconductor transmission lines. Various lumpedcircuit models of a single short segment of a transmission line are shown in Figure 2.4. In the limit of ~z > 0, any one of the circuit models of Figure 2.4 can be used to derive the fundamental transmission line equations. In the following, we use the simplest of these models (Figure 2.4b), shown in further detail in Figure 2.5.
2.2.1 Transmission Line Equations The section of line of length ~z in Figure 2.5 is assumed to be located at a distance z from a selected point of reference along the transmission line. We consider the total J>(z, t)
I I
Lfl.z Rfl.z Cfl.z Gfl.z
1
J>(z + fl.z, t)
I
+ °V(z + fl.z, t)
+ °V(z + fl.z, t)
z I
I
z
z + fl.z
Figure 2.5 Equivalent circuit of a short length of Az of a twoconductor transmission line.
30
Transient Response of Transmission Lines
Chap.2
voltage and current at the input and output terminals of this line section: that is, at points z and (z + ~z). The input and output voltages and currents are denoted as °V (z, t), !P (z, t) and °V (z + ~z, t), !P (z + ~z, t), respectively. Using Kirchhoff s voltage law, we can see that the difference in voltage between the input and output terminals is due to the voltage drop across the series elements R ~z and L~z, so we have
a!P(z,t) °V(z + ~z, t)  °V(z, t) = R~z !P(z, t)  L~z  at Note that we shall consider ~z to be as small as needed so that the lumped circuit model of the segment can accurately represent the actual distributed line. Upon dividing by ~z 11 and expanding °V (z + ~z, t) in a Taylor series, in the limit ~z > 0 we have
. °V(z 11m
+ ~z, t) 
~z+O
~z
°V(z, t)  a°V(z, t) ((, a!P(z, t) R.y(z,t)Laz at
or
a°V(z, t) az
R+L
a at
!P(z,t)
(2.1)
Similarly, using Kirchhoff s current law, the difference between the current at the input and output terminals is equal to the total current through the parallel elements G ~z and C ~z, so we have
!P(z
+ ~z, t) 
Upon dividing by we have: lim
!P(z
~z+0
~z
!P(z, t) =
G~z
and expanding °V(z
+ ~z, t) 
!P(z, t)
°V(z
+ ~z,t) 
+ ~z' t) C ~zat a°V(z
+ ~z, t) in a Taylor series, and taking ~z
a°V(z,t) = G °V (z, t)  C at
~z
> 0,
. 11m {~z (· · ·)}
~z+0
or
a!P(z,t) az
G+C
a at
°V(z,t)
(2.2)
Equations (2.1) and (2.2) are known as the transmission line equations or telegrapher's equations. We shall see in Chapter 8 that uniform plane electromagnetic wave propagation is based on very similar equations, written in terms of the components of electric and magnetic fields instead of voltages and currents. Most other types of wave phenomena are governed by similar equations; for acoustic waves in fluids, for example, one replaces voltage with pressure and current with velocity.
ll°V(z
+ ~z, t) =
°V(z, t)
+
avcz, t) az
~z
+
1 2!
a2vcz, t) az2
~z2 + ...
Sec. 2.2
31
Transmission Line Equations and Wave Solutions
2.2.2 TravelingWave Solutions for Lossless Lines Solutions of (2.1) and (2.2) are in general quite difficult and require numerical treatments for the general transient case when all of the transmission line parameters are nonzero. However, in a wide range of transmission line applications the series and parallel loss terms (R and G) can be neglected, in which case analytical solutions of (2.1) and (2.2) become possible. In fact, practical applications in which transmission lines can be treated as lossless lines are at least as common as those in which losses are important. Accordingly, our transmission line analyses in this chapter deal exclusively with lossless transmission lines. We now apply (2.1) and (2.2) to the analysis of the transient response of lossless transmission lines. For a lossless line we have R = 0 and G = 0, so that (2.1) and (2.2) reduce to av a!P =L(2.3) az at a!P av (2.4)  =C az at By combining (2.3) and (2.4) we obtain the wave equations for voltage and current, (2.5) or
a1!P
a1!P =LCaz2 at 2
(2.6)
Either one of (2.5) or (2.6) can be solved for the voltage or current. We follow the usual practice and consider the solution of the voltage equation (2.5), which can be rewritten as 2 az
1 v 
2 2·'
p ~
vp at
(2.7)
Note that once V(z, t) is determined, we can use (2.3) or (2.4) to find !P(z, t). The quantity Vp represents the speed of propagation of a disturbance, as will be evident in the fallowing discussion. For reasons that will become clear in Chapter 3, Vp is also referred to as the phase velocity; hence the subscript p. Any function 12 f (·)of the variable~ = (t  z/vp) is a solution of (2.7). To see that
V(z, t) =
f
z t 
=!(~)
Vp 12
An important function of (t  z /vp) that is encountered often and that we shall study in later chapters is the sinusoidal travelingwave function, A cos[w(t  z/vp)]. Depending on the location of the observation point z along the z axis, this function replicates the sinusoidal variation A cos(wt) observed at z = 0, except delayed by (z/vp) seconds at the new z. Thus, (z/vp) represents a time shift, or delay, which is a characteristic of the class of wave functions of the variable (t  z/vp).
32
Transient Response of Transmission Lines
Chap.2
is a solution of (2. 7), we can express the time and space derivatives of V (z, t) in ter1ns of the derivatives off (g) with respect to g:
since
ag /at
av af ag af at ag at ag
and
a2v a2f ag 2 at ag 2 at
= 1. Similarly, noting that
ag /az
= (1/vp), we have
av az
and
Substituting in (2. 7) we find that the wave equation is indeed satisfied by any function f (·) of the variable g = (t  z /vp). That an arbitrary function! (t  z/vp) represents a wave traveling in the +z direction is illustrated in Figure 2.6 for vp = 1 ms 1. By comparing f (t  z/vp) at two different times t = 2 and t = 3 s, we note that the function maintains its shape and moves in the +z direction as time t advances, as seen in Figure 2.6a. Similarly, by comparing! (t  z/vp) at two different positions z = 0 and z = 1 m, we note that the function maintains its shape but appears at z = 1 m exactly 1 s after its appearance at z = 0, as seen in Figure 2.6b. Figure 2.6c shows a threedimensional display off (t  z /vp) as a function of time at different points z1, z2, and z3. To determine the speed with which the function moves in the +z direction, we can simply keep track of any point on the
0.6 0.4 0.2
1
3
2
''
(a)
'
''
t
'
~~
f(t z/vp) 0.6
z=lm /
I
I
0.2
1
'
''
'
~
z=O
0.4
''
......
''
''
''
'
~~
z
'
'
''

''
2
t (s)
' ' .....
(c)
3
(b) Figure 2.6 Variation in space and time of an arbitrary function f (t  z/vp)· The phase velocity is taken to be equal to unity, that is, vp = 1 ms 1 . (a) f (t  z/vp) versus z at two different times. (b)f(tz/vp) versus tat two different locations. (c) Threedimensional display off (t  z /vp) as a function of time at different points z1, z2, and z3.
Sec. 2.2
Transmission Line Equations and Wave Solutions
33
function by setting its argument to a constant. In other words, we have t 
z
= const
>
Vp
dz dt
=
Vp
The speed of propagation of waves on a transmission line is one of its most important characteristics. It is evident from (2. 7) that Vp depends on the line inductance L and capacitance C. In the case of most (all except the microstrip line) of the commonly used twoconductor transmission lines shown in Figure 2.1, the phase velocity Vp in the absence of losses is not a function of the particular geometry of the metallic conductors but is solely determined by the electrical and magnetic properties of the surrounding medium. 13 When the medium surrounding the metallic conductors is air, the phase velocity is equal to the speed of light in free space, namely Vp = c. The propagation speeds for some other insulating materials are tabulated in Table 2.1. It is clear from the above analysis that any function of the argument (t + z /vp) is an equally valid solution of (2.7). Thus, the general solution for the voltage °V (z, t) is
°V(z, t) = TABLE 2.1
1+
t
z 
z t+ 
Vp
Vp
(2.8)
PROPAGATION SPEEDS IN SOME MATERIALS
Material Air Glass Mica (ruby) Porcelain Fused quartz (Si02) Alumina (Al203) Polystyrene Polyethylene Teflon Vaseline Amber (fossil resin) Wood (balsa) Water (distilled) Ice (pure distilled) Soil (sandy, dry)
Propagation Speed at"' 20°C (cm(ns) 1 at 3 GHz) 30 (315)* 12.9 (1013)* 15.4 10.1 18.8 20.0 20.7 20.4 18.6 27.2 3.43 16.8** 18.8
*Approximate range valid for most types of this material. **At 12°C. 13
This result will become evident in Chapters 4 and 6 as we determine the capacitance and inductance of selected transmission lines from first principles. That vp = (LC) 112 does not depend on the geometrical arrangement of the conductors can also be seen by considering the inductance and capacitance expressions given in Table 2.2, Section 2.7. For transmission lines that do not exhibit symmetry in the crosssectional plane, such as the microstrip line of Figure 2.le, the phase velocity depends in a complicated manner on the properties of the surrounding material, the shape and dimensions of the conductors, and the operating frequency. See Section 8.6 of S. Ramo, J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics, 3rd ed., John Wiley & Sons, New York, 1994.
34
Transient Response of Transmission Lines
Chap.2
wherel+(t  z/vp) andl(t + z/vp) represent waves traveling in the +z and z directions, respectively. Note that I+(·) and I(·) can in general be completely different functions. To find the general solution for the current !P (z, t) associated with the voltage °V(z, t), we can substitute (2.8) in (2.3) (or (2.4)), integrate with respect to time (or position), and then take the derivative with respect to z (or time) to find !P(z,t) =
1
Zo
1+
t
z z  I t+vp Vp
•
'
Zo
=
L
c
(2.9)
where Zo is known as the characteristic impedance of the transmission line. 14 The characteristic impedance is the ratio of voltage to current for a single wave propagating in the +z direction, as is evident from (2.8) and (2.9). Note that the current associated with the wave traveling in the z direction (i.e., toward the left) has a negative sign, as expected, since the direction of positive current as defined in Figure 2.5 is to the right. In other words, since the polarity of voltage and the direction of current are defined so that the voltage and current have the same signs for forward (to the right)traveling waves, the voltage and current for waves traveling to the left have opposite signs. The characteristic impedance of a line is one of the most important parameters in the equations describing its behavior as a distributed circuit. For lossless lines, as considered here, Zo is a real number having units of ohms. Since Zo for a lossless line depends only on L and C, and since these quantities can be calculated from the geometric shape and physical dimensions of the line and the properties of the surrounding material, Zo can be expressed in ter1ns of these physical dimensions for a given type of line. Formulas for Zo for some common lines are provided in Section 2. 7. Characteristic impedances for other types of transmission lines are available in other publications. 15 The fallowing example illustrates the meaning of the characteristic impedance of a line. An infinitely long and initially uncharged line (i.e., all capacitors and inductors in the distributed circuit have zero initial conditions) is considered, so there is no need for the I (t + z/vp) ter1n in either (2.8) or (2.9), which would be produced only as a result of the reflections of the voltage disturbance at the end of the line. Example 2.1: Step response of an infinitely long lossless line. As a simple example of the excitation of a transmission line by a source, consider an infinitely long lossless transmission line characterized by Land C and connected to an ideal step voltage source of amplitude Vo and source
14 To
find !Ji(z, t), we can also note that the wave equation (2.6) for the current is identical to equation (2.5) for voltage, so its solution should have the same general form. Thus, the general solution for the current should be
z
!Ji(z,t)=g+ t  
vp
z
+g t+ 
Vp
Substituting this expression for !Ji(z, t) and (2.8) into (2.3) or (2.4) yields g+ = z0 1f+ and g = z0 1f. 15 Reference Data for Engineers, 8th ed., Sams Prentice Hall Computer Publishing, Carmel, Indiana, 1993.
Sec. 2.2
r

35
Transmission Line Equations and Wave Solutions Rs


~
T
I I I I I I I I
" ~
j>s
+ °Vs
Infinite line
Zo
Zo


~
I
z
z =O
(a)
(V0Z 0)/ (Rs+ Z 0)
Rs
(b)
il
z
(c)
Vol(Rs + Zo)           
l
0
z
(d)
Figure 2.7 Step excitation of a lossless line. (a) Step voltage applied to an infinite lossless line. (b) Initial equivalent circuit seen from the source. (c) The voltage disturbance cy+ (z, t). ( d) The current disturbance .1> + (z, t).
resistance Rs, as shown in Figure 2.7a. Find the voltage, the current, and power propagating down the transmission line.
Solution: Before t = 0, the voltage and current on the line are identically zero, since the line is assumed to be initially uncharged. At t = 0, the step voltage source changes from 0 to Vo, launching a voltage cy+(z, t) at z = 0 propagating toward the right. In the absence of a reflected wave (infinitely long line), the accompanying current is !P+(z,t) = (Zo) 1°V+(z,t), as can be seen from equations (2.8) and (2.9). In other words, the characteristic impedance Zo = ,JI:/C is the ''resistance'' that the transmission line initially presents to the source, as shown in the equivalent circuit of Figure 2.7b. Accordingly, the initial voltage and current established at the source end (z = 0) of the line are +
VoZo °Vs(t) =°V (z =0,t) =  Zo + Rs
and ({,
({, +
Js(t) = J
Vo (z = O,t) =  Zo + Rs
36
Transient Response of Transmission Lines
Chap.2
The propagation of the voltage cy+ (z, t) and the current !P+ (z, t) down the line are illustrated in Figures 2.7c and d at t = l/vp as a function of z. Note that the flow of a current !Ps outward from a source producing a voltage Vo represents a total power of Po = !Ps Vo supplied by the source. Part of this power, given by !P~R 8 , is dissipated in the source resistance. The remainder, given by + + ~r+ ~r Pline = fP (0, t) v ( 0, t) = fP s v s =
~~ (Zo +Rs) 2
is supplied to the line. Because the line is lossless, there is no power dissipation on the line. Therefore, the power Ptne goes into charging the capacitances and the inductances 16 of the line, as discussed in connection with Figure 2.2. Note that as P tne travels down the line the amounts of energy stored respectively in the capacitance and inductance of a fully charged portion of the line of length dl are given by ~ ( C dl)°V; and ~ (L dl)!P;.
2.3 REFLECTION AT DISCONTINUITIES In most transmission line applications, lines are connected to resistive loads, other lines (with different characteristic impedances), reactive loads, or combinations of resistive and reactive loads. Such discontinuities impose boundary conditions, which cause reflection of the incident voltages and currents from the discontinuities, resulting in new voltages and currents that are launched in the opposite direction. In this section we consider the reflection process and also provide examples of step responses of transmission lines terminated with short and opencircuited terminations. Consider a transmission line terminated in a load resistance RL located at z = l, as shown in Figure 2.8, on which a voltage of (z, t) is initially (t = 0) launched by the source at z = 0. Note that for an ideal step voltage source of amplitude Vo and a source resistance Rs, as shown in Figure 2.8, the amplitude of Vi(z, t) is given by
Vi
v+co O) = 1 '
ZoVo Rs+ Zo
vt
In general, a new reflected voltage °VI (z' t) is generated when the disturbance (z' t) reaches the load position at time t = td, where td is the oneway travel time on the line, or td = l / Vp. In order to determine the amplitude of the reflected wave, we write the total voltage and current at the load position (i.e., z = l) at = (i.e., immediately after the arrival of the incident wave) as
t tt
°VL(t) = °Vi(Z,t) +°Vi(l,t) 1
+
1

!PL(t) = Zo °V 1 (l,t) Zo °V 1 (l,t)
16 ''Charging''
an inductance can be thought of as establishing a current in it.
(2.lOa) (2.lOb)
Sec. 2.3
Reflection at Discontinuit ies
37 l
....

..

.. +
+ Step Vo source
r
'• ~
"I ~
Z 0 , td = l/vP
°Vs
°VL


z =O
z= l
'
t>O
 v; (z, t) °V1 (z, t) +
°V2 (z, t)
Figure 2.8 A terminated transmission line. The load RL is located at z = l , while the source end is at z = 0.
Using (2.lOa) and (2.lOb) and the boundary condition V L(t) = !PL(t)RL imposed by the purely resistive termination RL, we can write >
Vi(l,t)
V!(l,t)
Zo
Zo
Vt (l, t)
+ V! (l, t) RL
(2.11)
From (2.11) we can find the ratio of the reflected voltage V!(l, t) to the incident one (l, t). This ratio is defined as the load voltage reflection coefficient, rL,
vt
(2.12) and it follows that (2.13) The reflection coefficient is one of the most important parameters in transmission line analysis. Accordingly, the simple expression (2.12) for rL should be memorized. For lines terminated in resistive loads, fL can have values in the range 1 :S fL :S + 1, where the extreme values of 1 and + 1 occur when the load is, respectively, a short circuit (i.e., RL = 0) and an open circuit (i.e., RL = oo ). The special case of rL = 0 occurs when 17 RL = Zo; meaning that the load resistance is the same as the characteristic impedance, that there is no reflected voltage, and that all the power carried by the incident voltage is absorbed by the load. It is important to note that expression (2.12) for the reflection coefficient was arrived at in a completely general fashion. In other words, whenever a (z' t) is incident on a load RL from a transmission line with characteristic voltage
vt
17 This
condition is referred to as a matched load and is highly preferred in most applications.
38
Transient Response of Transmission Lines
v;
Chap.2
vt
impedance Zo, the amplitude of the reflected voltage (l, t) is rL (l, t), with rL given by (2.12). The generality of (2.12) can also be used to determine the reflection of the new voltage V! (z, t) when it arrives at the source end of the line. Having originated at the load end at t = td, the reflected voltage V! (z, t) arrives at the source end (terminated with the source resistance Rs) at t = 2td. At that point, it can be viewed as a new voltage disturbance propagating on a line with characteristic impedance Zo that is incident on a resistance of Rs. Thus, its arrival at the source end results in the generation of yet another reflected voltage traveling toward the load. We denote this new voltage Vi (z, t), where the subscript distinguishes it from the original voltage Vi (z, t) and the superscript underscores the fact that it is propagating in the +z direction. The amplitude of the new reflected voltage Vi (z, t) is determined by the source reflection coefficient rs, which applies at the source end of the line and is defined as (2.14) Thus we have Vj:(z, t) = Vj:(O, t) = fsV!(O, t). Note that the voltage Vi (z, t) was created at t = 0 and still continues to exist. Thus the sourceend voltage and current at t = 2t are
;t
+ V! (0, t) + Vj: (0, t) =Vi (0, t)(l + fL + f sfL)
V s(t) = Vi (0, t)
1 + 1 !Ps(t) = V 1 (O,t)V 1 (O,t) Zo Zo =
1
Zo
+
(2.15a)
1 + + V 2 (O,t) Zo
V 1 (0, t)(l  fL + f sfL)
(2.15b)
The newly generated voltage Vi(z, t) will now arrive at the load end at t = 3td and lead to the creation of another new reflected voltage v;: (z, t), and this process will continue indefinitely. In practice, the stepbystep calculation of the successively generated voltages becomes tedious, especially for arbitrary resistive terminations. In such cases, the graphical construction of a bounce diagram is very helpful. We introduce the bounce diagram in the fallowing subsection.
2.3.1 Bounce Diagrams A bounce diagram, illustrated in Figure 2. 9, also called a reflection diagram or lattice diagram, is a distancetime plot used to illustrate successive reflections along a transmission line. The distance along the line is shown on the horizontal axis and time is shown on the vertical axis. The bounce diagram is a plot of the time elapsed versus distance z from the source end, showing the voltages traveling in the +z and z directions as straight
Sec. 2.3
39
Reflection at Discontinuities l ~
, '
~
Vt (z, t) 
+ ~
+

•
°Vs
°VL
~
~


..
°Vk(z, t)

Zo
Za
l
V!(z, t) I I
L
I I
1
I I I
______ L _____ _ I I I
°V1(z, t)
Po
°Vz(Z, t)
Vt(z, t)
I I I
°V3(Z, t) : I I I I I
t
'
Figure 2.9 Bounce diagram.
z
40
Transient Response of Transmission Lines
Chap.2
18 sloping
lines downward from left to right and right to left, respectively. Each sloping line corresponds to an individual traveling voltage and is labeled with its amplitude. The amplitude of each reflected voltage is obtained by multiplying the amplitude of the preceding voltage by the reflection coefficient at the position where the reflection occurs. The time sequence of events starting with the first application of the step voltage at the source end can be easily visualized from the bounce diagram. The application of the step voltage launches Vi (z, t) traveling toward the load. This voltage arrives at the load end at t = td, and its arrival leads to the generation of V! (z, t) = rLVi (z, t) propagating toward the source. This new voltage V! (z, t) arrives at the source end and leads to the generation of (z' t)' and this process continues back and forth indefinitely. Once constructed, a bounce diagram can be used conveniently to determine the voltage distribution along the transmission line at any given time, as well as the variation of voltage with time at any given position. Suppose we wish to find the voltage distribution V(z, to) along the line at t = to, chosen arbitrarily to be 2td < to < 3td. To deter1nine V (z, to), we mark to on the time axis of the bounce diagram and draw a horizontal line from t0 , intersecting the sloping line marked Vi (z, t) at point P0 , as shown in Figure 2.9. Note that all sloping lines below the point Po are irrelevant for V(z, to), since they correspond to later times. If we now draw a vertical dashed line through Po, we find that it intersects the z axis at zo. At time t =to, all points along the line have received voltages Vi (z, t) and V! (z, t). However, only points to the left of zo have yet received the voltage Vj: (z, t), so a discontinuity exists in the voltage distribution at z = zo. In other words, we have
vt
Ojf'(
v Z,
)
to =
Vi(z, t)(l + rL + rsrL) + V 1 (z, t) ( 1 + r L)
z < zo z > zo
Alternatively, we may wish to deter1nine the variation of voltage as a function of time at a fixed position, say Za· To deter1nine V(za, t), we simply look at the intersection points with the sloping lines of the vertical line passing through Za, as shown in Figure 2.9. Horizontal lines drawn from these intersection points, crossing the time axis at ti, t2, t3, t4, ... , are the time instants at which each of the new voltages Vi(z, t), Vl (z, t), Vj: (z, t ), V2 (z, t), ... , arrive at Z = Za and abruptly change the total voltage at that point. Thus, the time variation of the voltage at z = Za, namely V(za, t) is given as
0
0
Vi (0, 0) = 
Zo Vo Rs2 + Zo
Note that the negative sign in the sourceend boundary condition °Vs= Rs2!Ps is due to the defined direction of !Ps, with positive current coming out of the terminal of positive voltage. At t = td, the new voltage disturbance reaches the open end of the line, where a reflected voltage °V} (z, t) with amplitude °V} (l, td) =Vi (l, td) is produced, traveling toward the (z, t) source end. At t = 2td, °V} (z, t) arrives at the source end, and a reflected voltage with amplitude (0, 2td) = rs°V} (0, 2td) = (Rs2  Zo)°V} (0, 2td) I (Rs2 + Zo) is launched back toward the load. This process continues until a new steadystate condition is reached, when the line voltage eventually becomes zero. Figures 2.18b, c, and d show the bounce diagrams for three different values of Rs2, namely Zo/3, Zo, and 3Zo. Figures 2.18e, f, and g show the variation of the source and loadend voltages for all three cases as a function of time t.
Vi
Vi
The Transmission Line as a Linear TimeInvariant System. In some cases it is useful to think of the transmission line as a linear timeinvariant system, with a defined input and output. For this purpose, the input °Vin (t) can be defined as the voltage or current at the input of the line, while the output °V out(t) can be a voltage or current somewhere else on the line for example, the load voltage °VL (t) as indicated in Figure 2.19. Note that since the fundamental differential equations ((2.1) and (2.2)) that govern the transmission line voltage and current are linear, the relationship between °Vin (t) and
Sec. 2.4
Transient Response of Transmission Lines with Resistive Terminations R
"
((,
sl
J's
t=O
1
z=0
53
l
Z=
i======~0
+
2



(a)
rs2 =0.S
rL=l
z
3V0 /4 1 3V0 !4
rs2 = +O.S
rL=l
z
I
I I
........ ........
t Source end
Std .... .... ........
t Source end
Load end
t Source end
Load end
0.062S
3 I
0.12S 4
1
s
6 7
1

o.s
1
4
.
.
s
6
7
2
3
4
s
 7 0.12S
0.S
°V L(t)!Vo
t/td .
titd
6
°Vs(t)!Vo
3
Load end
0.2S
(e) R 82 = Zof3
2
Std
°V L(t)!Vo
0.2S
1
3td
(d) R 82 =3Z0
1 °Vs(t)!Vo

I I I I I I
V0 /16
(b) R 82 = Z 0/3
1
1
3td
I
2
I I
V018
1
3V0 !16 1
1
I td
V0 /4
I I
I
z
V0 !4 :
I td
I I I
3V0 /8
V0 !2 :
td
1
3V018
rL=l
1
2 3
t/td 4
s
6
7
(f) Rsz = Zo

°Vs(t)!Vo 1 0.7S
0.37S
0.187S
2
3
4
s
6
o.s 0.2S
t/td
L.
1
1
°VL(t)!Vo
7
1
2
3
4
t/td
0.12S
s

.
.
6
7
(g) Rsz = 3Zo
Figure 2.18 Discharging of a charged line. (a) A charged line connected to a resistor R s2 for Example 2.5. (b), (c), and (d) Bounce diagrams for R s2 = 'Zo/3, R s2 = 'Zo, and Rs2 = 3'Zo. (e), (f), and (g) Normalized source and loadend voltages as a function oft for the three different cases.
54
Transient Response of Transmission Lines
Chap.2
+ °Vout(t) = °VL(t)
°Vin(t) = °V(O, t) = °fs(t)
(a)
Transmission line system


°Vout(t) = °V(l, t) = °VL(t)
(b) Figure 2.19 The transmission line as a linear timeinvariant system. The input °Vin (t) to the system can be defined as the line input voltage °Vs (t), whereas the output °Vout(t) could be any voltage or current of interest anywhere on the line, such as the load voltage °VL(t).
is linear. In other words, for two different input signals °Vini and °Vm2 , which individually produce two different output signals °Vouti and °V out2 , the response due to a linear superposition of the two inputs, °Vin1+2 = ai °Vini + a2°Vin2 , is
°Vout(t)
Since the physical properties of the transmission line (L, C, td, Zo) do not change with time, the relationship between °Vin (t) and °Vout(t) is also timeinvariant. In other words, if the output due to an input °Vin1 (t) is °V out1 (t), then the output due to a timeshifted version of the input, namely °Vini (t  r), is simply a similarly shifted version of the output, namely °Vout1 (t  r). As with any linear timeinvariant system, the response of a transmission line to any arbitrary excitation signal can be deter1nined from its response to an impulse excitation. In the transmission line context, an input pulse can be considered to be an impulse if its duration is much shorter than any other time constant in the system or the oneway travel time td in the case of lossless lines with resistive ter1ninations. In most applications, however, it is necessary to determine the response of the line to step inputs, as were illustrated in Examples 2.3 through 2.5. For this purpose, it is certainly easier to determine the step response directly rather than to determine the pulse (or impulse) response first and then use it to deter1nine the step response. Treatment of a transmission line as a linear timeinvariant system can sometimes be useful in determining its response to pulse inputs, as illustrated in Example 2.6. Example 2.6: The transmission line as a linear timeinvariant system. Consider the transmission line system of Figure 2.20a, the step response of which was determined in Example 2.4. Determine the load voltage °V L (t) for an input excitation in the form of a single pulse of amplitude Vo and duration 0.5td (i.e., °Vm(t) = Vo[u(t)  u(t  td/2)]).
Sec. 2.4
Transient Response of Transmission Lines with Resistive Terminations
55
+ °Vout(t) = °VL(t)
RL = 9Zo
(a) °Vout1(t)!Vo
0.8 °Vin1(t)
Vo
0.6
+   
>0.63 0.702 " 0.45
0.4 0.2 ·• t/td 1
2
3
.•
t/td
1
°Vin2 (t)
1
1
2
3
4
• •
.•
5
6
°Vout2(t)!Vo
2
3
t/td
t/td
..
1
2
3
.
4
5
6
0.2 ..

0.4
"
0.6
"
0.8
0.45 0.63
0.702 >"
Figure 2.20 Pulse response of a transmission line. (a) The line configuration and source and load impedances. (b) The input pulse of amplitude Vo and duration td /2 is represented as a superposition of a °Vint and °Vin2 . (c) The response is computed as a superposition of the responses due to the individual step inputs. Note that °Vouti (t) was already computed in Example 2.4.
56
Transient Response of Transmission Lines
Chap.2
Solution: For the circuit of Figure 2.20a, it is convenient to define the input signal to be the excitation (source) voltage and the output as the load voltage, as indicated. As shown in Figure 2.20b, the input pulse of amplitude Vo and duration td /2 can be viewed as a superposition of two different input signals: a step input starting at t = 0, namely °Vini (t) = Vou (t), and a shifted negative step input, namely, °Vm2 (t) = Vou (t  td/2), where u(~) is the unit step function, u(~) = 1 for ~ > 0, and u(~) = 0 for ~ < 0. The output °Vout1 (t) due to the input °Vini (t) was determined in Example 2.4 and is plotted in the top panel of Figure 2.20c. Since the transmission line is a linear timeinvariant system, the response °Vout2 (t) due to the input °Vin2 (t) is simply a flippedover and shifted version of °Vouti (t), as shown in the middle panel of Figure 2.20c. The bottom panel of Figure 2.20c shows the superposition of the two responses, which is the desired pulse response. Note that our solution of this problem using the linear timeinvariant system treatment was simplified by the fact that the step response of the system was already in hand from Example 2.4.
2.4.2 Junctions between Transmission Lines We have seen that reflections from terminations at the source and load ends of transmission lines lead to ringing and other effects. Reflections also occur at discontinuities at the interfaces between transmission lines, connected either in cascade or in parallel, as shown, for example, in Figures 2.21a and 2.22a and as often encountered in practice. For example, consider the case of the two lossless transmission lines A and B (with characteristic impedances ZoA and ZoB) connected in tandem (i.e., in series) as shown in Figure 2.21a. Assume a voltage disturbance of amplitude °V{A (z, t) (with an associated current of !P iA (z, t) = °V{A/ZoA) to arrive at the junction between lines A and B (located at z = lj) from line A at t =to. A voltage °VIA (z, t) of amplitude °VIA (lj, to) = r AB ViA (lj, to) reflects back to line A, where the reflection coefficient r AB is given by r AB = (ZoB  ZoA) I (ZoB + ZoA)' since line B presents a load impedance of ZoB to line A. In addition, a voltage °V{B (z, t) of amplitude °V{B (lj, to) = ~AB°V{A (lj, to) is transmitted into line B, where ~AB is called the transmission coefficient, defined as the ratio of the transmitted voltage to the incident voltage, that is, ~AB °V{B (lj, to) /°V{A (lj, to). To find ~AB, we apply the boundary condition at the junction, which states that the total voltages on the left and right sides of the junction must be equal:
=
yielding ~AB = 1 + r AB = 2ZoB / (ZoB + ZoA). The transmission coefficient ~AB represents the fraction of the incident voltage that is transferred from line A to line B. Note that depending on the value of the reflection coefficient, the transmitted voltage can actually be larger in amplitude than the incident voltage, so that we may have ~AB > 1 (in those cases when r AB > 0). Similarly, if a voltage disturbance °V IB (z, t) of amplitude °V IB (lj, ti) (produced by reflection when °V{B (z, t) reaches the end of line B) arrives at the same junction between A and B from line B at t = t 1, a voltage °Vj:B (z, t) of amplitude
Sec. 2.4
Transient Response of Transmission Lines w ith Resistive Terminations
57
vtB (lj, t1) =
rBA°V lB (lj, t1) reflects back to line B and a voltage °V2A (lj, t1) = ~BA°V lB (lj, t1) is transmitted into line A, where rBA and ~BA are given by rBA =
ZoA ZoB ZoA +ZoB
= rAB
2ZoA ~BA = 1 + fBA =    ZoA + ZoB
ZoAar
~AB
ZoB
Example 2.7: Cascaded transmission lines. Consider the transmission line system shown in Figure 2.2la, where a step voltage source of amplitude l.S V and source resistance son excites two cascaded lossless transmission lines (A and B) of characteristic impedances son and 2Sn and lengths S cm and 2 cm, respectively. The speed of propagation in each line is 10 cmns 1 .24 The second line (B) is terminated with a load impedance of lOOn at the other end. Draw the bounce diagram and sketch the voltages °V8 (t) and °VL(t) as functions of time. Solution: With respect to Figure 2.2la, we note that lj = S cm and l = 7 cm. At t = o+, voltage °V{A (z, t) of amplitude °V{A (0, 0) = 0. 7S V is launched on line A. This voltage reaches the junction between the two lines at t = td 1 = S cm/(10 cm(ns) 1) = SOO ps, when a reflected voltage °V lA (z, t) of amplitude °V lA (lj, td1) = r AB°V{A (lj, td1) = (l/3)(0.7S) = 0.2S V, and a transmitted voltage °V{B (z, t) of amplitude °V{B (lj, td1) = ~AB °V{A (lj, td1) = (2/3) (0. 7S) = O.S V are created. The reflected disturbance arrives at the source end at t = 2td1 = 1 ns and is absorbed completely since rs = 0. The transmitted wave reaches the load at t = td1 + td2 = 700 ps, and a reflected voltage °V lB (z, t) of amplitude °V lB (l, 700 ps) = rL°V{B (l, 700 ps) = (0.6) (0.S) V = 0.3 V is launched toward the source. This reflected disturbance arrives at the junction from line B at t = td1+2td2 = 900 ps, and reflected and transmitted voltages ViB(z,t) and °V2A(z,t) of amplitudes respectively ViB (lj, 900 ps) = rBA°V lB (lj, 900 ps) = (1/3)(0.3) = 0.1 V and °V2A (lj, 900 ps) = ~BA°V lB (lj, 900 ps) = ( 4/3)(0.3) = 0.4 V are launched respectively toward the load and the source. The continuation of this process can be followed by means of a bounce diagram, as shown in Figure 2.2lb. The source and loadend voltages are plotted as a function of t in Figure 2.21 c.
Example 2.8: Three parallel transmission lines. Consider three identical lossless transmission lines, each with characteristic impedance Zo and oneway time delay td, connected in parallel at a common junction as shown in Figure 2.22a. The main line is excited at t = 0 by a step voltage source of amplitude Vo and a source resistance of Rs = Zo. Find and sketch the voltages °V8 (t), °VLl (t), and °VL2(t) for the following two cases: (a) RL1 = RL2 = Zo and (b) RL1 = Zo and RL2 = 00.
Solution: For any voltage disturbance °Vi (z, t) of amplitude Vi arriving at the junction from any one of the three lines, the parallel combination of the characteristic impedances of
24 Note from Table 2.1 that 10 cm(ns) 1 is approximately the speed of propagation in alumina (Al203), a ceramic commonly used for electronics packaging.
Transient Response of Transmission Lines
58
Chap.2
+ r AB= 1/3 °V1B •
r
Vo= 1.5 V

A. y

~
~ II
' '
+ °Vs 
" ~
son, tct1 = soo ps Line A
~

I
' '

~ ' '
' '
I 2sn, td2 = 200 ps + I LineB I °VL I~ rBA =+1/3 ~ ~

•
RL = 1oon ~

z = /.J = 5 cm
z=O


z = 1=7 cm
(a)
rBA =+1/3
r so
c_ +
rL=0.6
I
I I I I I
°VlA = 0.75 VI
I I I
+
z
°V1B = 0.5 V I I I I
1
1 ns +
°VzA = 0.4 V
°VzB = 0.1
0.7 ns
I I I
v:
1.2 1 0.8 0.6 0.4 0.2
+
I I I I
1.1 ns
1.8 ns +
1.5 ns
I I I
°V4B = 0.004 VI
°V4A = 0.016 V
I I I
: 1.9 ns
2.2 ns
0.9 v 0.75V o.sv t (ns)
1.2
1.5
1
2
°VL(t) (V)
vss = 1 v
0.8 0.6
I I 1
   ~ I
1         
°V3B = 0.02 V1
°V3A = 0.08 V
vss = 1 v
0.5
I I
1.4 ns
°V8 (t) (V)
 ;..=..::..::..;ir
0.8 v
0.96 v
0.4 0.2
t (ns)
0.5
1
1.5
(c)
I
t
(b) Figure 2.21 Cascaded transmission lines. (a) Circuit diagram for Example 2.7. (b) Bounce diagram. (c) Sourceend voltage °Vs and loadend voltage °VL as a function oft.
2
Transient Response of Transmission Lines with Resistive Terminations
Sec. 2.4
Line 1 Zo
+
Main line
Vg
r s =O
Zo
r.J = 113 _::::>
I I I I I I
°Vg(t)!V0 0.5 ~0.33
1/3 
td
~
0.33
1/3
1
r s =O
2
3
4
5
6
(b) RL1 = RL2 = Zo
r. = 113 r L2 = 1 J _:::>1 r.  1/3
:c..J
°Vg(t)!Vo 519 0.5 5 113· 4
°·
1
13!21
1
I
V0 !6
I
1
2td
I
2
I
3
I
I
I
4
5
6
°VL1(t)IV0
2Vol9
V019
4td V0!9
I I I I I I 1
I I I
519

0.5  11/3 13/27
t/td
.
1
2
.•
'
.•
'
3
4
5
6
°VL2(t)IV0 2/3 0.5          
 
14/27

  


419 2V0 /27
V0 /27
.
1
2
t/td
.•
'
.•
'
3
4
5
6
(c) RLl = Z 0 and RL2 = oo Figure 2.22 Three transmission lines connected in parallel at a common junction. (a) Circuit diagram for Example 2.8. (b) Bounce diagram and the variation of the °Vs and °VLI = °VL2 = °VL voltages as a function oft for RLI = RL2 = Zo. (c) Bounce diagram and the variation of the °Vs, °VLI, and °VL2 voltages as a function of t for RLI = Zo and RL2 = 00.
59
60
Transient Response of Transmission Lines
Chap.2
the other two lines acts as an equivalent load impedance at the junction. Once a voltage is incident at the junction, a voltage of amplitude Vr = rj Vi = ('Zo/2  'Zo) Vi/ (Zo/2 + Zo) = Vi/3 is reflected and a voltage of amplitude Vt = ;?lj Vi = (1 + rj) Vi = 2Vi/3 is transmitted to both of the other lines. Note that the reflection coefficient at the junction is denoted simply as rj, because all of the transmission lines are identical and the reflection coefficient at the junction is thus the same regardless of which line the wave is incident from. For the circuit shown in Figure 2.22a, the bounce diagram and the sketches of voltages °V8 , °VLt, and °VL2 for both cases are shown in Figures 2.22b and c. When RLi = RL2 = Zo, only one of lines 1 and 2 is shown in the bounce diagram, since what happens on the other is identical, as seen in Figure 2.22b. When RLi = Zo and RL2 = oo, only line 2 is shown in the bounce diagram in Figure 2.22c, since no reflection occurs on line 1.
2.5 TRANSIENT RESPONSE OF TRANSMISSION LINES WITH REACTIVE TERMINATIONS Up to now, we have studied only transmission lines with resistive terminations. In this section, we consider reactive loads. Reactive loads are encountered quite often in practice; in highspeed bus designs, for example, capacitive loading by backplanes (consisting of plugin cards having printed circuit board traces and connectors) often becomes the bottleneck when highspeed CPUs communicate with shared resources on the bus. Inductive loading due to bonding wire inductances is also important in many integratedcircuit packaging technologies. Packaging pins, vias between two wiring levels, and variations in line width can often be modeled as capacitive and inductive discontinuities. The capacitances and inductances of these various packaging components can range between 0.5 and 4 pF and between 0.1 and 35 nH, respectively. 25 For transmission lines with resistive loads, the reflected and transmitted voltages and currents have the same temporal shape as the incident ones and do not change their shape as a function of time. For a step excitation, for example, the reflected voltage produced by a resistive ter1nination remains constant in time, as discussed in preceding sections. However, in the case of capacitive or inductive terminations, the reflected and transmitted voltages and currents do not have the same temporal shape as the incident ones. The terminal boundary condition at the reactive termination must now be expressed as a differential equation whose general solution can be exceedingly complicated, whether the solutions are carried out in the time domain or by the use of Laplace transfor1nation. We illustrate the basic principles by considering a line terminated at an inductance, as shown in Figure 2.23. (z, t) (taken in Figures 2.23b, c as a constant When the traveling disturbance voltage Vo) and its associated current !Pi(z, t) (taken in Figures 2.23b, c as a constant current Io = Vo/Zo) first reach the end of the line, the inductive load acts as an open circuit, since its current cannot change instantaneously. Thus the disturbance is initially reflected in the same manner as an open circuit, the ter1ninal voltage jumping to 2 Vo
Vi
25 See
Chapter 6 of H. B. Bakoglu, Circuits, Interconnections, and Packaging for VLSI, AddisonWesley, Reading, Massachusetts, 1990.
Sec. 2.5
Transient Response of Transmission Lines w ith Reactive Terminations
c;t=Vo
61
z=l
Lossless Transmission Line
L
Zo
(a)
(b) V 0 + °V1 (z, t) "
1
~~~~~~~~~~ii~'~'~,~========::::::::::::=======~:I
 V1 (z, t)        
Io
I
I 0 + !fi 1 (z, ~ t)~1

~~/



(c) Figure 2.23 Reflection from a purely inductive load. (a) The source is assumed to be matched (i.e., no reflections back from the source) and to supply a constant voltage V0 . The distributions of the voltage and current are shown at two different times: (b) immediately after reflection when the inductor behaves like an open circuit and (c) later in time when the inductor behaves as a short circuit.
and the terminal current being zero, for that instant (Figure 2.23b). However, since a voltage of 2 Vo now exits across the inductor, its current builds up, until it is practically equivalent to a short circuit. At steady state, when the voltage across the inductance reduces to zero, the current through it becomes 2Io (Figure 2.23c), similar to the case of a shortcircuit termination (see Example 2.2). Between an initial open circuit and an eventual short circuit, the voltage across the inductive load goes through all intermediate values, and the reflected voltage
62
Transient Response of Transmission Lines
Chap.2
changes accordingly. To determine the analytical expression describing the variation of the voltage across the inductance, we need to simultaneously solve the transmission line equations (or the general solutions dictated by them, namely (2.8) and (2.9)) along with the differential equation describing the boundary condition imposed by the inductive load as VL(t) = Ld!P_L_(t_) dt
For a general incident voltage Vi (z, t), the load voltage V L(t) and current !PL(t) are given by v L(t) = vt cz, t)
+ v1 cz, t)
+
_
Vi(Z,t) !JL(t) =!JI (l,t) +!JI (l,t) =  Zo
V!(l,t)
Zo
where the location of the load (i.e., the inductive termination) is assumed to be at z = l. We thus have vtcz, t)
Zo or dV!(Z, t) dt
+
Zovcz t) = dVtcz, t) _ zov+cz t) L I ' dt L I '
which is a differential equation with V!(l, t) as the dependent variable, since Vi(Z, t) is presumably known, it being the incident voltage disturbance arriving from the source end of the line. The righthand side is therefore a known function of time, and the equation is simply a firstorder differential equation with constant coefficients. Note that we assume the source to be either far enough away or matched, so that the voltage V! (z, t) does not reach the source end and generate a reflected voltage Vi (z, t) before V!(l, t) reaches its ''steadystate'' value (Vo in the case when Vi(z, t) =Vo as shown in Figure 2.23c). To study the simplest case, let us consider an incident voltage with a constant amplitude Vo (i.e., Vi(Z, t) = Vo) launched by a step source reaches the inductor at t = 0. The above differential equation then simplifies to dV!(Z,t) dt
+
Zovcz t)=ZoVo L I ' L
The solution of this firstorder differential equation
26 The
26 is
solution can be found via Laplace transformation or as a superposition of the homogeneous solution and the particular solution; the validity of the solution can be shown by simply substituting it into the differential equation.
Sec. 2.5
Transient Response of Transmission Lines w ith Reactive Terminations
63
where the coefficient K needs to be determined by the known initial conditions as they relate to V!(l, t). At the instant of the arrival of the voltage disturbance (t = 0), the current through the inductance !PL(t = 0) = 0, and we thus have the incident voltage fully reflected, or V! (l, t = 0) = Vo. Thus, we must have K = 2Vo. The solution for the reflected voltage is then
which varies from its initial value of Vo to an eventual value of Vo, as shown in Figure 2.23c. Example 2.9: Lossy capacitive load. Consider the transmission line system shown in Figure 2.24a where a step voltage source of amplitude Vo and source resistance Rs = Zo excites a lossless transmission line of characteristic impedance Zo and oneway time delay td connected to a load consisting of a parallel combination of RL and CL. Find and sketch the source and loadend voltages as a function of time. Solution: At t = 0, an incident voltage Vi (z, t) of amplitude Vi (0, 0) = Vo/2 is launched at the source end of the line. This disturbance reaches the capacitive load at t = td, when a voltage V} (z, t) of initial amplitude V} (l, td) reflects toward the source. For t > td, the total load voltage V L(t) and current !PL(t) are given by VL(t) = Vi(Z,t) a.
a.+
J'L(t) = "'1 (l, t)
+ V}(l,t) a.
+ "'1 (l, t)
=
Vi(Z,t)
Zo
V}(l,t)
Zo
where Vi (l, t) =Vi (0, 0) = Vo/2. These two equations are related by the boundary condition imposed by the load; that is, a. ( )
J'L t =
V L(t) RL
+
C dVL(t) Ldt
Substituting the first two equations into the third equation yields
which is a firstorder differential equation for V} (t). Note that in deriving it, we have used the fact that the incident voltage is constant in time, so that dVi (l, t) / dt = 0. The solution of this firstorder differential equation can be found by noting that Vi (l, t) = Vo/2 and by writing the general solution as
Transient Response of Transmission Lines
64
~
.
+ °Vs 
0 r
"'
....
Chap.2
~
+
~
°VL
Zo, td
•
•
RL 
~
•
z=l
z =O (a)
                                RLVol(RL + Zo)
V01 2     
Time constant r = CLRLZ0 /(RL + Z 0)
t

           
~:.,::::..;;;..;
Time constant r = CLRLZ0 /(RL + Z 0)
t
(b) Figure 2.24 Step response of a capacitively loaded line. (a) Circuit diagram. (b) Time variation of the source and loadend voltages.
in which case the constants Ki and K2 can be determined by using the initial and final conditions. Note that the reflected voltage must vary exponentially from  Vi (l, t) at t = td to (RL Zo)°Vi(Z,t)/(RL + Zo) fort > oo. We thus have
RL  Zo v 1 (l, t) = v 1 (l, t) RL + Zo
Oj('
Ojf'+
2RL
RL + Zo
e [(RL +Zo)/(RLZoCL)](ttd)
Sec. 2.5
Transient Response of Transmission Lines with Reactive Terminations
65
which is valid for t > td. This behavior can be understood as follows: When the incident voltage reaches the capacitive load at t = td, the capacitor CL is initially uncharged and acts like a short circuit, resulting in V} (l, td) =  Vt (l, td) =  Vo/2. However, at steady state the capacitor is fully charged and acts like an open circuit, resulting in V} (l, oo) = (RL  Zo)Vt (l, oo)/(RL + Zo) = (RL  Zo)Vo/[2(RL + Zo)], as expected. Note also that the time constant of the exponential variation is r = RLZoCL/ (RL + Zo) = RTh CL, where RTh is the Thevenin equivalent resistance,27 as seen from the terminals of the capacitor. Substituting V} (l, t) into V L(t) yields
valid fort > td. When the reflected voltage reaches the source end at t = 2td, it is completely absorbed, since the source end of the line is matched (i.e., Rs = Zo). The voltage at the source end is given by Vs(t) =Vt (0, t) = Vo/2 fort < 2td, and
is valid fort > 2td. Sketches of Vs(t) and V L(t) are shown in Figure 2.24b.
Example 2.10: A lumped series inductor between two transmission lines. Microstrip transmission lines on printed circuit boards are often connected together with bonding wires, which are inherently inductive. In this example, we consider a typical model for such a connection, namely a lumped series inductor between two different transmission lines. The measurement of the bondingwire inductance is considered later in Example 2.13. Two microstrip transmission lines having equal line parameters of L = 4 nH(cm) 1 and C = 1.6 pF(cm) 1 and lengths lS cm and 10 cm are connected by a wire represented by a series lumped inductance of Lw = S nH, shown in Figure 2.2Sa. The end of the shorter line is matched with a son load, and the circuit is excited at t = 0 by a unit step voltage source of Rs =son. Find and sketch the variations with time of the source and loadend voltages. Assume lossless lines. Solution: Using the given line parameters Land C, the characteristic impedance and the phase velocity of the microstrip lines can be calculated as
Zo=
4 x 109 =Son 12 1.6 x 10
L
c
and vp
=
1 ~
=
1
J4 x
109 x 1.6 x 1012
9
= 12.S x 10 cms
1
Therefore, the oneway delay times of the two lines are td 1 = lS cm/(12.S cm(ns)1) = 1.2 ns and td2 = 10 cm/(12.S cm(ns) 1) = 0.8 ns, respectively, as indicated in Figure 2.2Sa.
27
In this case being simply equal to the parallel combination of the load resistance RL and the characteristic
impedance
Zo of the line.
66
Transient Response of Transmission Lines
tdl =
Lw=SnH + °Vind  A
1.2 ns
giind
son 12.S
(a) V0 =1V
+ °V
cm(ns) 1
A
td2 =
0.8 ns
+
son 12.S
Chap.2
°V 1 cm(ns)L
z = l2
z=O l1 = 1S cm, l 2 = 2S cm
1......_ _____ _ r = Lw/(2Z0) = SO ps
o.st I
(b)
2td1 =2.4 ns °VL(t) (V) 1
o.s _.___        ~r= SO ps t
Figure 2.25 A lumped series inductor between two different transmission lines. (a) Circuit diagram. (b) Time variation of source and loadend voltages for Example 2.10.
Vi
As in the previous example, at t = 0, an incident voltage (z, t) of amplitude (0, 0) = 0.5 V is launched from the source end of the line. Note that the amplitude of this incident voltage remains constant in time, as long as it is supplied by the source. When this disturbance arrives at the junction at t = td 1 = 1.2 ns, the uncharged inductor initially acts like an open circuit (i.e., it resists the flow of current), producing a reflected voltage V! (z, t) of initial amplitude V! (11, td1) =Vi (11, td1) = 0.5 V, the same as one would have if the two lines were not connected. Current flows into the second line through the inductor as the inductor charges exponentially and eventually behaves like a short circuit at steady state. The following equations apply for t > td 1 :
Vi
oir+ v
1 (11, t)
+
oirv
oir
1 (11, t) = v inct(t)
+
d !j ind (t) v A(t) = Lw dt
oir
({, + ZoJinct(t)
Sec. 2.5
Transient Response of Transmission Lines with Reactive Terminations
67
where °Vf (11, t) = O.S V and °V A(t) is the voltage at position A, at which the impedance seen looking toward the load is Zo =son. We also have
+ jind(t) = .9>1 (11, t)
+
°Vf (11,t) jl (11, t) =   Zo
°V}(l1,t) Zo
where s>inct(t) and °Vinct(t) are, respectively, the current through and the voltage across the inductor, as defined in Figure 2.2Sa. Substituting the second equation into the first and using d°Vf (11, t) I dt = o yield
which is a firstorder differential equation for °V} (11, t). The solution of this equation is 10 °V} (11, t) = °Vf (11, fdl)e(2Zo/Lw)(ttd1) = O.Se2x 10 (tl.2x 10 9) V which is valid fort > td 1 = 1.2 ns. Note that we have used the fact that °V} (11, td 1) = O.S and that the reflected voltage varies exponentially from °V} (11, td1) = °Vf (11, td1) at t = td1 (when the inductor initially behaves like an open circuit) to zero at t > oo (when the fully energized inductor eventually behaves like a short circuit). The time constant of the exponential variation is r = Lw/RTh =SO ps, where RTh = 2Zo = lOOn is the Thevenin equivalent resistance as seen from the terminals of the inductor. To satisfy the boundary condition at the junction, and noting that the inductor Lw is a lumped element, the current on both sides of the inductor must be the same. Thus, the inductor current s>inct(t) can be written as
which is valid for t>tdl· At t=td1, a transmitted voltage V1(z,t)=Zo.9>1(z,t) is launched at position A on the second line, where s>1(z,t) = s>inct(t). Therefore, the (z' t) can be written as transmitted voltage
v1
V1(z,t) = Zo.9>inct(t) = °Vf (l1,td1)[l  e td1. The voltage V1 (z, t) arrives at the load end at t = td1 + td2 = 2 ns, where it is completely absorbed, since RL = Zo =son. The load voltage is given by °VL(t) = °V1(l2,t) = °Vf (l1,td1)[l  e(2Zo/Lw)(t(td1+td2))] = O.S[l _ e2x1010 (t2x10 9)] y valid fort > (td1 + td2) = 2 ns. Note that °Vf (11, td1) = O.S V fort > td1, since the incident voltage remains constant unless the source changes. Similarly, the sourceend voltage is (0, t) = o.s v for t < 2td1 = 2.4 ns and given by °Vs (t) =
vt
1 °Vs(t) = °Vf (0, t)(l + e(2Zo/Lw)(t2tdl)] = 0.S[l + e2xl0 0(t2.4x10 9)] y fort > 2td 1 = 2.4 ns. Sketches of the time variations of source and loadend voltages are shown in Figure 2.2Sb.
68
Transient Response of Transmission Lines
Chap.2
In the preceding two examples we used timedomain methods to determine the step response of transmission lines with terminations or discontinuities involving reactive elements. The basis for our analyses was the simultaneous solution of (2.10), describing the voltage and current of the reactive element in terms of the transmission line incident and reflected voltage and current waves, together with the differential equations that describe the ter1ninal voltagecurrent relationship of the reactive load. These timedomain solutions were tractable partly because the excitation voltage was a simple step function and also because the reactive discontinuities that we analyzed involved only one energy storage element (i.e., a single capacitor or inductor). When the input voltage function is more complicated, or when the reactive discontinuity involves more than one energy storage element, it is often easier to use Laplace transfor1n methods to determine the response of the line. We demonstrate the use of the Laplace transform method in Example 2.11. Example 2.11: Reflections due to inductance of resistor leads. Consider the transmission line system shown in Figure 2.26, where the source voltage amplitude increases linearly from zero to Vo over a time28 of tr. The output resistance of the source is Rs = Zo, while the transmission line having a characteristic impedance Zo and a oneway time delay td is terminated in a reactive load consisting of a series combination of RL and LL. Find an expression for the reflected voltage at the load [i.e., °V! (l, t)] and determine its maximum value for RL = Zo.
Solution:
From Figure 2.26 we note that, starting at t = 0, an incident voltage is launched at the source end of the line, given by
+ Vo °V 1 (0, t) = [tu(t)  (t  tr )u(t  tr)] 2tr
R s= Z o z = O
Zo, td
+
z =l
°Vi (l, t) r = LJ(2Z0)
+ LL =?
0
Vg
0
°VL
r = Lif(2Z0)
RL = Zo

~
?

(a)

/
td
td + t,
t
(b)
Figure 2.26 Reflections due to inductance of resistor leads. (a) Circuit diagram of Example 2.11. (b) Time variation of the reflected voltage °Vi (l, t) for the case RL = Zo.
28 Note
that t, is not exactly the rise time discussed in Section 1.1, which was defined as the time required for the signal to change from 10% to 90% of its final value.
Sec. 2.5
Transient Response of Transmission Lines w ith Reactive Terminations
69
where u ( ·) is the unit step function. The Laplace transform of this voltage waveform is
+ Vo 1  etrs °V1 (s) = 2 2tr S This incident voltage propagates to the load end, and no reflected voltage exists until it arrives there at t = td. Fort> td, the total load voltage and current °VL(t) and !PL(t) are given by °VL(t) = Vi(Z,t) a.
J'L(t) =
a. + "'1
(l, t)
+ °V}(l,t) a.
+ "'1 (l, t)
=
Vi(Z,t)
Zo
°V}(l,t) Zo
and are related by the boundary condition imposed by the load,
Substituting the first two equations into the third equation yields
d°V} (l, t) Oj('d°Vi (l, t) Ojf' + LL + (RL +Zo) v 1 (l,t) =LL + (RL Zo) v 1 (l,t) dt dt Note that unlike the case of Examples 2.9 and 2.10 the derivative of the incident voltage is not zero, since Vi (l, t) is not constant in time during the time interval td < t < (td +tr). To solve this differential equation for the reflected voltage °V} (l, t) using the given functional form of°Vi(Z,t), we can take its Laplace transform,

+
(sLL + RL + Zo)°V 1 (s) = (sLL + RL  Zo)°V 1 (s) where
V~ (s)
is the Laplace transform of + °V 1 (s ), we then have
Vo 1 2tr
°V~(l, t).
etrs
s2
s + (RL  Zo)/LL
Vo 1 
s + (RL + Zo)/LL
2tr
which can be expanded into its partial fractions as
where the coefficients are and
Using the previously noted form of
etrs
s2
70
Transient Response of Transmission Lines
Chap.2
Taking the inverse Laplace transform29 yields
cy(l t) =Vo ZoLL {[1 e td, or t' > 0. A practical case of interest is that in which RL = Zo. When a microstrip is terminated at a matched load resistance to avoid reflections, the nonzero inductance of the resistor leads may nevertheless produce reflections. To determine the maximum reflection voltage due to the inductance of the resistor leads, we substitute RL = Zo in the solution for °V} (l, t) to find
The time variation of °V} (l, t) is plotted in Figure 2.26b, showing that the reflected voltage rises and falls exponentially with a time constant of L/ (2Zo). The maximum reflected voltage occurs at t = td +tr and is given by
[°Vj(l,t)Jmax
=Vo~~ tr
[l e(2Zo/LL)tr]
0
Note that in practice the maximum reflected voltage can easily be measured for example, by using a timedomain reflectometer (see Section 2.6), from which the value of the inductance LL of the resistor leads can be calculated, since Zo and tr are known in most cases.
2.6 TIMEDOMAIN REFLECTOMETRY In practice, it is often necessary to make a number of measurements on a given transmission line system to characterize its transient response. The quantities that need to be measured include the nature (capacitive, inductive, or resistive) of the load termination, the characteristic impedance of the line, the maximum voltage level at which the line can be used, and others of a more specialized character. A timedomain reflectometer30 (commonly abbreviated as TDR) is an instrument which is used to test, characterize, and 29 We
use the following Laplace transform pairs:
ea(tb)u(t 
b)  s +a bs
e (t  b )u (t  b) s2
where a and b are constants. 30 B. M. Oliver, Time domain reflectometry, HewlettPackard J., 15(6), February 1964.
TimeDomain Reflectometry
Sec. 2.6
71
Oscilloscope

r1 I I I
Step source 

I I I I 1
+ ~
Load
I
ZL
1
I
________ J


________ J
Figure 2.27 Timedomain reflectometry. Essential components of a typical TDR system.
model a system involving transmission lines and their accessories. In general, it consists of a veryfastrisetime (typically less than 50 ps) step pulse source and a display oscilloscope in a system that operates like a closedloop radar, as shown in Figure 2.27. The source produces an incident step voltage, which travels down the transmission line under investigation, and the incident and the reflected voltages at a particular point (typically the source end) on the line are monitored by the display oscilloscope using a highimpedance probe. The output impedance of the step source is typically well matched to the nominal characteristic impedance of the line to eliminate reflections from the source end. The most common use of timedomain reflectometry involves the measurement of the characteristics of an unknown load termination or a discontinuity on the line. The former application is illustrated in Example 2.12 for resistive loads. A discontinuity on a transmission line could, for example, be a point of breakage on a buried coaxial line, an unwanted parasitic capacitance on an interconnect, or the inductance of a bonding wire between two interconnects. The latter case is illustrated in Example 2.13.
Example 2.12: TDR displays for resistive loads. A TDR system (represented by a step pulse source of amplitude Vo and output impedance Rs = Zo) is connected to a transmission line of characteristic impedance Zo terminated with a resistive load RL, as shown in Figure 2.28. Three TDR waveforms monitored at the source end are shown for three different values of RL. Find the load resistance RL for each case.
72
Transient Response of Transmission Lines
Chap.2
+ Ys



z =O

(a) Ys(V)
Ys(V)
Ys(V)
0.3
0.2

0.2
0.1


0.2 I
t
1 ns
t
1 ns
t
Figure 2.28 TDR displays for resistive loads. (a) A TDR system connected to a transmission line terminated with an unknown load resistor RL. (b) RL = RL1. (c) RL = RL2· (d) RL = RL3·
Solution: The initial value (immediately after the application of the step input) of the sourceend voltage Vs (0) is equal to
V
8 (0) =
v+1 (z =
0, 0) = Vo = 0.2 V 2
from which the amplitude of the step voltage is found to be Vo = 0.4 V. At t = 1 ns, the reflected voltage arrives at the source end and is completely absorbed. So
Vs (lns)
=
Vi (0, 1 ns) + V~ (0, 1 ns) = Vi (0, 1 ns) (1 + rL)
where rL = (RL  Zo) I (RL + Zo). For RL = RL1, we have 0.2(1
+ rL1)
= 0.1
from which rL1 =  0.5, yielding RL1 = Zo/3. Similarly, we find RL2 = Zo and RL3 = 3Zo.
A simple summary of the TDR waveforms observed at the source end for purely resistive, capacitive, and inductive ter1ninations is provided in Figure 2.29. Note that the case of a resistive ter1nination was discussed in the preceding example, while a simple inductive termination was discussed in Section 2.5 and in connection with Figure 2.23. The result for the capacitive ter1nination case corresponds to that of Example 2.9 for RL = 00.
Sec. 2.6
TimeDomain Reflectometry
73
Termination TDR Waveform
R
+1

R>>Zo R>Zo R = Z0 R 2tdl
To find the area A under the glitch, we integrate (°V8 (t)  Vo/2) from t = 2td1 tot = oo: 00
A= 2tdl
v;
_oe(2Zo/Lw)(t2td1) dt = Vo 2
2
oo e(2Z()/Lw)t'
o
dt' =  Lw Vo e(2Z()/Lw)t' 4Zo
oo
LwVo
0
4Zo
where t' = t  2td1. Therefore, the bondingwire inductance Lw is given in terms of the area A as 4ZoA Lw=Vo
A simple summary of TDR signatures of purely resistive or purely reactive discontinuities on a transmission line is provided in Figure 2.31. The signatures of discontinuities involving combinations of reactive and resistive elements are dealt with in several problems at the end of this chapter.
Sec. 2.7
Transmission Line Parameters
75
Discontinuity
TDR waveform
~
Zo
R
1
) Z 1(2R + Z 0 0 ____/

R
Zo, td
Rl(R +2Z0)
1
Zo 
~..L
o1
L
Zo, td 1o
._
__
Zo
1 ~
o1
1
L
1o
I(
1o 1o
Zo, td
I c
._
o1
c
Zo, td
1 + e(t2td)2Z0!L
Zo
_
_._ _
2 e(t2td)!(2Z0 C) 1
._
o1 o1
Zo
1  e 2(t  2td)!(ZoC) 1

0
2td
t
Figure 2.31
TDR signatures produced by simple discontinuities. Sourceend TDR voltage signatures for shunt or series purely resistive, inductive, and capacitive discontinuities. In terms of excitation by a step voltage source of amplitude Vo and source resistance Rs, the TDR voltage waveforms shown are drawn for Vo= 2 V and Rs= Zo, and oneway travel time td from the source to the disconuity. (This figure was adapted from Figure 6 of B. M. Oliver, Time domain reflectometry, HewlettPackard J., 15(6), pp. 149 to 1416, February 1964. @HewlettPackard Company 1964. Reprinted with permission.)
2.7 TRANSMISSION LINE PARAMETERS
We have seen in previous sections that the response of a lossless transmission line to a given excitation depends on its characteristic impedance Zo and the propagation speed Vp (or the oneway travel time td = l /vp), which in turn depends on the line inductance L and capacitance C per unit line length. The response of lossy lines is additionally influenced by the values per unit line length of the series resistance R and shunt conductance G. In general, the values of these transmission line parameters depend on (1) the geometric shapes, physical dimensions, and proximity of the two conductors that for1n the line; (2) the electromagnetic properties of the material surrounding the conductors; and (3) the electrical conductivity of the conductors and the frequency of operation.
76
Transient Response of Transmission Lines
Chap.2
In later chapters, after we have introduced the governing electromagnetic equations, we will discuss methods by which the line capacitance, inductance, resistance, and conductance per unit length can be defined and determined from basic principles. In the case of the common transmission lines shown in Figure 2.32, we will be able to find convenient analytical expressions for the line parameters. For other, more complicated, structures R, L, C, and G can be either evaluated using numerical techniques or measured. Parameters 31 for many different transmission lines are also extensively available in handbooks. Approximate expressions for L, R, C and G parameters and for Zo for the common unifor1n transmission lines shown in Figure 2.32 are given in Table 2.2. The characteristic impedances (Zo) provided are for lossless lines (i.e., Zo = L/ C ). In Table 2.2, we have assumed the transmission line conductors to be made of copper and the surrounding medium to be air. Note that the parameters depend on the geometric shapes and the physical dimensions of the lines (d, a, and b). The line capacitance C and characteristic impedance Zo for the case when the surrounding medium is a nonmagnetic32 material other than air can be derived from those given in Table 2.2 by using the propagation speed Vp for these media as given in Table 2.1. Specifically we have c2 [ C]material
=
[C]air 2
and
[Zo] material =
VP
8 10
Vp
c
[Zo] air
1 ms .
where c is the speed of light in free space, or c r v 3 x The line inductance L remains the same, since it is governed by the magnetic properties of the surrounding material.
Twowire Coaxial Parallel plate
~1b~1 b
d
Air Air
Ta
_ _ _ J_ Air
Figure 2.32 Crosssectional view of three common uniform transmission lines. Expressions for the circuit parameters L, R, C, and G for these coaxial, twowire, and parallelplate lines are provided in Table 2.2.
31 Reference
Data for Engineers, 8th ed., Sams Prentice Hall Computer Publishing, Carmel, Indiana, 1993. 32 Magnetic properties of materials are discussed in Section 6.8. In the transmission line context, all materials can be considered nonmagnetic except for iron, nickel, cobalt, a few of their alloys, and some special compounds involving mixtures of magnetic materials with barium titanate.
Sec. 2.7
77
Transmission Line Parameters
TABLE 2.2 TRANSMISSION LINE PARAMETERS FOR SOME UNIFORM TWOCONDUCTOR TRANSMISSION LINES SURROUNDED BY AIR
Coaxial L
(µ,Hm 1 )
0.2ln(b/a)
C
(pFm 1)
55.6 ln(b/a)
Tuowire 0.4ln
G** (Sm 1)
+ b),Jf
d 2a
ln(b/a)
60 ln(b /a)
+
2
l.26a
1
2a
b
120ln
d
+
8.85b a
2
1
2a
8.3 x 108 ,JI
5.22 x 101 ,JJ
a
b
3.67 x 104
1.17 x 104 b
ab 7.35 x 104 ln
Zo (Q)
d
27.8 ln
4.15 x 108 (a
d 2a
Parallelplate*
d 2a
d
+
d 2a
1
2a
d
+
a
2
2a
2
1
377a b
*Valid for b >> a. **For polyethylene at 3 GHz.
The series resistance (R) is inversely proportional to the electrical conductivity of the particular metal that the conductors are made of, with the values given in Table 2.2 being relative to that of copper. The physical underpinnings of electrical conductivity are discussed in Chapter 5. For now, it suffices to know that it is a quantitative measure of the ability of a material to conduct electrical current and that the values of conductivity for different materials are tabulated extensively in various handbooks (see Table 5.1). A brief list of conductivities of some common metals relative to that of copper is provided in Table 2.3. The series resistance R is proportional to the square root of the frequency TABLE 2.3
Material Aluminum Brass Copper Gold Lead Magnesium Nickel
RELATIVE CONDUCTIVITIES OF METALS VERSUS COPPER
Relative Conductivity 0.658 0.442 1.00 0.707 0.0787 0.387 0.250
Material Silver Sodium Stainless steel Tin Titanium Tungsten Zinc
Relative Conductivity 1.06 0.375 0.0192 0.151 0.0361 0.315 0.287
78
Transient Response of Transmission Lines
Chap.2
because of the socalled skin effect, which results from the nonuniform distribution of electrical current in a metal at higher frequencies, and which is discussed in Chapter 8. With air as the surrounding medium, the shunt conductance G = 0, since air is an excellent insulator and leakage losses through it are generally negligible. In the case of other surrounding media for which leakage losses may not be negligible, the value of G depends on the geometrical layout of the conductors (as do the values of C, L, and R) but is more strongly deter1nined by the loss properties of the insulating medium surrounding the conductors and is, in general, a rather complicated function of the frequency of operation. Table 2.2 provides a representative expression for G for polyethylene as the surrounding medium at an operating frequency of 3 GHz. Highfrequency losses in insulating materials are discussed in Sections 8 .3 and 11.2. Example 2.14: RG58/U coaxial line. RG58/U is a commonly used coaxial line with an inner conductor of diameter 0.45 mm and an outer conductor of inside diameter 1.47 mm constructed using copper conductors and filled with polyethylene as its insulator. The line is to be used at 3 GHz. Find the line parameters (i.e., R, L, C, G, and Zo). Note from Table 2.1 that the propagation speed for polyethylene at 3 GHz is vp ~ 20 cm(ns) 1 = 2 x 108 ms 1 . Solution: We use the expressions provided in Table 2.2, except for the multipliers needed for C and Zo to correct for the fact that the filling is polyethylene rather than air. Note that we can also use the expression from Table 2.2 for G, since it was also given for polyethylene and for 3 GHz. We have rv
R 
L
~
4.15
108 (0.45
+ 1.47)
103 J3
X
109
3
(0.45 x 10 )(1.47 x 10 ) 0.2 ln
v2 p
1.47 0.45
~
55.6
b) 
ln ( a
0.237 µHm3

2
2
55.6 ln
( 1.47 )
4
G ~
X
3
c2 c~
X
7 .35 x 10ln
( 1.47 )
~ 6.21 x 10
rv

6 .6 Qm
1
1
~ 106 pFm
1
0.45
4
Sm
1
o.45
Zo '.: :'. ( ~) 60 In
~
'.:::'. 47 .4rl
Note that the value of Zo is close to the nominal 50ohm characteristic impedance of this coaxial line. Note also that the difference between 47.4Q and 50Q is within the range of accuracy (i.e., two digits after the decimal point) by which the physical quantities were specified (e.g., the radii of conductors, the value of vP, etc.).
2.8 SUMMARY This chapter discussed the fallowing topics:
• Transmission line parameters. A transmission line is commonly characterized by its distributed parameters R (in Qm 1), L (in Hm 1), G (in Sm 1), and C
Sec. 2.8
79
Summary 1 Fm ),
(in whose values are determined by the line geometry, the conductivity of the metallic conductors, the electrical and magnetic properties of the surrounding insulating material, and the frequency of excitation. Formulas for these parameters for three common transmission line geometries are provided in Table 2.2. • Transmission line equations. The differential equations governing the behavior of voltage and current on a lossless (R = 0 and G = 0) transmission line, and the wave equation for voltage derived from them, are
av az a!P az
a!P =Lat av =C at
)
• Propagatingwave solutions, characteristic impedance, and phase velocity. The general solution of the transmission line equations leads to mathematical expressions for voltage and current along the line that are wave equations in nature, depending on both distance and time. These are
V(z, t)
=
1+ t
z
+1 t +
Vp
1
!J(z,t) = Zo 1+ t
z Vp
z Vp
1 t +
z Vp
where Zo is the characteristic impedance of a lossless line, which is defined as the voltagetocurrent ratio of a single disturbance propagating in the +z direction and is given by Zo = L / C. The characteristic impedance is one of the most important quantities that determine the response of a transmission line. Formulas for calculating Zo are provided in Table 2.2. The velocity with which waves on a transmission line propagate is called the phase velocity and is given by Vp = (LC ) 112. For a lossless transmission line, the phase velocity is deter1nined by the properties of the material surrounding the transmission line conductors. For unifor1n transmission lines, Vp is a constant. Values of Vp for selected materials are tabulated in Table 2.1. • Transmission lines terminated in resistive loads, reflection coefficient. Transient response of a lossless transmission line to step or pulse excitation involves reflections from discontinuities along the line or from loads at its termination. Reflection effects are described in ter1ns of the reflection coefficient r, defined as the ratio of the reflected to the incident voltage at a given point. The reflection coefficients at the load and source ends of a transmission line are given by
RL Zo r L= RL+ Zo RsZo rs = Rs + Zo
Load end Source end
80
Transient Response of Transmission Lines
Chap.2
where RL is the resistance terminating the load end of the line, the line's characteristic impedance is Zo, and Rs is the source resistance. In the special case of a matched termination, we have RL = Zo, so rL = 0, and thus no reflection occurs from the termination. Similarly, when Rs = Zo, rs = 0, and no reflection occurs from the source end. In general, when a voltage disturbance is launched from the source end of a transmission line (e.g., due to a step change in input voltage), a sequence of reflections from both the load and source ends of the line occurs. The process of multiple reflections from the load and source ends of a transmission line can be described using a bounce diagram. • Transmission lines terminated in reactive loads. To determine the transient behavior of lossless lines terminated in reactive elements, it is necessary to solve the differential equations that determine the voltagecurrent relationships of the terminations subject to the appropriate initial conditions. The reflected voltage due to a step excitation is no longer a simple step function but, in general, varies continuously at a fixed position with respect to time depending on the nature of the reactive termination. PROBLEMS 2.1
Lumped or distributed circuit element? A uniform lossless transmission line with the L and C parameters and the length l given as shown in Figure 2.33 is excited by a unit step source with risetime equal to tr = 0.1 ns. Determine the following: (a) The characteristic impedance Zo and the oneway time delay td of the transmission line. (b) Whether it is appropriate or not to model the transmission line in this circuit as a lumped element. R 8 =50Q
1=5 cm j>L
.1>s
+
L=4nHcm1 C = 1.6 pFcm1
+ ~
t, = 0.1 ns
°VL
_j_


+ RL= lSOQ

Figure 2.33 Transmission line. Problem 2.1.
2.2
Lumped or distributedcircuit element? A uniform lossless transmission line is used to connect a periodic pulse waveform to a load as shown in Figure 2.34. Assuming the R 8 =50Q
l= 1 m
j>L
.1>s
+ f=l GHz SOo/o Duty cycle

+ lV 0


Z 0 =50Q Vp=2c/3
~
_j_

Figure 2.34 Transmission line. Problem 2.2.
+ °VL

RL=SOQ
Chap.2
81
Problems
characteristic impedance and phase velocity of the transmission line to be, respectively, Zo = son and Vp = 2c /3, determine the following: (a) The perunitlength parameters Land C of the transmission line. (b) The oneway propagation time delay td of the transmission line. (c) Whether this transmission line can be treated as a lumped element or not.
2.3
Opencircuited line. Consider the circuit shown in Figure 2.3S, with an ideal unit step source connected to a lossless line of characteristic impedance Zo = Son having an opencircuited termination at the other end. Assuming a oneway propagation delay across the line of td, use a bounce diagram to sketch the load voltage °VL(t) versus time for 0 < t < lOtd.
'
l
+

_L


Figure 2.35 Opencircuited line. Problem 2.3.
2.4
Step and pulse excitation of a lossless line. A uniform, lossless transmission line is excited with a step source as shown in Figure 2.36. (a) Provide an appropriate bounce diagram and use it to sketch both the sourceend voltage °Vs and the loadend voltage °VL as a function of time between 0 and 10 ns. Provide all the pertinent values on your sketches. (b) Repeat part (a) if the step voltage source was a pulse source given by 10 [u(t)  u (t  0.3 ns)] V. Z 0 = 90n, td = 1.8 ns Rs= ion ~~or=================~ + +
+
10 u(t)
(V)

RL=630n
Ys
_L


Figure 2.36 Step and pulse excitation of a lossless line. Problem 2.4.
2.5
Resistive loads. The circuit shown in Figure 2.37 consists of an uncharged transmission line connected to a load resistance RL. Assuming that the switch closes at t = 0, sketch the load voltage °VL(t) over the time interval 0 < t < 3 ns for the following load resistances: (a) RL = 2Sn, (b) RL = SOn, (c) RL = lOOn. t=
0
son, 0.5 ns +
3V

j_ 

Figure 2.37 Resistive loads. Problems 2.5 and 2.7.
82 2.6
Transient Response of Transmission Lines
Chap.2
Ringing. The transmission line system shown in Figure 2.38 is excited by a stepvoltage source of amplitude 3.6 V and source impedance lOQ at one end and is terminated with an open circuit at the other end. The line is characterized by the line parameters L = 0.5 µ,Hm1, C = 0.2 nFm 1, R = 0, G = 0, and has a length of l = 30 cm. Sketch the load voltage °VL(t) over 0 < t < 10 ns with the steadystate value indicated.
v
•
v
l
............... +
r _f~ 3.6V
0
~
_L



Figure 2.38 Ringing. Problem 2.6.
2.7
2.8
Discharging of a charged line. For the circuit of Problem 2.5, assume that the switch has been closed for a long time before it opens at t = 0. Sketch the load voltage °VL(t) over 0 < t < 3 ns for the same three cases. 1\vo step voltage sources. Consider the lossless transmissionline circuit which is excited by two step voltage sources, as shown in Figure 2.39. Use a bounce diagram to sketch the voltages °V 1 and °V 2 as a function of time. Note that the stepvoltage source on the left side turns off at t = 0.
+
°Vi


Figure 2.39 Pulse excitation. Problem 2.8.
2.9
Pulse excitation. The circuit shown in Figure 2.40 is excited by an ideal voltage pulse of 1 V amplitude starting at t = 0. Given the length of the line to be l = 10 cm and the propagation speed to be 20 cm(ns) 1; (a) sketch the voltage at the source end of the line, °V8 (t), for an input pulse duration of 10 ns; (b) repeat part (a) for an input pulse duration of 1 ns. Z 0 =75n
+

+
Figure 2.40 Pulse excitation. Problem 2.9.

Chap.2
2.10
83
Problems
Pulse excitation. The circuit shown in Figure 2.41 is excited with a voltage pulse of amplitude A and pulse width tw. Assuming the propagation delay of the line to be td, sketch the load voltage °VL(t) versus t for 0 < t < lOtd for (a) tw = 2td, (b) tw = td, and (c) fw = td/2.
+
+
J_

J_ = 

Figure 2.41 Pulse excitation. Problem 2.10.
2.11
Observer on the line. A transmission line with an unknown characteristic impedance Zo terminated in an unknown load resistance RL, as shown in Figure 2.42, is excited by a pulse source of amplitude 1 V and duration tw = 3td / 4, where td is the oneway flight time of the transmission line. An observer at the center of the line observes the voltage variation shown. (a) Determine Zo and RL. (b) Using the values found in (a), sketch the voltage variation (up to t = 4td) that would be observed by the same observer if the pulse duration were fw = l.5td.
I I
ctr
Rs = lOOQ . ~
, 'IC"
+
lV ., 0 '
Z 0 , td/2
°Vctr
Z 0 , td/2
• ~
'
'

'
ctr
0.5V
 
0.3 v
...._   


    

    

I 

I
0.5
1.0
titd
I
1.25
1.5
2.0
2.25
Figure 2.42 Observer on the line. Problem 2.11.
2.12
Cascaded transmission lines. For the transmission line circuit shown in Figure 2.43, sketch °V5 (t) and °VL(t) over 0 < t < 5 ns.
84
Transient Response of Transmission Lines SOQ, O.S ns
Rs= SQQ
~
+
Chap.2
7SQ, O.S ns
+
+
lV
~
0
_L


_L


Figure 2.43 Cascaded lines. Problem 2.12.
2.13 Timedomain reflectometry (TDR). A TDR is used to test the transmission line system shown in Figure 2.44. Using the sketch of °V8 (t) observed on the TDR scope as shown, determine the values of Zo 1 , 11 , and R 1 . Assume the phase velocity of the waves to be 20 cm(ns) 1 on each line. Plot °VRI (t) versus t for 0 < t < 4 ns.
Z 02 =7SQ
+
__.......,_ +
lV 0
_L




~(t)
o.sv i     0.37SV I I I
1 ns
t
Figure 2.44 Timedomain reflectometry. Problem 2.13.
2.14 Timedomain reflectometry (TDR). TDR measurements can also be used in cases with more than one discontinuity. Two transmission lines of different characteristic impedances and time delays terminated by a resistive load are being tested by a TDR, as shown in Figure 2.45. (a) Given the TDR display of the sourceend voltage due to a 3V, lOOQ step excitation starting at t = 0, find the characteristic impedances (Zo1 and Zo2) and the time delays (td1 and td2) of both lines, and the unknown load RL. (b) Using the values found in part (a), find the time and magnitude of the next change in the sourceend voltage °Vs (t), and sketch it on the display. 2.15 Timedomain reflectometry (TDR). Consider the circuit shown in Figure 2.46. The two line segments are of equal length l. Assuming the propagation speeds on the two lines are equal to 15 cm(ns) 1 each, find Zo 1 , Zo2, RL, and l using the TDR display of the source voltage °Vs (t), as shown. 2.16 Multiple lines. In the three lossless transmissionline circuit shown in Figure 2.47, the switch closes at t = 0. Assuming all the lines to be uncharged before t = 0, draw a bounce diagram and sketch voltages °V 8 , °VLl, and °VL2 between t = 0 and t = 10 ns using the impedance and time delay values indicated on Figure 2.47.
Chap. 2
85
Problems lOOQ
+
+
3V 0



~(t)
1.25 v
1.5 v

lV I
I
I
I
3
7.5
t (ns)
Figure 2.45 Timedomain reftectometry. Problem 2.14.
+ 5V
+
_r 
j_ 
j_ 
~(t)

5.82 v 3.6 v
3V
t (ns)
10
5
Figure 2.46 Timedomain reftectometry. Problem 2.15.
t =O
Z 01 = lOOQ, tdl = 3 ns
~'.3 ~ J.
+
000
~
4.5V
, i"
Cl'J ~< Os

_L

_L

+ RL2 = 100Q
VL2
Figure 2.47 Multiple lines. Problem 2.16.
86
Transient Response of Transmission Lines
Chap.2
2.17 Digital IC chips. Two impedancematched, inpackageterminated Integrated Circuit (IC) chips are driven from an impedancematched IC chip, as shown in Figure 2.48. Assuming the lengths of the interconnects to be 15 cm each and the propagation velocity on each to be 10 cm(ns) 1 , do the following: (a) Sketch the voltages °VLt and °VL2 for a time interval of 10 ns. Indicate the steadystate values on your sketch. (b) Repeat part (a) if one of the load IC chips is removed from the end of the interconnect connected to it (i.e., the load point A is left opencircuited).
son, lS cm
1oon, lS cm 1oon 
1oon, lS cm r~

Remove in part (b) A
I
+
1oon
1 vL2 I I
_ _ __
I____.~;_
IL __ _ _ _ _ _ _ _ _ _ _ _ _ _J
Figure 2.48 Digital IC chips. Problem 2.17.
2.18 Multiple lines. For the distributed interconnect system shown in Figure 2.49, and for Zo1 = Zo2 = 50Q, find and sketch the three load voltages °V1(t), °V2(t), and °V3(t) for a time interval of 5 ns. Assume each interconnect to have a oneway time delay of 1 ns. (b) Repeat part (a) for Zo1 = Zo2 = 25Q.
son
A
son
B
son
c
D RL3 = SOn
son
+
sv 0
_L

_L

_L


F
E

+
+
RL1 = SOn
RL2 = SOn
Figure 2.49 Multiple lines. Problem 2.18.

Problems
Chap.2
2.19
87
Reflections due to parasitic effects. The circuit shown in Figure 2.SO consists of a lowimpedance driver driving a distributed interconnect system that was intended to be impedancematched, with Zo = 120n. An engineer performs some tests and measurements and observes reflections due to parasitic effects associated with the two interconnects terminated at the 120n loads. Assuming that the effective characteristic impedances of these interconnects (i.e., taking parasitic effects into account) is such that we have Zo = 80n instead of 120n, find and sketch the voltages °V 1(t) and °V2(t) for 0 < t < 12 ns, assuming the oneway time delay on each interconnect to be 2 ns. Comment on the effects of the mismatch caused by parasitic effects. Assume the initial incident wave launched at the = 4 V. driver end of the 60n line to be
Vi
120Q
+

Driver 6012, 2 ns
LineA
_L

12012
Figure 2.50 Reflections due to mismatches. Problem 2.19.
2.20
2.21
Parallel multiple lines. The transmission line system shown in Figure 2.S 1 consists of three lines, each having Zo =son and a oneway propagation delay of 1 ns. (a) Find and sketch the voltages °V8 (t) and °V L(t) versus t for 0 < t < 10 ns. (b) Repeat part (a) when the opencircuited ends are terminated with a load resistance of son each. Optimized multiple lines. A multisection transmission line consists of three lossless transmission lines used to connect an ideal step source of SV amplitude and 6n output impedance
0

_L

_L

Figure 2.51 Parallel multiple lines. Problem 2.20.
88
Transient Response of Transmission Lines
Chap.2
to two separate load resistances of 66Q each, as shown in Figure 2.52. All three lines are characterized by line parameters L = 364.5 nHm 1, and C = 125 pFm 1 . To minimize ringing effects, the line lengths are optimized to be of equal length. If each line length is l = 40 cm, sketch the voltages °Vs and °VL versus t for 0 < t < 20 ns, and comment on the performance of the circuit in minimizing ringing.

0

_j_


Figure 2.52 Optimized multiple lines. Problem 2.21.
2.22 Transient response of a cascaded transmissionline circuit. Consider a circuit consisting of two cascaded lossless transmission lines with characteristic impedances and oneway time delays as shown in Figure 2.53. Assuming both lines to be fully discharged before t = 0, sketch the sourceend, junction and loadend voltages °Vs, °V1, and °VLas a function of time for the time interval 0 < t < lOtd for the following three cases: (a) Rs = Zo and RL = 3Zo, (b) Rs= Zo and RL = Zo, (c) Rs= Zo/3 and RL = 3Zo.
+
+


+ t =O

Figure 2.53 Cascaded transmission line. Problem 2.22.
2.23
Charging and discharging of a line. For the transmission line system shown in Figure 2.54, the switch S2 is closed at t = 2td (where td is the propagation delay of each line) after the switch S 1 is closed at t = 0. Find and sketch the voltage °VL versus t for 0 < t < 6td.
Chap.2
89
Problems
1sn
t=
0
lV 60Q




Figure 2.54 Charging and discharging of a line. Problem 2.23.
2.24
Digital IC interconnect. A circuit consists of one logic gate driving an identical gate via a 1ftlong, 50Q interconnect. Before t = 0, the output of the driver gate is at LOW voltage state and can simply be approximated as a 14Q resistor. At t = 0, the output of the driver gate goes from LOW to HIGH state and can be approximated with a 5V voltage source in series with a 14Q resistor. The input of the load gate can be approximated to be an open circuit (i.e., RL = oo). Assuming that a minimum load voltage of 3.75 Vis required to turn and keep the load gate on, (a) find the time at which the load gate will tum on for the first time. (b) Find the time at which the load gate will tum on permanently. (Assume a signal time delay of 1.5 nsft 1 along the interconnect for both parts.) 2.25 Terminated IC interconnects. The logic circuit of Problem 2.24 needs to be modified to eliminate ringing. Two possible solutions are to terminate the line in its characteristic impedance at either the source (series termination) or receiver (parallel termination) end. Both of these circuits are shown in Figure 2.55. (a) Select the value of the termination resistance RT in both circuits to eliminate ringing. (b) Compare the performance of these two circuits in terms of their speed and de power dissipation. Which technique is the natural choice for a design to achieve lowpower dissipation at steady state? 1 ft
son Driver gate
(a)
gate
1 ft
son Driver gate
gate
(b)

Figure 2.55 Terminated IC interconnects. (a) Series termination. (b) Parallel termination. Problem 2.25.
2.26
Digital IC gate interconnects. A disadvantage of the series termination scheme in Problem 2.25 is that the receiver gate or gates must be near the end of the line to avoid
90
Transient Response of Transmission Lines
Chap.2
receiving a twostep signal. This scheme is not recommended for terminating distributed loads. The two circuits shown in Figure 2.S6 have three distributed loads equally positioned along a 3ftlong son interconnect on a pc board constructed of FR4 material (take vp ~ 14.3 cm(ns) 1 ). Each circuit uses a different termination scheme. Assuming the driver and all the loads to be the same gates as in Problem 2.24, find the times at which each load gate changes its logic state after the output voltage of the driver gate switches to HIGH state at t = 0. Comment on the performance of both circuits and indicate which termination scheme provides faster speed. (Use some of the data provided in Problem 2.24.) A
1 ft
B
1 ft
son
1 ft
son
son
c
Driver gate (a)
1 ft
son
A
1 ft
son
Driver gate
B
1 ft
c
son
(b)
Figure 2.56 IC gate interconnects. (a) Series termination. (b) Parallel termination. Problem 2.26.
2.27 Digital IC circuit. For the digital IC circuit shown in Figure 2.S7, the driver gate goes from LOW to HIGH state at t = 0, and its Thevenin equivalent circuit (including the series termination resistance) can be approximated as a SV voltage source in series with a son resistor. If the time delay for all interconnects is given to be 2 nsft 1, find the time at which each receiver gate changes its state permanently. Assume each load gate to change state when its input voltage exceeds 4 V. Also assume each load gate to appear as an open circuit at its input. Support your solution with sketches of the two load voltages °V 1 and °V2 as functions of time for a reasonable time interval. 2.28 1\vo driver gates. Two identical logic gates drive a third identical logic gate (load gate), as shown in Figure 2.S8. All interconnects have the same oneway time delay td and characteristic impedance Zo. When any one of these driver gates is at HIGH state, its Thevenin equivalent as seen from its output terminals consists of a voltage source with voltage Vo in series with a resistance of value Rs = Zo. At LOW state, its Thevenin equivalent is just a resistance of value Rs = Zo. The input impedance of the load gate is very high compared with the characteristic impedance of the line (i.e., Zn >> Zo). (a) Assuming steadystate conditions before both driver gates change to HIGH state at t = 0, sketch the load voltage °VL as a function of time for 0 < t < 1td. What is the eventual steadystate value of the load voltage? (b) Assume steadystate conditions before t = 0 to be such that driver gate 1 is at HIGH state and gate 2 is at LOW state. At t = 0, gate 1 and gate 2 switch states. Repeat part (a). 2.29 Capacitive load. For the transmission line system shown in Figure 2.S9, the switch is closed at t = 0. Each of the two transmission lines has a oneway time delay of 2 ns.
Chap.2
91
Problems 2 ft
1 ft
son
1oon
Driver gate
1.S ft
1oon


Figure 2.57 Digital IC circuit. Problem 2.27.
Driver gate 1
gate
Driver gate2
Figure 2.58 Two driver gates. Problem 2.28.
1oon
t=
0
son, 2 ns
lOOn, 2 ns .~

.~~.
I
I
}
+
sv
.,r.,.
S pF
r

Figure 2.59
2.30


Capacitive load. Problem 2.29.
Assuming both transmission lines and the 5 pF capacitor to be initially uncharged, find and sketch the voltage °V 1 (t) across the resistor Ri. Unknown lumped element. Using the sourceend voltage waveform observed on the TDR display due to an ideal step voltage source exciting the double lossless transmission line
92
Transient Response of Transmission Lines
Chap.2
circuit connected as shown in Figure 2.60, determine the type of the unknown lumped element at the junction and its value.
son + ~
3V
?•
t=O



~ (V)
2.4
~~ 
1.5
r=2 ns
6.0
t (ns)
Figure 2.60 Unknown lumped element. Problem 2.30.
2.31
Unknown lumped element. The transmission line circuit has an unknown lumped element, as shown in Figure 2.61. With the sourceend voltage due to step excitation measured to be as plotted, determine the type of the unknown element, and find its value in terms of the shaded area A . ~
. •
+ I'
_f \..:
~
Vo

)
•
+ ~
?•

_j_

•
•
Unknown lumped element
'
J


~(t)
A V 0 /2
1
~
t
Figure 2.61
Unknown lumped element. Problem 2.31.
2.32 Sourceend TDR waveform. The sourceend TDR voltage waveform of a transmissionline circuit terminated with three unknown lumped elements is as shown in Figure 2.62. Using
Chap.2
93
Problems
this waveform, (a) determine the characteristic impedance and the oneway time delay of the line and (b) the types and the values of the three unknown elements.
z 0, td
SQQ ~
..,
?•
~
'
4u(t) r +" (V) \..
+
+
~ 
°VL

?•
?•

)


~(V)
10/3 r = 0.15 ns
2..,___ _ _____. 1.25
I r~
1.0
t (ns)
Figure 2.62 Sourceend TDR wavefor1n. Problem 2.32.
2.33
Unknown load. A 5 V de voltage source is used to excite a lossless transmissionline circuit terminated with two unknown lumped elements, as shown in Figure 2.63a. The switch at the source end is closed at t = 0. The transmission line parameters Zo, td, and length L are unknown. Based on the sourceend TDR voltage waveform shown in Figure 2.63b, determine the type (resistor, capacitor, or inductor) and values of the two unknown lumped elements. Assume the transmission line to be at steadystate condition at t = o. L
sv
?•
?•
t=O


(a)
~(V)
2.5 11 • 25 :
r = 0.8 ns
I
6.0 (b)
Figure 2.63
Unknown load. Problem 2.33.
t (ns)
94
2.34
Transient Response of Transmission Lines
Chap.2
Capacitive load. Two transmission lines of characteristic impedances 7Sn and son are joined by a connector that introduces a shunt resistance of lSOn between the lines, as shown in Figure 2.64. The load end of the son line is terminated with a capacitive load with a 30 pF capacitor initially uncharged. The source end of the 7Sn line is excited by a step function of amplitude 3.6 V and a series resistance of 7Sn, starting at t = 0. Assuming the total time delay of each line to be td 1 = 6 ns and td2 = 2 ns, respectively, find and sketch (a) the voltage °VL(t) at the load end of the son line and (b) the voltage °V 8 (t) at the source end of the 7Sn line. t dl =
7Sn
3.6 v 0
6 ns
td 2 =
Z 01 =7Sn
_r
r
Z 02 = SOn
1
+
+ Vg
+__.........
2 ns
30pF
lSOQ
lOOQ
1oon





_ _ _ _ _ _ _ _ _ _ .J
Load Figure 2.64 Capacitive load. Problem 2.34.
2.35 Reactive element at the junction. In the lossless transmissionline circuit shown in Figure 2.6S, find and sketch both the sourceend voltage °Vs and the loadend voltage °VL as a function of time. son, 2 ns ....___ _ _ _ _ ____.
son
0.27 µH
~
300n, 2 ns
.....
~~t0
+
+
6u(t)
300Q
lSOQ
(V)



Figure 2.65 Reactive element at the junction. Problem 2.35.
2.36 Transmission line terminated with a capacitive load. Consider the lossless transmissionline circuit shown in Figure 2.66 where the switch at the load end closes at t = 0, after t =O
Z 0 = Son, td = 3 ns
son
+ Vg
sv 

son

Figure 2.66 Transmission line terminated with a capacitive load. Problem 2.36.
Chap.2
2.37
95
Problems
being open for a long time. Find and sketch the sourceend and the loadend voltages °Vs(t) and °VL(t). Transients on a transmission line. Consider the lossless transmissionline circuit as shown in Figure 2.67. The switch is opened at t = 0, after being closed for a long time. Find and sketch the sourceend voltage °Vs (t) as a function of time. t =O Z 0 = SOQ, td = 3 ns lOOQ
+ 30nH
Ys
6V


300Q

Figure 2.67 Transients on a transmission line. Problem 2.37.
2.38
Transient response. Consider the lossless transmission line system shown in Figure 2.68. The 20 pF capacitor is initially uncharged. The switch S is initially open but is closed at t = 7 ns. Determine mathematical expressions for and sketch the capacitor voltage °Vc (t) and the sourceend voltage °V 8 (t) for 0 < t < 12 ns. Make assumptions as needed but clearly state and justify them. td1 =
SQQ
12u(t) (V)
I
s
td2 =
1 ns
z 01 =son
1 ns
25Q
+ 20pF°Vc
Z 02 = 2SQ
Switch S (initially open) Closed at t = 7 ns lOOQ
lOOQ
Figure 2.68 Transients on a transmission line. Problem 2.38.
2.39
Inductive load. Two transmission lines of characteristic impedances 50Q and 75 Q are joined by a connector that introduces a series resistance of 25 n between the lines, as shown in Figure 2.69. The load end of the 75Q line is terminated with an inductive load. The
son
son, s ns
2SQ
Load
7SQ, S ns
+
+
7SQ
Ys
20nH


________ J
Figure 2.69 Inductive load. Problem 2.39.
96
2.40
Transient Response of Transmission Lines
Chap.2
inductor is initially uncharged, and a step source with amplitude s v turns on at t = 0. Find and sketch the voltage °V L· First determine the initial and final values and accurately mark all points of your sketch. Capacitive load. A steptype incident voltage wave of 1V peak value arrives at a capacitive load at t = 0, as shown in Figure 2.70. After t = 0, find the approximate time at which the reflected voltage wave cy at the load position z = 0 is equal to zero. Assume the load capacitor to be initially uncharged.
!1 v I
+ Zo = lOOn
~
°VL ~ >300Q 
,
~
'
20pF
'
I
~z
z =O Figure 2.70 Capacitive load. Problem 2.40.
2.41 Step excitation. The circuit shown in Figure 2.71 is excited by a stepvoltage source of amplitude 5 V and source resistance Rs = 2Zo, starting at t = 0. Note that the characteristic impedance of the shorted stub is half that of the main line and that the second segment of the main line is twice as long, so its oneway time delay is 2td. (a) Assuming the load to be an open circuit (i.e., a very large resistance), sketch the load voltage °VL(t) versus t for 0 < t < lltd. (b) Repeat part (a) for the case when the input is a pulse of duration tw = 4td.
sv
+
+
0
Figure 2.71
2.42
Step excitation. Problem 2.41.
Capacitive load excited by two sources. For the transmission line system shown in Figure 2.72, find the mathematical expression for the capacitor voltage °Vc(t) and sketch it fort > 0. Assume the capacitor to be initially uncharged.
Chap.2
97
Problems lOOQ
tdl =
500 ps
td 2 =
1 ns
lOOQ
+ lOOQ
Su(t)
lOOQ
Su(t)
+
Figure 2.72 Capacitive load excited by two sources. Problem 2.42.
Zo
Zo
+
+
V 0 u(t)
Figure 2.73 Two sources. Problem 2.43.
2.43 2.44
2.45
1\vo sources. For the circuit shown in Figure 2.73, sketch the voltages °V sl (t) and °V8 2(t) for 0 < t < 1td. RG 8 coaxial line. A student buys an RG 8 lowloss coaxial cable for a VHF antenna project. He looks up the specifications of the RG 8 coax in a product catalog and finds out that its characteristic impedance is Zo = 50 n, its velocity factor is vp/c = 0.66, and its line capacitance is C = 26.4 pFft 1 . He then cuts a portion of this coax and measures the diameter of the inner conductor and the outer diameter of the dielectric to be approximately 2 mm and 7.5 mm, respectively. Using these values, find or verify the values of the unit length line parameters L, C, R, and G and the characteristic impedance Zo of this cable at 100 MHz. Note that the dielectric inside RG 8 coax is polyethylene and that the leakage conductance per unit length of a polyethylenefilled coaxial line at 100 MHz is approximately given by G ~ 1.58 x 105 /ln(b/a) Sm 1 . 1\vowire line. Calculate the perunitlength line parameters L, C, R, and G and the characteristic impedance Z0 of an airinsulated twowire line made of copper wires with wire separation of 2.1 cm and wire diameter of 1.2 mm at a frequency of 200 MHz.
This page intentionally left blank
_...tea y tate aves on Transmission Lines
In digitalintegrated electronics, computer communication, and many other applications, it is important to understand the response of transmission lines to steplike changes in their inputs, as we studied in detail in Chapter 2. We have seen that waves travel down a transmission line by successively charging the distributed capacitors of the line and establishing current in the distributed inductors. In this context, a wave is a function of both time and space but does not necessarily involve periodic oscillations of a physical quantity (e.g., the height of water in ocean waves). If a disturbance that occurs at a certain point and time causes disturbances at other points in the surrounding region at later times, then wave motion is said to exist. Because of the nonzero travel time along a transmission line, disturbances initiated at one location induce effects that are retarded in time at other locations. The natural response of transmission lines to sudden changes in their inputs typically involves wave motion, with disturbances traveling down the line, producing reflections at terminations, or discontinuities, which in turn propagate back to the source end, and so on. However, except in special cases, the natural response eventually decays after some time interval and is primarily described by the intrinsic properties of the transmission line (characteristic impedance and oneway travel time or length) and its termination (i.e., the load), rather than the input excitation. In many engineering applications, however, it is also important to understand the steadystate response of transmission lines to sinusoidal excitations. Electrical power and communication signals are often transmitted as sinusoids or modified sinusoids. Other nonsinusoidal signals, such as pulses utilized in a digitally coded system, may be considered as a superposition (i.e., Fourier series) of sinusoids of different frequencies. In sinusoidal signal applications, the initial onset of a sinusoidal input produces a natural response. However, this initial transient typically decays rapidly in time, while the forced
99
SteadyState Waves on Transmission Lines
100
Chap. 3
response supported by the sinusoidal excitation continues indefinitely. Once the steady state is reached, voltages and currents on the transmission line vary sinusoidally in time at each point along the line while also traveling down the line. The finite travel time of waves leads to phase differences between the voltages (or currents) at different points along the line. The steadystate solutions for voltage and current waves when reflections are present lead to standing waves, which also vary sinusoidally in time at each point on the line but do not travel along the line. The differences between traveling and standing waves will become clear in the following sections. In discussing the steadystate response of transmission lines to sinusoidal excitation, we take full advantage of powerful tools 1 commonly used for analysis of altematingcurrent phenomena in Jumped electrical circuits, including phasors and complex impedance. The phaser notation eliminates the need to keep track of the known sinusoidal time dependence of the various quantities and allows us to transform the transmission line equations from partial differential equations to ordinary differential equations. The magnitude of the complex impedance of a device or a load is a measure of the degree to which it opposes the excitation by a sinusoidal voltage source; a load with a higher impedance (in magnitude) at any given frequency requires a higher voltage in order to allow a given amount of current through it at the given frequency. The phase of the impedance represents the phase difference between the voltage across it and the current through it. In Jumped circuits, the impedance of a load is determined only by its internal dynamics (i.e., its physical make up as represented by its voltagecurrent characteristics, e.g., V = L dl / dt for an inductor). In transmission line applications, the impedance of a load presented to a source via a transmission line depends on the characteristic impedance and the electrical length2 (physical length per unit of wavelength) of the line connecting them. This additional length dependence makes the performance of transmission line systems dependent on frequencyto a greater degree than is typical in Jumped circuit applications. In this chapter, we exclusively consider excitations that are pure sinusoids. In most applications, transmission line systems have to accommodate signals made up of modified sinusoids, with the energy or information spread over a small bandwidth around a ce ntral frequency. The case of excitations involving waveforms that are other than sinusoidal can usually be handled by suitably decomposing the signal into its Fourier components, each of which can be analyzed as described in this chapter. In the first six sections we consider lossless (i.e., R = 0 and G = 0) transmission lines. Lossy transmission lines (i.e., R =f. 0 and G =f. 0) are discussed in Section 3.7. We shall see that the effects of small but nonzero losses can be accounted for by suitably modifying the lossless analysis. Note also that the Joss terms are truly negligible in many applications, so that the lossless cases considered in detail are of practical interest in their own right.
111le.c;e toolc; were first introduced by a famous electrical engineer, C. P. Steinn1etz, in hie; first book, Theory and Calcularion of 11C Pht•nomt•na, f\.1cGrawHill, I\1ew York, 189'7. 211le electricaJ
length of a transmission line is its physical length djvided by the wavelength at the frequency of operation. For ex3Jltple, the electrical length of a 1.5 mlong airtilled coaxial line operating at 100 /\.1Hz
=
(wavelength A 3 n1) is 0.5A, while the electrical length of the S3Jlte line operating at I J\.1Hz (~'avelength ), = JOO m) is 0.0051,.
Sec. 3.1
101
Wave Solutions Using Phasors
3.1 WAVE SOLUTIONS USING PHASORS The fundamental transmission line equations for the lossless case were developed in Chapter 2 (see equations (2.3) and (2.4)): av az a!P az
=L
a!P
=C
(3.1)
at av
(3.2)
at
Equations (3.1) and (3.2) are written in ter1ns of the spacetime functions describing the instantaneous values of voltage and current at any point z on the line, denoted respectively as V(z, t) and !P(z, t). When the excitation is sinusoidal, and under steadystate conditions, we can use the phasor concept to reduce the transmission line equations (3.1) and (3.2) to ordinary differential equations (instead of partial differential equations, as they are now) so that we can more easily obtain general solutions. As in circuit analysis, the relations between phasors and actual spacetime functions are as follows: V(z, t) = 9lle{V (z)eiwt}
(3.3a)
!P (z, t) = 9lle{l (z)eiwt}
(3.3b)
Here, both phasors V (z) and I (z) are functions of z only and are in general complex. We can now rewrite3 equations (3 .1) and (3 .2) in terms of the phasor quantities by replacing a/at withjw. We have dV (z) = j wLI (z)
(3.4)
dz and di (z) = j w CV (z)
(3.5)
dz Combining (3 .4) and (3 .5), we can write a single equation in terms of V (z), d 2 V(z)  2
dz
3 The

2
2
(j w LC)V(z) = 0
or
d 2 V(z)
 dz 2

2
(j {3) V (z) = 0
actual derivation of (3.4) from (3.1) is as follows: a°V az
= L aJi at
> a [ffie{V (z)eiwt}]
az
at      °V(z,t)
> ffie
= L!_ [ffie{/ (z)eiwt}]
eiwtdV(z)
m
i(z,t)
=ffie{L(jw)eiwt/(z)} > dV(z) = jwL/(z)
m
(3.6)
SteadyState Waves on Transmission Lines
102
Chap. 3
where fJ = (J),J[C is called the phase constant. Equation (3.6) is referred to as the complex wave equation and is a secondorder ordinary differential equation commonly encountered in analysis of physical systems. The general solution of (3.6) is of the form (3.7) where, as we shall see below, e i.Bz and ei.Bz represent wave propagation in the + z and  z directions, respectively, and where v + and v  are complex constants to be determined by the boundary conditions. The corresponding expression for the current I (z) can be found by substituting (3.7) in (3.4). We find (3.8) where Zi = V +/I + =  v / I  = ./L7C is the characteristic impedance of the transmission line. Using (3.3), and the expressions (3.7) and (3.8) for the voltage and current phasers, we can find the corresponding spacetime expressions for the instantaneous voltage and current. We have
"V'(z, t) = ~{V (z)ei"" ) = \i!te{v+e i.B' ei"" = v + cos(wt  {Jz ) + v  cos((JJt
+ v  e+i.B• ei"" )
+ {Jz)
(3.9)
where we have assumed4 v + and v  to be real. Similarly, we have
9'(z,t ) =
~[v+ cos((J)t  {Jz)  v  cos((J)t + {Jz )]
(3.10)
The voltage and current solutions consist of a superposition of two waves, one propagating in the + z direction (i.e., toward the load) and the other in the  z direction (i.e., a reflected wave moving away from the load). To see the wave behavior, consider the case of an infinitely long transmission line; in this case no reflected wave is present and thus v  = 0. The voltage and current for an infinitely long line are
"V'(z, t) = v + cos(wt  {Jz)
4
lf instead
v+ and v 
were complex, with
'V(z , t)
=IV +1cos(wr 
v+ =
1v +1J~+ and
v =
(3.11)
1 v 1&·~ . we would have
llz + ~+) + iv  1oos(w1 + /J z + ~)
Sec. 3.2
103
Wave Solutions Using Phasors
and
!P(z, t) =
v+ Zo
cos(wt  {3z)
(3.12)
Note that, everywhere along the line, the ratio of the voltage to current phasors is Zo; hence, Zo is called the characteristic impedance of the line. Note, however, that this is true not only for an infinitely long line but also for a line of finite length that is terminated at a load impedance ZL = Zo, as we discuss later. The solutions (3.11) and (3.12) are in the form of (1.5), which was introduced in Chapter 1 by stating that electromagnetic quantities with such spacetime dependencies are often encountered. Here we see that this for1n of solution is indeed a natural solution of the fundamental transmission line equations. 5 The spacetime behavior of the voltage wave given by (3.11) is illustrated in Figure 3.1. We note from Figure 3.la that the period of the sinusoidal oscillations (as observed at fixed points in space) is Tp = 2n /w. The voltage varies sinusoidally at all points in space, but it reaches its maxima at different times at different positions. Figure 3.lb indicates that the voltage distribution as a function of distance (observed at a fixed instant of time) is also sinusoidal. The distance between the crests of the voltage at a fixed instant of time is the wavelength A = 2n / f3. As time progresses, the wave propagates to the right (+z direction), as can be seen by observing a particular point on the waveform (e.g., the crest or the minimum) at different instants of time. The speed of this motion is the phase velocity, defined as the velocity at which an observer must travel to observe a stationary (i.e., not varying with time) voltage. The voltage observed would be the same as long as the argument of the cosine in (3.11) is the same; thus we have
wt  f3z = const.
)
Vp
=
dz dt
(1)

(1)

1
v1LC
(3.13)
where Vp is the phase velocity, which was also introduced in Sections 1.1.3 and 2.2 (see equation (2.7)). As discussed in Section 2.2, for most of the commonly used twoconductor transmission lines (Figure 2.1 ), the phase velocity Vp is not a function of the particular geometry of the metallic conductors but is instead solely determined by the electrical and magnetic properties of the surrounding insulating medium. When the surrounding medium is air, the phase velocity is the speed of light in free space, namely, vp = c ~ 3 x 108 ms 1 = 30 cm(ns) 1 . The phase velocity vp for some other insulating materials was tabulated in Table 2.1. Phase velocities for some additional materials are given in Table 3.1, together with the corresponding values of wavelength at a frequency of 300 MHz. Note that since A = 2n / {3, we have vp = w / f3 = Af, so that the phase constant f3 and wavelength A depend on the electrical and magnetic properties of the material surrounding the transmission line conductors as well as on the frequency of operation. 5 We
will encounter the same type of spacetime variation once again in Chapter 8, as the natural solution of Maxwell's equations for timeharmonic (or sinusoidal steadystate) electric and magnetic fields.
SteadyState Waves on Transmission Lines
104
Chap.3
°V(z, t)
.,,.,
1...,.......
'
I /
0.5 I
I
I
I
I
.,,., \
I
\
°V(z = Al4, t)
',, / \ \
I
0.5.
0.25
\
0.5
0.75
I
I
I
I
\
I
\
1
\
\
\
\
1.25
\
1.5
I
\
\
\
I \
°V(z = 0, t)
1
'
I/
' ,_.,,
I
I
ti~
I
I /
(a) °V(z, t)
Z=A
..... ,
1
'
I /
0.5 I
I
I
I
I
VP
0.25
',,/ \
\
\
I
/ °V(z, t = 0)
1
0.75
0.5
1
•
\
0.5
I /
°V(z, t = Tl4)
\
I
.,,.,
\
\
\
''
,_.,,/
I
I
I
I
I
I
I
I
I
'
\
\
I
1.25
\
\
\
\
\
1.5
I
zlA
(b) Figure 3.1 Wave behavior in space and time. (a) °V(z,t) = v+cos [2n(t/Tp) 2n (z/A.)] versus t/Tp for z = 0 and z = A./4. (b) °V(z, t) = v+ cos [2n(t/Tp)  2n(z/A.)] versus z /A. for t = 0 and t = Tp / 4. In both panels we have taken V + = 1.
TABLE 3.1
PHASE VELOCITY AND WAVELENGTH IN DIFFERENT MATERIALS
Material Air Silicon Polyethylene Epoxy glass (PC board) GaAs Silicon carbide (SiC) Glycerin
Wavelength (m at 300 MHz)
Phase Velocity (cm(ns) 1 at 300 MHz)
1 0.29 0.67 0.45 0.30 0.15 0.14
30 8.7 20.0 13.5 9.1 4.6 4.2
Sec. 3.2
Voltage and Current on Lines w ith Short or OpenCircu it Terminations
105
3.2 VOLTAGE AND CURRENT ON LINES WITH SHORT OR OPENCIRCUIT TERMINATIONS Most sinusoidal steadystate applications involve transmission lines terminated at a load impedance ZL. Often, voltages and currents near the load end are of greatest interest since they determine the degree of matching between the line and the load and the amount of power delivered to (versus that reflected from) the load. A portion of a lossless transmission line terminated in an arbitrary load impedance ZL is shown in Figure 3 .2. We can use this setup to explore the concept of reflected waves on transmission lines, a fundamental feature of distributed circuits in general. Assume that a forwardpropagating (+z direction) wave of the for1n v+ej.Bz produced by a source located at some position z (z < 0) is incident on load ZL located at z = 0. Contrary to the case of an infinitely long transmission line, here the ratio of the total voltage V (z) to the total current I (z) at any position z along the line is not equal to Zo. For example, at the load position (z = 0), we must satisfy the boundary condition [V(z)/l(z) ]z=O = ZL, where ZL is in general not equal to Zo. Thus, since v + jJ+ = Zo, and in general Zo i= ZL, a reversepropagating ( z direction) reflected wave of the form v e+j .Bz with the appropriate value for v must be present so that the load boundary condition is satisfied. The total voltage and current phasors, V (z) and I (z), at any position on the line consist of the sum of the forward and reverse waves as specified by (3. 7) and (3.8), namely V(z) = v+ej.Bz I(z) =
+ ve+j.Bz
(3.14)
1 [V+e1',B z  ve+J',B Z] Zo
(3.15)
where V + and v are in general complex constants to be deter1nined by the boundary conditions. When a transmission line has only a forwardtraveling wave with no reflected wave (e.g., in the case of an infinitely long line), the ratio of the total voltage to the current is the characteristic impedance Zo, as was discussed in Section 3.1. When the line is terminated
I(z)
 z Source
z

•
+
+ V(z)

Zo
VL

Z(z)
z=O Figure 3.2 A terminated lossless transmission line. For convenience, the position of the load is taken to be z = 0.
SteadyState Waves on Transmission Lines
106
Chap.3
so that, in general, a reflected wave exists, the ratio of the total line voltage V (z) to the total line current l (z) at any position z the line impedance is not equal to Zo. The line impedance is of considerable practical interest; for example, the impedance that the line presents to the source at the source end of the line (called the input impedance of the line, denoted by Zin) is the line impedance evaluated at that position. The source, or the generator, does not know anything about the characteristic impedance of the line or whether a reflected wave exists on the line; it merely sees that when it applies a voltage of Vs to the input terminals of the line, a certain current ls flows, and thus the source interprets the ratio of Vs/ls as an impedance of a particular magnitude and phase. The line impedance as seen by looking toward the load ZL at any position z along the line (see Figure 3.2) is defined as Z( ) z
= V(z) l(z)
(3.16)
In general, the line impedance Z (z) is complex and is a function of position z along the line. From electrical circuit analysis, we know that a complex impedance Z (z) can be written as Z(z) = R(z) + jX(z) where the real quantities R(z) and X (z) are the resistive and the reactive parts of the line impedance, respectively. The following example considers the case of a matched load, defined as a load impedance equal to the characteristic impedance of the line, or ZL = Zo. Example 3.1: Matched load. A lossless transmission line is terminated with a load ZL = Zo, as shown in Figure 3.3a. Find the magnitude of the reflected wave v and the line impedance Z(z).
z =O

I
T
v+e jf3z (V= 0)
rc Z(z) Z =
I I I I I
Zo • •
0
(a)
I I I
_ ___.. z
~1~~~~~~~,l Zo, the reflection coefficient rL is purely real with lfr = 0 and 0 s p s 1.
Example 3.5: A Yagi antenna array driven by a coaxial line. To increase the geographic coverage area of a broadcast station, four Yagi antennas, each having a feedpoint impedance of son, are stacked in parallel on a single antenna tower and connected to the transmitter by a son coaxial line, as shown in Figure 3.14a. (a) Calculate the load reflection coefficient rL. (b) Calculate Vmax' Vmin' Imax' and Imm along the line, assuming v+ = 1 V. (c) Sketch IV(z)I and l/(z) I as functions of z /A., taking the position of the antenna array terminals to be at z = 0.
SteadyState Waves on Transmission Lines
120
Chap.3
Solution: (a) The total load impedance seen by the coaxial line is a parallel combination of the four son impedances, resulting in 50 ZL = = 12.5Q 4
The load reflection coefficient is then given by
r
Zo ZL + Zo
= ZL L
so that p = 0.6 and
= 12.5  50 = 0 6 = 0 6ei180o
12.5 + 50
.
.
1/1 = 180°.
(b)
vmin = 1v+1c1 
1v+1c1 + p) = 1.6 V
Vmax =
IV+ I (1 + p)
Imax =
= 32 mA
/min=
Zo
1v+1 Zo
p) =
o.4 v
(1  p) = 8 mA
Yagi array

son
Transmitter
son coax
son
I I I
z=O (a) IJ(z)I (mA) IV(z)I
32
I J(z)I ~
....
I /
24
'
\
I
\
\
1
I I
\
\ \ \
J
\
I
0.7S
I
\
I
0.S
0.8
I
\ \
8
1.2 \
\
16
IV(z)I (V) 1.6
I '/
0.2S
0.4 0
z/A. (b)
Figure 3.14 Yagi array driven by a coaxial line. (a) Array of a stack of four Yagi antennas fed by a coaxial line. (b) Voltage and current standingwave patterns.
Sec. 3.3
121
Lines Terminated in an Arbitrary Impedance
(c) The voltage reflection coefficient at any position z along the line is given by r(z) = pt!
p=
s 1 s+1
(3.23)
Note that S varies in the range 1 ::::; S ::::; oo. Example 3.6: UHF blade antenna. A UHF blade antenna installed in the tailcap of a small aircraft is used for communication over the frequency band 225400 MHz. The following table provides the measured values of the feedpoint impedance of the antenna at various frequencies: 9
/(MHz)
225 300 400
22.5  j51 35  j 16 45  j2.5
A 50Q coaxial line is used to connect the communication unit to the antenna. Calculate the load reflection coefficient rL and the standingwave ratio S on the line at (a) 225 MHz, (b) 300 MHz, and (c) 400 MHz.
Solution: (a) At 225 MHz, the load reflection coefficient is given by (equation (3.19))
rL
ZL  Zo 22.5  j51  50 57.9ej 1180 _ 83 2o = rv r v 0 6 54e 1 . 0 ZL + Zo 22.5  j51+50  88.6ej35.t  .
The standingwave ratio S at 225 MHz can then be obtained using (3.23) as follows:
s
9 R.
1+ p = 1 p
~
1+0.654 1  0.654
~
4.78
L. Thomas, A Practical Introduction to Impedance Matching, Artech House, Inc., Dedham, Massachusetts, 1976.
SteadyState Waves on Transmission Lines
122
Chap.3
(b) Using the same equations, at 300 MHz, we have rL =~ 0.254ej 1220 and S ~ 1.68. (c) At 400 MHz, we have rL =~ 0.0588ej 1520 and S ~ 1.13. We see that the reflections on the coaxial line are quite significant near 225 MHz but are much reduced near 400 MHz.
Another quantity that can sometimes be measured in experimental settings is Zmin, or the distance from the load to the first minimum of the voltage standingwave pattern. 10 From (3 .20) we have
Since
IV (z) I =
Vmin when ej(1ff+2f3z) = 1, or
1/f + 2f3Zmin =
s
(2m
+ l)n
m = 0, 1, 2, 3, ...
s
where Jr 1fr < Jr and Zmin 0. For any given frequency, measuring the wavelength (by measuring the distance between successive minima) provides a means to determine the phase velocity Vp = f A. At the location of the first minimum we have
1/f + 2f3Zmin
=
Jr
>
1fr = n 
2f3zmin
(3.24)
We see that Zmin is directly related to the phase 1fr of the reflection coefficient rL, whereas s determines its magnitude through (3.23). Once rL is known, the load can be fully specified (assuming the characteristic impedance Zo is known), or Zo can be found (if ZL is known). Thus, the two measurable quantities, S and Zmin, completely characterize the transmission line ter1ninated in an arbitrary load impedance. It is often useful to rewrite (3 .20) and (3 .21) in terms of the load voltage VL and load current IL. Using the fact that VL = V (z) lz=O and IL= I (z) lz=O, and after some manipulation, we have V (z) = VL cos(f3z)  jlLZo sin(f3z)
(3.25)
I (z) =IL cos(f3z)  j VL sin(f3z)
(3.26)
Zo
The voltages and currents at the source end (z = l) can be found from (3.25) and (3.26) by substituting z = l. Note that for a general complex load impedance ZL, the load voltage VL and current IL are in general complex, so that equations (3.25) and (3.26) do not necessarily constitute a decomposition of V (z) and I (z) into their real and imaginary parts. 10 In
practice, it may often be difficult to actually measure the first minimum; however, if the location of any of the minima can be measured, the location of the first minimum can be deduced by using the fact that successive minima are separated by 'A /2.
Sec. 3.3
Lines Terminated in an Arbitrary Impedance
123
In general, the voltage and current standingwave patterns on a terminated line depend on the nature of the load. Typically what is plotted is IV (z) I, as was shown in Figure 3.13 for a purely resistive load RL = 2Zo. In the general case, with a complex load ZL, the reflection coefficient rL is complex, with 1fr # 0. From (3.20) we have
V (z) = v+[cos(f3z)  j sin(f3z) + p cos(lfr
+ f3z) + j p sin(lfr + f3z) ]
and IV(z) I = 1v+ 1J [cos(f3z)
+ p cos(lfr + f3z) ] 2 + [sin(f3z) + p sin(lfr + f3z) ] 2
(3.27)
which is the quantity plotted in various figures as the voltage standingwave pattern. For 1fr = 0 or Tr (i.e., load is purely resistive) (3 .27) reduces to (3.28) where the lower signs correspond to the case for 1fr = Tr. Similar expressions can also be written for II (z) I. Voltage and current standing wave patterns for different types of load impedances are shown in Figures 3.15 and 3.16. The interpretation of some of these patterns will become clearer after the discussion of line impedance in the following subsection. In general, for purely resistive loads (ZL = RL + j 0), the load position is a point of a voltage maximum or minimum, depending on whether RL > Zo or RL < Zo, respectively. This behavior is apparent from Figure 3.15 and can also be seen by considering (3.20) and (3.21). For RL > Zo, 0 < r L :::; 1 and IV(z = 0) 1 = Vmax = 1v+ 1c1 + p), whereas for RL < Zo, 1:::; rL < O and IV(z = 0) 1=Vmm= 1v+ 1c1  p). The standingwave patterns in Figure 3 .16 for ZL = Zo ± jZo illustrate specific cases of the general behavior for loads with a reactive (capacitive or inductive) component. In general, the sign of the reactance (positive or negative) can be determined by inspection of the voltage standingwave pattern. For ZL = RL + jXL, XL is negative (i.e., the load is capacitive) when the first minimum is at a distance smaller than one quarter of wavelength from the load (i.e., Zmin < A/4) and XL is positive (inductive) when the first minimum is at a distance greater than one quarter of a wavelength from the load (i.e., A/4 < Zmin < A/2), as illustrated in Figures 3.l 7a and b. The behavior illustrated in Figures 3. l 7a and b can be understood upon careful examination of (3 .19) and (3 .20). For a general complex load impedance, (3 .19) can be rewritten as
The phase angle 1fr of the reflection coefficient is 0 < 1fr < Tr if the load impedance is inductive (XL > 0) and Tr < 1fr < 0 if the load impedance is capacitive (XL < 0). The magnitude of the voltage along the line can be written from (3.20) as
SteadyState Waves on Transmission Lines
124
c:=====================================:=Jo, 2
....  .... /
/
I
IV(z) I
''
Z 0 1J(z) I .... .... /
''
/
' \
I
II
\
I
/
z/A. 0.75
\ \
I \
I
1
\
I \
I
\
I
\
I
' \
I \
I
1.5
I \
I
1

'/
\
I
0.5
\
'
0.5
0.25
0 2
.... .... /
RL = 2Z0
 .... ',

.... .... .... .... ;
/
' ' ' .... _....
;
1.5
Z 0 1J(z) I
IV(z) I
''
''
/
1
' ....
0.5
z/A. 1
0.75
0.5
0.25
0 2
.... '
Z 0 1J(z) I
IV(z)I ;
'
1.5
....
/
' ' ' .... _....
1
/
;
0.5
z/A. 1
....
0.5
0.75
Z 0 1J(z) I
I
\
2
IV(z)I
\
I \ \ \

I \
I \
I
I
0.75
I \
I
0.5
1
I
\
I \
I \
I \
1
0
' '\ \
z/A.
0.25
' ..... '
I
0.25
0.5
0
Figure 3.15 Voltage and current standingwave patterns for different purely resistive loads. Magnitudes of the voltage and current phasors (i.e., IV (z) I and Zo I/ (z) I) are shown as functions of electrical distance from the load position z / /..., for V + = 1 and for RL =
5Zo, 2Zo, Zo/2, and Zo/ 5.
Chap.3
Sec. 3.3
125
Lines Terminated in an Arbitrary Impedance
2 Z 0 1J(z) I
IV(z) I
1.5 1
0.5
1
0.75
0.5
0.25
0
2
/
... .... "" .....
/
I
''
'
1
' 
/
'
.....
''
1.5 '
I I I \
I
0.75
 ....
I
\
I
""
/
\
I \
/
\
I \
Z 0 1J(z) I
IV(z) I
0.5
' ,_.,,,.
I /
0.5
0.25
0
Figure 3.16 Voltage and current standingwave patterns for complex loads. Magnitudes of the voltage and current phasors (i.e., IV (z) I and Zo I/ (z) I) are shown as functions of electrical distance from the load z/A. for load) and ZL = Zo  }Zo (capacitive load).
v+ =
1 and for ZL
= Zo +JZo
(inductive
where z ::::; 0. Noting that iv+ I is a constant, consider the second ter1n and its variation with z (note that z decreases as one moves away from the load (at z = 0) along the transmission line). This ter1n is the magnitude of the sum of two numbers, one being the real number 1 and the other being a complex number, f(z) = pej(lfr+2/3z), which has a constant magnitude p (0::::; p ::::; 1, as determined by ZL and Zo) and a phase angle 1/f + 2f3z that decreases with decreasing z (corresponding to clockwise rotation of this complex number on a circle with radius p centered at the origin on the complex plane). The two different cases of capacitive and inductive load are shown respectively in Figures 3.l 7c and d. For an inductive load, we see from Figure 3.l 7d that as we move away from the load (i.e., starting at z = 0 and rotating clockwise), IV (z) I (which is proportional to 11 + f(z) I) first increases, reaches a maximum (at 1/f + 2f3z = 0), and then decreases, consistent with the variation of IV (z) I for the inductive load as shown in Figure 3.17b. Similarly, for a capacitive load, we see from Figure 3.17c that as z decreases (starting with z = 0), IV (z) I first decreases, reaches a minimum (at 1/f + 2{3z = n), and then increases, consistent with Figure 3.l 7a.
SteadyState Waves on Transmission Lines
126 fllll"' .....
IV(z) 1
1
,,'
',
I/
'
I I I
14 Zmin
''
''
/
''
I .. I 1
< A/4 1
,,,,....
Imaginary axis

.........
.......
/
''
I
'
I /
I
'
\
\
I I 1
I

Chap.3
1
I I /
•
I 1
/ 7
~~
'
'
\
ZL = RL + jXL
'
//
z=O
  I,___ ___,
..... O; inductive)
(b)
\
1+flt.)
//
I
/
I I I
/ I
\
/
d/ ~~•
\
1
I '
''
Real • axis
I .......
......
 
/
/
/
/
(d)
Figure 3.17 Variation of the voltage standingwave pattern in the vicinity of the load for inductive or capacitive loads. In descriptive terms, starting from the load (i.e., z = 0), the standingwave voltage at first increases (decreases) as one moves away from the load (i.e., clockwise in the diagrams shown) for an inductive (capacitive) load. (a) IV(z) I for a capacitive load. (b) IV(z) I for an inductive load. (c) IV(z)I = 11 + r(z) I as the sum of two complex numbers, 1 and r(z), for a capacitive load ( :rr < 1/1 < 0). (d) IV (z) I = 11 + r(z) I as the sum of two complex numbers, 1 and r(z), for an inductive load (0 < 1/1 < :rr). Note once again that v + = 1.
3.3.2 Transmission Line Impedance An important property of a transmission line is its ability to transform impedances. In Section 3.2, we saw that the input impedance of a short or opencircuited transmission line segment can be made equal to any arbitrary reactive impedance by simply adjusting its electrical length (l /A). The input impedance of a transmission line terminated in an arbitrary load impedance ZL is similarly dependent on the electrical length of the line or the distance from the load at which the impedance is measured. For the transmission line shown in Figure 3.12, the impedance seen looking toward the load ZL at any position z along the line (l s z s 0) given by (3.22) can be rewritten as
Z( ) = V(z) z I(z)
ZL  jZo tan(f3z)
=ZoZo  jZL tan(f3z)
(3.29)
Sec. 3.3
Lines Terminated in an Arbitrary Impedance
127
using (3.19), (3.20), and (3.21). Expression (3.29) for the line impedance is often written as ZL cos({Jz)  jZo sin({Jz) Z (z ) = Z o        Zo cos({Jz)  jZL sin({Jz) Note that since f3 = 2n /A, the impedance varies periodically with electrical distance (z /A) along the line, with the same impedance value attained at intervals in z of ±A /2. At the load, where z = 0, we have
as expected. In particular, the input impedance seen by the source at the source end, where z = l, is
~n
ZL + jZo tan({Jl) = [Z(z)]z=l = ls = Zo Zo + jZL tan({Jl)
Vs
(3.30)
For example, for a shortcircuited line (ZL = 0), the input impedance is Zin
= jZo tan(f3l)
and for an opencircuited line (ZL = oo), it is Zin = jZo cot(f3l)
as was shown in Sections 3.2.1 and 3.2.2. Example 3.7: Input impedance of a line. Find the input impedance of a 75cm long transmission line where Zo = 70Q, terminated with a ZL = 140Q load at 50, 100, 150, and 200 MHz. Assume the phase velocity vP to be equal to the speed of light in free space. Solution: (a) At f = 50 MHz, we have 'A ~ (3 x 108 )/(5 x 107 ) = 6 m, so the electrical length of the line is l/'A ~ 0.75/6 = 0.125. Noting that fJl = 2nl/'A, we then have, from (3.30), 140 + j70 tan(2n x 0.125) . 70 56 142 Zn ~ 70+j140 tan(2n x 0.125) = Q since tan(2n x 0.125) = 1. Note that the input impedance of the line at 50 MHz is capacitive. (b) Atf = 100 MHz, we have 'A~ 3 m and l/'A ~ 0.75/3 = 0.25. From (3.30) we have Zin ~
140 + j70 tan(2n x 0.25) (70) 2 70 = = 35Q 70+j140 tan(2n x 0.25) 140
since tan(2n x 0.25) = oo. Note that the input impedance of the line at 100 MHz is purely resistive.
SteadyState Waves on Transmission Lines
128
Chap.3
(c) At f = 150 MHz, A.~ 2 m and l/A. ~ 0.75/2 = 0.375. Again using (3.30), we have Zin ~ 56 + j42Q, which is an inductive impedance. (d) Atf = 200 MHz, we have A.~ 1.5 m and l/A. ~ 0.75/1.5 = 0.5. Using (3.30), we have Zin ~ 140Q. Note that the input impedance of the line at 200 MHz is purely resistive and is exactly equal to the load impedance.
Normalized line impedance.
In transmission line analysis, it is often convenient and common practice to normalize all impedances to the characteristic impedance Zo of the transmission line. Denoting the normalized version of any impedance by using a bar at the top, we can rewrite (3 .29) to express the nor1nalized line impedance Z (z) in ter1ns of the nor1nalized load impedance ZL
Z (z)


_i
= ZL  j tan(f3z) = ZL cos(f3z) sin(f3z) 1  jZL tan(f3z) cos(f3z)  jZL sin(f3z)
(3.31)
where
Using (3 .23) and (3 .24), we can further write
1 + jS tan(/3Zmin)
S
+j
tan (/3 Zmin)
+ jS sin(/3Zmin) S cos(/3Zmin) + j sin(/3Zmin) cos(/3Zmin)
(3.32)

which expresses the normalized load impedance ZL in terms of the measurable quantities S and Zmin· Sometimes, it is useful to express the real and imaginary parts of the load impedance ZL = RL + jXL explicitly in ter1ns of Zmin and S:
RL RL=
Zo
s S 2 cos 2 (/3Zmin)
+ sin
2
(/3Zmin)
2
XL (S  1) cos(/3Zmin) sin(/3Zmin) XL=  2 2 2 Zo S cos (/3Zmin) + sin (/3Zmin) The relationship between the polarity of XL (i.e., inductive versus capacitive) and the distance to the first minimum, as depicted in Figure 3 .17, can also be deduced by careful consideration of the preceding equation for XL.
Example 3.8: Unknown load. Determine an unknown load ZL from Sand Zmin measurements. The following measurements are carried out on a lOOQ transmission line terminated with an unknown load ZL, as shown in Figure 3.18. The voltage standingwave ratio S is 5, the distance between successive voltage minima is 25 cm, and the distance from ZL to the first voltage minimum is 8 cm. (a) Determine the load reflection coefficient rL. (b) Determine the unknown load impedance ZL. (c) Determine the location of the first voltage maximum with respect to the load.
Sec. 3.3
129
Lines Terminated in an Arbitrary Impedance

•
I
Z 0 =100Q S=S

ZL
unknown
I
Figure 3.18 Transmission line ter1ninated in an unknown impedance.
Solution: (a) Using (3.23) and (3.24), we have = P
s  1 = 5  1 ~ 0.667
S+l
A. =25 cm 2
5+1
A.= 50 cm
>
2n 1/f = n  2fJZmin = n + 2 50 rL
= ptJl/I ~
(8) = 0.36n rad or  64.8°
80 64 0.667ej ·
(b) We have (fJZmin) =
n
+ 0.36n 2
= 0.32n rad
>
tan(fJZmin)
~
1.58
and so using (3.32), ZL = Zo 1 + jS tan(fJZmin)
S + j tan(fJZmin)
~ 1001  j5(1.58) ~ 63.4  j 137.6Q (5  j 1.58)
(c) The location of the first voltage maximum is A./ 4 away from the location of the voltage minimum. Thus, we have Zmax
= Zmin  A./4 = 8  12.5 = 20.5 cm
Transmission line admittance. In Sections 3.5 and 3.6, when we discuss impedance matching and the Smith chart, it will be useful at times to work with the line admittance rather than the impedance. From (3 .29), we can find the expression for line admittance as YL  jYo tan(fiz) 1 1  rLe+j 2,Bz (3.33) Y (z ) = = Yo .2 = Y o      Z (z) 1 + fLe+1 ,Bz Yo  jYL tan(fiz) where Yo= (Zo)can be written as
1
.
Similar to (3.31), the normalized line admittance Y(z) = Y(z)/Yo
y (z) =


_j
YL  j tan(fiz) = YL cos(fiz) sin(fiz) 1  jYL tan(fiz) cos(fiz)  jYL sin(fiz)
(3.34)
SteadyState Waves on Transmission Lines
130
The load reflection coefficient
rL
Chap.3
can also be written in ter1ns of admittances as
Line impedance for resistive loads. The variation with z of the real and imaginary parts of the nor1nalized line impedance is illustrated in Figure 3. l 9a for the case of a resistive load with ZL = RL = 2Zo. The voltage and current standingwave patterns (from Figure 3.13) are also shown for reference in Figure 3.19b. Note from Figure 3. l 9a that, as viewed from different positions at a distance z from the load, the real part of the normalized line impedance varies between R(z) = 2 and R(z) = 0.5, with the distance between successive maxima being A/2. The line impedance is purely real
Zo

2.5 

ffie[Z(z)]
=
R(z)
2
1.5 1
;, ,1
;
,1
............
z /A_ I
....
0.5
',
......................~
0.5
',0.25
I
~~~~~~'...,_~~~~~~~~~._,_~~.~ro
1
0.75 ' ... ,
,' '
I ' ........ .;1
' ... , 

.9>m[Z(z)] = X(z) Z1 :::::::
I
0.5
I
'
1
0.0985A.
(a) 2
;
,,
;
/
/
/ .;;;
/
IV(z) I ....
''
;
''
Z 0 1I(z) I .... .... .... ~

;
' ' __ ....
/ ; ;
/
/
1.5
'
''
'' _ '
1
....
0.5 z/A.
1
0.75
0.5 (b)
0.25
0
Figure 3.19 Impedance along a line tenninated with ZL = RL = 2Zo. (a) The real and imaginary parts of the normalized line impedance Z (z) are shown as functions of z /A.. (b) Magnitudes of the voltage and current phasors (i.e., IV(z) I and Zo l/(z) I) are shown as functions of electrical distance from the load z /).. for V + = 1.
Sec. 3.3
Lines Terminated in an Arbitrary Impedance
131
at the load and at distances of integer multiples of A/ 4 from the load. These positions also correspond to the positions of voltage maxima and minima along the line. For z < A/4, the line impedance Z (z) is capacitive (i.e., its imaginary part is negative; X (z) < 0), reminiscent of the behavior of the opencircuited line 11 (see Figure 3.9). For A/4 < z < A/2, the line impedance Z(z) is inductive (X(z) > 0), and continues to alternate between capacitive and inductive impedance at intervals of A/4. An interesting aspect of the result in Figure 3.19a is the fact that ffie{Z(z)} = 1 at z1 ~ 0.0985).. 12 If the imaginary part of the line impedance at that position could somehow be canceled (as we shall see in Section 3.5), the line would appear (from all positions at locations z < 0.0985).) as if it were matched (i.e., terminated with an impedance Zo). For example, such cancellation can in principle be achieved by introducing a purely reactive series impedance that is opposite in sign to the reactive part of Z (z) at that position, as will be discussed in Section 3.5. The following example illustrates the determination of the point at which ffie{Z (z)} = 1 for a specific complex load impedance. Example 3.9: An invertedV antenna. A 50Q coaxial line filled with tefion (vp ~ 21 cm(ns) 1) is connected to an invertedV antenna represented by ZL, as shown in Figure 3.20. Atf = 29.6 MHz, the feedpoint impedance of the antenna is approximately measured to be ZL ~ 75 + j25Q. 13 Find the two closest positions to the antenna along the line where the real part of the line impedance is equal to the characteristic impedance of the line (i.e., Zo). Solution:
The line impedance at any position z is given by (3.29):
(75 + j25)  j50~ 3 + j(l  2~) Z(z) = 50 = 50   50  j(75 + j25)~ (2 + ~)  j3~ where ~ = tan(,Bz), ,B = 2n/A., and A.= vp/f ~ (2.1 x 108 )/(29.6 x 106 ) ~ 7.09 m. Multiplying both the numerator and the denominator with the complex conjugate of the denominator, we can extract the real part of Z (z) as
ffie{Z(z)} = ffie 50
3 + j (1 
2~)
(2+~)j3~

(2 + ~) + j3~ · (2+~)+j3~
= 50 3(_ 2 _+_ ~_ ) _3~_(_1__ 2~_) (2 + ~)2 + (3~)2
•
I
SQQ coax vp ~ 21

cm/ns
I
z=O
Z(z)
Figure 3.20 A coaxial line connected to an antenna.
11 Note
that this makes sense because in Figure 3.19a the load resistance is larger than the characteristic impedance (RL > Zo), which is also the case for the open circuit. 12The value of z can be read roughly from Figure 3.19a or accurately evaluated from (3.31) by letting 1 

Z(z) = 1 + jX(z). 13 R.
Dean Straw (Ed.), The ARRL Antenna Book, 17th ed., American Radio Relay League, pp. 2729, Newington, Connecticut, 19941996.
SteadyState Waves on Transmission Lines
132
Chap.3
The value of z for which we have ffie{Z(z)} = Zo = 50Q can then be found as ffie{Z(z)} = Zo
503(2 + ~)  3~(1  2~) = 50 (2 + ~)2 + (3~)2
>
2~
>
2
+ 2~
Using tan(fJz) =tan
2nz
~tan
A.
~1, ~2 ~ 1.37, 0.366
>
 1= 0
2nz 7.09
=~
and noting that z < 0, we find
z1
~
0.149A.
~
1.06 m
and
z2
~
0.444A.
~
3.15 m
as the locations at which the real part of the line impedance is equal to the characteristic impedance of the line.
Some aspects of the behavior of the real and imaginary parts of the line impedance shown in Figure 3 .19 for ZL = 2 can be generalized. For example, the line impedance seen at the positions of voltage maxima (minima) is always purely real and has maximum (minimum) magnitude. To see this, consider the voltage along the line at the position of a voltage maximum (i.e., z = Zmax) given by
with a maximum magnitude of
IV(z = Zmax) lmax = Vmax = IV+ l(l occurring at
l/f + 2,Bzmax = m2n
+ p)
m = 0, 1, 2, 3, ...
where TC s l/f < TC, Zmax s 0, and where m = 0 does not apply if TC same position, the current is equal to
I(z = Zmax) =
v+
s l/f
< 0. At the
.
eJ.BZmax(l  p)
Zo
with a minimum magnitude given by II (z = Zmax) Imin = lmin =
1v+ 1 Zo
(1  P)
so that the line impedance Z (Zmax) = V (Zmax) /I (Zmax) is clearly purely resistive and has a maximum magnitude given by IZ (Zmax) Imax = Rmax =
Vmax /min
= Zo
1+ p
1 P
= SZo
(3.35)
Sec. 3.3
133
Lines Terminated in an Arbitrary Impedance

ffie[Z(z)] 4 I~
I
z/')..
I I I
'
I I I
', ',
........
0.75 ........ ,
1
'
'
''
.... .... .... 0.25 ........ ,
0.5 '
I ' , 1 im[Z (z)]
'
1
z I A.
1
;   ;;:::.... :::: ...."'.'...,.... ........
0.75
/
,,
\2,'

0.25
.... _  . 0.5 .9lm[Z(z)]
.... ....
........ __ o
2
ffie[Z(z)]
1 ' ....   0.75
I
........ ........ ...  ~:::

z/A.
0
2
0.5 .... _   0.25
...  >c:::::......1
:">o.c~~
0
.9lm[Z(z)]
2

ffie[Z(z)]
I ' I ', I '
4
I I I
..
'
2
'
',
z/A._.,c:::::::::::::::::____~,~==~'~~;i:::::::::::::::.~_;_'~=~'~. ~!i{_ . ........ . 1 ............ , 0.75 0.5 .... , 0~25 0
''
.I

''"' .9lm[Z(z)]
',
I ' ....I
2
Figure 3.21 Line impedance for different purely resistive terminations. The real and imaginary parts of the normalized line impedance Z (z) are shown as functions of electrical distance z/'A along the line for ZL equal to (a) 5Z0 , (b) 2Z0 , (c) Zo/2, and (d) Zo/5.
whereas at the voltage minima (Zmin = Zmax  A./4), the line impedance Z(Zmin) is purely resistive with a minimum magnitude given by IZ(Zmin) lmin = Rmin =
Vmin Imax
1 p = Zo  1+ p
Zo s
(3.36)
Also, for purely resistive terminations (ZL = RL), the load is at a position of either minimum or maximum for the voltage and therefore, the load impedance RL is either equal to the minimum (RL = Zo/S when RL < Zo) or maximum (RL = SZo when RL > Zo)
SteadyState Waves on Transmission Lines
134
Chap.3
magnitude for the line impedance. Note that we have
S=l+_P 1p
for
RL > Zo
for
RL < Zo
Line impedance for complex load impedances. For general load impedances that are not purely resistive (i.e., ZL = RL + jXL), the behavior of the line impedance Z(z) = R(z) + jX(z) is similar to that for purely resistive loads, in that its real part R(z) varies between a maximum value of SZo and a minimum value of Zo/S, and the imaginary part X(z) alternates sign at intervals of 'A/4. (SZo occurs at the positions of the voltage maxima when the line impedance is purely resistive and therefore is also the maximum magnitude of the line impedance. Zo/S occurs at the positions of the voltage minima when the line impedance is also purely resistive and therefore is also the minimum magnitude of the line impedance.) However, the maxima and minima of the magnitudes of either the voltage or the line impedance are not at

3

Re[Z(z)] 2 ~
,
....
I
I
.... ....
z/A.
I
....
1
.... ....
I
.... 0.75 .... .... ....
I I I
I
0.25 ....
I I
ZL =Zo + jZo
' ' .... .,,,.I
,_/ Im[Z(z)]
0
I
~

1
I ,
0.5

I
1
2 3 
Re[Z(z)] ,
I
z/).
I
2
I
I
I
'
II
'
t~~'~~~'"'~t~+1"~~t'~~~+ 0
1
I
'
I
0.75
'
I
',,0.5 I _ 0.25 ' , ...,.;/ Im[Z(z)]
'...,,
,
' , 1
2 Figure 3.22 Line impedance for two different complex load impedances. The real and imaginary parts of the normalized line impedance Z (z) are shown as functions of electrical distance z/). along the line for (a) ZL = Zo + jZo and (b) ZL = Zo  jZo.
Sec. 3.3
135
Lines Terminated in an Arbitrary Impedance
the load position. Figures 3.21 and 3.22 show plots of the real and imaginary parts of the nor1nalized line impedance Z (z) as functions of z/A. for selected load impedances. Example 3.10: Reflection coefficient, standingwave ratio, and maximum and minimum resistances. A radio transmitter is connected to an antenna having a feedpoint impedance of ZL = 70 + j lOOQ with a 50Q coaxial line, as shown in Figure 3.23. Find (a) the load reflection coefficient, (b) the standingwave ratio, and (c) the two positions closest along the line to the load where the line impedance is purely real, and their corresponding line impedance values.

I
Radio transmitter

•
Coax
...
Antenna ZL = 70 + jlOOQ
Z 0 =50Q
.. 
I
z=O
Figure 3.23 Transmission line terminated in an antenna. In general, the feedpoint impedance of an antenna is complex.
Solution: (a) The load reflection coefficient is
r
= ZL  Zo = 70+j100  50 =~ 0.653d38.90 L ZL + Zo 70+j100 + 50
(b) The standingwave ratio is
s=
1 + p ~ 1 + 0. 653 ~ 4. 7 6 1 p 1  0.653
(c) The maximum voltage position is the position Zmax at which the line impedance is purely real and the magnitude of the line impedance is a maximum; note that the position of the first voltage maximum is closer to the load than that of the first voltage minimum because the load impedance is inductive. To find the maximum voltage position, use 1/1 + 2f3Zmax = 0 > Zmax ~ 0.054A.:
SZo
Rmax =
~
(4.76)(50) = 238Q
The minimum voltage position is the position Zmin at which IZ (z) I is minimum; note that this is the next closest position where the line impedance Z (z) is real. To find the minimum voltage position, use
1/1 + 2fJZmin
=
n
>
Zmin ~ 0.304A.
Note that as expected, Zmin = Zmax  A./4. We then have Rmin =
Zo/S
~
50/(4.76)
~
10.5Q
SteadyState Waves on Transmission Lines
136
Chap.3
3.3.3 Calculation of v+ Up to now, we have primarily focused on the line impedance and the variation of the voltage and current along the line without particular attention to the source end of the line. The source that excites the transmission line shown in Figure 3.12 is a voltage source with an opencircuit phasor voltage Vo and a source impedance Zs. Using equation (3 .20), the voltage Vs at the source end of the line (z = l) is
As seen from the source end, the transmission line can be represented by its input impedance, Zin. We can thus also express the sourceend voltage phasor Vs in ter1ns of the source parameters Vo and Zs by noting the division of voltage between Zin and Zs, namely,
By equating the two preceding expressions for Vs, we can solve for the constant v+:
Note that the knowledge of V + and the wavelength A = 2rr / f3 completely specifies the transmission line voltage and current as given in (3.20) and (3.21), as for any given transmission line with characteristic impedance Zo and load ZL (and hence r L).
Example 3.11: Coaxial line feeding an antenna. A sinusoidal voltage source of °Vo(t) = 10 cos (5n x 107 t) V and Rs = 20Q is connected to an antenna with feedpoint impedance ZL = lOOQ through a 3m long, lossless coaxial transmission line filled with polyethylene (vp = 20 cm(ns) 1) and with a characteristic impedance of Zo = 50Q, as shown in Figure 3.24a. Find (a) the voltage and current phasors, V (z) and I (z), at any location on the line and (b) the corresponding instantaneous expressions °V(z, t) and !P(z, t). Solution: (a) At f = w / (2n) = 25 MHz, the wavelength in a polyethylenefilled coaxial line is A= vp/f = (20 cm(ns) 1)/(25 MHz)= 8 m The electrical length of the 3m line is then l /"A = 3 /8 = 0.375, so we have {Jl = 2n(0.375) = 3n /4 and tan({Jl) = 1. The input impedance seen at the source end can be calculated using (3.30):
z. Ill
=so 100 + jSO(l) = 40 ·3on 50 + j 1QQ( 1) +j
Sec. 3.3
137
Lines Terminated in an Arbitrary Impedance
z.
lil
l = 3m
Rs = 200
v0 = 10ejo (V) 25MHz
+
son coax filled
+
vs
with polyethylene vp == 20 cm(ns)1
VL
Antenna ZL = 1000
z =O
z = 3m (a)
z = 3m Equivalent Circuit (b)
Figure 3.24 Coaxial line feeding an antenna. (a) Circuit configuration. (b) Thevenin equivalent circuit seen from the source end.
Using the equivalent circuit shown in Figure 3.24b, we have . Vs = Zn Vo= 40 + 130 (10) Rs + Zin 60 + j30
~
.36.9° 5e1 (10) 3J5ej26.6°
~ 7.45dl0.3o V
We can also write an expression for Vs by evaluating V (z) at z =  3 mas
where
rL
is the load reflection coefficient given by (3.19): 100  50 100 + 50
1 3
Equating the two expressions for Vs, we can determine the complex constant •
Vs = v+ei 3n/4 1 + }_
3
~
30 10 7.45d ·
>
v+
~ 7.07ej 1430
so that the voltage phasor at any position z from the load is given as V(z) ~ 7.07ej143oejnz/4 (1 + jeinz/2) V
and the corresponding current phasor is /(z) ~ 0.141ej143oejnz/4 (1  jeinz/2) A
v
v+
SteadyState Waves on Transmission Lines
138
Chap.3
(b) Using °V(z, t) = ffie{V (z)ejwt} and !P(z, t) = ffie{/ (z)ejwt}, we find
°V (z, t) ~ 7.07 cos ( 5n 107 t  n z  143°) 4
+ 2.36 cos (sn 107 t + n z 4
143°) V
and
3.4 POWER FLOW ON A TRANSMISSION LINE In practice, the primary purpose of most steadystate sinusoidal transmission line applications is to maximize the timeaverage power delivered to a load. The power and energy flow on a transmission line can be determined from the line voltage and line current; the product of the instantaneous current !P (z, t) and instantaneous voltage °V (z, t) at any point 14 z is by definition the power that flows into the line at that point. In most applications, the quantity of interest is not the rapidly varying instantaneous power but its average over one sinusoidal period Tp, namely, the timeaverage power, which is given by Tp
1
Pav(z) = °V (z ' t) !P (z ' t) dt Tp o
where Tp = 2n / w. The timeaverage power can also be calculated directly from the voltage and current phasors: 1 Pav(z) = ffie{V (z)[I (z)]*}
2
Consider the general expressions for the voltage and current phasors along a lossless unifor1n transmission line: V(z) =
I(z) =
y+ ej,Bz
+ fL y+ ~,Bz
v+ j,Bz  e Zo
forward
v+ _;,Bz rL e......
wave
Zo
...,,...,
reverse wave
We denote the timeaverage power carried by the forward and backward traveling waves asp+ and p, respectively, and we evaluate them directly from the phasors. The timeaverage power carried by the forward wave is
14
Note that the product °V(z, t)i(z, t) represents power flow into the line, rather than out of the line, due to the defined polarity of the current I (z) and voltage V (z) in Figure 3.12.
Sec. 3.4
139
Power Flow on a Transmission Line
Note that although v+ is in general complex, v+cv+)* = The power carried by the reversepropagating wave is 1
(rL y+ ej /3Z)(rL y+ ej /3Z)*
2
Zo
p = ffie
2 1v+ 1
is a real
15 number.
The fact that p is negative simply indicates that the backward wave carries power in the opposite direction with respect to the defined polarity of I (z) and V (z) in Figure 3.12. The net total power in the forward direction is then given by +
_
Pav= P + P =
2
Iv + 1 2Zo

2
P
2
Iv+ 1 2Zo
_
2
IV + 1 ( 1 _ 2Zo
2)
P
(3.37)
Thus, the net timeaverage power flow on a transmission line is maximized when the load reflection coefficient rL is zero, which, according to (3.19), occurs when ZL = Zo. Note that when rL = 0 and thus ZL = Zo, the standingwave ratio S = 1, as is evident from (3.23). Since the transmission line is assumed to be lossless, all of the net power flowing in the +z direction is eventually delivered to the load. The same result can also be obtained by examining the power dissipation in the load, which is given by 16
Noting that we have
= V(z)lz=o = v+ + rLv+ = v+c1 + rL) v+ v+ v+ /L = I(z)lz=O =  rL = (1  rL)
vL
Zo
Zo
Zo
and substituting in the preceding expression for PL, we find
(3.38)
If y+ =A+ jB, then v+cv+)* =(A+ jB)(A  jB) = A 2 + B 2 =IA+ jB l2 = 1v+1 2 . 16 Note that PL can also be written in terms of the load voltage VL and load admittance YL as 15
1 2 1 2 PL= IVLI ffie{YL} = IVL I Gr. 2 2
140
SteadyState Waves on Transmission Lines
Chap.3
which is identical to the total net forward power Pav as derived in (3.37). We see that Pav = PL as expected, since, for the case of a lossless line as assumed here, all of the net power traveling toward the load must be dissipated in the load. The same result can further be obtained by evaluating P (z) at any point along the line using the total voltage and current phasors (rather than separating them into forward and reverse traveling wave components). In other words, 1 Pav(z) =  ffie{V (z)[I (z)]*}
2
where V (z) and I (z) are given by (3.20) and (3.21), respectively. (This derivation is left as an exercise for the reader.) In summary, the total net power propagating in the +z direction is 2
Pav (z) =
Iv+ 1 2Zo
2
(3.39)
(1  p )
and the fallowing observations concerning power flow on a lossless transmission line can be made: • For a given v+' maximum power is delivered to the load when ZL = Zo, rL = 0 (i.e., p = 0), and S = 1. Noting that Zo is a real number, this condition is realized when the load is purely resistive, that is, when ZL = RL = When RL = Zo, the load is said to be matched to the line and all of the power P + is delivered to the load. Detailed discussion of impedance matching is given in Section 3.5. • To deliver a given amount of power (say, PL) when the line is not matched (i.e., S > 1) requires higher wave power in the incident wave with correspondingly higher voltages (P + = IV+ 12 / (2Zo)). The higher voltages are undesirable as they may cause breakdown 17 of the insulation between the two conductors of the line. • The power efficiency achieved by matching can be assessed by considering the ratio of the power PL that is dissipated in a given load to the forward wave power P + that would be delivered to the load if the line were matched:
.zo.
PL 2 4S =1p =   p+ (1 + S) 2
The variation of PL/P+ with S is plotted in Figure 3.25. We see that PL/P+ = 1 for S = 1 and monotonically decreases to zero as S gets larger. Note that for voltage standingwave ratios S < 1.5, which are relatively easy to achieve in practice, more than 90 percent of the power in the forward wave is delivered to the load. In other words, it is not necessary to strive for S very near unity to attain maximum power transfer to 17 Electrical
breakdown of insulating materials will be discussed in Section 4.10.
Sec. 3.4
141
Power Flow on a Transmission Line
1.2
0.8 0.6 0.4 0.2
s 1.5
2
2.5
3
3.5
4
Figure 3.25 Power efficiency as a function of standingwave ratio S.
the load. Usually the more important issues are ensuring that the value of S is not so large as to make the line performance highly sensitive to frequency (see Section 3 .5), and that the design of the line can accommodate large reactive voltages and currents that accompany a large value of S. The degree of mismatch between the load and the line is sometimes described in terms of return loss, which is defined as the decibel value of the ratio of the power carried by the reverse wave to the power carried by the forward wave, given as S+l Return loss =  20 log 10 p = 20 log 10  S 1 If the load is perfectly matched to the line (p = 0), the return loss is infinite, which simply indicates that there is no reverse wave. If the load is such that p = 1 (i.e., a shortcircuited or opencircuited line, or a purely reactive load), then the return loss is 0 dB. In practice, a wellmatched system has a return loss of 15 dB or more, corresponding to a standingwave ratio of rvl.43 or less. Example 3.12: A 125MHz VHF transmitterantenna system. A VHF transmitter operating at 12S MHz and developing Vo= 100ej v with and source resistance of Rs= son feeds an antenna with a feedpoint impedance18 of ZL = 100  j60 through a son, polyethylenefilled coaxial line that is 17 m long. The setup is shown in Figure 3.26a. (a) Find the voltage V (z) on the line. (b) Find the load voltage VL. (c) Find the timeaverage power absorbed by the VHF antenna. (d) Find the power absorbed by the source impedance Rs. 00
18 R.
Dean Straw (Ed.), The ARRLAntenna Compendium, Vol. 4, p. 56, The American Radio Relay League, Newington, Connecticut, 19951996.
SteadyState Waves on Transmission Lines
142
R 8 =50n •
A y
•

y
+
VHF transmitter r " "V V 0 = lOOejO V "" (125 MHz)
.•


.• 
+
Coax filled with polyethylene Zo=50n
Vs
~
l=17m

Chap.3
VL
VHF antenna ZL = l00j60Q
•
.I
(a)
+ V 0 = lOOejO V (125 MHz) "V
z.
lil
(b) Figure 3.26 A 125MHz VHF transmitterantenna system. (a) Circuit diagram. (b) Equivalent circuit seen by the source.
Solution: (a) First we note that for a polyethylenefilled coaxial line, the wavelength at 125 MHz is (using Table 3.1 and assuming vp at 125 MHz is the same as that at 300 MHz) 8 "A= vp/f = (2 x 10 m/s)/(125 MHz) = 1.6 m. The length of the line is then l = 17 m = 10.625"A = 10.5"A + 0.125"A
Noting that tan({Jl) = tan[(2n/"A)(0.125"A)] = tan(n/4) = 1, and the input impedance of the line seen from the source end is then (equation (3.30)):
z. in
(100  j60) + j50tan(n/4) = 50 50 + j (100  j 60) tan(n /4)
~
22. 6  1·25.1n
The equivalent circuit at the source end is as shown in Figure 3.26b. The sourceend voltage Vs is then
22.6  j 25 .1 _;oo r v j 28 .9o Vs = Vo . lOOe _ 44.0e V Rs+ Zin 50 + 22.6  ]25.1 Zn
rv
But we can also evaluate Vs from the expression for the line voltage V (z) as
Sec. 3.4
143
Power Flow on a Transmission Line
where we have used the facts that ej 21 ·25:rr = d 1·25:rr = ejo.75:rr, that the load reflection coefficient rL is r
ej 42 .5:rr
=
ejO.S:rr,
and
50  j 60 ~ 0.483ej28.4o 150  j60
_ 100  j60  so L100j60+50
Equating the two expressions for Vs, we can detennine the unknown voltage V + as Vs ~
v+
ej 3:rr/4(0.770
90 28  j0.425) ~ 44.0ej ·
>
v+
=
sod 3:rr14 v
Thus the expression for the line voltage is V (z) ~
40 28 + 0.483ej( ·  5rrz/2)) V
50d 3:rrf4 ejS:rrz/4 (l
(b) The voltage at the load end of the line is VL = V (z = 0) ~
4 3 50d :rr/ (1
40 28 + 0.483ej · )
~ 5od 3:rr/4(1.43  j0.230) ~ 72.2ej 1260 V
(c) Using the value of VL, the timeaverage power delivered to the VHF antenna can be calculated as VL 2 RL ZL
~
(72.2) 2(100) 4 2(1.36 x 10 )
~
19.2 W
2 2 2 4 2 where IZL l = 1100  j601 = (100) + (60) = 1.36 x 10 . Note that PL can also be found using the sourceend equivalent circuit (Figure 3.26b). Since the line is lossless, all of the timeaverage power input to the line at the source end must be absorbed by the antenna. In other words, P =
L
!
2 Vs
2 Zin
R·
2
~!
44 C ·0) (22.6) 2 [(22.6) 2 + (25.1) 2]
m
~ 19.2 W
(d) Noting that Rs = 50Q and Zm ~ 22.6  j25.1Q, the current at the source end (again considering the lumped equivalent circuit shown in Figure 3.26b) is jQO
ls= _Vi_o_ Rs+ Zin
rv
lOOe ~ l.30dt9.10 A 50 + 22.6  j25.1
Using the value of ls, the timeaverage power dissipated in the source resistance Rs is then 1 2 PRs = lls l Rs 2
~
1 2 (1.30) (50) 2
~
42.3 W
Therefore, the total power supplied by the VHF transmitter is Ptotal = PRs +PL 61.5 w.
~
SteadyState Waves on Transmission Lines
144
Chap.3
Example 3.13: Parallel transmission lines. Three lossless transmission lines are connected in parallel, as shown in Figure 3.27. Assuming sinusoidal steadystate excitation with a source to the left of the main line, find the reflection coefficient on the main line and the percentage of the total net forward power that is absorbed by the two loads ZL2 and ZL3 for the following cases: (a) Zo1 = Zo2 = Zo3 = ZL2 = ZL3 = lOOn, (b) Zo1 = SOn, Zo2 = Zo3 = ZL2 = ZL3 = lOOn, and (c) Zo1 = Zo2 = Zo3 = lOOn, ZL2 = ZL3 =SO+ jSOn. Solution: (a) Zo1 = Zo2 = Zo3 = ZL2 = ZL3 = lOOn Since the two parallel branches are both matched, the input impedance seen at the terminals of each branch is independent of its line length and is simply lOOn. Thus, the line impedance ~ seen from the main line is the parallel combination of two 1oon impedances, or Zj =son. The reflection coefficient at the junction (as seen from the main line) is then
so  100 so+ 100
1 3
In other words, the power efficiency, defined as the percentage of total power that is delivered to the loads versus that of the forward wave on the main line, is PL p+ = (1 
2
1rj 1) x 100 ~ 88.9%
Since each line presents the same impedance at the junction, each load absorbs half of the total power delivered, or approximately 44.4% of the total power of the incident wave.
A.14
z.J
Source end
Main line Zo1
A.12
Figure 3.27 Parallel transmission lines. A main line with characteristic impedance
Zo1 drives two other lines of lengths A./4 and A./2, with characteristic impedances Zo2 and Zo3, respectively.
Sec. 3.4
145
Power Flow on a Transmission Line
(b) Zo1 = SOn, Zo2 = Zo3 = ZL2 = ZL3 = lOOn The line impedance at the junction seen from the main line is once again Zj =son. However, the reflection coefficient is now zero since r·= J
~
 Zo1
=
~ + Zo1
SO  so so + so
= o
In other words, 100% of the power of the forward wave in this case is delivered, with each load absorbing SO%. (c) Zo1 = Zo2 = Zo3 = lOOn, ZL2 = ZL3 =SO + jSOn We now need to evaluate the input impedances of each of the two parallel sections at the junction as seen from the main line. Note that the length of line 2 is A./ 4, so that using (3 .30), its input impedance is
2 (100)    = 100  ·100 n so + jSO J whereas line 3 has length A./2 and thus presents ZL3 at the junction; in other words, Zm3 = ZL3 = SO + j SOn. The line impedance ~ at the junction as seen from the main line is then the parallel combination of Zin2 and Zin3 , namely, (100  jlOO)(SO + jSO) 6 . z. = = o+1 2on J 100  j 100 + so + j so and the reflection coefficient at the end of the main line is
r·= J
~
 Zo1
~ + Zo1
=
60 + j20  100 60 + j20 + 100
~
_; 1460
o.211e
The percentage of the incident power delivered to the two loads is thus PL
2
p + = (1  1rj1 ) x 100 ~ 92.3% Since the two impedances Zin2 and Zin3 appear at the junction in parallel, they share the same voltage. In other words, we have and We can thus calculate the ratio of the powers delivered to the two loads as
Therefore we have PL2
~
1
3
x 92.3% ~ 30.8%
and
PL3
~
2 3
x 92.3%
~
61.S%
146
SteadyState Waves on Transmission Lines
C hap . 3
Example 3.14: Cascaded transmission tines. An antenna of measured feedpoint impedance of 72 + j36Q at JOO MHz is to be driven by a transmiuer through two cascaded coaxial lines with the following characteristics:
c::: 30 cm(ns) 1
Zo1 = 120Q
11 = 3.75 en airfilled
vp 1 =
Zo2 = 60Q
12 = 1.75 en polyelhylenelillcd
vp2 ::: 20
cm(ns) 1
where we have used Table 3.1 for Uie phase velocity vp2 for lhc polyethylenefilled coaxial tine assuming it to be approximately lhc same at 100 MHz. (a) Assuming both lines to be lossless and assuming a source voltage of Vo= 100& 0 V and resisLancc of R, = 50Q for the transmiuer, find the timeaverage power delivered LO lhc load. (b) Repeal part (a) with 11 = 4.5 m. The setup is shown in Figure 3.28a. Solution: (a) At 100 MHz, lhc wavelengths for Lile. two coaxial lines are i 1 = vp 1If ::: 3 x 108 / 108 = 3 en and i2 = vp 2 /f = 2 x 108/ 108 = 2 m, respectively. The lengths of the lines arc lhcn 11 = 3.75 en ::: l.25)q = ). 1 + 0.25>. 1 12
= 1.75 en = 0.875>.2 = 0.5,l.2 + 0.375!2
Note lhat lhc corresponding phase constants are /! 1 = 2rr/ i 1 and fh = '2Jr/i2. The impedance Z, seen looking toward the load at lhc interface between Uie two coaxial
Zin
11 =3.75 m
R,  500 Transmitt~r
r;:...

+ + ...__
Vo= lOOeJO ~
v,
(V) lOOMHz
Air
Zo1= 1200
vP, = 30 cm(nsr' ,.
(a)
Polyethylene 2 02 =600 VL v1, 2 =20cm(nst'
Antenna
ZL=72 +j360
I 12 =L75 m
Zx
R, +
v,
Vo= IOOeiO
(b)
F"igure 3.28 Cascaded transmission line.~. (a) Circujt diagram showing an airfilled line of in1pedance 120Q cascaded "'ith a polyethylenefiUed line of impedance 60Sl. (b) Equivalent circuit seen by the source.
Sec. 3.5
147
Impedance Matching lines is, using (3.30), ZL + j:Z02 tan(f32l2) (72 + j 36) + j 60( 1) . Zx = :Z02 = 6O = 36 + J 12Q Zo2 + jZL tan(f32l2) 60 + j (72 + j 36) ( 1) The input impedance is then
Zx + jZ01 tan(f3ol1) ZJi Zin = Zo1 = Zo1 + jZx tan(f3ol1) Zx
~
2 (120) 36+j12
.
= 360  J 120Q
With reference to the equivalent circuit in Figure 3 .28b, we have
Vs =
Zin Vo = 360  ~ 120 (100) Rs+Zin 4101120
~ 88.8ej2.120 V
Thus the power delivered to the antenna is PL= Pin=
!2
2
Vs Zin
2 88 8 ' ) (360) ffie{Z· } = Ill 2 (1.44 x 105)
! (
~ 9.86 W
(b) With 11 = 4.5 m = l.5A. 1, we have Zin= 36+j12
>
so that PL
~
v.
= 36+j12 (100) s 86+j12
1 (43.7) 2 (36) 2 (1440)
~
~ 43.7ei10.50 v
23.9 W
which is a significant improvement in power delivered, achieved simply by making the first line segment longer. This result indicates that the amount of power delivered to a load sensitively depends on the electrical lengths of the transmission lines.
3.5 IMPEDANCE MATCHING We have already encountered the concept of impedance matching in previous sections, in connection with standing waves on transmission lines. It was shown that if the characteristic impedance Zo of the line is equal to the load impedance ZL, the reflection coefficient rL = 0, and the standingwave ratio is unity. When this situation exists, the characteristic impedance of the line and the load impedance are said to be matched, that is, they are equal. In most transmission line applications, it is desirable to match the load impedance to the characteristic impedance of the line in order to reduce reflections and standing waves that jeopardize the powerhandling capabilities of the line and also distort the infor1nation transmitted. Impedance matching is also desirable in order to drive a given load most efficiently (i.e., to deliver maximum power to the load), although maximum efficiency also requires matching the generator to the line at the source end. In the presence of sensitive components (lownoise amplifiers, etc.), impedance matching improves the signaltonoise ratio of the system and in other cases generally reduces amplitude and phase errors. In this section, we examine different methods of achieving impedance matching.
SteadyState Waves on Transmission Lines
148
C hap . 3
3.5.1 Matching Using Lumped Reactive Elements
The simplest way to match a given transmission line to a load is to connect a lumped reactive element in parallel (series) at the point along the line where the real part of the line admittance (impedance) is equal to the line characteristic admittance (impedance). 19 This method is useful only at relatively low frequencies for which lumped elements can be used. The method is depicted in Figure 3.29, which shows a shunt (parallel) lumped reactive element connected to the line at a distance l from the load. Since the matching element is connected in parallel, it is more convenient to work with line admittance mther than line impedance. The normalized admittance Y(z) seen on the line looking toward the load from any position z is given by (3.33):



In Figure 3.29, matching requires that !'2 = Y1 + Ys _ 1. Thus ~e first need to_ choose the position l along the line such that Y(z =  l ) = Y1 = l  jB, that is, ~{Yi} = l. Then we c~oose !,he lumped shunt ele,!llent !O be purely reactive with an appropriate value such that Ys = j B, which results in Y2 = Y1+ y, = 1. Substituting z =  l , we have l  l Le i 2ftl .Y1 = Y (z =  l) = ·i pt = 1  J B l + i Le J .
from which we can solve for the position I of the lumped reactive element, its type (i.e., capacitor or inductor), and its normalized susceptance  iJ. Since the load reflection
Y:i =Yi +Y, Y,
I
~
Yo
Y,
I
L' Yo
YL
c:====t~L.,1::::===================>:,J z  1 z0 l'"igure 3.29 1\tatching by a lumped shunt e.Jeme.nt. The shunt element Y." is connected at a distance I from the load such that the line admjttance Y2 (.t =  I )= Y1 (z =  l ) + Y."' = Yo. 19Tuis (X>SSibility was noted earUer in Section 3.3.2 in connection "''ith Figure 3. 19, "''here 0'te {Z(z ) } = I 1 at z1 ~  0.0985>.. Note that, in general, the reaJ pan of the line admittance is equal to unity at some other point z2 ¢ z1.
Sec. 3.5
coefficient
149
Impedance Matching
rL
is, in the general case, a complex number given by
we have 1  pejl/Jej 2f3 1 1 + pejl/J ej 2f3 1
(3.40)

where () = l/f  2{31. By multiplying the numerator and the denominator by the complex conjugate of the denominator,20 we obtain 1  p2
_
Yi= l
\,,,,
2 p sin ()
.
2 + 2p cos() + p __.,,..___
(3.41)
1~~~~~
1 + 2p cos()
+ p2
ffie{Y1 }= 1
Since

ffie { Y1}
= 1, we can write
1 p2
= 1 1 + 2p cos() + p 2
which yields () =
l/f  2{31 = cos 1 (p)
(3.42)


In other words, the distance l from the load at which Y (z = l) = 1  j B is given by
lfr  () l/f  cos  I ( l = = 2/3 2/3
p)
=
A 4rr
[l/f
 cos
1 (
)
 p ]
(3.43)
Note that, in general, () = cos 1 (p) (with p > 0) has two solutions, one in the range rr /2 s ()I s rr and the other in the range rr s e2 s Jr /2. Also, if (3.43) results in negative values for l, then the corresponding physically meaningful solution can be found by simply adding A./2. 21 To find B, we substitute cos()= p and sin()= ±Jl  p 2 (where the plus sign corresponds to rr /2 ::::; e1::::; rr and the minus sign corresponds to rr s e2 s rr /2) in the imaginary part of Y1 given by (3.41), resulting in
B
=
.9im{Y1} [cosO=p
=
2pJ 1  p 2 ± 2 2 1  2p + p
2p
= ± J1 
2o[l + peie]* = 1* + (peie)* = 1 + peje 21 For negative l, we have 2rr < 2{Jl < 0. By adding A./2, we have 2{J(l 2{Jl + 2rr > 0, which lies between 0 and 2rr.
p2
(3.44)
+ A./2) = 2{Jl + 2(2rr/A.)(A./2) =
SteadyState Waves on Transmission Lines
150
Chap.3
where the plus and minus signs correspond to a shunt capacitor (B1 > 0) and a shunt
inductor (B2 < 0), respectively. 22 This susceptance also determines the value of the lumped reactive element Ys, which must be connected in parallel to the line in order to cancel out the reactive part of Y1. In particular, we should have Ys = +j B so that the total admittance Y2 seen from the left side of Ys in Figure 3 .29 is 

Y2 = Y1


+ Ys =
1  jB

+ jB =
1
When matching with series lumped reactive elements, similar equations can be derived for the distance l away from the load at which Z1 = Z (z = l) = 1  j X, 1
l/f 
l =
cos p
2/3
A ( = l/f 4:rr

cos
_1
p
)
(3.45)
and the nor1nalized reactance of the series lumped element that would provide matching is given by 
2p
X=±  
Jl p2
Example 3.15: Matching with a single reactive element. An antenna having a feedpoint impedance of 1 lOQ is to be matched to a 50Q coaxial line with vp = 2 x 108 ms 1 using a single shunt lumped reactive element, as shown in Figure 3.30. Find the position (nearest to the load) and the appropriate value of the reactive element for operation at 30 MHz using (a) a capacitor and (b) an inductor. Solution:
The load reflection coefficient is
rL
ZL = ZL +
Zo Zo
110  50 0 375 = 110 + 50 = ·
Using (3.44), the reactive admittance (or susceptance) at the position of the shunt element is
jj = ±
2 x 0.375
Ji  (0.375)2
~ ±0.809
(a) For B 1 ~ + 0.809, the shunt element must be a capacitor. The nearest position of the capacitor with respect to the load can be found as
11 = 
81
2{3
~

A.
cos 4n ...
_1
(0.375) V"
~

A.
4n
(1.955)
~
0.156A.
rr /2 ~ • ZL= llOn
20 cm(ns)1
.• 
• •
son at30MHz
(b) Figure 3.30 Matching with a single reactive element. The two solutions determined in Example 3.15. (a) Using a shunt capacitor. (b) Using a shunt inductor.
Since 11 < 0, we add A./2 to get 11 ~ 0.156A. + 0.5A. = 0.344A.. Using A. = vp/f = 2 x 108 /(30 x 106) ~ 6.67 m, the actual position of the shunt capacitor is 11 ~ 0.344 x 6.67 ~ 2.30 m. To determine the capacitance Cs, we use
(j wCs) (Zo) = j B 1
>
[j (2n x 30 x 106 Cs)(50)] ~ +j0.809
>
Cs
~
85.8 pF

(b) For B2 ~ 0.809, the shunt element must be an inductor. Similarly, the nearest position of the inductor is
12 = 
82 2~
~

A.
4n
cos
_1
(0.375)
~

A.
4n
(1.955)
~
0.156)..
n
~
0.156 x 6.67
[j/(2n x 30 x 10 Ls)J(50) ~ j0.809 6
Ls
~
0.328 µ,H
~
1.04
SteadyState Waves on Transmission Lines
152
Chap.3
3.5.2 Matching Using Series or Shunt Stubs In Section 3.2, we saw that short or opencircuited transmission lines can be used as reactive circuit elements. At microwave frequencies, it is often impractical or inconvenient to use lumped elements for impedance matching. Instead, we use a common matching technique that uses single open or shortcircuited stubs (i.e., transmission line segments) connected either in series or in parallel, as illustrated in Figure 3 .31. In practice, the shortcircuited stub is more commonly used for coaxial and waveguide applications because a shortcircuited line is less sensitive to external influences (such as capacitive coupling and pickup) and radiates less than an opencircuited line segment. However, for microstrips and striplines, opencircuited stubs are more common in practice because they are easier to fabricate. For similar practical reasons, the shunt (parallel) stub is more
l
Yo
Parallel (shunt) 1 s stub
I I I
Open or : shorted 1
I I
l I
Zo
Zo
~I~. '
~'
I
' '
' '
Series stub

I >~
~
Open or shorted Figure 3.31 Matching by shunt or series open or shortcircuited stubs.
Sec. 3.5
153
Impedance Matching I Yz I I I l I ...~~~~~~~~~ I ...~~~~~~~~~~~.
l0
Figure 3.32 Matching with a single parallel (shunt) shortcircuited stub.
convenient than the series stub; the discontinuity created by breaking the line may disturb the voltage and current in the case of the series stub. The principle of matching with stubs is identical to that discussed in Section 3.5.1 for matching using shunt lumped reactive elements. The only difference here is that the matching admittance Ys is introduced by using open or shortcircuited line segments (or stubs) of appropriate length Zs, as shown in Figure 3.32. In the following, we exclusively consider the case of matching with a shortcircuited shunt stub, as illustrated in Figure 3.32. The corresponding analysis for opencircuited stubs is similar in all respects and is left as an exercise for the reader. With the required location Z and the normalized admittance B of the stub as determined from (3.43) and (3.44), we need only to find the length of the stub Zs necessary to present a normalized admittance of Ys = +j B at the junction. For this purpose, we can use expression (3.17) from Section 3.2 for the nor1nalized input impedance of a shortcircuited line of length Zs and set the corresponding nor1nalized admittance equal to +j B. Recalling that for a shortcircuited line Zin = j tan(fJZs), we have 1 Y.  +1·B s  j tan(fJZs) 
or 1
tan(fJZs) =  =B 
(3.46)
The value of B determined from (3.44) can be used in (3.46) to find the length Zs of the shortcircuited stub. Note that in (3.46), we have assumed the characteristic impedance of the shortcircuited stub to be equal to that of the main line.
SteadyState Waves on Transmission Lines
154
Chap.3
l + mA.12

ZL = Zo(0.5  j0.5)
2
ls
~lmax
, ..... /
I I I
 .... '
/
' '
z/A.
Stub location
1
0.75
Z 0 1J(z)I / /
''
,
 .... ' '
1.5
'
/
''
/ /
.... _.. / '
/
IV(z) I
0.5
0.25
0.5
Figure 3.33 Voltage standingwave pattern on a transmission line with singlestub matching. The standingwave ratio S is unity to the left of the stub. The stub location is at a distance of D..lmax from the nearest voltage maximum toward the load end. The particular case shown is for ZL = 0.5  }0.5. Note that, as usual, we have assumed v + = 1.
In practice, singlestub matching can be achieved even if the load impedance ZL is not explicitly known, by relying on measurements of S to determine p and measurements of the location of the voltage minimum or maximum to determine l/f. To see this, consider that the stub location l can be measured relative to the position Zmax of the nearest voltage maximum toward the load end so that l  ll.lmax = lzmax l < A/2, as shown in Figure 3.33. Using (3.42), we can then write () =
l/f  2{31
=
l/f + 2f3Zmax  2{3 ll.Zmax
= m27r  2{3ll.lmax
noting that m = 0 does not apply if Jr we have
s l/f
m = 0, 1, 2, ... < 0. Using the preceding expression for(),
(3.47) Thus, p can be directly determined from the measured standingwave ratio, and (3.47) deter1nines the stub location with respect to the measured location of the voltage maximum, as shown in Figure 3.33. Figure 3.33 also illustrates the fact that, although the proper choice of the stub location l and its length Zs achieves matching so that the standingwave ratio on the main line on the source side of the stub is unity (and so V (z) = Zol (z)), a standing wave does exist on the segment of the main line between the stub and the load. Note also that
Sec. 3.5
155
Impedance Matching
some current is shunted to the matching stub, resulting in the discontinuity in the plot of Zo II (z) I at the stub position along the main transmission line. Example 3.16: Singlestub matching. Design a singlestub system to match a load consisting of a resistance RL = 200Q in parallel with an inductance LL = 200 / n nH to a transmission line with characteristic impedance Zo = 1OOQ and operating at 500 MHz. Connect the stub in parallel with the line. Solution:
y = L
At 500 MHz, the load admittance is given by
1
1
RL
1 1  . =  . = 0.005  . 0.005 J WLL 200 J [2n(500 x 106)(200/n)(l0 9)] J
s
The reflection coefficient at the load is 0.01  (0.005  j0.005) 0.01 so that p
~
l~
0.447 and
+ (0.005 
j0.005)
1+j 40 63   ~ 0.447d · 3j
1/f ~ 63.4°. Using (3.43), we have
1.11  cos 1 (0.447) 2(2n/A.)
~
1.11=r=2.034
4n
>
11 ~ 0.073).. 12 = 0.25A.
where the first solution with a negative value of l can be realized by simply adding 0.5A. so that the stub position is between the load and the source. Thus, the stub position for the first solution is 11 ~ 0.073A. + 0.5A. = 0.426A.. Note that we have used () = cos 1 ( 0.447) ~ ±117° ~ ±2.034 radians. Using (3.44), the normalized susceptance of the input admittance of the shortcircuited shunt stub needed is
iJ
~±
J
2(0.447) ~ ±l 1  (0.447) 2
where the plus sign corresponds to the stub position 11 and the minus sign corresponds to 12. From (3.46), the length 18 1 of the shortcircuited stub at a distance 11 from the load is A. 1 1 18 1 = tan  =2n B
~
A. tan 1 (1) = 0.125A. 2n
>
0.375A.
and similarly, the stub length 182 needed at position 12 is A. 1 ls2 = tan (1) = 0.125A. 2n Both of the alternative solutions are shown in Figure 3.34. Since Solution 2 gives a stub position closer to the load and a shorter stub length, it would usually be preferred over Solution 1. In general, standing waves jeopardize powerhandling capabilities of a line and also lead to signal distortion. Thus, it is desirable to minimize the lengths of line over which the standingwave ratio is large. In the case shown in Figure 3.34, more of the line operates under matched (S = 1) conditions for Solution 2.
SteadyState Waves on Transmission Lines
156
Chap.3
12 = 0.25A
Solution 2
1oon
YL = (1/RL)j(llwLL)
1oon
RL = 2000, LL= (200/7r) nH
/ 82 =
0.125A

YL = 0.5  j0.5
at500MHz
Solution 1
1oon
1oon
YL
= (1/RL) j(llwLL)
RL = 2000, LL= (200/7r) nH
1oon / 81
=0.375A
Figure 3.34 Two alternative singlestub matching solutions. Solution 2 would in general be preferred since it has shorter segments of line (and a shorter stub) over which the standingwave ratio differs from unity.
The frequency dependence of the various designs is important in practice. A comparison of the frequency responses of the two alternative solutions for the previous example is given in Figure 3.35. Note that the load admittance as a function of frequency f = w/(2n) is given by 9 1 10 YLif) = 200  j 400f Assuming that the phase velocity along the line is equal to the speed of light c, we have {31 = 2nfl /c. The line admittance seen just to the right of the shortcircuited stub is given as a function of frequency:
YL + jYo tan(2nfl/c) Yi if) = Y o      Yo + jYL tan(2nfl/c) If we also assume f3ls = 2nfl8 /c, the total admittance seen from the source side of the shortcircuited stub (see Figure 3.32) is
jYo Y2 if) = Ys +Yi = tan(2nfls/ c)
YL + jYo tan(2nfl/c) + YoYo + jYL tan(2nfl / c)
Sec. 3.5
157
Impedance Matching Standingwave ratio
5 I
I 1 1 Solution
1
I
4
I I I
I I
3
I I I I I
........ ....
400
2
........
I
I
Solution 2
I I
....
/
/
500
600
Frequency (MHz) Figure 3.35 Frequency sensitivity of singlestub matching. The standingwave ratio S2 versus frequency for the two alternative solutions given in Figure 3.34.
The load reflection coefficient f2 and the standingwave ratio S2 on the main line to the left of the shortcircuited stub are then given as
Yo  Y2if) r 2 if) ·  Yo + Y2 if) ' The quantity S2 if) is plotted in Figure 3 .35 as a function of frequency between 400 23 and 600 MHz. Note that the bandwidths of the two designs are dramatically different. Which solution to choose depends on the particular application in hand, although in most cases minimizing reflections (e.g., S2 < 2) over a wider frequency range is desirable. Note that, for the case shown, Solution 2 provides matching over a substantially broader range of frequencies than does Solution 1.
3.5.3 QuarterWave Transformer Matching A powerful method for matching a given load impedance to a transmission line that is used to drive it is the socalled quarterwave transfor1ner matching. This method takes advantage of the impedance inverting property of a transmission line of length l = A/4, namely, the fact that the input impedance of a line of length l = A/4 is given by Z· m
23
_ Zo Z_L_+_jZ_o_ta_n_(fJ_l_) L=J../4 Zo + jZL tan(f3l) l=J../4
Defined as the frequency range over which the standingwave ratio S2 is lower than a given amount, for example, S2 < 2. The maximum value of S2 used depends on the application in hand.
SteadyState Waves on Transmission Lines
158
Chap.3
•
~I
, I __
•
Main line
Figure 3.36 Quarterwave transfor1ner.
t~I
A load ZL to be driven by a transmission line of characteristic impedance Zo connected to the line via a quarterwavelengthlong line of characteristic impedance ZQ.
,__I
or in terms of normalized impedances, we have 
Zin l=A/4 =
~n
Zo
L=A/4
Zo ZL
1
(3.48)
hence the ter1n ''impedance inverter." Thus, a quarterwave section transforms impedance in such a way that a kind of inverse of the terminating impedance appears at its input. Consider a quarterwavelength transmission line segment of characteristic impedance ZQ, as shown in Figure 3.36. In the general case of a complex load impedance ZL = RL + jXL, we have
so that a load consisting of a series resistance (RL) and an inductive reactance (XL > 0) appears at the input of the quarterwave section as an admittance consisting of a conductance RL/Z6_ (or a resistance Z6_/RL) in parallel with a capacitive susceptance XL/Z6_ > 0. Similarly, if the load were capacitive (XL < 0), it would appear as a conductance RL/Z6_ in parallel with an inductive susceptance XL/Z6_ < 0.
Purely resistive loads. The practical utility of the impedanceinverting property of a quarterwavelength transmission line becomes apparent when we consider a purely resistive load. Any arbitrary purely resistive load impedance ZL = RL is transfor1ned into a purely resistive input impedance of Z6_/RL. Thus, by appropriately choosing the value of the characteristic impedance ZQ of the quarterwavelength line, its input impedance can be made equal to the characteristic impedance Zo (a real value for a lossless line) of the main line that is to be used to drive the load. This property of the quarterwave line can be used to match two transmission lines of different characteristic impedances or to match a load impedance to the characteristic impedance of a transmission line. Note that the matching section must have a characteristic impedance of (3.49)
Sec. 3.5
159
Impedance Matching
where Ri and R1 are the two purely resistive impedances to be matched. Note that in the case shown in Figure 3.36, Ri = Zo and R1 = RL. Alternatively, R1 could be the characteristic impedance of another transmission line that may need to be matched to the main line (Zo) using the quarterwave section. Example 3.17: Quarterwave transformer for a monopole antenna. Design a quarterwavelength section to match a thin monopole antenna of length 0.24A. 24 having a purely resistive feedpoint impedance of RL = 30Q to a transmission line having a characteristic impedance of Zo = lOOQ.
Solution:
With reference to Figure 3.37 and according to (3.49), the A./4 section must match two impedances R2 = RL and Rt = Zo = lOOQ and thus must have a characteristic impedance ~ of
The standingwave ratio is unity beyond the quarterwave section. However, note that S 1.83 within the A./4 section.
~
Af 4
'
I
Monopole antenna
Z 0 =54.8Q = 1.83
Z 0 =100Q
s
S=l
< RL=30Q
Main line I
'
'
'
·~~·

Figure 3.37 Quarterwave transformer. The load is a monopole antenna of length 0.24).., which has a purely resistive impedance of 30Q.
Complex load impedances. In using quarterwave transformers to match a complex load impedance to a lossless transmission line (i.e., where Zo is real), it is necessary to insert the quarterwave segment at the point along the line where the line impedance Z (z) is purely resistive. As discussed in previous sections, this point can be the position of either the voltage maximum or minimum. In most cases, it is desirable to choose the point closest to the load in order to minimize the length of the transmission line segment on which S # 1 because the presence of standing waves jeopardizes powerhandling capabilities of the line, tends to reduce signaltonoise ratio, and may lead to distortion of the signal transmitted. 24
The reactive part of the impedance of such monopole antennas with length just shorter than a quarterwavelength is nearly zero. Monopole or dipole antennas with purely resistive input impedances are referred to as resonant antennas and are used for many applications. (See Section 14.06 of E. Jordan and K. Balmain, Electromagnetic Waves and Radiating Systems, PrenticeHall, New York, 1968.)
SteadyState Waves on Transmission Lines
160
Chap.3
Example 3.18: Thinwire halfwave dipole antenna. A thinwire halfwave dipole antenna25 has an input impedance of ZL = 73 + j 42.50Q. Design a quarterwave transformer to match this antenna to a transmission line with characteristic impedance Zo = lOOQ.
Solution:
We start by evaluating the reflection coefficient at the load
rL = so that
_;.1,
pe 't'
ZL  Zo
73 + j42.5  100 _; 1090 = = ::'.'. 0.283eZL + Zo 73 + j42.5 + 100
1/f ::'.'. 109° ::'.'. 1.896 radians. Note that the standingwave ratio is
s=
1 + p ::'.'. 1+0.283 ::'.'. 1.79 1 p 1  0.283
From previous sections, we know that, for an inductive load, the first voltage maximum is closer to the load than the first voltage minimum (see Figure 3.17). The first voltage maximum is at
1/f + 2fJZmax
= 0
>
1/f Zmax 
rv 
l.896A rv 0 . 151 A
4n
2fJ
Thus the quarterwave section should be inserted at z ::'.'. 0.151A., as shown in Figure 3.38. Noting that the normalized load impedance is ZL = 0.73 + j0.425 and that we have tan[fJ(0.151A.)] ::'.'. 1.392, the normalized line impedance Z(z) seen toward the load at z ::'.'. 0.151A. is
( ) I Z Z Z:::'.0. 151)..
ZL  j tan (fJ z) = . 1  J ZL tan({Jz)
0.73+j0.425+j1.392 ::'.'. 1.79 1 + j (0.73 + j0.425)(1.392)
z:::'.0.151A.
l = 0.151A I
•
Z 0 = 100Q S=l
Halfwave dipole antenna ZL = 73 + j42.SQ
Zo= 100Q s = 1.79
z 0 =133.7Q s = 1.34 •
1
Main line I
z= 0 Z(z = 0.151A) Figure 3.38 Quarterwave matching of the halfwave dipole antenna in Example 3.18.
25 See,
for example, Section 14.06 of E. C. Jordan and K. Balmain, Electromagnetic Waves and Radiating Systems, PrenticeHall, New York, 1968.
Sec. 3.5
Impedance Matching
161

Note that Z(z ~ 0.151A.) ~ 1.79 = S, as expected on the basis of the discussion in Section 3.3.2 (i.e., Rmax = S ~ 1.79). The characteristic impedance ZQ of the quarterwave section should thus be
~=
Jz(z
~ 0.151A.)Zo ~ J(lOO)(l.79)(100) ~ 133.7Q
Note that in general, as in this specific example, we have Z (z = Zmax) = S, and thus ZQ = ZoJS. Note also that the standingwave ratio in the quarterwave section is S ~ 1.34, as can be calculated by using Z (z ~ 0.15 lA.) and ZQ.
Frequency sensitivity of quarterwave matching. The frequency sensitivity of a quarterwave transfor1ner is a serious limitation since the design is perfect (i.e., provides S = 1) only at the frequency for which the length of the transformer segment is exactly A/4. The bandwidth of the transformer can be assessed by plotting the standingwave ratio S versus frequency, as was done in Section 3.5.2 for the singlestub tuning example and as is shown in the fallowing example. Example 3.19: Multiplestage quarterwave transformers. A resistive load of RL = 75Q is to be matched to a transmission line with characteristic impedance Z0 = 300Q. The frequency 6 8 of operation is fo = 300 MHz (which corresponds to A.0 = vp/fo ~ 3 x 10 /(300 x 10 ) = 1 m, assuming an airfilled coaxial line). Design multiplecascaded quarterwave matching transformers and at 300 MHz compare their frequency responses between 200 and 400 MHz. Solution: Three different designs are shown in Figure 3.39. Note that the choice of the impedance ZQ 1 for the singletransformer case is straightforward. For the doubletransformer case, the condition for exact quarterwave matching is ZQ2 = ZQ1 JZo/ RL, allowing for different choices of ~2 and ZQ1 as long as the condition is satisfied. The design shown in Figure 3.39b is one that provides a standingwave ratio in the first and second quarterwave sections of S ~ 1.40 and S ~ 1.43, respectively. For the triple transformer case the condition for exact quarterwave matching can be shown to be ZQiZQ3 = ZQ2v'ZOJIL = 150ZQ2· Once again, many different combinations of ZQ1, ~2, ZQ3 satisfy this condition, and other performance criteria (such as minimizing S in the quarterwave sections) must be used to make particular design choices. One design approach26 is to require the characteristic impedance of the second quarterwave segment to be the geometric mean of the two impedances to be matched, namely ZQ2 = J(300)(75) = 150Q. The choices of ZQ 1 and ZQ3 shown in Figure 3.39 is one that provides a relatively low value of S Z1
Z1
(a)
0
0 0 0
Mz
(b)
Figure 3.41 Tapered impedance transformer and a mechanical analogy. (a) A gradual taper provides wide bandwidth; many small reflections from a series of small incremental steps add with different phases to produce very small net total reflection. (b) Mechanical analog of impedance matching to provide better power absorption at a termination.
The transfer of energy between Mi and M1 can be improved by the insertion of a third mass between them, as shown in Figure 3.4lb. The energy transfer is optimum if the third mass M is the geometric mean of the other two, that is, if M = ,JMiM2. Further improvement in the energy transfer can be achieved by using several bodies with masses varying monotonically between Mi and M1.
3.6 THE SMITH CHART Many transmissionline problems can be solved easily with graphical procedures, using the socalled Smith chart. 28 The Smith chart is also a useful tool for visualizing transmissionline matching and design problems. Many aspects of the voltage, current, and impedance patterns discussed in previous sections can also be interpreted and visualized by similar means using the Smith chart. One might think that graphical techniques are not as useful in this age of powerful computers and calculators, but it is interesting to note that some commonly used pieces of laboratory test equipment have displays that imitate the Smith chart, with the line impedance and standingwave ratio results presented on such displays. In this section, we describe the Smith chart and provide examples of its use in understanding transmissionline problems.
28 P.
H. Smith, Electronics, January 1939. Also see J. E. Brittain, The Smith Chart, IEEE Spectr., 29(8), p. 65, August 1992.
Sec. 3.6
165
The Smith Chart
3.6.1 M apping of Complex Impedance to Complex r
The S1nith chart is essentially a conveniently parameterized plot of the normalized line impedance Z (z) of a transmission line and the generalized voltage reflection coefficient f (z) as a function of distance from the load. To understand the uti lity of the Smith chart, we need to understand the relationship between Z (z) and f (z). From (3.22), we can write the normalized line impedance as Z (z )
=
I + f (z ) I  f (z )
(3.50)
where we note that f (z ) = pejlo/1+2P') is the voltage reflection coefficient at any position z along the line. Denoting f (z ) simply as r , while keeping in 1nind that it is a function of z, we can rewrite (3.50) as
~
(3.51 )
t.:..8J 


where Z = R + j X and r = 11 + j v are both complex numbers, so (3.51) represents a mapping between two complex numbers. Note that if the load is g iven, then we know rL, and therefore r (and thus Z = R + jX , from (3.51)), at any position at a distance  on the r z from the load. The Smith chart conveniently displays values of Z (or R,X) (or 11 , v) plane for graphical calculation and visualization. From (3.51 ) we have Z

R
+J·x 
I+
( 11
I  ( 11
+ jv) [ I + (11 + j v))[ I  (u  j v))  + jv) ( I  11)2 + v2
(3.52)
Equating real parts in (3.52) and rearranging, we have
(3.53)


which is the equation of a circle in die 11v plane centered at 11 = R/( I + R), v = 0, and having a radius of 1/( 1 + R). Examples of such circles are shown in Figure 3.42a. Note d1at R = I corresponds to a circle centered at 11 = v = 0, having a radius and passing through the origin in the 11v plane. Similarly, by equating the imaginary parts in (3.52) and rearranging, we find
i·
t.
(3.54) which is the equation for a circle in the 11v plane centered at u = I, v = l/ X, and having a radius of I/ X. Exan1ples of such circular segments are shown in Figure 3.42b. Note d1at X = ± I corresponds to a circle centered at 11 = I, v = ± I, having a radius of I, and tangent to the v axis at v = ± I.
SteadyState Waves on Transmission Lines
166 v
Chap.3
v
1
1
u=1
X=2 u=1

'X=1 \
X=0.5
\
0.5 ,.  .... ; _.,,. ' I/ R = 1 ,_......... I R =2

R = 0.5
1
0.5
0.5

X=O
\
''
' ' ................ ,.
0.5
0.5
\
\
;
\
'0.5 '.... .... __ ,.,.
o.5""
I I I I I I I
/
;
/
/
I I
I
I /_ I X=1
U
I

X=2
x = 0.5 1
1
1
(b)
(a) Figure 3.42 Contours
of constant R

or X. (a) The circles in the uv plane are centered
at [R(l + R) 1, O], with radius (1 + R) 1. Note the R = 1 circle (dashed lines) passes through the origin (i.e., u, v = O); this circle is centered at (i_,0) with radius Sb) The circles in the uv plane are centered at (1,X 1), with radius x 1. Note that for X = ±1 we have two circles (dashed lines) with unity radii and centered at (1, ± 1).
i.
The voltage reflection coefficient r = u + jv is defined on the complex uv plane so that the locus of points of constant Ir I = IrLI = p are circles centered at the origin. Once p is known (the value of which is set by ZL and Zo), motion along the line (i.e., variation of z) corresponds to motion around this r circle of fixed radius p. To see this, consider (3.55) As illustrated in Figure 3.43, motion away from the load (i.e., decreasing z) corresponds to clockwise rotation of r around a circle in the uv plane. Since f3 = 2rr /A, a complete rotation of 2f3z = 2rr occurs when z decreases by J./2, which is why a complete cycle of line impedance (or admittance) is repeated every J./2 length along the line. A circle
v z =O \
\
, 2/3z I I
u
I I
/
Figure 3.43 Complex reflection coefficient r. The complex number r is shown in the uv plane, together with its variation with z.
Sec. 3.6
167
The Smith Chart
of constant radius p (corresponding to a given load) also corresponds to a fixedvoltage standingwave ratio S, since S = (1 + p)(l  p) 1. Although all r values along the line terminated with ZL lie on the circle of radius p, each value of r corresponds (through (3.51)) to a different value of Z = R + jX, which is the normalized line impedance seen at that position. On a typical Smith chart, shown in Figure 3 .44, the contours of constant R or X are plotted and labeled on the uv plane so that the line impedance at any position along the constant p (or S) circle can be easily read from the chart. A summary of various Smith chart contours and key points is provided in Figure 3.45. 0.11
o.>9
0.12
0.13
0.38
0.37
0.14 0.36
90
100
1, ' t1rR = 1
/
I
I
R = O
X =O ~ (short circuit)
I I


X =0 1
p
Rm{Z (z)} > 0) in the approximate range A/ 4 < z < 0, attains a large real value (but not quite an open circuit) at z = A/4, is capacitive in the approximate range A/2 < z < A/4, but does not quite return to zero at z = A /2. The peak in the real part of the impedance at z =  3A / 4 is considerably smaller than that at z = A/4. In general, the maxima and minima of the imaginary part of Z (z) both approach zero as z attains larger and larger negative values, while the maxima and minima of the real part of Z (z) both approach Zo. At sufficient distances from the load, for example, for z <  l .5A in the case of a = 1.5 np /A, the line impedance Z (z) ~ Zo, just as if the line were infinitely long or matched.
Sec. 3.7
Sinusoidal SteadyState Behavior of Lossy Lines
4
''
3
189
I a= 1 np/,,\ I
'
3
._.,
2
/ IV(z) I z/,,\ 1.5
0.5
4
[a= 1.5 np/,,\ I
''
'
3
1
1
0.5
1 0
4
Ia= 2 np/,,\ I
3
\
'" z/,,\ 1.5
z/,,\ 1.5
0
2
'
1 1
4
0.5
0
2
2
1
1
1
z/,,\ 1.5
0.5
0
Figure 3.57 Voltage and current standingwave patterns on a terminated lossy transmission line. The magnitudes of the voltage and current phasors (for current, the quantity plotted is I/ (z) IZo) of a lossy line terminated in ZL = 5Zo are shown for values of the attenuation constant a = 0.5, 1, 1.5, and 2 np/A., where).. = 2rr / {J.
For a lossy line terminated in an open circuit (ZL = oo ), expressions for V (z), I (z), and Zin can be obtained in a manner analogous to the preceding discussion for a shortcircuited line. This straightforward procedure is left as an exercise. The general behavior of the line voltage, current, and impedance for other terminations is quite similar, as illustrated in Figures 3.57 and 3.58 for a resistive load impedance of ZL = 5Zo. Results are shown for four different values of the attenuation constant. For simplicity, we have once again assumed the phase of the characteristic impedance 0.5 and when the line is long. Example 3.29: A highspeed microstrip interconnect. Consider a highspeed microstrip transmission line of length 20 cm used to connect a 1V amplitude, 1GHz, 50n sinusoidal voltage source to a digital logic gate having an input impedance of 1 kn, as shown in Figure 3.59. Based on measurements, the transmission line parameters of this interconnect at 1 GHz are 32 See
Section 63 of R. K. Moore, Traveling Wave Engineering, McGrawHill, New York, 1960.
Sec. 3.7
Sinusoidal SteadyState Behavior of Lossy Lines
191 lkQ
Microstrip
+ +
VL
..     20 cm    ..

==
Vs
.l



(a)
+ V 0 = 1e j 0.......
(V) "\J 1 GHz'""'"

(b)

Figure 3.59 A lossy highspeed microstrip interconnect. (a) The microstrip transmission line connected to a 1kQ load and driven by a 1GHz source. (b) Thevenin equivalent circuit as seen by the source, where Zin is the input impedance at the source end of the microstrip.
approximately given by R = 5Q(cm) 1, L = 5 nH(cm) 1, C = 0.4 pF(cm) 1, and G = 0, respectively. (a) Find the propagation constant y and characteristic impedance Zo of the line. (b) Find the voltages at the source and the load ends of the line. (c) Find the timeaverage power delivered to the line by the source and the timeaverage power delivered to the load. What is the power dissipated along the line?
Solution: (a) The propagation constant is given by y
=a + j fJ
= J(R
+ jwL)(G + jwC)
(500 + j2n x 109 x 500 x 109 )(j2n x 109 x 40 x 1012 ) 50 85 ~ 28.3ei · ~ 2.23 where a ~ 2.23 npm 1 and by
Zo =
fJ
+ j28.2
~ 28.2 radm 1. The characteristic impedance is given
R + jwL ____ G + jwC
,........,
~ 112.5ej 4 ·5220n
500 + j3142 ,........, j0.251
3181ej80.96o 0.251ej 9QO
SteadyState Waves on Transmission Lines
192
Chap.3
(b) The reflection coefficient at the load end can be found as 5220 4 ZL  Zo 1000  112.5ej · . 0 r = rv r v 0 798e1 l.03 L ZL + Zo  1000 + 112.5ej4.522o  . The reflection coefficient at any position along the line is given by r(z) = rLe2yz ~
~
0.798dl.029°e4.458zd56.37z
O. 798 e4.458z d (56.37z+0.018)
The input impedance of the line is given by Z· = Z(z) I  02 Ill z .
ID
1 2 _+_ r _ = Zo (_ 0_. _) 1  r(0.2)
We first find r(0.2) as r(0.2)
~
0.798e4.458(0.2)ei[56.37(0.2)+0.018] ~ 0.327d75.05°
We now use r(0.2) to find
Zin·
We have
·75 05° rv j4.5220 1 + 0.327 el . Zin _ 112.5e .75 05 0 1  0.321eJ · 310 790 5220 35 30 4 ~ (112.5ej · )(1.166d · ) ~ 131.2d · n Based on voltage division in the equivalent circuit of Figure 3.59b, the sourceend voltage Vs is Zin 13 l .2ej30.79o v.s = Zn+ Zs Via r v 112.7 + j67.16 + 50 (1) rv
Using (3.74a),
v+
13 l .2ej30.79o ~ 0.745d8.3610 V 176ej22.43o
can be written as V(z)
v+=erz [1 + r(z)] At z = 0.2 m, we have 3610 8 V (z = 0.2) = Vs ~ 0.745d · V and ey(0.2)
~
e0.446ej37.010
Using these values, the value of
v+ ~
v+
and
r(0.2)
~
0.327d75.050
can be found as "8.361°
. o.1;se1 . 0 ~ o.423ei29.120 v e0.446e1 31.01 (1 + 0.327eJ75.05 )
Sec. 3.8
193
Summary
The voltage at the load end of the line is given as vL =
v (z
= O) = v +[1 +
120 29 rLJ ~ (0.423ei · )(l
570 0290 29 1 + 0.798ei · ) ~ o.760ei · v
(c) The timeaverage power delivered to the line is given by 2 1 Vs Ps = Rin 2 Zin
~
1 0.745 2 (112.7) W 2 131.2
~
1.82 mW
Similarly, the timeaverage power delivered to the load can be found as p
=
L
! IVLl2 ~ ! (0.760)2 2 RL
2
W
~ 0.289 mW
1000
Thus, based on conservation of energy, the power dissipated in the lossy line is P1ost = Ps 
PL~
1.82  0.289
~
1.53 mW
3.8 SUMMARY This chapter discussed the following topics: • Transmission line equations. When a transmission line is excited by a sinusoidal source of angular frequency w at steady state, the variations of the line voltage and current can be analyzed using the phasor form of the transmission line equations, which for a lossless line are dV (z) = j wLI (z) dz dl(z) = jwCV(z) dz
where L and C are the perunit length distributed parameters of the line, and V (z) and I (z) are the voltage and current phasors, respectively, which are related to the actual spacetime voltage and current expressions as follows: !P(z, t) = ffie{l (z)eiwt}
• Propagatingwave solutions, characteristic impedance, phase velocity, and wavelength. The solutions of the lossless transmission line equations consist of a superposition of waves traveling in the +z and z directions. The voltage and current phasors and the corresponding spacetime functions have the for1n (assuming v+ and v are both real) v (z) = v+ ej ,Bz I(z) =
v+
Zo
+ ve+j,Bz;
. v . e1,Bz e+1,Bz;
Zo
°V(z,t) = v+cos(wt f3z) + vcos(wt !P(z, t) =
v+
Zo
cos(wt  f3z) 
v
Zo
+ /3z)
cos(wt + f3z)
SteadyState Waves on Transmission Lines
194
Chap.3
The characteristic impedance Zo of the line is the ratio of the voltage to the current phasor of the wave propagating in the +z direction (or the negative of the ratio of the voltage to the current phasor of the wave traveling in the z direction) and, for a lossless line, is given by Zo = ,JIJE. The phase velocity and the wavelength for a lossless line are given as Vp
= lj,JIE;
A = 2nI f3 =
Vp
If
Note that the phase velocity of a lossless line is independent of frequency. • Input impedance of short and opencircuited lines. The line impedance of a transmission line seen looking toward the load at any position along the line is defined as
The input impedances of short or opencircuited transmission lines of length l are purely imaginary and are given by Zsc = jZo tan(f3l)
shortcircuited line
Zoe = jZo cot(f3l)
opencircuited line
Since any arbitrary reactive impedance can be realized by simply adjusting the length l of open or shortcircuited stubs, these stubs are commonly used as reactive circuit elements for impedance matching and other applications. • Reflection coefficient. It is common practice to treat steadystate transmission line problems by considering the wave traveling in the +z direction (toward the load) as the incident wave and the wave traveling in the z direction (away from the load and toward the source) as the reflected wave. The ratio of the reflected to the incident voltage phasor at any position z along the line is defined as the reflection coefficient, represented by r (z). The reflection coefficient at the load end of the line (where z = 0) is given by
The case of ZL = Zo is referred to as a matched load, for which there is no reflected wave, since rL = 0. The reflection coefficient f (z) at any other location z (where z < 0) on a lossless line is given by
r (z) _
vejf3z
+ '{3 V e1 z
= fL ~;2{3z = p ~; (1/1+2{3z) e'
e"
• Standingwave pattern. The superposition of the incident and reflected waves constitutes a standingwave pattern that repeats every A/2 over the length of the
Sec. 3.8
195
Summary
line. The standingwave ratio S is defined as the ratio of the maximum to minimum voltage (current) magnitude over the line and is given by
S=l+_p 1p where p = lfL I· The standingwave ratio S has practical significance because it is easily measurable. The value of S varies in the range 1 s S s oo, where S = 1 corresponds to p = 0 (i.e., no reflection case) and S = oo corresponds to p = 1 (i.e., the load is either open or short circuit or purely reactive).
• Transmission line as an impedance transformer. The line impedance of a lossless transmission line ter1ninated in an arbitrary load impedance, defined as the ratio of the total voltage to current phasor at position z, is in general complex and is a periodic function of z, with period of )... /2. The line impedance is purely real at locations along the line where the total voltage magnitude is a maximum or • • rmn1mum. • Power flow. The net timeaverage power propagating toward the load on a lossless transmission line is given by P(z) =
and is equal to the power PL delivered to the load. For a given value of 1v+ 1, the power delivered to the load is maximized under matched conditions, or p = 0. The degree of mismatch between the load and the line can be described in terms of return loss, given as
S+l Return loss= 20log 10  S 1 • Impedance matching. In most applications it is desirable to match the load impedance to the line in order to reduce reflections and standing waves. In singlestub matching, a short or opencircuited stub is placed in shunt or series at a location z = l along the line at which the normalized line admittance or the impedance is given as 

Y1 (z) lz=l = 1  j B;


Z1 (z) lz=l = 1  jX
The matching is then completed by choosing the length Zs of a short or opencircuited stub so that it presents an admittance or impedance at z = l of Ys = j B or Zs = jX. In quarterwave matching, it is first necessary to determine the location l along the line at which the line impedance is purely real, that is, where
Z(z) lz=l = R
+ jO
Matching to a line of impedance Zo is then completed by using a quarterwavelength long line of characteristic impedance ZQ = ,JZOR.
SteadyState Waves on Transmission Lines
196
Chap.3
• Smith chart. The fact that the impedance Z (z) and the reflection coefficient r (z) on a lossless line are both periodic functions of position z along the line makes it possible to analyze and visualize the behavior of the line using a graphical display off (z), S, and Z (z) known as the Smith chart. The Smith chart provides a convenient means of analyzing transmission line problems to determine values of impedance and reflection coefficient (or standingwave ratio). The Smith chart is also a useful tool for matching network design. • Lossy transmission lines. The solutions for voltage and current propagating in the z direction on a lossy transmission line have the fortn (assuming v+ is real) °V(z' t) = v+ eaz cos(wt  {Jz) !P(z, t) =
v+
eaz cos(wt  {Jz  z)
IZo I
where a and f3 are the real and imaginary parts of the propagation constant y =a+ j f3 = j(R + jwL)(G + jwC), R, L, G, and Care the perunit distributed parameters of the line, and w is the angular frequency of the excitation. The characteristic impedance for a lossy line is in general complex and is given by
Zo = IZo leiz
=
+ jwL G + jwC R
For terminated lines, the general expressions for line voltage and current are V(z) = v+eazejf3z[l
/(z) =
+ f(z)]
1 v+eaze1'{3 z[l  r(z)] Zo
where r (z) = rLe az ei f3z' with rL = (ZL  Zo) I (ZL + Zo) being the complex load voltage reflection coefficient. The line voltage and current exhibit a standingwave pattern near the load, but the differences between the maxima and minima become smaller as distance from the load increases. At sufficient distances from the load, the magnitudes of the line voltage and current do not vary significantly with distance, as if the line were matched. The impedance of a lossy transmission line is given by 2
2
The real and imaginary parts of Z (z) exhibit maxima and minima near the load, similar to that of a lossless line. However, at sufficient distances from the load, Z (z) approaches Zo, as if the line were matched.
Chap.3
Problems
197 PROBLEMS
3.1
Transmission line capacitor. (a) Design an opencircuited son air transmission line with the shortest length that will provide the impedance of a 4 nF capacitor at 10 GHz. (b) Redesign the same capacitor using a shortcircuited son air transmission line. (c) Which design yields the shortest length and why? Transmission line inductor. (a) Design an opencircuited son lossless microstrip transmission line with shortest electrical length that will provide the impedance of a S nH inductor at S GHz. (b) Repeat the same design using a shortcircuited son microstrip line. (c) Which design resulted in a shortest length line and why? Input impedance of a transmission line. Consider a shortcircuited son lossless transmission line as shown in Figure 3.60. Find the shortest electrical length of the line such that (a) Zin = j Son, (b) Zin = j lSOn, (c) Zin = oo, and (d) Zin = 0.
3.2
3.3
. =?. l rmn
Z 0 =50Q Figure 3.60 Shortcircuited transmission line. Problem 3.3.
z.
lil
3.4
Terminated transmission line. A son lossless transmission line is terminated with a capacitive load having a load impedance of ZL = 100  j lOOn, as shown in Figure 3.61. (a) Find the load reflection coefficient, rL. (b) Find the standing wave ratio S on the line. (c) Find the input impedance of the line, Zin, for four different line lengths: 11 = 0.12SA., 12 = 0.2SA., 13 = 0.37SA., and 14 = O.SA., respectively. l l
•
•
ZL = 100 jlOOQ
Z 0 =50Q l
z.
lil
3.5 3.6 3.7

• •
• •

Figure 3.61 Tenninated transmission line. Problem 3.4.
Input impedance. Find the input impedance Zin of the two cascaded lossless transmission line system shown in Figure 3.62 for (a) 12 = 0.2SA., (b) 12 = O.SA., and (c) 12 = 0.12SA.. Source and loadend voltages. For the lossless transmission line circuit shown in Figure 3.63, calculate the phasorform sourceend and loadend voltages Vs and VL. Capacitive termination An air transmission line with Zo = son is terminated with a capacitive load impedance of SO  j90n. If the line is operated at 100 MHz, calculate (a) the load reflection coefficient rL, (b) the standingwave ratio S on the line, (c) the distance from the load to the first voltage minimum, and (d) the distance from the load to the first voltage maximum.
SteadyState Waves on Transmission Lines
198
Chap.3
z.
1Il •
'
'
2oon
z 02 =1oon
z 0 1 =son
.
1
.~~~~~~~~~
·~~~~~~~~~

Figure 3.62 Two cascaded transmission lines. Problem 3.5.
z.
1Il
son
SA./ 8
Zo = SOn
Figure 3.63
3.8
1son
Source and loadend voltages. Problem 3.6.
A wireless communication antenna. The following table provides the approximate values at various frequencies of the feedpoint impedance of a circularly polarized patch antenna used in the wireless industry for making cellular phone calls in difficult environments, such as sport arenas and office buildings:
/(MHz)
ZL(Q)
800 850 900 950 1000
21.5  j 15.4 38.5 + j2.24 43.8 + j9.74 55.2  j 10.2 28.8  j7.40
If this antenna is directly fed by a 50Q transmission line, find and sketch the standingwave ratio S as a function of frequency. 3.9 Resistive line impedance. A 1OOQ transmission line is terminated with a load impedance of 30 + j 60Q at z = 0. Find the minimum electrical length l /A. of the line at which the line impedance (i.e., Z (z =  l)) is purely resistive. What is the value of the resistive line impedance? 3.10 Real Zin position. Consider a 50Q transmission line terminated with a load impedance of 50 + j140Q. (a) Find the load reflection coefficient rL and the standing wave ratio S on the line. (b) Find the electrical position on the line nearest to the load where the line impedance is purely real. (c) Find the value of the line impedance at the position found in part (b).
Chap.3
3.11
Problems
199
Cascaded transmission lines. Two lossless transmission lines are cascaded as shown in Figure 3.64. (a) Find the standingwave ratio S on each line. (b) Find the timeaverage power delivered to the load.
z =O R 8 = 50n

,
&
r
7A./4 
..
.• 
z02 = 1oon
"I
rv "
Z 01 = SOQ
~

z
..
ZL = 200  j200Q
.• •
•
Figure 3.64 Inductive termination. Problem 3.11.
3.12
Inductive termination. A 75Q lossless transmission line is terminated with an inductive load given by 45 + j60Q, as shown in Figure 3.65. Calculate (a) the load reflection coefficient rL, (b) the standingwave ratio S on the line, (c) the percentage timeaverage incident power that is absorbed by the load, and (d) the Vmax and Vmin positions that are nearest to the load.
.. 
·
ZL = 45 + j60Q
Z 0 = 75n
·
•
z z =O
3.13
Figure 3.65 Inductive ter1nination. Problem 3.12.
Input impedance. The input impedance of an 80Q transmission line terminated with an unknown load ZL and having an electrical length of 3 /8 at frequency f 1 is measured to be 64  j48Q. Find the new value of Zin at frequency !2 =2f1. Assume ZL and Zo values to stay the same with frequency.
3.14
Resistive load. A lossless line is terminated with a resistive load of 120Q. If the line presents an impedance of 48 + j36Q at a position 3A./8 away from the load, what is the characteristic impedance Zo of the line?
3.15
Input impedance. For the lossless transmissionline system shown in Figure 3.66, find Zin for the following load impedances: (a) ZL = oo (open circuit), (b) ZL = 0 (short circuit), and (c) ZL = Zo/2. A.14
Zo
A.14
2Z0 Figure 3.66 Input impedance. Problem 3.15.
200
SteadyState Waves on Transmission Lines
Chap.3
3.16 Input impedance. Find the input impedance Zin of the two cascaded lossless transmission lines as shown in Figure 3.67, where Zp = jSOn is a lumped impedance connected at the junction. l = 5A./8
l = 7A./4  •  •
• •
(
z.
.•

• •
Zp
Z 01 =50Q (

• •
• •

ZL = 100jSOQ
Z 02 =50Q 
.• 
• •
Zp=j50Q
lil
Figure 3.67 Cascaded transmission line. Problem 3.16.
3.17 Input impedance. Repeat Problem 3. lS for the circuit shown in Figure 3.68.
A.14
A.14
2Z0
z.
lil
Figure 3.68 Input impedance. Problem 3.17.
3.18 Unknown load. Consider a son transmission line terminated with an unknown load. If the standingwave ratio on the line is measured to be 4.2 and the nearest voltage minimum point on the line with respect to the load position is located at 0.21A., find the following: (a) The load impedance ZL. (b) The nearest voltage maximum and the next voltage minimum positions with respect to the load. (c) The input impedance Zin at each position found in part (b). 3.19 Input impedance. For the lossless transmissionline system shown in Figure 3.69, what is the ratio Zo1/Zo2 if Zin = 22Sn?
A.14
A.14
lOOQ
z.
lil
Figure 3.69 Input impedance. Problem 3.19.
3.20 Unknown termination. Consider a transmission line with Zo unknown load impedance ZL. (a) Show that
=son
terminated with an
1  jS tan(fJlmin) ZL = Z o      S  j tan (fJ lmin) where lmin is the length from the load to the first voltage minimum and S is the standingwave ratio. (b) Measurements on a line with Zo = son having an unknown termination
Chap.3
201
Problems
show that S = ,J3, lmin = 2S mm, and that the distance between successive minima is 10 cm. Find the load reflection coefficient rL and the unknown termination ZL. Unknown load. A son air transmission line with a standing wave ratio of 3 has its first and second voltage maximums nearest to the load located at 0.1 m and 0.3 m respectively. Calculate (a) the operating frequency f and (b) the unknown load impedance ZL. Unkown load. A 7Sn air transmission line with a standing wave ratio of 6 has a voltage maximum and an adjacent voltage minimum position located with respect to the load at 3 m and 4.S m, respectively. Calculate (a) the operating frequency f and (b) the unknown load impedance ZL. Distance to the first maximum. Derive a formula similar to that in Problem 3.20 in terms of lmax, where lmax is the distance from the load to the first voltage maximum. Terminated transmission line. A son air transmission line is terminated with an inductive load impedance given by ZL = 100 + j lSOn and excited by a sinusoidal voltage source as shown in Figure 3.70. Calculate: (a) the load reflection coefficient rL, (b) the standing wave ratio S, (c) the timeaverage power delivered to the load, and (d) the first two Vmax and the first two Vmin positions on the transmission line nearest to the load.
ZL
3.21
3.22
3.23 3.24
l = lOm
son 
V 0 = lOejO V (/= 300MHz)
.., .., A

~
' '
+ r
" rv '
..
Air line Zo = SOn
~
~

' '
ZL = 100 + jSOn

~
' '
' '

Figure 3.70 Terminated transmission line. Problem 3.24.
3.25
Terminated transmission line. A son, 10.Sm long air transmission line terminated with a load impedance of ZL = 70 + j 1on is excited by a sinusoidal voltage source, as shown in Figure 3.71. (a) Calculate the load reflection coefficient rL and the standing wave ratio S on the line. (b) Find all the Vmax and Vmin positions (in actual lengths) on the line. (c) Find all the Imax and /min positions on the line. (d) Find the input impedance Zin seen at each Vmax and Vmin position. (e) Find the line impedance Z (z) seen at the source end of the line and draw the equivalent lumped circuit with respect to the source end. (t) Find the phasor voltages V8 , v+, v, and VL. (g) Find the Vmax and Vmin values. (h) Find the Imax and /min values. (i) Find the timeaverage powers p + , p, PRs' PL, and Psource· What percentage of the power carried by the incident wave reflects back towards the source? G) Repeat parts (a) through (i) for a load impedance of ZL = lS  j3Sn.
son ~
~
y
V 0 = lOejO V (/= 300MHz)
l = 10.5 m 
' '
+ r
" rv '~

+
Air line Zo = SOn
~
VL

Figure 3.71
' '
ZL = 70 + j10n
' '
' '

Terminated transmission line. Problem 3.25.
202
SteadyState Waves on Transmission Lines
Chap.3
3.26 Power dissipation. For the lossless transmission line system shown in Figure 3.72, with Zo = lOOQ, (a) calculate the timeaverage power dissipated in each load. (b) Switch the values of the load resistors (i.e., RLi = 200Q, RL2 = 50Q), and repeat part (a). A.12
A.14
2sn Z0 = 1oon
RL1 =500
Zo=lOOn RL2=200Q
10ejO (V)
Figure 3.72 Power dissipation. Problem 3.26.
3.27 Power dissipation. Consider the transmission line system shown in Figure 3.73. (a) Find the timeaverage power dissipated in the load RL with the switch S open. (b) Repeat part (a) for the switch S closed. Assume steady state in each case.
2sn 10cos(wt) (V)
A.14
3A./2
s Zo=SOn
1oon
Zo=SOn
Figure 3.73 Power dissipation. Problem 3.27.
3.28 Power dissipation. Repeat Problem 3.27 if the characteristic impedance of the transmission line on the source side is changed from son to 25J2n. 3.29 Cascaded transmission lines. Two cascaded lossless transmission lines are connected as shown in Figure 3.74. (a) Find the standingwave ratio on each line. (b) Find the timeaverage powers delivered to impedances Z1 and Z2.
'
'
'
Figure 3.74 Cascaded transmission lines. Problem 3.29.
3.30 1\vo antennas. Two antennas having feedpoint impedances of ZL1 = 40  j 30Q and ZL2 = 100 + j50Q are fed with a transmission line system, as shown in Figure 3.75. (a) Find S on the main line. (b) Find the timeaverage power supplied by the sinusoidal source. (c) Find the timeaverage power delivered to each antenna. Assume lossless lines.
Chap.3
Problems
203
,,\/2 ,,\/ 4
lOOQ Z 0 = SOQ
Main line
3.A/ 4
Figure 3.75 Two antennas. Problem 3.30.
3.31
Power dissipation. For the transmission line network shown in Figure 3.76, calculate the timeaverage power dissipated in the load resistor RL. ,,\/ 4
2sn ~
TT
20e jO , ... '\J.. (V) '"
•
,,\/ 4 •
son
•
•
....
~
T
:>
z 01 = son
T
RL = 200n
20e jO '\J ... .. (V) ~
Z 02 = lOOQ
...
H...
Figure 3.76 Power dissipation. Problem 3.31.
3.32
Three identical antennas. Three identical antennas Al, A2, and A3 are fed by a transmission line system, as shown in Figure 3.77. If the feedpoint impedance of each antenna is ZL = 50 + j 50Q, find the timeaverage power delivered to each antenna. lOOQ
13.A/ 4
1oon
S.A/4
lOOQ
S.A/ 4
lOOQ
A3
A2 Figure 3.77 Three identical antennas. Problem 3.32.
3.33
Power delivery. The transmission frequency f such that w = 2nf = power supplied by the source and three different 25Q load resistances
line system shown in Figure 3.78 is to be used at a 0.5 x 109 rads 1 . Determine the total timeaverage also the timeaverage power supplied to each of the (connected to the ends of lines 2, 3, and 4).
SteadyState Waves on Transmission Lines
204
Af 4
son 5 cos(wt) 'V (V)
z 01 =son
Af 4
25n
Zo2=SOn
2sn
Chap.3
2SnH
.80pF
Figure 3.78 Power delivery. Problem 3.33.
3.34 Power delivery. Consider the transmission line system shown in Figure 3.79. (a) Calculate the timeaverage power dissipated in the load resistances RL1 =son and RL2 =son at f = !1. (b) Calculate the timeaverage power dissipated in the load resistances RL1 =son and RL2 =son at f = 2f1. (c) Calculate the timeaverage power dissipated in the load resistances RL1 =SOn and RL2 =SOn atf = l.Sf1.
Zo=SOn
son
10cos(27rft) (V)
Zo=SOn
Zo=SOn
Figure 3.79 Power delivery. Problem 3.34.
3.35 Matching with a single lumped element. The transmission line matching networks shown in Figure 3.80 are designed to match a lOOn load impedance to a son line. (a) For the network with a shunt element, find the minimum distance l from the load where the unknown shunt element is to be connected such that the input admittance seen at BB' has a conductance part equal to 0.02 S. (b) Determine the unknown shunt element and its element value such that the input impedance seen at AA' is matched to the line (i.e., ZAA' =son) at 1 GHz. (c) For the matching network with a lumped series matching element, find the minimum distance l and the unknown element and its value such that a perfect match is achieved at 1 GHz. Assume vp = 30 cm(ns) 1. 3.36 Matching with series stub. A load impedance of 13S  j90n is to be matched to a 7Sn lossless transmission line system, as shown in Figure 3. 81. If A. = 20 cm, what minimum length of transmission line l will yield a minimum length 18 for the series stub?
Chap.3 A
B
son
A
l

205
Problems

son
?•
son

I
I
A'
B'
B
l
son I
I
A'
B'
(b)
(a)
Figure 3.80 Matching with a single lumped element. Problem 3.35.
l
7Sn
.. ZL = 13Sj90n
75n

•
I
)
•
•
a
ls
'Ir)
r:
~ 

Figure 3.81 Matching with a series shorted stub. Problem 3.36.
3.37 Series stub matching. A seriesshortedstub matching network is designed to match a capacitive load of RL = 50n and CL= 10/ (3n) pF to a lOOn line at 3 GHz, as shown in Figure 3.82. (a) The stub is positioned at a distance of A./ 4 away from the load. Verify the choice of this position and find the minimum electrical length of the stub to achieve a perfect match at the design frequency. (b) Calculate the standingwave ratio S on the main line at 2 GHz. (c) Calculate S on the main line at 4 GHz.
Main line lOOn
1oon
RL = SOn
  CL = 10/(37r) pF .~~~~~~~~~~ri'
~~~~~~~~~~~~~~}6
A.14 Figure 3.82 Series stub matching. Problem 3.37.
3.38 Quarterwave transformer. (a) Design a singlesection quarterwave matching transformer to match an RL = 20n load to a line with Zo = 80n operating at 1.5 GHz. (b) Calculate the standingwave ratio S of the designed circuit at 1.2 and 1.8 GHz. 3.39 Helical antenna. The feedpoint impedance of an axialmode helical antenna with a circumference C on the order of one wavelength is nearly purely resistive and is approximately
SteadyState Waves on Transmission Lines
206
Chap.3
given33 by RL ~ 140(C /A.), with the restriction that 0.8A. < C < l.2A.. Consider a helical antenna designed with a circumference of C = A.o for operation at a frequency fo and corresponding wavelength A.o. The antenna must be matched for use with a son transmission line atfo. (a) Design a singlestage quarterwave transformer to realize the design objective. (b) Using the circuit designed in part (a), calculate the standingwave ratio S on the son line at a frequency 1S% above the design frequency. (c) Repeat part (b) at a frequency 1S% below the design frequency. 3.40 Quarterwave matching. Many microwave applications require very low values of S over a broad band of frequencies. The two circuits shown in Figure 3. 83 are designed to match a load of ZL = RL = 400n to a line with Zo = SOn, at 900 MHz. The first circuit is an airfilled coaxial quarterwave transformer, and the second circuit consists of two airfilled coaxial quarterwave transformers cascaded together. (a) Design both circuits. Assume ZQiZQ2 = ZoZL for the second circuit. (b) Compare the bandwidth of the two circuits designed by calculating S on each line at frequencies 1S% above and below the design frequency.
,,\/4
RL = 4000
Z 0 = 50n
,,\/4
,,\/4
RL = 4000
Z 0 = 50n
Figure 3.83 Quarterwave matching. Problem 3.40.
3.41 Quarterwave matching. A 7Sn coaxial line is connected directly to an antenna with a feedpoint impedance of ZL = 1S6n. (a) Find the loadreflection coefficient and the standing wave ratio. (b) An engineer is assigned the task of designing a matching network to match the feedpoint impedance of the antenna (1S6n) to the 7Sn coaxial line. However, all he has available to use for this design is another coaxial line of characteristic impedance S2n. The engineer uses the quarterwave matching technique to achieve the match. How?
3.42 Lsection matching networks. A simple and practical matching technique is to use the lossless Lsection matching network that consists of two reactive elements. (a) Two Lsection matching networks marked A 1 and A2, each consisting of a lumped inductor and a capacitor, as shown in Figure 3.84, are used to match a load impedance of ZL = 60 j 80n to a lOOn line. Determine the L section(s) that make(s) it possible to achieve the
33 See
Chapter 7 of J. D. Kraus, Antennas, 2nd. ed., McGrawHill, New York, 1988.
Chap.3
207
Problems
design goal, and calculate the appropriate values of the reactive elements at 800 MHz. (b) Repeat part (a) for the two Lsection networks marked Bl and B2, consisting of two inductances and two capacitors, respectively. Al
Bl
I loon A2
Cz ~ t.......0.
B2
I loon [

(b)
(a)
Figure 3.84 Lsection matching networks. Problem 3.42.
3.43
Variable capacitor. A shunt stub filter consisting of an airfilled coaxial line terminated in a variable capacitor is designed to eliminate the FM radio frequencies (i.e., 88108 MHz) on a transmission line with Zo = lOOn, as shown in Figure 3.8S. If the stub length is chosen to be 2S cm, find the range of the variable capacitor needed to eliminate any frequency in the FM band. Assume the characteristic impedance of the stub to be also equal to lOOn.


loon
loon

Figure 3.85 Variable capacitor. Problem 3.43.
3.44
Impedance matching network design. Consider the transmissionline circuit as shown in Figure 3.86. As a design engineer, your task is to determine the electrical lengths of the two shortcircuited stubs (each son) connected at the load position to match the load impedance ZL = 2S  j 7sn to the line characteristic impedance Zo =son such that 11 + 12 is minimum. 3.45 Matching with lumped reactive elements. Two variable reactive elements are positioned on a transmission line to match an antenna having a feedpoint impedance of 100 + j lOOn to a Zo = lOOn airfilled line at S GHz, as shown in Figure 3.87. (a) Determine the values of the two reactive elements to achieve matching. (b) If the reactive elements are to be replaced by shorted son airfilled stubs, determine the corresponding stub lengths. 3.46 Standingwave ratio. For the transmission line shown in Figure 3.88, calculate S on the main line at (a) 800 MHz, (b) 880 MHz, and (c) 960 MHz. 3.47 Quarterwave transformer design. Consider the transmissionline circuit, as shown in Figure 3.89. Design the quarterwave matching network shown (find its electrical position l /A. from the load and its characteristic impedance ZQ) to match the load impedance ZL to Z 0 under the condition ZQ > Z 0 .
SteadyState Waves on Transmission Lines
208
Main line

Source end
Zo = SOn
ZL = 25j75Q

Figure 3.86 Impedance matching network design. Problem 3.44. 3,,\/ 16
Zo = lOOn
L~
c~
1oon
ZL = 100 + jlOOQ
Figure 3.87 Matching with lumped reactive elements. Problem 3.45.
f = BOOMHz
Main line Zo = SOn
Figure 3.88 Standingwave ratio. Problem 3.46. Main line

Z 0 = 6on
,,\/4 •
l •
Zo
•
•
Z 0 = 60n
ZL = 30j18Q
Figure 3.89 Quarterwave transfonner design. Problem 3.47.
Chap.3
Chap.3
3.48
209
Problems
Impedance matching. A load of ZL = 36 + j 40n is to be matched to a main line of characteristic impedance Zo = lOOn using a circuit as shown in Figure 3.90. (a) The opencircuited stub is positioned at a distance of 11 =A./ 4 away from the load. Determine the characteristic impedance ZQ of this quarterwave segment and the stub length ls which would result in a match. (b) For the values found in part (a), calculate the standing wave ratio S on the quarterwave line segment with the characteristic impedance ZQ. Also find the magnitude and the phase of the V1 at the opencircuited end of the stub. lOOQ
10 cos(wt) '\J
(V)
,,\/4
Main Line Z 0 =100Q
ZL=36+j40Q
Figure 3.90 Impedance matching. Problem 3.48.
3.49
Unknown feedpoint impedance. A son transmission line is terminated with an antenna that has an unknown feedpoint impedance. An engineer runs tests on the line and measures the standingwave ratio, wavelength, and a voltage minimum location away from the antenna feed point to be at 3.2 cm, 20 cm, and 74 cm, respectively. Use the Smith chart to find the feedpoint impedance of the antenna. 3.50 A lossy highspeed interconnect. The perunit line parameters of an IC interconnect at S GHz are extracted using a highfrequency measurement technique resulting in R = 143.Sn(cm) 1, L = 10.1 nH(cm) 1, C = 1.1 pF(cm) 1 , and G = 0.014 S(cm) 1, respectively. 34 Find the propagation constant y and the characteristic impedance Zo of the interconnect at S GHz. 3.51 Characterization of a highspeed GaAs interconnect. The propagation constant y and the characteristic impedance Zo at S GHz of the GaAs coplanar strip interconnects considered in Example 3.27 are determined from the measurements to be y ~ 1.1 np(cm) 1 +j3 rad(cm) 1 and Zo ~ 110j40n, respectively. Using these values, calculate the perunit length parameters (R, L, G, and C) of the coplanar strip line at S GHz. 3.52 A lossy highspeed interconnect. Consider a highspeed microstrip transmission line of length 10 cm used to connect a 1V amplitude, 2GHz, son sinusoidal voltage source to an integrated circuit chip having an input impedance of son. The perunit parameters of the microstrip line at 2 GHz are measured to be approximately given by R = 7.Sn(cm) 1, L = 4.6 nH(cm) 1 , C = 0.84 pF(cm) 1 , and G =0, respectively. (a) Find the propagation constant y and the characteristic impedance Zo of the line. (b) Find the voltages at the source and the load ends of the line. (c) Find the timeaverage power delivered to the line by the source and the timeaverage power delivered to the load. What is the power dissipated along the line?
34
W. R. Eisenstadt and Y. Eo, S parameterbased IC interconnect transmission line characterization, IEEE Trans. Comp. Hybrids Manufact. Technol., 15(4) pp. 483489, August 1992.
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The . . . . tatic Electric Fiel
Following our discussion of transmission line behavior in Chapters 2 and 3, we begin in this chapter a more traditional study of electric and magnetic fields, following the historical development of ideas that led to the general laws of electromagnetics known as Maxwell's equations. The voltage and current waves on twoconductor transmission lines are but special cases of electromagnetic waves, which can propagate in empty space and in material media as well as being guided by a variety of conducting or insulating structures. Kirchhoff's voltage and current laws, which were used in Chapter 2 to derive the basic transmission line voltage and current equations, are special cases of Maxwell's equations, which in general are expressed in terms of electric and magnetic fields. This chapter is the first of four chapters in which we undertake the development of these fundamental laws of electromagnetics, culminating in the establishment of the full set of interconnected laws in Chapter 7. Our bases for the development of Maxwell's equations are experimentally established facts of nature, as verified by physical observations. The interaction between electric and magnetic fields is at the root of electromagnetic wave propagation the basis for all telecommunications, from the telegraph to today's satellite, wireless, and optical fiber networks. The interaction between electric and magnetic fields is also responsible for the behavior of electric circuit elements and networks as well as the workhorses of our industrial society: electrical motors and machinery. Our interest in electromagnetics stems from a need to understand the behavior of practical devices and systems, to describe such devices and systems mathematically, to predict their perfor1nance, and to design systems for particular applications. To achieve these goals requires an understanding of the physical bases of the fundamental laws of electromagnetics, which are developed in Chapters 4 through 7 and stated compactly as Maxwell's equations in Section 7.4. The branch of electromagnetics dealing with electric charges at rest, namely static electricity or electrostatics, involves the study of the first,
211
212
The Static Electric Field
Chap.4 1
and one of the most important, of these fundamental laws, known as Coulomb's law. Coulomb's law is based on physical observation and is not logically or mathematically derivable from any other concept. The experimental basis of this law, the mathematical for1nulations that it leads to, and its broad range of applications and implications are the subjects of this chapter. The study of electrostatics constitutes a relatively simple first step in our quest to understand the laws of electromagnetics. In electrostatics we deal only with electric charges that are at rest and that do not vary with time. Yet, mastering the behavior of static electric fields and the techniques for the solution of electrostatic problems is essential to the understanding of more complicated electromagnetic phenomena. Furthermore, many natural phenomena and the principles of some important industrial and technological applications are electrostatic in nature. Lightning, corona discharge, and arcing are natural phenomena that involve very strong electrostatic fields that cause ionization in the surrounding medium. Lightning discharges involve currents of tens of kiloamperes that flow for a few hundred microsec10 2 onds and represent release of energies of up to 10 joules. Applications of electrostatics in industry encompass diverse areas cathode ray tubes (CRTs) and flat panel displays, which are widely used as display devices for computers and oscilloscopes; ink jet printers, which can produce goodquality printing at very fast speeds; and photocopy machines are all based on electrostatic fields. Electrostatic technologies are extensively used for sorting charged or polarized granular materials; 3 applications of this technology are reflected in hundreds of patents, extending from mineral beneficiation and seed conditioning to recycling of metals and plastics from industrial wastes. In Microelectromechanical Systems (MEMS), where small mechanical devices are fabricated together with integrated circuits, electrostatic forces are often utilized by the integrated digital electronics as a means to interact with the surrounding environment (see Section 4.14). Among other applications, electrostatic spraying and painting, electrostatic precipitators and filters, electrostatic transducers, and electrostatic recording are utilized in numerous industrial and household applications. Our discussion of electrostatics in this chapter also serves to bring us closer to a full understanding of the underlying physical basis of the transmission line behavior discussed in Chapters 2 and 3. One of the important physical properties of a transmission line is its distributed capacitance, which comes about as a result of the separation of charge induced on the two conductors that constitute a transmission line. In Chapters 2 and 3, we took it for granted that a twoconductor system exhibits capacitance, and we presented formulas (in Table 2.2) for the distributed capacitances of a few common transmission line
1That
electric charges attract or repel one another in a manner inversely proportional to the square of the distance between them. C. A. de Coulomb, Premiere memoire sur l' electricite et magnetisme (First memoir on electricity and magnetism), Histoire de l'Academie Royale des Sciences, p. 569, 1785. 2 See Sections 1.3 and 1.8 of M.A. Uman, The Lightning Discharge, Academic Press, San Diego, California, 1987. 3 A. D. Moore (Ed.), Electrostatics and Its Applications, John Wiley & Sons, 1973; F. S. Knoll, J. E. Lawver, and J. B. Taylor, Electrostatic separation, in Ullman's Encyclopedia of Industrial Chemistry, 5th ed., VCH, Weinheim, New York, 1988, Vol. B2, pp. 2012011.
Sec. 4.1
Electric Charge
213
structures. In this chapter we define the concept of capacitance and discuss how the capacitance of different conductor configurations can be determined using Coulomb's law. In addition to starting our discussion of the fundamental underpinnings of electromagnetic theory, we also present in this chapter the important concepts of vector algebra. Although some of the most basic aspects of vector algebra are provided in Appendix A, we introduce important concepts such as gradient and divergence along with the relevant physical laws. This approach ensures that the physical significance of the vector operations can be best understood in the electromagnetic context. Throughout this book, vector quantities are often written using boldface symbols (e.g., G), although a bar above the symbol is also used (e.g., C§). 4 In either case, the vector in question is understood in general to have three components, Gx, Gy, and Gz, and it is often written as G = X.Gx + yGy + zGz, where x, y, and z are the unit vectors in the x, y, and z directions, respectively. The ''hat'' notation is always used to represent a unit vector, so we " " can also write the vector G as G = GG, where G is a unit vector in the direction of G and G is the magnitude of G, G = IGI. Points in threedimensional space are identified by means of their position vectors, which point from the origin of the coordinate system to the point in question. For example, the position vector for point P located at (xp, Yp, Zp) 1s rp = XXp + YYp + zzp· •
A
A
A
4.1 ELECTRIC CHARGE Our experiences with electricity date back to ancient times 5 and have their roots in the observation that, for example, a piece of glass and a piece of resin (or rubber) attract one another if they are first rubbed together and then separated. Also, if a second piece of glass is rubbed with another piece of resin, the two pieces of glass or resin repel one another, while each glass piece attracts each piece of resin. Various manifestations of 6 electrification by friction are encountered in our daily experiences. 4 This
alternative notation is used starting in Chapter 7 in order to distinguish between real physical quantities and their corresponding complex phasors, necessary for sinusoidal (or timeharmonic) applications. 5 For example, Thales of Miletus [640540 B.C.] wrote that a piece of amber rubbed in silk attracts pieces of straw. 6 In his text, Electricity and Magnetism (Cambridge Press, 1907), Sir James Jeans gives the following amusing account of Robert Symmer's [1759] observations: He was in the habit of wearing two pairs of stockings simultaneously: a worsted wool one for comfort and a silk pair for appearance. In pulling off his stockings, he noticed that they gave a crackling noise, and sometimes they even emitted sparks when taken off in the dark. On taking two stockings off together from the foot and then drawing the one from inside the other, he found that both became inflated to reproduce the shape of the foot and exhibited attractions and repulsions at a distance as much as a foot and a half. ''When this experiment is performed with two black stockings in one hand, and two white in the other, it exhibits a very curious spectacle; the repulsion of those in the same colour, and attraction of those of different colours, throws them into an agitation that is not unentertaining, and makes them catch each at that of its opposite colour, and at a greater distance than one would expect. When allowed to come together they all unite in one mass. When separate, they resume their former appearance, and admit of the repetition of the experiment as often as you please, till their electricity, gradually wasting, stands in need of being recruited."
214
The Static Electric Field
Chap.4
4.1.1 Electrification by Friction, Induction, and Conduction These electrical phenomena of attraction and repulsion that come about due to friction and that are part of our daily experiences7 are understood in ter1ns of electric charge. Electric charge is said to be acquired by the material as a result of rubbing. In actual fact, when glass and resin, for example, are rubbed together, a small amount of charge is transferred from one to the other, causing each material to become nonneutral, that is, charged. The glass becomes positively charged, while the resin acquires negative charge. The different behavior of glass and resin indicates that there must be two different types of charge. Materials that behave upon electrification like glass are said to be positively charged, and those that behave like resin are said to be negatively charged. Other materials also acquire charge as a result of being rubbed, although this property may be less apparent for some than others. Metallic materials, such as brass, do not retain electricity for a sufficient time (after they are rubbed to another object) for us to observe it. However, a brass rod with a glass handle becomes electrified to a marked degree on rubbing. Once it is electrified, such a brass rod with a glass handle loses all of its electricity if it comes in contact with water, flame, or the human body. Conversely, the same rod retains its power if it comes in contact with hard rubber, a piece of silk, or wood. We understand such behavior in ter1ns of the classification of materials into conductors of electricity and insulators. Metals are excellent conductors of electricity. Solutions of salts, acids, and other electrolytes are also conductors. Examples of good insulators are oils, waxes, silk, glass, rubber, and plastics. Gases are ordinarily good insulators, but flames or other ionized gases are good conductors. Distilled water is almost a perfect insulator, but any other type of water contains impurities, which in general cause it to conduct reasonably well. Being made largely of water, the human body conducts reasonably well and is generally a bad insulator. Although electrification by friction is part of our daily experiences, material bodies can also acquire electric charge through induction. If we suspend an uncharged metallic (e.g., brass) object by silk threads, as shown in Figure 4.la, and bring near one end of it a positively (or negatively) charged rod, the nearby end of the suspended object becomes negatively (or positively) charged, while the other end becomes primarily positive (or negative). If the charged rod is removed, the suspended object loses its electrification. However, if the suspended object is cut in the middle before the inducing charge is removed, each of its parts remains electrified, one positive and the other negative, as shown in Figure 4.lb. When a charged conductor is connected to another conductor (charged or not), for example by means of a metal wire, the total charge on both bodies is shared between the two. The second body is then said to have become electrified via conduction through the 7 In
our everyday lives, we experience frictional electrification every time we rub our feet across a wool carpet, pull clothes out of a dryer, or comb our hair with a plastic comb. On a dry summer day, it is not uncommon to receive a shock on touching a doorknob, hear an accompanying crackling noise, and sometimes see a spark. We see sparks in such cases because large amounts of charge create electric fields that cause local breakdown of air.
Sec. 4.1
215
Electric Charge
+ + + +
+ y... y... y...
+ + + +
y...
+
+ +
y...
y...
y...
y... y...
y... y...
y... y...
y...
(a)
(c)
I I I I I
+, + +
+ ..
+
+ + + +
+ +
+ +
+ +
+ + +
y... y... y...
y...
y... y...
y...
y... y...
(b)
(d)
Figure 4.1 Electrification by induction. (a) A charged rod near a suspended neutral object induces opposite charge on the surface closer to it. Since the suspended object is initially neutral, an equal amount of positive charge remains on its far surface. Since the negative charge is closer to the charged rod, the attractive force is stronger than the repelling force, and the suspended object deflects toward the charged rod. (b) If we cut the suspended object in the middle while the charged rod is near it, each of its two parts remains electrified, one positive, one negative. This is an example of electrification by induction. (c) If the charged rod is brought in physical contact with the neutral body, the excess positive charge on the rod is shared between the two objects and the body is now positively charged. This is an example of electrification by conduction. (d) The two positively charged objects now repel one another.
metal wire. Such conduction would clearly not occur if the two bodies were connected with a silk thread. Once again we see a basis for classifying materials as conductors (e.g., metals) and insulators (e.g., silk). Electrification by conduction is essentially what occurs when we place the charged rod in contact with the neutral body as in Figure 4.lc; the charge on the rod is now shared between the two objects, which now repel one another, as shown in Figure 4. ld.
216
The Static Electric Field
Chap.4
4.1.2 Faraday's GoldLeaf Electroscope Long series of experiments carried out by M. Faraday, 8 many years after the establishment of Coulomb's law, were instrumental in bringing about a physical understanding of electrical phenomena and demonstrated the principle of conservation of electric charge. In assessing the quantity of electrification or the quantity of charge associated with any body, it is useful9 to think in terms of a goldleaf electroscope, as shown in Figure 4.2. Under normal conditions the gold leaves in Figure 4.2a hang flat side by side. When an electrified body (e.g., a charged rod) touches (conduction) or is brought near the brass rod of the goldleaf electroscope (induction), the two gold leaves separate because of electrostatic repulsion (Figure 4.2b), so that the electroscope can be used to examine the degree to which a body is charged. Now consider a metal vessel placed on top of the brass rod (Figure 4.2c), closed but having a lid attached to a silk thread so that it can be opened or closed without touching it. When a charged glass ball is inserted into the vessel and the lid is closed, opposite charges are induced on the inner surface of the vessel. Since the metal vessel and the gold leaves were originally neutral, unbalanced positive charge appears on the outside of the vessel, some of which is on the gold leaves, which repel one another and diverge. The separation of the gold leaves, which indicates the degree of electrification (i.e., amount of charge), remains exactly the same if we do the experiment by placing the electrified ball at different positions inside the vessel, as long as it does not come into contact with the vessel or other conductors. The separation of the gold leaves is also independent of any changes in shape of the glass piece (e.g., a thin rod versus a round ball or a straight rod versus a bent one) or any changes in its state (e.g., it might be heated 10 or cooled). It thus appears that the separation of the gold leaves is due only to a quantity of electricity, or electric charge, associated with the glass ball. Now imagine two balls, one glass (A) and one rubber (B), electrified by rubbing against one another (Figure 4.2d). If we introduce A and B separately into the vessel, the gold leaves diverge by the same exact amount either way. If we introduce both A and B together into the vessel, we find that the leaves stay flat; that is, no electrification occurs outside the vessel. From this we conclude that the process of electrification by friction is not one that creates charge; rather, rubbing merely transfers charge 11 from one object to another, slightly disturbing the neutrality of each. If we insert two charged glass balls A and A' into the vessel, the induced charge on the vessel would be the algebraic sum of what it would be when each ball was introduced separately. If, while A and A' were inside the vessel, we connect them with a conducting wire, the induced charge on the vessel would not change, indicating once again that the quantity of total charge remained constant. 8 M.
Faraday, Experimental Researches in Electricity, Vol. 3, Art. 3249, Bernard Quatrich, London, 1855. 9 We do not pretend here that Faraday's experiments can be carried out under modem conditions to verify electrical laws accurately or that a goldleaf electroscope would be used today to measure charge. The discussion in this section should simply be viewed as a set of thought experiments that illustrate the properties and constitution of electricity remarkably well. 10 Not with a flame, however; as was mentioned before, flames are good conductors because they consist of ionized matter. 11 In actual fact, electrons are transferred, in this case from glass to rubber.
Sec. 4.1
217
Electric Charge
+ + + +
+ + +
+ +
+ + + + + + + .. +
 
I I
 
+ + + + + + + ... + + +
 +,.. + + .... + +"" + + ....   
(b)
D
~
+ + + + + + r + + ~ + ++ + ++ + +
I
+ + +~ + f F ·+
~
\...
(a)
I
A r
~
1

..
..
r
'" 
•

B
"
~
~
\...
(c)
...
(d)
Figure 4.2 Faraday's goldleaf electroscope. (a) An actual unit might consist of a glass vessel, through the top of which a metal (brass) rod is passed, to which are attached two gold leaves. (b) A charged object is brought near the brass rod; the gold leaves repel one another and diverge. (c) A closed metal vessel with a lid is placed on top of the brass rod so that a charged glass ball can be inserted within it. The gold leaves deflect by the same amount, regardless of the position of the charged ball within the vessel. (d) If two equally but oppositely charged bodies are inserted into the vessel, the gold leaves stay flat.
4.1.3 Electric Charge and the Atomic Structure
Our goal in this book is to develop and apply the fundamental laws of electrical and magnetic phenomena that are valid in the macroscopic, nonatomic realm. The physical experiments performed to verify Coulomb's law, and most applications of electrostatics, involve the use of ponderable macroscopic objects. The same is true for the other laws of electromagnetics that we shall study in later chapters. A brief review of modern notions concerning the atomic nature of matter is helpful in clarifying these macroscopic principles. Matter is composed of atoms with dimensions of order of one angstrom, or 10 8 cm, each containing a nucleus of dimensions rv 10 12 cm and electrons that move about the nucleus. It is the electrons' electrical influence on those of other atoms that
218
The Static Electric Field
Chap.4
effectively defines the atomic dimensions. Regardless of where they reside, electrons are all alike, each bearing a charge qe ~ 1.602 x 10 19 coulombs. 12 Most of the mass of 13 an atom is contained in its nucleus, which bears a positive charge of Na lqe I, where Na is the atomic number14 of the particular element. Thus, there essentially are two kinds of electrical charge in matter: positive and negative. Each atom or molecule in its normal state has as many electrons as it has units of positive charge in its nucleus or nuclei, and thus is electrically neutral. Nearly all of the macroscopic effects of electricity arise from the fact that electrons may be separated from their atoms in some circumstances, leading to separation of positive and negative charges by distances appreciably greater than atomic dimensions. At present, the universe is believed to contain equal amounts of positive and negative electricity, with the result that an excess of one polarity of charge at one place implies the presence elsewhere of an equal but opposite charge. Furthermore, net amounts of electric charge can neither be created nor destroyed (the law of conservation of electric charge). The atomic nature of matter is inherently quantized in that electric charges are made up of integral multiples of the electron charge qe. However, any charged object of macroscopic size (i.e., much larger than atomic dimensions) typically holds so many electrons (or, if positively charged, would have a deficiency of so many electrons) that it is considered to have a ''continuous'' distribution of charge. Since we restrict our attention here to the macroscopic realm, all quantities that are dealt with, whether they be charges, fields, or potentials, should be understood to be macroscopic in nature. For example, the discussion of Coulomb's law in the next section describes the electrostatic force between two point charges residing at rest at two different locations in vacuum. In the context of our discussion, we implicitly understand that these point charges may actually occupy physical space of many atomic dimensions in extent, but that the size of these regions is nevertheless negligible on a macroscopic scale.
4.2 COULOMB'S LAW Although the preceding discussion introduces us to the fundamentals of electrostatic principles, the law of action between electrified bodies needs to be specified for mathematical formulation of these concepts. This law of action, known as Coulomb's law, is the quantitative expression of the experimental observations discussed above. Coulomb's law states that the electric force between two point charges 15 Q 1 and Q 2 is proportional to the product of the two charges QiQ2 and inversely proportional to the square of the 12 The
most precise value of the charge of an electron is qe = 1.60217733(49) x 10 19 C, as specified in The CRC Handbook of Physics & Chemistry, 16th ed., CRC Press, Boca Raton, Florida, 19951996. 13 An electron is physically quite light, having a mass of me ~ 9.11 x 1031 kg, much smaller than the mass of a proton ("''1.66 x 1027 kg) or a neutron, the two types of particles that together form the nucleus. 14 The atomic number of an element is defined as the number of protons present in its nucleus. 15 By ''point charges'' we mean charges that occupy a macroscopic region of space much smaller in extent than the distance between the charges.
Sec. 4.2
Coulomb's Law
219
distance between them. In vector for1n, the forces Fi 2 and F2i felt by charges Q2 and Qi are given as kQiQ2 (4.1) Fi2 = F2i = R  A
R2
where R = r2  ri, with ri and r2 being the position vectors i 6 for points Pi and P2 (where charges Qi and Q2 reside), respectively, shown in Figure 4.3. As for any other vector, we can write R as R = RR, where R is the unit vector in the direction of R and R is the magnitude of R. Note that R is the vector pointing from point Pi to P2 and that R = R/R = (r2  ri)/(lr2  ril) and R = lr2  ril. The force as given in (4.1) is repulsive (as shown in Figure 4.3) if the charges are alike in sign and is attractive if they are of opposite sign (i.e., QiQ2 < 0, so that the directions of both Fi2 and F2i in Figure 4.3 are reversed). The proportionality constant k is equal to (4n Eo)i, with the 4n ''rationalization'' factor included so that a 4n factor does not appear in Maxwell's equations, which are more commonly used than (4.1 ). A
A
A
R
Figure 4.3 Coulomb's law. Forces on point charges Q 1 and Q2 when the two charges are alike in sign.
Using the SI (Le Systeme International d'Unites, International System of Units) units of force, charge, and distance respectively, newtons (N), coulombs (C), and meters (m) the value of the quantityi 7 Eo, referred to as the permittivity of free space, is is Eo ~ 8.854 x 10i 2 farads per meter (or Fmi ). Since the proportionality constant k is needed in numerical evaluations, it is helpful to note that in SI units, k = (4nEo)i ~ 9 x 109 mFi. A coulomb can be defined as the quantity of charge that flows through a given cross section of a wire in 1 second when there is a steady current of 1 ampere (A) flowing in the wire. i 9 The following example introduces the application of Coulomb's law. Example 4.1: Tuo point charges. Two point charges of Q 1 = +37 nC and Q2 = + 70 nC are located at points (1, 3, 0) m and (0, 0, 2) m, respectively, as shown in Figure 4.4. Find the force exerted on Q2 by QI· 16 As
defined earlier, the position vector for any point P located in threedimensional space is the vector pointing from the origin to the point P. 17 It will later become evident that the precise value of this quantity is Eo = (4rr x 107 c 2 ) 1, where c is the speed of light in vacuum. 18 The physical meaning of the dimensions of Eo will become clearer in later sections, when we discuss capacitance. 19 The definition of an ampere will be discussed in Chapter 6 in connection with Ampere's law.
The Static Electric Field
220
Chap.4
z
Q2 + (0, 0, 2)
3 ri
/
y
/
        Ql+ (1, 3, 0)
Figure 4.4 Two point charges. Configux
ration of two point charges for Example 4.1.
Solution:
The repulsive force Fi2 on charge Q2 exerted by Qi is
where
12z
R = lr2  ril =
since r2 =
2z and ri
2 2 2 2 ]i1  x  3yl = [(1) + (3) + 2 = ,J14 m A
=
x + 3y. The unit vector R in direction of Fi2 is
R=
R = r2 ri R lr2  ri I
2z  x 3y ,J14
Therefore, we have
9 x 109 mFi x 37 x 109 C x 70 x 109 C Fi2 = 14 m2
x3y+ 2z
,J14
'V"·R"
~
1.66 x 10
_6
x 
3y + 2z
f1A
v14
~
445( x

3y + 2z) nN
It is obvious that the repulsive force F2i on charge Qi exerted by Q2 is F2i = Fi2.
To acquire a feel for the magnitude of the electric force as represented by (4.1 ), (e.g., in comparison with the gravitational force), consider the force between two protons in the nucleus of a helium atom, which are rv 10 15 m apart. 20 The gravitational force between the two protons is rvl.84 x 1034 N, but the electrical force from (4.1) is rv230 N! The electrical force, in other words, is rvl036 times larger. Were it not for the nuclear force that keeps the nucleus together, the initial acceleration that each of the protons would acquire due to the electrical force would be rv 1028 g ! 20 Approximate
radius of helium nuclei.
Coulomb's Law
Sec. 4.2
221
To appreciate the enormity of one coulomb of charge, note that two 1 C charges 1 m apart would exert an electrical force upon one another of rv9 x 109 newtons! In nor1nal air, a single 1 C charge would cause electrical breakdown21 at distances of rv50 m. If there are more than two charges present, (4.1) holds for every pair of charges, and the principle of superposition can be used to determine the net force on any one of the charges due to all others. For this purpose, the force due to each charge is determined as if it alone were present; the vector sum of these forces is then calculated to give the resultant force, as illustrated in Example 4.2. Example 4.2: Three point charges. Consider two point charges of Qi = + 1 /LC and Q2 = + 2 /LC located at (1, 0) m and (1, 0) m, respectively, as shown in Figure 4.5. (a) What is the magnitude and direction of the electrical force felt by a third charge Q3 = + 1 nC when placed at (0, 1) m? (b) At what point(s) must the third charge Q3 = + 1 nC be placed in order to experience no net force? y F23 F13
(a) Q3 + (0,1)
Ri3 r3 Q2 + (1, 0) r2
rl
+
Qi
(1, 0)
x
(x1, 0)
(b)
I I
Q2 F13
+ Q3 F23
Ql
+
Figure 4.5 Three point charges. Configx
urations of the three point charges Qi, Q1, and Q3 for Example 4.2.
Solution:
(a) Note from Figure 4.5a that r3 = y, r 1 = x, and r2 = x. Using Coulomb's law along with the superposition principle, the repulsive forces exerted on charge Q3 by charges Qi and Q2 are, respectively,
where
21 Breakdown
of air occurs when the electric field intensity in air exceeds 3 x 106 Vm 1, at which time atmospheric electrons and ions are accelerated to high energies as a result of the Coulomb force. As these energized particles have inevitable collisions with neutral air molecules, many other electrons and ions are knocked out of the neutral air molecules, resulting in arcing and corona discharges.
222
The Static Electric Field A
and the unit vectors
Chap.4
A
Ri3
and
R23
are
Therefore we have
yx A
A
v2
9 x 109 mF 1 x 106 C x 109 C 4.5 2m2 =y2(i+y)
µ,N
and similarly
µ,N The total electrical force on charge Q3 is thus
(b) Since the forces exerted by the two charges on Q3 must be equal and opposite in order to have no net force, Q3 must be placed somewhere along the straight line between Qi and Q2 (i.e., on the x axis), as shown in Figure 4.5b. Let us assume that Q3 is located at (x1, 0) and find x1. Referring to Figure 4.5b, we have
Solving for x 1, we find x1
~
0.172 m.
It is important to remember that Coulomb's law as stated above is defined for ''point'' charges. To see this, consider two spheres A and B at a distance R from one another; sphere A is charged by an amount +QA while sphere B is uncharged, QB = 0. Direct application of Coulomb's law specifies the force between the two bodies to be
IFI =
1 (+QA)(O) = 0 4nEo R2
However, based on the discussion in connection with Figure 4.1, we know that the force is not zero, because sphere A will induce negative charge on the side of sphere B closer to it and will be attracted toward it. For this induction effect to be negligible, we must have b r' or R > 0). The 42 ''Faraday,
in his mind's eye, saw lines of force traversing all space where the mathematicians saw centres of force attracting at a distance: Faraday saw a medium where they saw nothing but distance: Faraday sought the seat of the phenomena in real actions going on in the medium, they (i.e., the mathematicians) were satisfied that they had found it (i.e., the seat of the phenomena) in a power of action at a distance impressed on electric fluids." From J. C. Maxwell, A Treatise on Electricity and Magnetism, Clarendon Press, Oxford, 1892.
Sec. 4.3
The Electric Field
229
resolution of this apparently unphysical result lies in our original premise (Section 4.1.3) of restricting our attention to effects on macroscopic scales, thus ensuring that the distance R = Ir  r' I between source and observation points is much larger than atomic dimensions. Example 4.3: Electric field. Find the electric field at P(3, 1, 0) due to a point charge Q = +80 nC located at (2, 0, 2) mas shown in Figure 4.8. Assume all distances to be in meters. Solution: Ep =
Note from the geometry that r = 3x + y and r' = 2x + 2z. Using (4.5) we have
x+ y 
80 nC
(9 x 109 ) 1 [(3  2)2
+
2z '"'/ 49(x + 12 + (2)2]3/2 
y
2z) Vm 1
Note from Figure 4.8 that Ep points predominantly downward, consistent with the fact that the magnitude of the z component is twice as large as that for the other two components.
z
2
1 / /
y
/
r
/
/
/
/ /
r r'
/
v
x
Ep
/
P(3, 1, 0)
Figure 4.8 Electric field. Electric field Ep at point P(3, 1, 0) due to a positive point charge Q at (2, 0, 2) m.
4.3.1 Electric Field Resulting from Multiple Charges The electric field concept is most useful when large numbers of charges are present, since each charge exerts a force on all the others. Assuming that for a particular test charge the forces resulting from the different charges can be linearly superposed,43 we can evaluate 43 We
are able to assume linearity simply because, at the macroscopic level, its use produces accurate results in many different experiments and applications involving groups of charges and currents. We note here that some material media can behave in a highly nonlinear manner and that such behavior often leads to interesting applications or consequences, such as in magnetic materials, nonlinear optics, crystals, and electrostatic breakdown phenomena (e.g., lightning discharges). As far as fields in vacuum are concerned, linear superposition is remarkably valid even at the atomic level, where field strengths of order 1011 to 1017 Vm 1 exist at the orbits of electrons. For further discussion, see Section 1.3 of J. D. Jackson, Classical Electrodynamics, Wiley, New York, 1975.
230
The Static Electric Field
r{
Chap.4
Qz+ rr2
r2
p
•
~
Ep
• Figure 4.9 Electric field due to multiple charges. The electric field at the point P at
r r'n
position r is the sum of the electric fields due to the point charges Q 1, Q1, ... , Qn, located at source points r~, r'2 , •.. , r:i, respectively.
r'n
the electric field at a given point simply as a vector sum of the field contributions resulting from all charges. For a set of discrete point charges Q1, Q1, ... , Qn located at points with position vectors r~, ~' ... , r~, as shown in Figure 4.9, the electric field at a point P with position vector r is given as n
E p=
1
4nEo
LR"
k
k=l
Qk
Ir  r'k 12
(4.6)
,...
where Rk is the unit vector pointing in the direction from the source point P~ (~) at which ,... charge Qk resides to the observation point P(r). Noting that Rk = (r  r~)/(lr  r~ I), (4.6) can be written as (4.7)
Example 4.4: Tuo point charges. Consider two point charges Qi = + 5 nC and Qz = 5 nC located at points (1, 0, 0) m and (1, 0, 0) m as shown in Figure 4.10. (a) Calculate the electric field at points Pi (0, 0, 0), P2(l, 1, 0), and P3(0, 0, 1), respectively. (b) Repeat for Qz = +5 nC. Assume all coordinates to be in meters.
Solution:
Using (4.7), the electric field at point P with a position vector r resulting from point charges Qi and Qz located at positions r~ = x and r~ = x, respectively is given as
(a) For point Pi (0, 0, 0) we have r = 0. Therefore, E Pi
9 = 5 x 10 (x) (9 X 109)i 13
9
5 x 10 (+x) = _ ox Vmi 9 (9 X 109)i 13
Sec. 4.3
231
The Electric Field
z
P 3 (0, 0, 1)
E
,' I I I
2
' Qz //
Ei, Ez e (1 0 0) '
/
I I
,'
/
P1
/
I I I
'
'
y
Ez//
I
/
Figure 4.10 Two point charges. The electric field vectors E1 and E2, due to charges Qi= +5 nC and Q1 = 5 nC, respectively, are shown at points P1, P2, and P3.
(1, 0, 0) +      / E1 Qi P2 (1, 1, 0)
x+ y so that 5 x 1o 9 x+ y  x 5 x 1o 9 x+ y + x 
For point P2(l, 1, 0), we have r = Ep2 =
(9 x 109) 1
,.. = 45 y 
,.. ) 18 ,.. 9 (5 ,JS  1) ,.. x+y = x+ y
9 (2,..
,JS
For point P3(0, 0, 1) we have r =
E
_ P3 
5 x 109 (9
X
(9 x 109) 1 [22 + 12]3/2
13
,JS
,JS
v m1
z. Thus,
z x
9
109)1 [12 + 12]3/2
z x
5 x 10+ = _ 45 (9 x 109)1 [12 + 12]3/2 v'2
x Vml
(b) When Q2 changes sign, all E2 vectors reverse direction. As a result, we have
Ep 1 =0 At P2
Ep 2
_ 18 ,.. 
v'sx +
9(5,JS + 1) ,.. V _1
,JS
y
m
45 ,.. 1 Ep3 = v'2z Vm
4.3.2 Electric Field Resulting from Continuous Charge Distributions The atomic theory of matter has shown us that matter is quantized and that electric charges are made up of integral multiples of a certain minimum electric charge, which is known as the charge of an electron, namely qe ~ 1.602 x 10 19 C. Although we know that charge is so quantized, it is often useful to ignore that charges come in packages of electrons and protons and to instead think of continuous distributions of charge. Such a view is valid because the ''graininess'' of electrical charge is not significant over the macroscopic scale sizes that are considered here. We can consider charge as if it were ''continuous'' because we deal with very large numbers of charged particles that are extremely close (typically, confined to material bodies of macroscopic size). As an example, the current that flows
232
The Static Electric Field
Chap.4
V'
p R=rr' ~+,...;::.._...:........:.......+11~.1•~ dEp
p(r')
Figure 4.11 Electric field due to a continuous charge distribution. The field gen
r
erated at point P (position vector r), due to the continuous charge distribution p (r') in the source volume V ', is found by integrating the contributions of infinitesimal volume elements such as that shown.
in a wire connected to an ordinary 110volt, 100watt light bulb is about 1 ampere (or coulomb per second), so that rv6 x 1018 elementary charges flow per second through the crosssectional area of the wire. We can describe such ''continuous'' distributions of charge in ter1ns of density of charge, much like representing matter in terms of its mass density rather than counting the number of molecules. The volume charge density, p (x, y, z) can in general be a function of all three coordinates of space and has units of 3 Cm . In other cases, it might be more convenient to represent the charge distribution in terms of a surface charge density Ps or a line charge density Pt. Consider electric charge to be distributed over a volume V' with a volume density p (r') as shown in Figure 4.11. Since a differential element of charge behaves like a point charge, the contribution of a charge element dQ = p (r') dv' located in a differential volume element44 dv' to the electric field at point P in Figure 4.11 is
,.. p (r') dv' dEp=R2
4nEo lr  r' l
(4.8)
The total electric field at point P due to all the charge in V' is then given as 1 Ep=
Rp(r') dv'
4nEo v 1
Ir  r' l2
(4.9)
Note once again that (4.9) can alternatively be written as
1 Ep=4nEo
V'
(r  r')p(r') dv' Ir  r' l3
(4.10)
If the electric charge were instead distributed on a surface S' with a surface charge density Ps (having units of Cm2 ), then the integration would need to be carried out only over the surface. Thus, 1 Ep=
4nEo s
44
RPs(r') ds' 1
Ir  r' l2
(4.11)
Note that although dv' can be made very small compared with the macroscopic dimensions of the problem in hand, it is nevertheless taken to be large enough compared with atomic scales so that it still does contain many discrete charges.
Sec. 4.3
233
The Electric Field
or, alternatively, 1 (r  r') Ps (r') ds' Ep=Ir  r' l3 4nEo s'
(4.12)
Surface charge distributions arise commonly in electromagnetics, especially when metallic conductors are placed in electric fields. As we shall see later, freely mobile electrons in a metallic conductor redistribute quickly (within '""'10 19 s) under the influence of an applied electric field by establishing distributions of charge confined to infinitesimally (on macroscopic scales) thin regions at the conductor surfaces. For a line charge distribution Pt (having units of Cm 1) we have 1 Ep=4nEo
RP1(r')~l
v
Ir  r
1
(4.13)
12
or, alternatively, 1 Ep=4nEo
L'
(rr')p1(r')dl' Ir  r' l3
(4.14)
where L' is the line along which the charge is distributed. Note that L' need not be a straight line, just as S' need not be a flat surface. The concept of a line charge distribution becomes convenient when the objects on which the charges reside have significant extent only in one dimension. A good example is a thin wire with negligible crosssectional dimensions. Example 4.5: Line charge of finite length. Consider a cylindrical rod of charge of length 2a centered at the origin and aligned with the z axis, as shown in Figure 4.12, where the diameter d of the rod is much smaller than any other dimension in the system, so we can represent45 the charge distribution in terms of a line charge density Pl. Determine the electric field at a point P(r, (r) = 00
Pl
2nEo
r .
oo ~
(rd~)
= 
Pl
2nEo
[ln(~)]~
= oo !
We can understand this seemingly unphysical result by recalling the meaning of electrostatic potential at any point as the work that needs to be done to bring a unit positive test charge from infinity to that point. With an infinitely long line charge, the electric field apparently does not fall rapidly enough with distance to allow moving a test charge from oo to any point r without requiring infinite energy.
Sec. 4.4
245
The Electric Potential
Note, however, that the difference in electrostatic potential between any two points r = a and r = b, such that 0 < b < a, is straightforward to evaluate; we find b
Pl a*b = (b)  (a) =   
2nEo
a
dr

r
=
Pl
2nEo
a In 
b
which is the amount of work that we need to do to move a unit positive test charge closer to the line charge from r =a to r = b. Note that for a < b, we have ln(a /b) < 0, so a*b < 0, indicating that the field does work (i.e., work is extracted from the field) when the unit positive test charge moves away from the positive line charge.
4.4.2 The Electrostatic Potential and the Electric Field Since forces on charges are determined by the electric field E, the introduction of the additional concept of electrostatic potential may at first appear unnecessary. One advantage of the potential concept is that it is a scalar, which is relatively easier to evaluate in the vicinity of complicated charge distributions than the vector electric field, and from which the electric field can be readily calculated by simply taking derivatives, as described in the following paragraphs. Another advantage of the potential concept is that it allows us to link field concepts with the more familiar circuit concepts, which are couched in ter1ns of potential (or voltage) differences between different points in a circuit and voltage drops across circuit elements. We can now establish a differential relationship between the electrostatic potential and the electric field E. Consider two points at x and (x + ~x) with the same y and z coordinates. The differential amount of work done (dW) in moving a test charge q between these points is directly proportional to the potential difference between them. Hence, a (dW)x = q(x + ~x,y,z)  q(x,y,z) = q ~x ax If the two points also had slightly different y, dW =q
z coordinates, we then have
a a a ~x+ ~y+ ~z ax ay az
But the work done by an external force against the electric field is also given by dW = q E · ~I, where ~I is the differential length element given by ~I = (x~x + y~y + z~z). Thus we have a a a  ~x+ ~y+ ~z=E·~l ax ay az a x ax
a + y ay
a + z az a x ax A
. (x~x
+ y~y + z~z) =
a + y ay A
a + z az
· ~I =
E. ~I  E · ~I
246
The Static Electric Field
Chap.4
from which it follows that
E=
(4.21)
Equation (4.21) indicates that the electric field E at any given point will have its largest component in the direction opposite to that along which the spatial rate of change of electrostatic potential is the largest. In other words, the electric field at any given point 1 is the negative gradient5 of the electrostatic potential at that point. It is often customary to represent the gradient of a scalar by using the del operator, sometimes also called the grad or nab la operator, represented by the symbol V. For this purpose, the del operator is defined in rectangular coordinates as V
a a a =x ax +y ay +zaz A
A
A
(4.22)
so (4.21) can be compactly written as (4.23) As an example of gradients of potential functions that vary in two dimensions, consider the function shown in Figure 4.20. The function itself is illustrated using a threedimensional surface plot, whereas the corresponding negative gradient (i.e., the electric field) is shown by means of a vector field plot. The plots illustrate that the electric field magnitudes (i.e., size of the arrows) are highest in regions of highest spatial slope of the potential functions and that the electric field has no components in directions along which there is no potential variation. In problems for which a system of charges is specified and where it is necessary to determine the resultant electric field due to the charges, it is often simpler first to find the electrostatic potential (x, y, z) and then determine E(x, y, z) by finding the negative gradient of the potential. This roundabout way of finding E is simpler because the electric field is a vector quantity, whereas is a scalar and can be found as an algebraic (rather than vector) sum of the potentials due to each system of charge. In simple problems, there may be little advantage in using the potential method; however, in more complicated problems, the use of the scalar potential results in real simplification, as in the case of the electric dipole discussed in Section 4.4.3.
51 The
gradient of a scalar function at any given point is the maximum spatial rate of change (i.e., the steepest slope) of that function at that point. Gradient is thus naturally a vector quantity because the maximum rate of change with distance must occur in a given direction. The direction of the gradient vector is that in which the scalar function (e.g., temperature) changes most rapidly with distance. A good analogy here is the gravitational potential. A marble placed on the slope of a mountain rolls down (i.e., its velocity vector is oriented) in the direction opposite to the gradient of the gravitational potential, and its speed (i.e., the magnitude of its velocity) is determined by the spatial derivative in that direction.
Sec. 4.4
247
The Electric Potential E(x, y) " ""\" ,
~
~
t
, '
"' " ' \ \ \ ' t f I
'11 I
I/
~y=2
'/ /
"""''\\\~''I//'/
' '
'' ' .............
.... .......
" " , ' , ' '
' , , ' , ' , ,
.......
/
/
,, y
1~
0.5
(x, y)
o  0.5
' ' '
y
~
"
x 2
2
x=2
x=2
x
(b)
(a)
Figure 4.20 An example of a twodimensional potential function and its gradient field. (a) The function shown is (x,y) = sin(x 2 + y 2 ). (b) The corresponding electric field is E = V = x2x cos(x 2 + y 2 )  y2y cos(x 2 + y 2 ). The units for the x and y axes are in (radians) 112 for both panels.
Gradient in other coordinate systems. The preceding definition of the concept of the gradient of scalar potential in ter1ns of the motion of a test charge in the presence of an electric field was presented in a rectangular coordinate system. We now consider gradient in other coordinate systems (see Appendix A). If we consider the motion of the test charge in a cylindrical (r, , z) coordinate system, we realize that in moving by a differential amount d
and the electric field E at a point P along its axis.
z P(O, , z) Ps 27rr' dr'
Figure 4.26 Disk of charge. Figure for Example 4.10.
Solution: We work in the cylindrical coordinate system (with the disk on the xy plane and centered at the origin). Consider an annular ring of radius r' and a differential width dr', as shown in Figure 4.26. The contribution of the surface charge residing on this ring to the electrostatic potential at point P along the z axis is dct>p(r = O) =
Ps2nr' dr'
4nEoJz 2 + (r') 2 where the 2n term comes from an integration along the ring (i.e., (r =0) = 
a
r' dr'
;::::::========= 2Eo o Jz2 + (r')2
+C
=
p s
2Eo
[(a
2
2 112 ) 
+z
lzIJ + C
where C is a constant and we retain lz I = # in order to ensure that ct> decreases53 as z > ±oo (or lz I > oo) and to ensure that ct> (z) = ct> ( z), as is necessary in view of the 53
At large distances from the disk the electrostatic potential should vary like that of a point charge.
Sec. 4.4
255
The Electric Potential
symmetry in z. Since we view electrostatic potential at a point in reference to infinity, we need to have C = 0 so that ct> > 0 as z > oo. The electrostatic potential at point Pis thus
ct> (r = 0) = Ps [J a 2 + z 2 2Eo

lz I]
2 2 2 Note that for z >>a we can make the approximation Ja +z ~ z +a /(2z), so that
we have
ct>(r = 0)
~
Ps 
2Eo
z
+
a2 2z z
1
Q
4nEo z
z>O
where Q = PsJt a 2 is the total charge on the disk. For z < 0 we would have ct> ~ Q/(4nEoZ) since for the positively charged disk (i.e., Ps > 0) the potential ct> must be positive. To see this, we need to remember that electrostatic potential at a given point is the energy required to bring a unit positive test charge from infinity to that point. We note that, as expected, at large distances the potential due to the disk charge Q behaves like that of a point charge. To find the electric field, we must in general take the negative gradient of ct>, which means that we need to know the variation of ct> as a function of r, ¢, z. However, because of the symmetry of the problem, the electric field at a point on the axis can only be in the z direction. Thus, the electric field at point P is related only to act>/ Bz. In other words, even though ct> in the vicinity of point P may in general depend on r and ¢, we are only concerned with its z dependence as determined above. Thus we have
[Ez]P = 
dct> dz
+ z2)1f2]
for z > 0
+ z (a2 + z2)1f2]
for z < 0
Ps [1  z (a2
=
2Eo
Ps [1
2Eo
Note that Ez (z) = Ez (z), as is expected on physical grounds.
Example 4.11: Finitelength line charge. In Example 4.5, we found the electric field at a point equidistant from the end points of a thin cylindrical rod charge, which had a component only in the r direction due to the symmetry involved. For points off the central axis, such as point P(r, ¢, z) as shown in Figure 4.27, it is clear that the field would have both Er and Ez components. This example illustrates the usefulness of the electrostatic potential concept by determining ct> at P via direct scalar integration and subsequently determining E from E =  V ct>. Assume that a >> d, so that the rod of charge can be represented by a line charge density Pl. As shown in Figure 4.27, we consider an elemental charge dQ = Pl dz' located at (0, 0, z'). The distance between this source element and point P is Ir  r'I = 2 + (z  z') 2 . The total potential ct> at Pis then given by
Solution:
Jr
ct>=
+a a
Pl dz' 4n Eo [r 2 +
Pl
= ln (z  z') 2] 112 4nEo
z  a+ [r 2 + (z  a) 2] 112 z + a + [r 2 + (z + a ) 2 ] 112
A threedimensional plot of ct> (r, z) is shown in Figure 4.27. For the purpose of this display, the normalized value of the electrostatic potential has been limited to 3 units, since for a < z < a, the potential increases without limit as r > 0.
256
The Static Electric Field
Chap.4
z a=l a
3
z
(r, z)
+a
2
pz dz'
z'
· .. d
r
z
r
r
_.., dE,
P(r, , z)
a
1
dEp
t
dE z
Figure 4.27 Electrostatic potential in the vicinity of a straight rod of charge. The amplitude range of the display is limited to 3 units since the potential becomes arbitrarily large as r + 0.
The electric field components can now be found from E =  V ct>; since ct> depends only on z and r, only E, and Ez components exist and are given by
act> Ez =  Bz act> E,=Br
1
Pl
4nEo
[r2
+ (z
1
_a )2]1/2
[r2
+ (z +a )2] 1;2
p1r
1
4nEo
r2 + (z  a)2 + (z  a)[r2 + (z  a)2]1/2
1 +~~~~~~~~~~~~~~~ r2 + (z + a)2 + (z + a)[r2 + (z + a)2]1/2
It is useful to check the behavior of the field and potential in limiting cases. For r z >> a, the potential approaches that of a point charge of Q = Pl (2a), namely, lim ct> = Pl (2a) 4nEor
r+oo
>> a
or
lim ct> = Pl (2a)
and
z+oo
4nEoZ
For z = 0, the solution boils down to that worked out in Example 4.5. For r = 0, and at points lz I > a, we have E, = 0 and _ Ez
Pl
1
1
4nEo
lz  al
lz +a l
±p1a
where the''+'' and'''' correspond to the cases of z > a and z < a, respectively. The result in Figure 4.27 serves to give a better understanding of the relationship between the electric field and the electrostatic potential. Since E =  V ct>, the direction of the
Sec. 4.5
Electric Flux and Gauss's Law
257
electric field can be easily deduced from the threedimensional surface plots of the potential. An analogy with gravitational potential is particularIy useful here; if the quantity plotted, say, in Figure 4.27, were the gravitational potential, its gradient at any point is in the direction opposite to that in which a marble placed at that point rolls down (note that the top of the surface in Figure 4.27 looks flat only because of the limited amplitude range of the display). The electric field at any point specified by its r and z coordinates thus points in the direction of the most rapid change of slope of the potential distribution. The magnitude of the electric field is higher at points where the slope is higher that is, closer to the disk charge.
4.5 ELECTRIC FLUX AND GAUSS'S LAW The electric field line plots that we have seen suggest some sort of ''flow," or flux, of electric energy emanating from positive charges and terminating at negative charges. If the field lines represented the velocity of fluid flow, we could indeed think of flux of fluid along the field lines. Although, in fact, electric field lines do not represent the flow of anything material, it is helpful to think of them as describing the flux of something that, like a fluid, is conserved and that emanates from the charges into the surrounding space. 54 In such a picture, each field line represents a unit of flux of that something, from each unit of positive charge to a unit of negative charge.
4.5.1 Electric Displacement and Flux Density D Michael Faraday's experiments, carried out in 1837, well after Coulomb, and many years after his own work on induced electromotive force, 55 have established our present concepts of electric displacement and flux density. A schematic description of Faraday's experiments is shown in Figure 4.28. Basically, Faraday used two concentric spheres. First, the inside sphere was charged by a known quantity of electricity (say, Q); then a larger outer sphere (uncharged) was placed around it56 with a dielectric (insulating material) in between. The outer sphere was then grounded57 momentarily; then the two spheres were separated and the charge remaining on the outer sphere was measured. Faraday found that this charge was equal in magnitude and opposite in sign to that on the inner sphere, regardless of the size of the sphere and for all types of dielectric materials filling the space between the spheres. He thus concluded that there was a sort of ''displacement'' from the charge on the inner sphere 54 0ne
possibility is to think of small ''bullets'' shot out of charges and to require that none of the bullets could disappear once they were produced; however, one needs to be very careful in carrying such a model any further than simply a thought exercise [R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics, Addison Wesley, Reading, Massachusetts, 1964]. 55 See Chapter 7 for discussion of Faraday's law. 56 Actually the outer sphere consisted of two hemispheres that could be firmly clamped together. Faraday prepared shells of dielectric material to occupy the entire volume between the spheres. 57 In other words, connected to the ground, so that it can access as much charge as it required. In this context, ''ground'' is an uncharged object at zero potential that has essentially an infinite number of electrons (and positive nuclei) and that can accept or provide any amount of excess charge from or to any object connected to it.
258
The Static Electric Field
Chap.4
Insulator
+
+
Charged sphere (+Q)
+
+
Figure 4.28 Faraday's experiments on electric displacement. The apparatus used in Faraday's experiments consisted of two concentric spheres separated by an insulating material.
through the insulator to the outer sphere, the amount of this displacement depending only on the magnitude of the charge Q. This displacement is the closest we can come to identifying the something that must flow along the field lines, and we adopt this notion as our definition of the electric flux. In SI units, the electric displacement (or electric flux) is equal in magnitude to the charge that produces it; namely, it has units of coulombs (C). Consider now the case of an isolated point charge as shown in Figure 4.29a; all other bodies, such as the outer sphere of Figure 4.28, can be considered to be at infinity. The density of electric displacement, or flux density vector D, at any point on a spherical surface S of radius r centered at the isolated point charge Q is defined as D
=r4nr Q
2
Cm 2
(4.30)
Note that the electric displacement per unit area, D, depends on the orientation of the area; hence, it is a vector quantity. In simple materials, we can write (4.31) with the dielectric constant E being a simple constant58 so that the vector D is in the same direction as the electric field E. Note, however, that Dis in units of Cm 2 , while E is in Vm 1. In view of the simple proportionality of D and E, our previous plots showing electric field lines also represent electric flux lines. In the following, we confine 58
We discuss dielectric materials and their behavior under applied electric fields in Section 4.10. At this point, it suffices to note that, in general, the relationship between D and E can be quite complex. In addition to the possibility of the value of E not being a constant and depending instead on the magnitude of the electric field (IEI), x, y, z, and t, in socalled anisotropic materials E might also depend on the direction of the applied electric field; for example, an electric field in the x direction (i.e, Ex) may produce an electric flux density in the y direction (i.e., Dy).
Sec. 4.5
Electric Flux and Gauss's Law
 ........ I
/
259
''
'\ s
I
I
\ \
I I
~~.~{+Q l~4'~
\ \
I I \
I \
 
,. .... ....
\
/
I I I I \ \ \
''
/ I
' ' .... .... _
....... .... ,
\
I '
I
 ' \ sb
I
\
I
I/
/
....
(a)
  .... 
....
(b)
sd ,.
....
'
I
: Sa1
Q
II
\
\ I I
''
''
''
\
\ \ \ I I I I
Sc I
I
'
I I I I \
''
\
\
\
\ I I
\
\ I I I
Sc
I
I
I
'
(c)
........
,. ....
/
I
I
I I
/
 
........
,. ....
/
(d)
Figure 4.29 Electric field lines around point charges. (a) A positive point charge, (b) two opposite equal point charges, (c) two point charges +2Q and Q, and (d) two point charges +4Q and Q.
ourselves to the consideration of simple materials with constant E unless otherwise stated. In fact, until dielectric materials are discussed in a later section, we exclusively consider free space, in which D and E differ through the constant Eo, (i.e., D = EoE). Thinking of each unit of flux as a unit of charge enables us to have a deeper understanding of the distribution of electric field lines around charge configurations. Consider, for example, the flux lines around two opposite charges shown again in Figure 4.29b, and the closed surfaces Sa, Sb, Sc, and Sd. If a closed surface encloses a charge of +Q (as is the case for Sa), all flux lines emanating from charge Q must go through it. Thus, by counting all the flux lines passing through this closed surface we should be able to know the amount of charges enclosed, independent of the properties of the material surrounding the charges. For this purpose, we would adopt the convention that lines outward (inward) represent positive (negative) enclosed charge. Since equal numbers of lines enter and leave from surface Sb, the charge enclosed by it is zero, whereas surface Sc encloses negative charge, equal in magnitude to that enclosed by surface Sa, since all lines of flux coming out of surface Sa eventually enter surface Sc. Similarly, the large surface Sd encloses both charges +Q and Q, so equal numbers of field lines enter and
260
The Static Electric Field
Chap.4
leave, and the net total flux through it is zero, just as the net amount of charge enclosed in it is zero. Consider the other two cases shown in Figures 4.29c and d. In the case of Figure 4.29c, the positive charge is twice as large as the negative charge. Accordingly, twice as many (i.e., 24, versus 12 for the flux line density shown in Figure 4.29c) lines cross through surface Sa as do through surface Sc. Also, the large surface Sd now encloses a net charge of 2Q  Q = +Q, so that it is crossed by the flux lines emanating from the +2Q charge that do not connect back to the Q charge (i.e., 12 out of the total 24 lines 59 in Figure 4.29c), but instead terminate on charges located at infinity. We thus see that, indeed, the number of flux lines is an accurate measure of the amount of charge. Similar observations can be made in connection with Figure 4.29d, in which case the positive charge is four times as large as the negative charge; accordingly, four times as many flux lines (i.e., 24 in Figure 4.29d) cross through surface Sa as do through surface Sc (i.e., 6 in Figure 4.29d). The net number of outward flux lines (i.e., 18, Figure 4.29d) through Sd is determined by the net enclosed charge ( +4Q  Q = +3Q).
4.5.2 Gauss's Law Consider now the region of space around a single positive point charge and an arbitrary closed surface S as shown in Figure 4.30a. If the electric field (E) is indeed like a flow, the net flow out of this truncated conical box should be zero. Let us find the surface integral of the component of the electric flux density vector D nor1nal (perpendicular) to the surface. Using (4.30), we have D · d s = 0 on the side faces, whereas on the radial faces we have D · ds = ±IDlds, plus on the outer surface Sb and minus on the inner surface Sa. Since the magnitude of the electric flux density D due to the positive point 2 2 , charge decreases as r and the surface areas of the radial faces increase as r , the total flow (or flux) inward through surface Sa completely cancels the flow outward though surface Sb. Since the total flow through the entire closed surface S is the integral of D · ds over S, we have :fs D · d s = 0. Note that we would not have arrived at this conclusion if the electrostatic force law depended on distance in any other way than r 2 . Considering Figure 4.30b, we conclude that the result we have just obtained is true even if the two end surfaces Sa and Sb were tilted with respect to the radial distance. Note 1 that when the surface is tilted at an angle (), the area is increased by a factor (cos() ) , but the component of D nor1nal to the surface is decreased by the factor cos (), so the product D · ds is unchanged. Thus, the net total electric flux out of the closed surface is still zero. It is thus clear that the net flux out of any closed surface (that does not enclose any charge) must be zero, because, as shown in Figure 4.30c, any such surface can be constructed out of infinitesimal truncated cones such as those in Figures 4.30a and b.
59 We
note that the particular number of lines (e.g., 12) is determined entirely by the resolution of the field line plot. In the case of Figure 4.29, we have chosen a particular density of field lines that conveys the essential aspects of the field distributions. Too high a density of lines (e.g., 100 lines emanating from +Q instead of 12) would clutter the figure, while too few lines (e.g., 2 instead of 12) would not be sufficient to convey the underlying symmetries and to illustrate the field intensity quantitatively in the regions surrounding the charges.
Electric Flux and Gauss's Law
Sec. 4.5
261
I Eb
/ /
/
/
/
/
s
/
/
/
/
/ /
/
s
/
/ / /
/ /
/
,
,'o ,
E
Sb
(b)
(a)
s E
Q ; ~

/
(c)
(d)
Figure 4.30 The flux of E out of a surface is zero. This figure is adapted with permission from R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics, AddisonWesley, 1964. Copyright 1964 by California Institute of Technology.
Now consider the surface shown in Figure 4.30d. We see that the magnitudes of the electric flux through the two surfaces Sa and Sb are once again equal (because the flux density decreases as r 2 but the area increases as r 2 ); however, they now have the same sign, so the net flux out of this surface is not zero. To determine the net outward (or inward) flux from a surface enclosing a point charge, consider the general surface S shown in Figure 4.31, shown for simplicity as a twodimensional crosssection. Emanating from the point charge Q are cones of flux, which might cross the surface one or more times as shown. Also shown is another small spherical surface S' around the point charge Q. Note that the volume enclosed between S and S' has no enclosed charge, so that the total flux emanating from this volume is 60 zero, by the arguments given above. Thus, the net total flux through the larger surface S must be identical to that through S'. In other words, the flux emanating from the end (near F) of the truncated cone between points E and Fis precisely that which leaves the surface S' in the same solid angular range. Note that the above argument holds for a 60 Basically,
the net flux emanating from any of the truncated cones, such as those between points A and B, B and C, or C and D, is zero, since these surfaces are just like those in Figures 4.30a and b and do not enclose any charge.
262
The Static Electric Field
Chap.4
ds
s B E I I I I I I
I
.... .... .... ........ .... .... ........   ........ .... G
H
~=
............
ds
Figure 4.31 Gauss's law. The total flux emanating from the closed surface S' surrounding the charge Q is equal to that emanating from any other closed surface of arbitrary shape (i.e., S), as long as it encloses no other charges. This figure was inspired by Figures 8 and 9 of J. H. Jeans The Mathematical Theory of Electricity and Magnetism, 4th ed., Cambridge Press, 1920.
surface S' of any shape; we choose S' to be a sphere centered at the point charge Q for convenience so that we can easily calculate the flux through it. Taking the radius of the sphere S' to be r, the electric flux density everywhere on its surface is given by (4.30). Since S' is a spherical surface with surface area 4nr 2 , we have Q D · ds = 2
4nr
As expected, the net flux out of the spherical surface S' is independent of the radius of the sphere. Since all of the flux out of S' must go out of the closed surface S, and since we imposed no restrictions whatsoever on the shape of S, we have arrived at a rather general result: D · ds = Q any closed surfac.e enclosing Q
To illustrate the point further, a second point charge Q' that lies outside the closed surface S is shown in Figure 4.31. It is clear from the foregoing discussion in connection with Figure 4.30 that any solid angular range (i.e., conical tube of flux) emanating from Q' cuts the closed surface S an even number of times (twice, four times, etc.), so the net flux from any of the pairs of truncated conical surfaces (such as G and H) is zero. Although the result we have obtained is for a single point charge, it is obvious that 61 it can be generalized to any free charge distribution since any continuous or discrete 61 In
Faraday's experiments on electric displacement, he measured the total amount of mobile (or free) charges on the inner and outer conducting spheres. The flux density D is a measure of this free charge. We will see in Section 4.10 that in the presence of an electric field, the dielectric material that fills the space between the inner and outer spheres contains an induced bound polarization charge density.
Sec. 4.5
Electric Flux and Gauss's Law
263
distribution of charges can be subdivided into a smaller number of elements, each of which can each be treated as a point charge. Using the superposition principle, the total flux out of any closed surface S is the sum of that from each individual enclosed charge element. We are thus ready to write down the most general f or1n of Gauss's law: fl
(J
.. s
fl
D · ds =
. v
p dv =
(4.32)
Qenc
where V is an arbitrary volume enclosed by the closed surface S, p (x, y, z) is the volume density of free electric charge inside the volume V, and Qenc is the total charge in volume V enclosed by the surface S. This general result is the most convenient method at our disposal for expressing the electrostatic inverse square action law of Coulomb. Gauss's law is a direct consequence of Coulomb's law: that electrostatic force between point charges varies with distance as r 2 . It does not provide us with any additional infor1nation about the way static charges interact; it simply allows us to state Coulomb's law in a way that may be more useful in the solution of some electrostatic problems, especially those that exhibit some sort of symmetry. Gauss's contribution was actually not in originating the law but in providing the mathematical framework for this statement.
4.5.3 Applications of Gauss's Law We now consider examples of the use of Gauss's law, as expressed in (4.32), to deter1nine the electric flux density D (and hence E = D/E) in cases where the charge distribution is known. Easy solutions of this integral equation (where the unknown D is under the integral) are possible only if we can take advantage of the inherent symmetries of the problem to identify a socalled Gaussian surface S such that D is everywhere either tangential or normal to the closed surface, so that D · d s is either zero or simply Dn ds, with Dn = const. on the surface. Gauss's law is thus useful only for symmetric charge distributions, which lead to symmetric distributions of E and D. The following examples illustrate the use of Gauss's law for a few different charge distributions. Example 4.12: Line charge. Consider an infinitely long line charge in free space with a line charge density Pl coulombs per meter, as shown in Figure 4.32. Find the electric field at a radius r from the line charge. Solution: To find the field at a radius r, we can use Gauss's law (equation (4.32)) and integrate the flux density D over a coaxial cylindrical surface S of length l and radius r, as shown in Figure 4.32. Due to symmetry, the electric field has a component only in the r direction, so there is no contribution to the integral over the closed surface from the end surfaces of the cylinder (Ez = 0). Thus, we can write 2n
D·ds =
EoE · ds = Eo
Erlr d¢ = 0
Qenc =
P1l
>
Er=
Pl
2nEor
264
The Static Electric Field
Chap.4
Pz
Figure 4.32 Line charge. A cylindrical Gaussian surface S is appropriate since by symmetry E, is constant on the side surface of the cylinder.
E,
When there is appropriate symmetry, the use of Gauss's law is clearly much simpler than direct use of Coulomb's law to find the electric field, as in Example 4.5.
Example 4.13: Spherical cloud of charge. Consider a spherical cloud of charge with radius a, as shown in Figure 4.33a. The electric charge is uniformly distributed over the spherical volume, with volume charge density p, so that the total charge in the cloud is Q = ~na 3 p. (a) Find the electric field E(r), for both r a. (b) Find the electric potential ct:>(r). Solution: (a) We first observe that, due to the spherical symmetry, the electric flux density D and hence E are both in the radial direction (i.e., E = rE,) and depend only on r. In
E, Q(47rcoa2)1

r 2 S2
"'
,
.,,.,,,.
  
/
S1
I
.
I I I \ \
......
'
....
....
''
a \
\
\ ' r I
r (b)
\ I I I
I
I
\
I
\
' .... .... , _____ , ,,. "' (a)
/
3Q(87rc0 a) 1 Q(47rc a) 1
(a2 ,2)
r 1
0
a
r (c)
Figure 4.33 Electric field and potential due to a spherical cloud of uniform charge. (a) The spherical cloud of charge and Gaussian surfaces S 1 and S2 . (b) Electric field. (c) Potential.
Sec. 4.5
265
Electric Flux and Gauss's Law
other words, D = constant on any spherical surface that is concentric with the spherical charge cloud. We consider two spherical Gaussian surfaces S 1 and S2, respectively, inside and outside the cloud of charge, and apply Gauss's law (equation (4.32)) to both surfaces:
ra Thus, we have
E=
Q r" 4nEor 2 Qr r" 4nEoa 3
r>a r= 
00
(rEr) · (rdr) =
a
00 00
r>a
4nEor 2
r Qr dr
Qdr 4nEor 2
a
4nEoa 3
r =
4nEor Q 4nEoa
r>a
+
Q 4nEoa 3
a2 
2
r2
r is plotted in Figure 4.33c. The relationship between E and ct>, namely, E =  V ct> (in this case, Er = d ct>/ dr ), is apparent.
Example 4.14: Infinite sheet of charge. Consider the planar sheet of charge of infinite extent62 located in the xy plane (i.e., z = 0), as shown in Figure 4.34. The surface charge density is given as Ps. Determine the electric field E and the electrostatic potential ct>. Solution: We observe from the symmetry of the problem that the electric field can only be in the z direction: E = zEz for z > 0 and E = ZEz for z < 0. We choose to work with cylindrical (r, b. It is clear that the net charge enclosed by the surfaces S 1 and S3 is zero, and since the electric field must be constant over the side surfaces (due to symmetry), from Gauss's law we have Er = 0. We now consider the surface S2 and apply Gauss's law, noting that the surface elements on the top, bottom, 63 This
type of a charge distribution is in fact what occurs for a coaxial line, consisting of concentric coaxial conductors. The surface charge densities as assumed in this problem are established on the surfaces of the conducting cylinders when a potential difference is applied between them. The field and potential as determined in this example will be directly useful in the treatment of coaxial capacitances and coaxial transmission lines.
268
The Static Electric Field
Chap.4
and side surfaces are ds = zrdrd 0, and also noting that the line integral of the electrostatic field around any closed path is identically zero (4.20), we have Et dl = Et ~ w = 0
E · dl =
(4.41)
da
Thus, the tangential component of the electric field at the surface of a conductor is identically zero. Electric field lines ter1ninate perpendicularly on conductor surfaces. In Chapter 7, when we examine Faraday's law, we shall see that, for timevarying fields, the line integral of the electric field around a closed loop is not zero (i.e., :f E · dl =f. 0) 72 0nly
a spatial averaging of the fields is necessary, since in any macroscopically significant region there are so many nuclei and electrons that temporal fluctuations are completely washed out by averaging over space. To appreciate this, note that any macroscopic amount of ordinary matter (with a lowest onedimensional extent of say '""10 6 cm or volume of '""10 18 cm 3 ) has of order rvl06 nuclei and electrons. For further discussion, see Section 6.7 of J. D. Jackson, Classical Electrodynamics, Wiley, New York, 1975. 73 In metals, the lowest energy bands are only partially filled, with many vacant levels. For a brief elementary discussion, see Section 9.6 of R. L. Sproull, Modem Physics, The Quantum Physics ofAtoms, Solids, and Nuclei, Wiley, New York, 1963. For more detail, see V. F. Weisskopf, Am. J. Phys., 11:1943. 74 This proportionality of vd and Eis the basis for Ohm's law, which we will discuss in Chapter 5.
280
The Static Electric Field
Chap.4
/\
n
a d
,
b
c
__
/ /
Conductor (E = 0)
Conductor (E = 0)
(b)
(a) External source charges
/).s I /\
I
 E 2n
n
+
+
+ ~ I I I
~ I I I
I I I
I I I
+ ~
+ ~ I I I I I I
I I I I I I
~ I I I I I I
~ I I I I I I
n
I
+
+
/\
I
fl~
+++++++ F" =t \
Conductor (E
+ '
/
/
=
0)
Conductor (E = 0)
(c)
(d)
Figure 4.40 Field at the surface of a conductor. (a) Contour abcda used to show that E1 = 0. (b) Application of Gauss's law to relate normal electric field to surface charge. (c) Electric fields at the conductor surface; E1n is the field due to the surface charge Ps, whereas E2n is the field due to the external ''source'' charges of opposite polarity. (d) The fields at the surface of the conductor arise from two systems of charges.
but instead is given by the negative time rate of change of the magnetic flux passing through the surface enclosed by the closed loop. Nevertheless, the boundary condition on the conductor surface (i.e., the fact that Et = 0) will be shown to be still valid in general. Using Gauss's law it is possible to determine a relationship between the surface charge density on a conductor at equilibrium and the electrostatic field at the surface. Consider a very small portion of any charged conducting surface, as shown in Figure 4.40b. An infinitesimal pillboxshaped Gaussian surface is visualized as shown, with half of it above and the rest below the surface. In applying Gauss's law, we note that no flux crosses the lower surface, since the field inside the conductor is zero. No flux leaves through the sides either, since this would require that E have a component tangential to the surface. Furthermore, we lose no generality by considering the limit ~h > 0, so that the total surface area of the closed surface is primarily due to the top and bottom faces.
Sec. 4.7
281
Metallic Conductors 75
On the top surface, a nor1nal component of E exists. The net outward flux from this top surface is EoE · ~s and must equal the charge within it. Denoting the surface charge density on the conductor surface as Ps, and the electric field nor1nal to the surface as En, we have
En
~s
=
Ps
~s
)
En = :fi . E = Ps
EQ
EQ
where :fi is the unit vector outwardly perpendicular to the conductor surface, as shown. Note that since D = EoE, this condition can also be written as I fi · D = Dn = EoEn = Ps I
(4.42)
This more general relationship is also valid if the conductor is surrounded by an insulating material other than vacuum, except that in (4.42), Eo would have to be replaced with a different constant E to account for the electrical behavior of the particular material (see Section 4.10). If we think of the conductor surface as an infinite plane surface with charge density Ps, the result just obtained (i.e., En = Ps / Eo) seems to contradict the electric field that we found for an infinite sheet of charge in Example 4.14, which was En = Ez = Ps / (2Eo). The reason for this difference is that the field at the surface of a charged conductor actually arises from two systems of charges: the local surface charge and the ''source'' charges, which are remote from the conductor and that are physically required by the ''charged'' state of the conductor. 76 In the simplest geometry, we could think of these distant charges as lying unifor1nly on a distant parallel planar surface, as shown in Figure 4.40c. By examination of Figure 4.40c, we can see that the field solved in Example 4.14 represented only the partial field caused by the surface charge (i.e., Ein) and that the total field can be found only if the external source charges of opposite polarity are included. In other words, the conducting surface can have net induced positive charge only if these negative external source charges are present, typically because a potential difference (e.g., with a battery) is applied between the conducting surface and some other conductor (on which the external negative charges reside). The field E1n due to the external charges cancels out E 1n inside the conductor and doubles it outside the conductor, so the net field nor1nal to the surface of the conductor is En = Ein + E1n = Ps / Eo. We can also understand the difference between the field near a sheet of charge and that at the surface of the conductor by considering Gauss's law with a cylindrical pillbox surface, as shown in Figure 4.40d. Since the field inside the conductor is zero, and since we take ~h > 0, the only nonzero contribution to the surface integral is from the top surface. Thus, we have EoEn ~s = Ps ~s, which gives En = Ps / Eo. However, in the case of the sheet of charge (Example 4.14), the field is nonzero on both the top and 75 Presumably
this field is the reason the conductor is charged; that is, the electric field emanates from (or converges on) the positive (negative) surface charges and connects to (emanates from) other charges of negative (positive) sign, which constitute the source of the ''charging'' of the conductor. 76 Considering the common situation in which the conductor is charged by applying a potential difference between it and another conductor, these distant charges are the charges of opposite sign induced on the other conductor.
282
The Static Electric Field
Chap.4
bottom sides of a similar pillbox surface, so we have 2EoEn !::,.s = Ps !::,.s, which gives En = Ps / (2Eo). Both of these points of view (i.e., as discussed in relation to Figures 4.40c and d) are equally valid.
Example 4.20: Surface charge on a conductor. It is interesting to consider how small an excess or deficiency of electrons can be expected to occur on the surface of a metallic conductor in practice. Normal air breakdown occurs (see Section 4.10.3) when the magnitude of the electric field in air reaches approximately Emax = 3 x 106 Vm 1. (a) Find the charge density Ps on the surface of a metallic conductor that would produce such a field. (b) How many excess electrons would such a surface density represent in a square pillbox volume of surface area 1 µ,m x 1 µ,m and thickness one angstrom? Note that a surface density at which the field immediately outside the conductor would be at the breakdown level for air represents essentially the maximum surface charge that can be placed on a conductor in air before one sees corona breakdown. 77
Solution: (a) The maximum surface charge density can be found as
Ps =
EoEmax =
8.854 x 1012 Fm 1 x 3 x 106 Vm 1 ~ 2.66 x 105 Cm2
(b) Now consider the surface region of area 106 x 106 m 2 and a thickness of 10 10 m. Such a region has a volume 106 x 106 x 10 10 = 1022 m 3 , with on average about 8 r v 10 free electrons and nuclei, assuming one free electron per atom and atomic dimensions of r v 1 angstrom3 = 1o 3o m 3 . The number of excess electrons represented by the charge density Ps determined above, in the same volume just described, can be found as
PsA lqe I
~
2
2 ·66 x los Cm (lo 6 x 106 m 2 ) 1.6 x l019 Ce2
~ 1.66 x
102
b
(b) The electrostatic potential ct> with respect to a point at infinity can be found by taking the line integral of E (equation (4.17) ). For points outside the spherical shell, we have r
Er dr = 
=
Q
r
00
Q
dr = 4nEor 2 4nEor
00
r>b
Since the conductor is an equipotential, the electrostatic potential ct> remains constant in the region a < r < b, namely b
cl> = 
r
Er dr 
Q
Er dr
4nEob
b
00
a
=0
Sec. 4.9
297
Capacitance
z
(r) = 
C1
r
+ C2
where C 1 and C2 are integration constants to be determined by the boundary conditions. One boundary condition is that (r =a) = Vo, which gives C1
 +C2 =Vo a
Voa ( (r) = r +C2
>
a)
1;:
To determine C2, we need a second boundary condition. For a sphere of finite size, we would expect the potential (r) to vanish as r > oo, which in turn implies that C2 = 0. Our solution is then
V0 a
(r) = 
r
>
E=r
B
ar
Voa
=r
r2
This solution can be compared with what can be found using Gauss's law, if we were to assume that raising the sphere to a potential Vo deposits a charge Q on it. In such a case, the electrostatic potential for r > a would be (r) = _Q_ 4n Eor
For the two solutions to be identical, we must have
Q = Vo(4nEoa) We shall see in the next section that the ratio of Q/Vo represents the capacity of the sphere to hold charge per unit applied voltage.
4.9 CAPACITANCE The relation between induced charges on a set of conductors and the resulting potentials in their vicinity depends only on the geometric arrangement of the conductors. If we
298
The Static Electric Field
Chap.4
double the surface charge densities at every point on the conductors (presumably by doubling the applied potential difference between the conductors), the configuration (i.e., spatial variation) of the electric field remains unchanged, but the magnitude of the field is doubled, and hence the work required to take a unit positive test charge from one conductor to another (i.e., potential difference between them) is also doubled. Consider a single isolated conductor of any shape. If we place a charge Q on it, the charge is distributed in some equilibrium pattern on the surface of the conductor. If we then increase the surface charge density Ps everywhere by the same factor, we still have an equilibrium arrangement, since the electrostatic potential is everywhere greater by the same factor. This constant chargetopotential ratio of an isolated conductor is its capacitance. In other words, we define Q
C=
(4.49)
where is the potential of the isolated conductor. The units of capacitance are coulombs per volt (Cv 1), or farads (F). In particular, for a spherical conductor of radius a, far away from any other charges, conductors, or ground planes, we have (r = a) = = Q/(4:rrEoa) (see Example 4.25), so that we have (4.50)
C = 4:rrEoa For example, a sphere of radius a = 10 cm would have a capacitance of C = 4:rr (8.854 x 10 12) (10 x 102 ) ~ 11.1 pF L=~v'~~~...
faradsm 1
m
In Example 4.20 it was found that a surface charge density of Ps ~ 2.66 x 105 Cm2 on a metallic conductor would cause electrical breakdown of the surrounding air. Such a charge density unifor1nly distributed over the surface of a sphere of radius a = 10 cm corresponds to a total charge of Q = Ps4:rra 2 ~ 3.34 x 106 C. Using the capacitance value of C ~ 11.1 pF, we find that placement of that much charge on a 10cm radius sphere requires = Q/C ~ (3.34 x 106 )/(ll.l x 10 12) ~ 3 x 105 V, or ~300 kilovolts! In the case of an isolated conductor, the electric flux that leaves the conductor can be thought to terminate at infinity. If the amount of charge on the conductor is increased, the flux pattern remains the same, but the flux density D proportionally increases. If we have a pair of conductors and if these conductors are given equal and opposite charges (by applying a potential difference 12 = 2  1 between them94 ), then all of the flux 94 When
a battery is connected between the two conductors, each conductor becomes an extended battery terminal. Within rv 10 19 s following the connection, electrons flow from the conductor connected to the positive terminal through the battery and connecting wires and to the conductor connected to the negative terminal, and equilibrium is established. As mentioned before, the time involved here (rvl0 19 s) is the time in which charge is rearranged in a material, which is in the range of 10 18 to 10 19 s for most metallic conductors. The subject of redistribution of charge is discussed in Section 5.5.
Sec. 4.9
299
Capacitance
emanating from the positively charged conductor terminates on the negatively charged one. If the potential difference is increased, the magnitude of charge on both conductors, as well as the flux density D, also increases, without a change in the distribution of flux. Once again, we see that the amount of charge and the potential difference are proportional to one another, and we define their ratio to be the capacitance of the twoconductor configuration: Q C=2  1
Q
(4.51)
12
In (4.51), Q is the amount of charge on one of the conductors, usually taken to be the positive one. Note that the case of the single isolated conductor can be considered as a special case with the second conductor being at ''infinity," so that 2 = 0. In summary, capacitance is the measure of the ability of a conductor configuration to hold charge per unit applied voltage between the conductors (i.e., their capacity to store charge). In the following examples, we evaluate the capacitance of some simple twoconductor configurations, including the parallelplate and coaxial transmission line configurations that were considered in Section 2.7, the capacitance expressions for which were given in Table 2.2. Although the capacitance expressions for these configurations are derived on the basis of electrostatic formulations, they are nevertheless quite valid up to optical frequencies. 95 Example 4.26: Parallelplate capacitor. Consider two parallel conducting plates separated by free space as shown in Figure 4.48. Determine the capacitance C.
_..Plate 2 A ~___,.
+ d
+
+
+
+
+
Plate 1
z
+ E
y
Q x
Figure 4.48 Parallelplate capacitance.
Solution: We start by assuming that the plates are charged so that the upper plate has a total charge of +Q while the lower plate has Q. The charge on both plates is uniformly distributed on the inner surfaces, supporting an electric field pattern as shown. Note that
95 The
basic reason for this is the extremely small (rv 10 19 s) time within which charges can redistribute in metallic conductors, as mentioned in Section 4.7 and the previous footnote and further discussed in Section 5.5.
300
The Static Electric Field
Chap.4
when the smallest linear dimension of the plates is sufficiently larger than the plate separation d, we can neglect ''fringing'' fields on the sides and assume the electric field between the plates to be identical to that in Example 4.23. If the area of the plates is A, the surface charge density is Ps = Q /A, positive on the upper plate and negative on the lower one. Based on the results of Section 4. 7 .2 and Example 4.23, the electric field between the plates is
The potential difference between the plates can be found from d
= 
cf>12
0
,. . Ps
d
Ps
Qd
z
z · (zdz) = Eo
EQ
EoA
0
The capacitance is then
c
=
Q
=
Q ~~~Qdj(EoA)
cf> 12
We note that the capacitance is a function only of the geometry of the configuration (i.e., A and d). If the region between the conductors is an insulating material other than air, then the constant Eo is replaced by another constant characteristic to the material (Section 4.10). Thus, capacitance is also a function of the material medium around the conductors.
Example 4.27: Coaxial capacitor. Consider the coaxial capacitor, consisting of two concentric cylinders separated by free space, as shown in Figure 4.35 in connection with Example 4.15. Find the capacitance per unit length. Solution: We can find the capacitance of this configuration by the usual method: (1) Assume a uniform surface charge density Ps on the inner conductor (i.e., a total charge of Q = Ps(2nah), where h is the height of the cylinder); (2) find E in terms of Ps; (3) integrate E to find the potential difference between the inner conductor (r = a) and the outer conductor ( r = b ), denoted ct> ba, in terms of Ps ; and (4) take the ratio of Q and ct>ba. However, in this case we already have the solution of the problem worked out in Example 4.15, where two different expressions were found for ct>(r): Psa
ct>(r) = 
Eo
r ln 
and
ct> (r) = ct>
b
ln(r / b) ba ln(a/b)
Equating the two expressions for ct>(r) we find

Psa
Eo
r ln  = b
ln(r/b) ct>ba _ __ ln(a/b)
>
Q
cf>ba
2nEoh
ln(b/a)
Hence we have
c
=
Q cf>12
2nEoh ln(b /a)
>
Cu = C =
h
2rrEo ln(b/a)
rv
55.6 pFml ln(b/a)
which agrees with the expression listed in Table 2.2 as the perunitlength capacitance of a coaxial line.
Sec. 4.9
301
Capacitance
As a numerical example, consider a practical coaxial line such as RG11, which has an innerconductor diameter of '""'1.21 mm and the inner diameter of its outer conductor of '""'7 .24 mm. Assuming the conductors to be separated by free space, the capacitance per meter of this line is Cu~ 55.6/[ln(7.24/1.21)] ~ 31 pFm 1 .
Example 4.28: Capacitance of concentric spheres. Consider a spherical capacitor consisting96 of two concentric conducting spheres as shown in Figure 4.49a. Find the capacitance.
Connecting • wrre
I I I
b
I I I I I ..... .......
,I
a
....... ....... .......
a
.......
.....................
I
.......
I
b .......
....... .......
I
Dielectric spacers
(b)
(a)
Figure 4.49 A spherical capacitor. (a) The geometry of a spherical capacitor consisting of two concentric conductor spheres of radii a and b. (b) In practice, the inner conductor would need to be supported by insulating spacers and can be accessed by means of a wire through a small hole as shown. The surface charges on the inner and outer spheres are not shown to avoid cluttering the figure.
Solution:
If we assume a charge +Q on the inner sphere and Q on the outer one, the electric field between the two spheres is
E = r"Er = r"
Q 4n Eor 2
aba
ab = 4nEob  a
Note that if we let b > oo, we get the capacitance of an isolated sphere discussed earlier. In such a case, the flux lines originating on the positively charged inner conductor terminate at ''infinity.''
In general, regardless of the conductor configuration, and with reference to Figure 4.50, we can define capacitance as Q :fs D · ds C=  12   fLE . dl
(4.52)
where S is any surface enclosing the positively charged conductor and L is any path from the negative (lower potential) to the positive (higher potential) conductor. The general procedure for determining the capacitance of any twoconductor configuration can be summarized as follows: (1) Choose a coordinate system appropriate to the geometrical layout and shapes of the two conductors. (2) Assume charges +Q and Q on the conductors. (3) Find E from Gauss's law, by direct integration or by other methods; the result will be proportional to Q. (4) Integrate E along any path between the two conductors to determine 12; result will be proportional to the assumed charge Q. (5) Find C by taking the ratio Q / 12; note that Q will cancel out and the result will depend only on the geometry of the conductors. Example 4.29: Capacitance of the twowire line. A rather practical transmission line configuration used in telephony, radio engineering, and electric power distribution is the twowire line, consisting of two parallel cylindrical conductors, each of radius a, separated by a distance d,
Sec. 4.9
Capacitance
303
.x I
+pz
#2 d  _I
~da
tE2 I I
d
Ei ~ a
o#1 '" (a)
pz
(c) I
I
d/2 E
(d)  = O E
I
I
I
I
E
(b)
(e) Figure 4.51
Twowire transmission line and a single wire above ground. (a) The
twowire line. (b) Electric field configuration. (c) Representation in terms of two line charges. (d) Single wire above ground. (e) Field configuration for single wire above ground.
as shown in Figure 4.51a. General solution of the capacitance for this configuration is more involved than that for the coaxial cable, since, in general, the charge distribution on the conductor surfaces is not uniform, because of the Coulomb attraction between the induced surface charges on the two conductors. This socalled ''proximity effect'' causes the charge density to be larger (smaller) on the sides of each conductor facing toward (away from) the other. The resultant distribution of the electric field lines and equipotentials is shown in Figure 4.51b. Closedform analytical expressions for the electric field and potential can be found using various methods. Find a simplified expression of the capacitance of the twowire line, valid for cases where d >>a. Note that this approximation is indeed quite valid in most cases, including the wellknown examples of the power transmission lines (for which a is a few cm, whiled is many meters) and the TV antenna wire (a < 1 mm and d r v 1 cm).
Solution: Ford >>a, the nonuniformity of the charge distribution on the conductor surfaces can be neglected and thus the electric field outside the conductors is the same as what would be produced by two line charges of opposite polarity (pz and  pz). From Example 4.11 we know that the electric field at a distance r from a line charge of linear charge density pz
304
The Static Electric Field
Chap.4
is given as Er = Pl/ (2n Ear). Taking the +Pl and  Pl line charges to be located at r = d and r = 0, respectively, the electric field as a function of distance x along the x axis (i.e., y = 0 and z = 0) is given as Ex(x,0,0) =
p1
2nEox
Pl
+ 2nEo(x  d)
To find the capacitance using its general definition (4.52), we need to integrate the electric field along any path from one conductor to the other. It is obvious from Figure 4.51c that one convenient integration path97 is along the x axis, from x = a to x = d  a. We find da
1
12 =   2nEo =
+
+P1
x +xd 
a
Pl
p1
dx =
2nEo
a
da ln d  (d  a) 2nEo
Pl
2nEo
[ lnx  ln(d  x) ]da =
Pl
2nEo
a
x dx
[ ln
a ln da
Pl
::2:'.
da
1 1 + x dx
dx
]~a
Pl 2ln
2nEo
da a
Pl In d
nEo
a
since d >> a. Noting that the total charge on one of the conductors of unit length is Q = Pl, the per unit length capacitance Cu of the twowire line is
_
Cu 
Pl
rv
'lrEQ
rv
ln(d/a)
l12I
27.8
_ __
ln(d/a)
pFm 1
As an example, consider the capacitance per kilometer between two Draketype steelreinforced aluminum conductors (ACSR) of a 115kV power transmission line, separated by d = 3 m. For Drake ACSR, we have a ::2:'. 1.407 cm, so that the approximation d >>a is valid. Using the above, we find Cu ::2:'. 5.19 nF(km) 1 . The above expression is valid for d >> a. A more accurate analysis for the general case yields Cu=
In[d/2a
+
JrEo J(d/2a) 2
27.8
rv

1]
ln[d/2a
2 + J~ (d/2a_)_ 1]
pFm 1
which is the expression listed in Table 2.2 for the perunitlength capacitance of a twowire line.
It is clear from (4.52) that the capacitance of a twoconductor configuration is primarily dependent on the electric field distribution. This realization allows us to simply infer the capacitance of some conductor configurations from others for which we have derived expressions. A good example is the case of a single cylindrical conductor above a ground plane, as shown in Figure 4.51d. We can simply obtain this configuration by placing an infinitely large conducting sheet (i.e., ground) coincident with the flat equipotential 97 Note
that the electric field at points other than along the x axis in general has components in both the x and y directions, as can be determined by a vector superposition of E1 and E2. However, since 12 can be found by integrating the total electric field along any path between the two conductors, it is convenient to choose this path along the x axis, where the electric field has only one nonzero component (i.e., Ex).
Sec. 4.10
305
Dielectric Materials
plane of the twowire line configuration of Figure 4.51 b. Such a conducting plane does not influence the existing field distribution, since the electric field lines are already perpendicular to it at all points. For a given value of Pt, the electric field everywhere is the same for the singlewire aboveground case as that for the twowire line. However, if the charge density of Pt corresponded to a potential difference of Vo between the two wires, the potential difference between a single conductor and the ground plane (for the same Pt) is Vo/2. In other words, the capacitance per unit length of the configuration shown in Figure 4.5 ld is Cu,single wire above ground
= 2Cu,twowire line
2rrEo ~ In (d /a)
(4.53)
Another way to analyze the problem of the single wire above a ground plane is the socalled method of images, which involves the replacement of boundary surfaces by appropriately placed ''image'' charges instead of a for1nal solution of Laplace's equation. In the case in hand, the conducting plane can be replaced by an image line charge of line charge density  Pt at a distance d /2 below the plane so that = 0 on the ground plane. The electric field at any given point is then simply a superposition of the field due to the original line charge (single wire) and the image line charge, which of course is entirely equivalent to the case of the twowire line. The method of images is a powerful method for finding the fields produced by charges in the presence of dielectric and conducting boundaries with certain symmetries. 98
4.10 DIELECTRIC MATERIALS An insulator is a substance whose electrons and ions cannot move over macroscopic distances under the influence of an applied field. Although no material is a perfect insulator, in this section we deal with materials within which the motion of charges over macroscopic distances is negligible. Insulators are said to be polarized when the presence of an applied field displaces the electrons within a molecule away from their average positions. When we consider the polarizability of insulators, we refer to them as dielectrics. 99 Although all substances are polarizable, the effects of polarizability can be readily observed only when the material does not conduct electricity that is, when it is an insulator. 98 See,
for example, Section 4.4 of D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., AddisonWesley, Reading, Massachusetts, 1989. 99 The word dielectric was coined by Faraday, who extensively studied the behavior of different insulators placed between electrodes and found that the capacitances of spherical capacitors filled with dielectrics were higher. His results and insights were documented in Experimental Researches in Electricity, Vol. 1, Section 12521270, B. Quatrich, London, 1839. For a historical overview and for an excellent and thorough discussion of dielectric materials see R. S. Elliott, Electromagnetics, IEEE Press, Piscataway, New Jersey, 1993. Another excellent reference on the topic of dielectrics is A. von Hippel (Ed.), Dielectric Materials and Applications, Artech House, Boston, 1995.
306
The Static Electric Field
Chap.4
4.10.1 Polarizability In Section 4.7 we saw (e.g., Example 4.20) that redistribution of charges in a metallic conductor involves the transfer of a very small percentage of the amount of free charge in a conductor over macroscopic distances. Polarization of dielectrics, on the other hand, involves the displacement of one or more electrons per atom over subatomic or microscopic distances. The atoms of dielectric materials have their outer1nost electron shells almost completely filled. As a consequence, the electrons are tightly bound, and only a negligible number of electrons are available for conduction of electric current. A measure of the conducting ability of a material is the charge rearrangement time, which as we shall see in Chapter 5 is ~io 19 s for metallic conductors. For a good dielectric, such as fused quartz, the charge rearrangement time is ~50 days. Thus, free charge deposited inside quartz stays in place for all practical purposes and can be considered as bound charge. The difference in the electrical behavior of dielectrics versus conductors is essentially due to the different numbers of free versus bound charges, with semiconductors exhibiting intermediate behavior. When a dielectric material is placed in an electric field, the electrons respond by shifting with respect to the nuclei in the direction opposite to the applied field, essentially establishing many small electric dipoles, as shown in Figure 4.52. Note that electric dipoles aligned with the field are produced in both polar and nonpolar materials. Nonpolar materials consist of molecules that do not possess a per1nanent dipole moment; the external field both induces the dipoles and orients them as shown in Figure 4.52a. E
Se
/,
I I I I I I I I I I I ,
0 00 0 0 0 0 0
 \ + I I I

I I • 4

\..;
+
(a) I
+
'
I

,.
/
I I
I I  , +
+ ;
+
 1+ I I

\ I
+
'
+
  ,.,,.
 + 
+
+

+ +

+ 
+
+
+
E
(b)
+ 
+
+

+
Figure 4.52 Dipoles oriented in an applied field. (a) Nonpolar molecules with no permanent dipole moment; the external field induces the dipoles and orients them along the field. (b) Polar molecules have permanent dipole moments randomized by thermal agitation; the applied field aligns the dipoles.
Sec. 4.10
307
Dielectric Materials
The molecules of polar materials (e.g., NaCl) have a permanent dipole moment, but the individual molecular dipoles are usually randomly oriented due to ther1nal agitation, as shown in Figure 4.52b; when an applied field is present, the individual dipoles tend to align in the direction of the field. It is important to note that polarization does not produce net charge inside the dielectric. Any interior volume of macroscopic dimensions 100 (e.g., that enclosed by the surface Si in Figure 4.52a) contains equal amounts of positive and negative charge. However, a net amount of surface polarization charge does appear on the surface of the polarized dielectric, as shown in Figure 4.52a for the volume enclosed by the surface Se. We see that a layer of charge adjacent to the boundary remains unneutralized and appears as polarization charge with bound surface charge density Psp· We shall see that the amount of surface charge is a direct indication of the degree of polarization of a material and that Psp is proportional to the magnitude of the electric field.
Electronic polarizability. Figure 4.53a shows a very simplified model of the atom, consisting of a positively charged nucleus surrounded by a negatively charged spherical electron cloud. 101 With an applied steady field the nucleus and the electron cloud are displaced (as illustrated in Figure 4.53b) until their mutual attractive force is just equal to the force due to the applied field. We can thus calculate the amount of displacement r by equating the two forces. We can treat the electron cloud with radius a as a spherical cloud of charge much like that in Example 4.13, with uniform volume charge density p = 3qe/(4rra 3 ), as shown in Figure 4.53a. Note that a is the effective atomic radius. The electric field due to such a spherical cloud of total charge Q at a distance r from its center was derived in 3 Example 4.13 to be Er = Qr /(4rrE 0 a ) for r ::::; a, which for the case in hand (replacing Q by lqe I) gives Er=
lqelr 4rrEoa 3
r rsine 8
V · (EoE + P) = p
This suggests that we can use a displacement density or flux density vector defined as (4.64) which per1nits us to write the familiar form of the differential form of Gauss's law (equation (4.36)) as (4.65) V · D= p Note that in materials for which (4.57) holds we have
[ D = EoE + EoXeE = (1 109 Free
conductor.
+ Xe)EoE =
EE [
(4.66)
charge includes all charge that is not the result of polarization, such as freeflowing electrons in a
314
The Static Electric Field
Chap.4
where the parameter E = (1 + Xe)Eo is called the electrical permittivity or the dielectric constant of the material. Typically we write (4.67) where Er is a dimensionless quantity called the relative permittivity or the relative dielectric constant. Note that (4.66) holds only for a class of dielectrics under certain conditions. Since (4.66) originates in (4.57), it depends primarily on the linear relationship between the polarization P and the electric field E. An important example of nonlinear behavior occurs when the applied electric field is intense enough to pull electrons completely out of their bound locations, causing dielectric breakdown, discussed in Section 4.10. Equation (4.66) also requires that the material be isotropic; that is, Xe must be independent of the direction of E. In anisotropic materials (e.g., crystals), electric field in one direction can produce polarization (and thus D) in another direction. Accordingly, the relation between D and 110 E must be expressed as a matrix, more commonly called a tensor. Equation (4.66) is valid for both homogeneous (where E is the same everywhere in the material) and for inhomogeneous materials as long as we allow for E to vary throughout the material as a function of position, expressed as E (x, y, z). Some authors refer to the charge density in (4.65) as the ''free'' charge density and use the symbol Pf to distinguish this quantity from the bound polarization charge density Pp· However, the distinction between bound charge distributions and all other charge density terms is usually clear from context alone. Since the definition of D includes the polarization vector P, the physical effects of bound charges, as determined through  V · P, are already included in the specification of D and E. Comparing (4.65) with (4.63), we see the simplification introduced by the notion of electric flux density D. From Coulomb's law and the definition of the electric field, E is deter1nined by all charges, including both free charges and bound charges, the latter resulting from a polarization density P. Hence, we see that the charge densities (volume, surface, or line) in equations (4.9), (4.11), (4.13), and thus (4.29) refer to the total charge densities, which include bound charges. By extension, the general form of Gauss's law (equation (4.63)) and Poisson's equation (equation (4.44)) also depend on both the bound and free charges. With the introduction of D and the concept of electrical permittivity, we can simply express Gauss's law in ter1ns of free charges only, represented by p (4.65). Similarly, for linear media (where D =EE), by rewriting (4.43) as V · E = p/E, Poisson's equation can be rewritten as (4.68) where, again, p now excludes any bound polarization charges. 11°For
a brief discussion, see Section 3.3 of D. H. Staelin, A. W. Morgenthaler, and J. A. Kong, Electromagnetic Waves, PrenticeHall, New Jersey, 1994. For further discussion of electromagnetics of anisotropic media, see Section 13.8 of S. Ramo, J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics, 3rd ed., Wiley, New York, 1994.
Sec. 4.10
315
Dielectric Materials
While there are versions of Gauss's law and Poisson's equation that involve D and the free charge distribution only, explicit equations for E and in ter1ns of the total charge distribution, such as equations (4.9), (4.11), (4.13), and (4.29), that depend on the fact that V x E = 0 for static electric fields, do not have equivalents for D. In general, even for static fields, we may have V x D =j:. 0, since V x P =j:. 0 may be true. The introduction of the permittivity concept greatly simplifies our analysis and understanding of electrostatic phenomena in the presence of dielectric materials. The polarizability and local molecular electric field concepts were introduced in Section 4.10.1 so that we can relate microscopic physical behavior to macroscopic fields. For most future problems we shall describe materials by means of their per1nittivity E, since this constant is easily measured. If E is known, we can find the polarization from the relation
IP= EoXeE = D 
EoE
= (E 
(4.69)
Eo)E I
The total polarizability aT is thus E  Eo = EoXe =
3Eo(E  Eo)
NaT +)
1  N aT/(3Eo)
aT=    
N(E
+ 2Eo)
Permittivities of selected materials. The relative permittivities of some selected materials are listed in Table 4.1. The values given are at room temperature and for static or lowfrequency applied electric fields; the electrical permittivity of most materials depends on frequency and can in general be quite different from the lowfrequency values, especially in the vicinity of resonances (see Section 11.2.1). The value of Er for most materials listed lies in the range 125, with the exception of distilled water, titanium dioxide, and barium titanate. The high permittivity of distilled water arises from the partial orientation of per1nanent dipole moments of its polar molecules. However, other polar materials such as mica and quartz do not have unusually large values of Er, so the substantially higher per1nittivity of distilled water is largely a result of the specific microscopic molecular structure of this ubiquitous material. Titanium dioxide is also a polar material but is also an anisotropic crystalline substance whose permittivity depends on the direction of the applied electric field. The value of Er for Ti02 is 89 when the applied field is in the direction of one of its crystal axes and 173 when the applied field is perpendicular to this axis. Other materials such as quartz also exhibit anisotropic behavior but with a narrower range of per1nittivities (e.g., 4.75.1 for quartz). Titanates (combinations of Ti02 with other oxides) also belong to a class of materials known as ferroelectric materials, which exhibit spontaneous polarization just as ferromagnetic materials (e.g., iron) exhibit per1nanent magnetization (see Chapter 6). Some of these materials have relative per1nittivities at room temperature of 5006000; a 111 with Er ~ 1200. good example is barium titanate (BaO·Ti02 or BaTi03 ), 3 is also anisotropic, with Er ~ 160 along its principal ferroelectric axis and Er ~ 5000 along the perpendicular axis at room temperature. For an aggregation of randomly oriented BaTi03 crystals, the permittivity can be taken to be Er ~ 1200. 11 1BaTi0
316 TABLE 4.1
The Static Electric Field
Chap.4
RELATIVE PERMITTIVITY AND DIELECTRIC STRENGTH OF SELECTED MATERIALS
Material Air Alumina (Al203) Amber Bakelite Barium titanate (BaTi03) Freon Fused quartz (Si02) Gallium arsenide (GaAs) Germanium (Ge) Glass Glycerin Ice Mica (ruby) Nylon Oil Paper Paraffin wax Plexiglass Polyethylene Polystyrene Porcelain Rubber Rutile (Ti02) Silicon (Si) Silicon nitride (Si3N4) Sodium chloride (NaCl) Styrofoam Sulphur Tantalum pentoxide (Ta20s) Teflon (P'IF'E) Vaseline Water (distilled) Wood (balsa)
Relative Permittivity (€7 ) (at Room Temperature)
Dielectric Strength (MVm 1) (at Room Temp. and 1 atm)
1 "'8.8 2.7 "'4. 8 1200 1 3.9 13.1 16 "'49 50 3.2 5.4 "'3.64.5 2.3 1.54 2.1 3.4 2.26 2.56 "'59 "'2.43.0 100 11.9 7.2 5.9 1.03 4 rv25 2.1 2.16 81 1.4
rv3
25 7.5 rv8 '"'' 1000 rv40 '"'' 10 rv30
200 15 15 30
20 11 25 rv30 '"'' 1000
Such materials can be used to construct compact generalpurpose capacitors with very small dimensions but relatively high capacitances, in the range of 5 pF0.1 µF. The permittivity of the titanates depends sensitively on temperature and also on the applied electric field strength. The latter property can be very useful in nonlinear circuit applications. Other crystals that have similar structure to BaTi03 and show similar behavior include SrTi03, NaTa03, and LaFe03. Other examples of ferroelectric materials include lithium selenite LiH3(Se03)2 and related salts, ammonium cadmium sulfate (NH4)2Cd2(S04)3, and lead metaniobate Pb(Nb03)2.
Sec. 4.10
Dielectric Materials
317
Further discussion of electrical properties of different dielectric materials can be found elsewhere. 112
4.10.3 Dielectric Breakdown The linear relationship (4.66) between E and D holds only for relatively small values of the applied electric field E. If the applied field is sufficiently high, a dielectric material is suddenly transfor1ned from a good insulator into an extremely good conductor, causing substantial current to flow. Dielectric breakdown is the ter1n used for this sudden loss of the insulating property of a dielectric under the influence of an electric field. The electric field strength at which dielectric breakdown occurs is referred to as the dielectric strength and is denoted by EBR· The values of EBR for selected materials are given in Table 4.1. The dielectric strength of a given material may vary by several orders of magnitude depending on the conditions under which it is used. For gases, EBR is proportional to pressure. For solids, microstructural defects, impurities, the shape of the dielectric, the manner in which it was prepared, and its environment are all factors that may strongly affect EBR· Under nor1nal conditions, dielectric breakdown does not per1nanently affect gaseous or liquid dielectrics. In solids, on the other hand, the breakdown leads to the for1nation of highly conductive channels, leaving behind a characteristic damage to the texture of the material, typically in the for1n of a channel of molten material, jagged holes, or a treelike decomposition pattern of carbonized matter. We now comment briefly on the nature of dielectric breakdown in gaseous and solid dielectrics.
Breakdown in gases. Consider the parallelplate capacitor of Figure 4.48 to be filled with a gas and connected to a constant voltage applied across its plates, and arrangements made so that the current flowing through the dielectric region between the plates can be measured. As the applied voltage is slowly increased, the current initially rises to a small value of a small fraction of a microampere and levels off to a saturation level. As the voltage is increased further, the current remains nearly constant until a certain critical voltage, VBR = EBRd, where VBR is called the breakdown voltage and d is the separation of the plates, is reached. When a voltage greater than VBR is applied, the current through the dielectric quickly rises (typically in a time interval of the order of "'"'10 8 s) to the maximum value allowed by the power supply. This behavior was first analyzed by Townsend, 113 who introduced the concept of electron avalanche breakdown. Avalanche breakdown occurs when the applied field is large enough to accelerate a free electron (a few of which are always present in any gas due to cosmic ray ionization) to sufficiently large energies so that it can liberate a new electron by impact ionization of a neutral gas molecule, leaving behind a positively charged gas ion. Both the initial electron and the newly produced electron can then be subsequently accelerated by the field again and produce more electrons by impact ionization. 112See
A. von Hippel (Ed.), Dielectric Materials and Applications, Artech House, 1995. 113 J. S. Townsend, Electricity in Gases, Oxford University Press, Oxford, UK, 1914.
318
The Static Electric Field
Chap.4
This process thus develops into an avalanche of impact ionization, rapidly creating many free electrons and positive ions in the gas, and accounts for the highly conducting behavior of the gas. The acceleration of the initial electron to energies at which it can impactionize the neutrals must occur between two successive collisions of the electron with gas molecules. Since the density of gas molecules (i.e., the average distance between them) is larger at higher pressures, the dielectric strength for gases is proportional to pressure; 114 in other words, pressurized gases can, in general, withstand stronger electric fields. This property of gases, and the fact that a gas recovers its insulating properties if the applied field is removed after breakdown, accounts for the frequent use of pressurized gases as dielectrics, especially in highvoltage applications. In most cases, the electric field distribution in a gas is not unifor1n, as was implicitly assumed to be the case for the gasfilled parallelplate capacitor just described. For example, in a singlewireaboveground configuration as shown in Figure 4.51e, the electric field is strongest on the surface of the conductor facing the ground. Similarly, the electric field in a coaxial line is strongest on the inner conductor, decreasing with distance as r 1. In general, on any charged conductor the electric field is higher in the vicinity of sharp points or conductor surfaces with the lowest radius of curvature. These high fields can sometimes locally exceed the dielectric strength of air and produce a breakdown in the vicinity of the highfield region although, further away from the conductor, the field is not sufficiently strong to sustain the breakdown. This type of a local discharge is referred to as a corona discharge. Consider, for example, a singlewire power transmission line above ground as shown in Figure 4.51e, with a typical height above ground of tens of meters. As the air breaks down in the vicinity of the conductor, it is ionized and behaves like a conductor. At night, the ionized air is apparent as a reddish (or sometimes blueyellow) glow that crowns the conductor, hence the term ''corona.'' To maximize power transmission, power lines typically operate very close to their highest voltage ratings. On days of bad weather, increased humidity and other effects may lower the dielectric strength of air, leading to corona discharges. Corona also comes about because of accumulation of rust and dirt on the lines, which create local inhomogeneities, resulting in large local electric fields in their vicinity. Corona discharge on power transmission lines must be avoided because it substantially increases losses and emits electromagnetic waves that can interfere with certain types of nearby communication systems. The reader may have experienced the effects of corona discharge while driving under a transmission line with the radio on. It was mentioned above that corona discharge occurs when the field is above the breakdown level in the vicinity of a conductor but not away from it. If the applied voltage is increased even further, a continuously ionized path in the for1n of a bright luminous arc is for1ned, extending to the nearest point of opposite polarity. An intense current (typically hundreds to thousands of amperes) flows through the gas, and an arc 114
This important result was first demonstrated by F. Paschen in 1889 and is valid at moderate pressures. At higher pressure levels, the breakdown voltage rises with increased pressure since too many collisions take place and an excessive amount of energy is wasted in various excitation processes. For a discussion and references, see A. von Rippel (Ed.), Dielectric Materials and Applications, Artech House, 1995; A. H. Beck, Handbook of Vacuum Physics, Macmillan, Pergamon Press, New York, 1965.
Sec. 4.10
Dielectric Materials
319
Figure 4.55 Sprites. Spectacular luminous glows occurring in clear air at 5090km altitudes above large active thunderstorms are called sprites. This image was taken from Fort Collins, Colorado, using a lowlightlevel television camera pointed toward the east. The event occurred above a thunderstorm in northern Kansas.
discharge is said to occur. The giant arcs of lightning are the best known examples of an arc discharge. 115 Rather spectacular examples of dielectric breakdown phenomena in air are the recently discovered luminous glows that occur at high altitudes above thunderstor1ns, referred to as sprites. 116 An example of sprites observed in the midwestern United States is shown in Figure 4.55. Sprites are believed to occur as a result of dielectric breakdown of air under the influence of quasistatic electric fields that appear at high altitudes following intense cloudtoground lightning discharges.
Breakdown in solids. The processes by which electrical breakdown occurs in liquids and solids are quite complex and depend on the particular periodic lattice structure of the material, the number of electrons in its conduction band, as well as imperfections due to the presence of foreign atoms. The externally measured voltagecurrent characteristics are quite similar to those observed for gases. When a pure, homogeneous solid dielectric is placed between suitable electrodes across which an increasing 115 M.
A. Uman, The Lightning Discharge, Academic Press, Orlando, 1987. 116 D. D. Sentman, E. M. Wescott, D. L. Osborne, D. L. Hampton, and M. J. Heavner, Preliminary results from the Sprites94 aircraft campaign, 1: Red sprites, Geophysical Research Letters, 22(10), pp. 12051208, 1995. For two descriptive articles, see Heaven's new fires, Discover, pp. 100107, July 1997, and Lightning between earth and space, Scientific American, pp. 5659, August 1997.
320
The Static Electric Field
Chap.4
voltage is applied, a small (microampere) current flows and levels off to a saturation value until the applied voltage exceeds a certain critical voltage. At this point, the current through the material suddenly (within ""'10 8 s) increases to the maximum value allowed by the supply. The dielectric strength of solids may vary by several orders of magnitude depending on the purity of the material, the shape of the material, the manner in which it was manufactured, the ambient temperature, and the duration of the applied field. In general, it is not possible to determine EBR for solids on purely theoretical grounds; practical experimental tests under precisely defined conditions are generally required to specify the dielectric strength of the material reliably in the context of a particular application. Three basic mechanisms of breakdown can be identified. Socalled intrinsic breakdown is electronic in nature and depends on the presence of conductionband electrons capable of migration through the lattice. The underlying process is an avalanche process similar to what occurs in gases; however, the microphysics can be understood only on the basis of the band theory of electronic structure and by including electronlattice interactions topics well beyond the scope of this book. The electric field levels at which intrinsic breakdown occurs typically represent the theoretical upper limits for EBR for a solid dielectric, reached only by extremely pure and homogeneous materials. More typically, impurities and structural defects cause the material to break down at lower EBR values. The second type of breakdown is called thermal breakdown and arises from the fact that small leakage currents flowing through a dielectric lead to resistive losses, which are dissipated as heat in the material. If the local heat is generated faster than it can be dissipated, the temperature of the material rises and breakdown occurs, typically by melting or decomposition. The dielectric strength EBR due to this breakdown mechanism depends on the duration of the application of the field as well as the ambient temperature. The third type of breakdown mechanism is often referred to as discharge breakdown and depends on the presence of voids or cracks within the material. The presence of air in such cracks or voids produces a reduction in the dielectric strength. Partial discharges in the vicinity of such imperfections can in time lead to arcing, localized melting, and chemical transfor1nations, creating conducting channels. Even in relatively pure and homogeneous materials, microscopic cracks can initially occur because of temperature or mechanical stresses, oxidation, or other aging (progressive degradation) effects.
Example 4.30: Parallelplate capacitor with sandwiched dielectrics. A parallelplate capacitor is constructed using two separate dielectric materials, stacked between two conductor plates, as shown in Figure 4.56. (a) Find the capacitance of this configuration. (b) If glass (Ert = 6 and (EBR)t = 30 MVm 1) and mica (Er2 = 5 and (EBR)2 = 200 MVm 1) are chosen to be used with 2 cm2 crosssectional areas and equal thicknesses of 5 mm each (i.e., d 1 = d2 = 5 mm), find the capacitance and voltage rating of the capacitor, using a safety factor of 10. Solution: (a) As usual in capacitance problems, we start by assuming that a potential difference is applied between the plates and that charges +Q and Q are induced on the lower and upper plates, respectively. The surface charge densities on the plates are Pst = Q/A and
Sec. 4.11
321
Electrostatic Boundary Conditions
.,. \. n
di
Mica
Ez
A'
dz
, ~~
El
Glass
Figure 4.56 Dielectric strength. Parallelplate capacitor with two sandwiched dielectric layers.
S1
Q/A, respectively, where A is the area of the plates. Taking a Gaussian surface S 1 as shown, we have Ps2 =
>
where we arbitrarily assume the positive direction for the electric field to be upward. Similarly, using the Gaussian surface S2, we can show that E2 = Ps2/E2 = E1E1/E2. The magnitude of the total potential drop between the plates is
and the capacitance is then
c
=
Q
I I
Using the permittivity values given with di = d2 = 5 mm, we find C ~ 0.966 pF. (b) The voltage rating of the capacitor is determined by the dielectric layer that breaks down first. We have (EBR) 1 = 30 MVm 1 and (EBR) 2 = 200 MVm 1. For a safety factor of 10, we must have (EBR)1 1 (E1)max = lO = 3 MVm
and
(EBR)2 1 (E2)max = lO = 20 MVm
But since E2y = E1E1y / E2 = 6E1y /5, the voltage rating of this capacitor is determined by the glass layer since the glass layer breaks down first. Thus, the voltage rating is
= (3
MVm 1 )(5
x
103
m) 1 +
6 5
=33 kV
Note that if the dielectric were filled uniformly with mica, the breakdown voltage with a safety factor of 10 would be llmax = (E2)max(d1 + d2) = (20 MVm 1)(0.01 m) = 0.2 MV. Thus, the presence of the glass layer lowers the voltage rating of the capacitor.
4.11 ELECTROSTATIC BOUNDARY CONDITIONS
In dealing with electrostatic problems, it is often necessary to relate the polarization charges induced in dielectric media to electric fields produced by external charges (e.g.,
322
The Static Electric Field
Chap.4
on conducting bodies). This is facilitated by relations such as (4.66), which are called constitutive relations. In most electrostatic applications, we also deal with interfaces between two or more different dielectric media. The manner in which D and E behave across such interfaces is described by the boundary conditions. The boundary conditions are derived from fundamental laws of electrostatics, which we reiterate below. Starting with the experimental facts as expressed in Coulomb's law, namely that the electric force between charges is a function of the distance between them 117 and that its magnitude is proportional to the inverse square of the distance, we have derived the fundamental laws of electrostatics 118 as
VxE=O V ·D =p
c s
Conservative field
E · dl = 0 D · ds =
v
Coulomb's law
pdv
The dielectric properties of different physical media are accounted for by the permittivity E of the material, and so, as stated earlier, the charge density p excludes any bound polarization charge density. For linear (polarization P is linearly proportional to E) and isotropic (polarization P does not depend on the direction of E) media, D and E are related via (4.66), namely D = EE, which is valid even for inhomogeneous materials where E is a function of position expressed as E (x, y, z). A very common type of inhomogeneity occurs in practice at the interface between two electrically different (i.e., characterized by different permittivity values) materials. To establish a basis for solving such problems, we now study the behavior of D and E at the boundary between two different materials and derive the conditions that these vectors must satisfy at such interfaces. These conditions are referred to as the boundary conditions. The boundary conditions must be derived using the integral forms of the fundamental electrostatic laws, because the differential forms apply only at a point. For the normal component of the electric flux density, we consider a pillboxshaped closed differential surface as shown in Figure 4.57 a. Since the contributions from the sides can be made infinitesimally small by taking ~h > 0, we have
s
D · ds = (fi · D1)~s  (fi · D2)~s =
Ps~s
>
I Din  D2n = Ps I
(4.70)
where the minus sign in front of the second term on the lefthand side is due to the unit vector normal to the surface of the pillbox being fi, and Ps is the free surface charge density at the interface. Hence, the nor1nal component of the electric flux density D is discontinuous through a surface that has free surfacecharge density present; however, at 117 It
is important to remember that our arguments concerning the conservative nature of the electrostatic forcethat :fc E · dl = 0, or that fc E · dl is pathindependent (Section 4.4)were based only on the fact that the force from a single charge was radial and spherically symmetric. The fact that the force was proportional to r 2 was not important in this regard; any other r dependence would give the same result. Gauss's law, on the other hand, is based entirely on the fact that the electric force is proportional to r 2 . 118 We do not discuss the definition and physical meaning of the notation V x E until Chapter 6, but it follows from :fc E · dl by Stokes's theorem, also to be introduced and discussed in Chapter 6.
Sec. 4.11
323
Electrostatic Boundary Conditions
interfaces between two dielectrics, the free surface charge density Ps is usually zero unless a surface charge density is externally placed at the interface. For Ps = 0, (4.70) becomes (4.71) In other words, the normal component of D is continuous across a chargefree dielectric boundary. The nor1nal component of E is then not continuous across the boundary, since we have (4.72) The discontinuity in fi · E or En can be understood physically as follows. The change in the nor1nal component of the polarization vector P at the boundary can be related to the surface polarization charge by integrating (4.60) over the differential pillbox shown in Figure 4.57a as ~h > 0. Noting that P · ds = P · fids = Pnds, we have
v
V · Pdv = 
s
P · ds =(Pin 
P2n)~s
=
v
Ppdv
Since ~h > 0, the volume integral of Pp selects only the surface polarization charge density at the boundary:
v
PpdV = Pp~h~s = Psp~S
a d
El
n
D 1n ll.s
n
El
\
Ps
)
E1.
324
The Static Electric Field
Chap.4
Simplifying, we can write (4.73) But, using (4.64), we have
so that
=0 from (4.71)
Using (4.73), we find the discontinuity in the nor1nal component of the electric field at the interface in terms of the bound polarization surface charge as Psp E1n E2n = Eo
Thus, at the surface of the dielectric the normal component of the electric field E is in general discontinuous at the boundary by an amount Psp/ Eo, which can be interpreted as if Psp is a surface layer of free charge at that location. The surface polarization charge represents the amount of charge on the ends of dipoles in medium 2 that are not canceled by the charge of opposite polarity on the ends of the dipoles in medium 1 (see Figure 4.52a). In practice, surface polarization charge does not need to be taken into account explicitly, since the usual methodology involves finding suitable solutions for E and D in the two dielectric regions and then adjusting the magnitudes of the solutions to satisfy the boundary conditions. Using the different permittivities in the two regions fully accounts for all the requirements placed on the fields due to the different microphysical electrical behavior of the materials. For the tangential components of the electric field, we consider the line integral of the electric field E around a closed differential contour such as that shown in Figure 4.57b. Since the contributions from sides ab and cd can be made infinitesimally small by taking ~h > 0, we have a
c
E2·dl=(E1tE2t)~w=O
E1·dl+ d
(4.74)
b
Thus the tangential component of the electric field is continuous across the interface between the two materials. Since D = EE, the tangential component of the electric flux density D is not continuous across a boundary. Instead we have D1t
D2t
EI
E2
A consequence of (4.70) and (4.74) is the change in orientation of electric flux lines across material interfaces, as illustrated in Figure 4.57 c. Consider an electric field
Sec. 4.11
325
Electrostatic Boundary Conditions
E1 that is oriented at an angle 81 from the normal in medium 1. From (4.70) and (4.74) we have E1sin81 = E2 sin 82 E1E1cos81 = E2E2 cos 82 where E1 and E2 are the magnitudes of E1 and E2 and where we have assumed ps = 0. Therefore, we can find
and
The above relationships indicate that electric field lines are bent further away from the normal to the boundary in the medium with the higher permittivity.
Example 4.31: Spherical dielectric shell. A positive point charge Q is at the center of a spherical dielectric shell of an inner radius a and an outer radius b. The relative dielectric constant of the shell is Er. Determine E, cf>, D, P, and the polarization charge densities Pp and Psp· Solution: This example is very similar to Example 4.21 except that the conducting shell has now been replaced by a dielectric shell (see Figure 4.58). In view of the spherical symmetry, we apply Gauss's law to find E and Din three regions: (a) r > b, (b) a < r < b, and (c) r < a. The electric potential cf> is then determined from the negative line integral of E, and polarization P follows from the relation P = D  EoE = Eo(Er  l)E
Note that the E, D, and P vectors each have only radial components. (a) For the region r > b, E and cf> are exactly the same as in Example 4.21. We have
Q Er=   2 4n Eor
r>b
cf> =
Q
r>b
4n Ear
From equations (4.64) and (4.66) we obtain
r>b
Pr= 0
Q
Dr= EoEr =  2 4n r
r>b
(b) For the region a < r < b (i.e., inside the dielectric shell), E can be found using Gauss's law (equation (4.32)) using the spherical Gaussian surface Sd as
a i2
+ q3 cf:> i3 + q3 cf:>23
= q2
qi
4n EoR i2
+ q3
qi
q2
4nE0Ri3
+ q3   4nEoR23
where we have Ri2 = 1, Ri3 = 1 and R23 = 2. If qi = q2 = q3 = q, we then have 5q2 We=8nEo
Note that the same result could have been obtained using (4. 78), which is simply another way of writing (4.76): We=
1
2
qi
q2 4nE0Ri2
+
q3 4nE0Ri3
+ +
1 q2
qi
2
4nE0Ri2
1
qi 4nE0Ri3
q3 2
+
q3 4nEoR23
+
q2 4nEoR23
2 which upon adding terms with values substituted also gives We = 5q /(8nEo).
4.12.2 Electrostatic Energy in Terms of Fields It is interesting to consider where the energy associated with the charge distribution is stored. This question is analogous to one we might ask in mechanics when we have two masses attached to the ends of a compressed spring. The stored energy might be considered to reside in the masses or in the stressed state of the spring. The first viewpoint is the same as that expressed by (4.78), where the stored energy is linked to the charges and the electric potentials at their positions. This view ascribes physical reality only to the charges and their spatial distributions. On the other hand, if the field concept is to be complete, it should be capable of expressing the stored energy without recourse to a description of the charges producing it. The energy should be describable in ter1ns of the ''elastic'' quality of the electric field, just as the spring stores the energy in the preceding mechanical example. Another mechanical example is the flowerpot on the windowsill, which possesses potential energy because if it is pushed over the edge, it gains kinetic energy. Is the potential energy stored in the windowsill, in the flowerpot, or in the gravitational field? It turns out that the energy associated with a charge distribution can indeed be expressed solely in ter1ns of the fields E and D. We can generalize (4.78) to the case of a continuous distribution of charges in a volume V by making the following transformations and taking the limit as N > oo, N
qi
> p(r)dv,
i
> (r)
and
L
> v
i=I
which gives (4.81)
332
The Static Electric Field
Chap.4
where is now a continuous function of position and is the potential distribution due to all charges. It is no longer necessary to exclude the contribution due to p 8v located at the point where is evaluated because in the limit 8v > 0, this contribution is zero anyway. Equation (4.81) can be interpreted as the volume integral over V of an electric energy density We = ip, which exists in regions where p # 0. From Section 4.10, we know that a charge distribution can in general include both free charges and bound polarization charges. Let us first consider the case where the 'p' in (4.81) represents only the free charge density (this scenario is certainly the more practical consideration since free charges are the ones that we can directly control). In this case, as the free charge distribution is assembled, the resulting electric field induces polarization in the dielectric, represented by P. The induced polarization weakens the electric field in the dielectric due to the shielding effect120 of the effective bound polarization charge densities. Thus, considering only the free and induced polarization charge distributions, it appears that the emergence of the polarization charge reduces the energy required to assemble the free charge distribution. However, while bringing the free charges into their final configurations, no net work is done on the bound charges, since they are bound to their final positions. By calculating (4.81) for free charges only, we implicitly include an extra energy ter1n: the energy necessary to polarize the dielectric material itself. 121 Writing out all the energy terms explicitly,
(4.82) where Wf is the energy associated with the free charges, Wb is the energy associated with the bound charges, and Wp is the energy required to polarize the dielectric material. Since Wb =  Wp, by calculating Wf, we actually account for all energy terms, including the energy required to polarize the dielectric material. Thus, if we use the free charge density as the 'p' term in (4.81), we calculate both Wf and We. Hence p should indeed be solely the free charge density. If we instead include both the bound and free charge densities in (4.81), we calculate only Wf + Wb and leave out the final term Wp in the equation for the total stored energy of the charge system. Since the charge distribution in question is the free charge represented by p distributed in volume V, to convert (4.81) into an expression involving D and E, we can substitute V · D for p: We= fv(V · D)dv
i
By using the vector identity V · (D) we find
We = 120 As
=(V · D) + D · V
i f v (V · D 
D · V ) dv
illustrated in Figure 4.52, since the induced polarization is oriented in the same direction as E for isotropic materials, the electric field from the induced effective bound charge densities has a destructive effect on the electric field generated by externally applied charges inside the dielectric (see also Example 4.31). 121 This energy may be thought of as a potential energy equivalent to stretching or compressing a mechanical spring. It is the energy required to ''stretch'' the bound charges from their equilibrium states.
Sec. 4.12
333
Electrostatic Energy
Since for electrostatic fields E =  V , we have
where S is the surface that encloses volume V, and fi is the outward unit nor1nal vector of the differential surface element ds. The surface S, and thus the volume V, must be large enough to enclose all the charges in the region of interest. Since there are no other restrictions on the choice of the surface S, we can extend it to infinity. By doing so, the surface integral ter1n in We reduces to zero, since the dependencies of the various quantities at large distances from the charges (at infinity the entire charge distribution looks just like a point charge) are
1 rv

r'
1 D"'"'r2
and
ds
rv
2 r
We thus have lim D · fi ds
r+oo
so the electrostatic energy We reduces to simply the volume integral ter1n, (4.83) Since we let S extend to infinity, we must have V > oo, meaning that the integral in (4.83) has to be carried out over all space in which the electric field is nonzero. Based on the above, we can define the volume energy density for the electrostatic field as (4.84) We have thus found a way of expressing the stored energy of a charge distribution solely in ter1ns of the electric field. Note that the two alternative expressions we have found for electrostatic energy density, namely, ip from (4.81) and iEE 2 from (4.84) are quite different. The for1ner implies that energy exists where free charges exist and is zero where p = 0. However, the latter indicates that electrostatic energy exists wherever the fields exist. Both points of view have merit, and it is neither necessary 122 nor possible to determine which one is more ''correct." The dilemma here is the same as that in the mechanical examples discussed earlier: The two masses at the ends of a compressed spring and the flowerpot 122 For
an interesting discussion, see Vol. II, Section 8 of The Feynman Lectures on Physics, AddisonWesley, Reading, Massachusetts, 1964. A more advanced discussion is given in Chapter 4 of W. B. Cheston, Elementary Theory of Electric and Magnetic Fields, Wiley, 1964. It is possible to formulate the subject of electrostatics such that the electrostatic energy and the fields produced by a system of charges are fundamental quantities and the force between charged particles is a derived conceptan approach more fruitful for treatment of quantum phenomena. In the context of the experimentally based approach adopted here, the electric force, as defined in Coulomb's law, is taken to be the fundamental quantity.
334
The Static Electric Field
Chap.4
on the windowsill. It is not possible to ''localize'' energy or to decide whether it is associated with the charge or the field. Thus, the quantities ip and iEE 2 represent energy density only to the extent that their volume integral over space yields the total potential energy. We can also write We in ter1ns of the polarization P = D  EoE as (4.85) In other words, to establish a given electric field E in a dielectric medium rather than in free space, additional energy needs to be supplied (by the source that sustains the field). This additional energy is given by the second integral in the last equation above and is the energy required to polarize the dielectric. We can also interpret (4.85) in the context of (4.82). The first term, i fv EoE 2dv, calculates the energy required to assemble all of the charges (free and polarization charge densities) as if they were all free charges assembled in free space. Using the notation in (4.82), we have wf + wb = i fv EoE 2dv. Hence the final term in (4.85) is the energy required to polarize the dielectric material: Wp = f v P · E dv.
Energy stored in a capacitor. Equation (4.81) or (4.84) can be used to derive the expression for energy stored in a capacitor, which we know from circuit theory to 2 be We = CV . Consider any configuration of two conductors constituting a general capacitor structure as shown in Figure 4.50. For simplicity, assume conductor 1 to be at potential Vo/2 and conductor 2 to be at potential Vo/2, with reference to zero potential at infinity. Since the charge density is zero everywhere inside the conductors and is nonzero only on their surfaces, (4.81) reduces to
i
We = where 2
1 2
fs cl Ps l 1 ds
+ fsc2 Ps22 ds
and Sc2 are the surfaces of the two conductors. Using = Vo/2, we have Sci
w.e =
1 2 1
2
1
= Vo/2 and
r Ps i d S + TVo f Sc2 Ps2 dS
Vo 2 JSc1
(4.86)
where Q and Q are the total surface charges residing on conductors 2 and 1, respectively, and where we have used C = Q/Vo. Example 4.33: Energy of a parallelplate capacitor. Consider the dielectric filled parallelplate capacitor shown in Figure 4.60. The parallel plates, separated by a distanced and each with area A, are held at a potential difference Vo. Calculate the total energy stored in the capacitor. Solution: Neglecting fringing effects and using the coordinate system shown in Figure 4.60, the electric field between the capacitor plates is given by (see Example 4.23)
Sec. 4.12
z=d

335
Electrostatic Energy
_ _ _ _ _ _ _.._._ _ _ _ __ A
z Figure 4.60 Energy in a capacitor. Parallelplate capacitor filled with a dielectric.
E = zVo/d. Using (4.83), the total energy stored, which equals the energy required to assemble the free charges on the conducting plates (Wf), is 1
We = 
2 v
1 EE · Edv = A
2
v.2
d
E
0
O dz=
d2
1 v.2 AEo o
d
2
where E = ErEo. Compared with the same capacitor filled with free space, using (4.69), the power supply responsible for maintaining the constant voltage Vo supplies an additional energy in the amount (see (4.85) and subsequent discussion) 1 Wp= 
2 v
1 p . Edv = A
2
v.2
d
Eo (Er 
1) ~ dz =
d
0
1 v.2 AEo o
2
d
(Er 
1)
We can also directly calculate the total energy Wb required to assemble the final distribution of bound charges as if they were free charges assembled in a vacuum. From the induced polarization P, there are two layers of bound surface charge: a negative layer at the top of the dielectric (at z = d) and a positive layer at the bottom of the dielectric (at z = 0). Using (4.61), these surface charge densities are Psp lz=d = P · (Z)lz=d =  :o Eo(Er  1) Psplz=O = p · (Z)lz=O = :o Eo(E,  1) Using (4.19), the electric potential ct>(z) with respect to the lower plate is given by z
ct>(z) = z1=0
, . . Vo , . . ' z · zdz
=
d
Hence, substituting Pp for p in (4.81) and converting the integration into a surface integral, we have
s
1 Ppds = APsp(Z = d)ct>(z = d) = 
2
1 v.2 AEo o
2
d
(Er 
1)
where we used ct>(z = 0) = 0. As expected, the energy Wb associated with the bound charges is equal and opposite to the extra energy WP required to polarize the dielectric. We note that
The Static Electric Field
336
Chap.4
the energy required to assemble the free and bound charges, excluding the energy required to polarize the dielectric, is 1 v.2  AEo o 2 d
We could have arrived at this result more directly by simply evaluating (again with reference to (4.85) and subsequent discussion) 2
v
EoE dv
Example 4.34: Energy of a sphere of charge. Consider a sphere of radius b having uniform charge of volume density p in free space. Determine the electrostatic energy of this charge configuration using (a) (4.81) and (b) (4.84). Solution: 4.13:
The potential ct>(r) and the electric field Er(r) were determined in Example
1 rp Er=  
3
r(r)= 6 Eo
r>b
ct>(r) =
EQ
1 pb 3 Er=   3 Eor 2
1 pb 2
+ 2 Eo
~ pb3 3 Ear
where we have made the appropriate substitutions of a
> b and p
> Q/[(4n/3)a 3 ].
(a) Using (4.81), the stored energy is
1 1 pct>(r) dv =  4n We= 2 v 2
2np2
b
2
pct>(r)r dr = 0
b5
b5
6
30
4n p 2 b 5 15
EQ
EQ
Note that p is finite only where r < b, so that we only need to evaluate ct> for r < b. (b) Since Dr = EoEr, the stored energy from (4.84) is 1 We= 2
00
0
1
D · Edv = 4n 2
oo
0
2np 2 1 EQ
9
2np2 1 DrErr dr = 2
EQ
9
0
b
4np 2 b 5
15Eo
As expected, this result is identical to the stored energy as calculated from (4.81 ).
Example 4.35: Electrostatic energy around a charged conducting sphere. Consider a conducting sphere of radius a carrying a total charge of Q in free space. Determine the electrostatic energy stored around this sphere using (a) (4.86) and (b) (4.84).
Sec. 4.13
337
Electrostatic Forces
Solution: The electrostatic potential for such a conducting sphere was determined in Example 4.25. The variations with r of the electric field E(r) and the potential ct> (r) are Q 4nEor Q 4nEoa
ct:>(r) =
r>a
r"
E(r) = ra
4nEor 2
r(r =a), the energy stored is given by 1 1 We=  QVo =  Q
2
2
Q
Q2
4nEoa
8nEoa
(b) We can also determine the stored energy by integrating the electric field over the entire space within which it exists (i.e., it is nonzero), by using (4.84). We have 1 We= 
=
2
1 EoE dv = 2
2 v 1
2n
2
00
(4n)
Eo a
00
Eo 0
Q
0
2
4nEo
a
1 dr= 2
r
2
Q
r 2 sin() dr d() d
dWe lQ Fy=  dy
(4.89)
We note that in the constantcharge configuration, where there are no external sources of energy, the internal electrostatic force acts in a direction to minimize the internally stored
340
The Static Electric Field
electrostatic energy . to obtain
124
Chap.4
We can simplify (4.89) by direct substitution of We IQ= Q
dWe lQ dC Q2 dC F   ydC dy  2c2 dy
2
/
(2C)
(4.90)
We return now to the parallelplate capacitor example depicted in Figure 4.61a. Substituting C = EoA/y into (4.90) and again noting that EoEA = Q, we have Q2 Fy = 2C2
EoA y2

which matches our result above.
Constant voltage. Figure 4.61b depicts a parallelplate capacitor held at a constant potential Vo. In this configuration, when the plates are separated by an increment ~y due to an external force Fext, the energy associated with the removal of charge by the battery from the plates to maintain the same potential must also be considered. The total energy increase in the electric field of the capacitor is equal to the mechanical work supplied by the external force moving the plates against the internal electrical force, plus the energy supplied by the battery to deposit an additional ~Q charge against the voltage Vo: ~We l v = Fext ~Y
+ Vo~Q
(4.91)
where We Iv represents the energy stored in a capacitor at a fixed potential. Thus, in a constantvoltage configuration, the internal electrostatic force Fy is related to the change in stored energy as (4.92) In terms of differentials,
Fydy = dWe Iv
+ VodQ
>
dW lv dQ Fy = + Vody dy
(4.93)
Using (4.86) and Q = CVo, (4.93) becomes
d Fy=dy
1
2 CV0
2
(4.94)
Substituting Vo = Q/C into (4.94) yields (4.90), which was derived using the constantcharge configuration. 124
This result may be generalized to a force acting in three dimensions by considering a virtual displacement in the x, y, and z directions:
Sec. 4.13
341
Electrostatic Forces
4.13.2 Electrostatic Forces in the Presence of Dielectrics If metallic conductors are in the presence of a dielectric of finite extent (i.e., instead of being immersed in a dielectric per1neating all space), then any virtual displacement of the dielectric is also expected to change the stored energy, depending on the geometries involved. Since change in stored energy under virtual displacement indicates the presence of mechanical force, it follows that dielectric bodies in electrostatic fields will experience a net force. The magnitude of this force can be calculated using the principle of virtual work, in a manner similar to the case of conducting bodies. We illustrate this technique in the fallowing example. Example 4.36: Force on a dielectric slab. Consider a uniform dielectric slab of permittivity E partially inserted between the plates of a parallelplate capacitor of width w, and separation d, as shown in Figure 4.62 and having depth L (not shown). The capacitor plates are held at a constant potential difference Vo. Determine the electrostatic force acting on the dielectric slab.
Fringing fiel~ /
Q
I
Eo
E
E
E \
\
~
7
y
Fx
\
'.....
d
'
x
x w
Figure 4.62 Force on a dielectric slab between capacitor plates.
Solution: Suppose that the portion of the dielectric between the plates is of length x. In this case, the deflection coordinate is in the x direction. Assuming a constantvoltage configuration as shown in Figure 4.62, equation (4.94) becomes (4.95) where we used Vo= Eyd. To evaluate the total capacitance C, we can take the usual approach of evaluating the total charge Q on the capacitor plates under the assumed potential difference Vo. Using Gauss's law (equation (4.32)), the total charge residing on the capacitor plate adjacent to the dielectric slab is E EyLx. Similarly, the total charge on the plate adjacent to free space is EoEyL(w  x ). Hence the total capacitance is
Q
EEyLx EoEyL(w  x) ELx EoL(w  x) C  Vo  Eyd + +Cf·r1ng1ng ·  d + +Cf·r1ng1ng · Eyd d where we have noted the effect of the fringing field on the capacitance explicitly. As long as the ends of the slab are not too close to the plate edges, the fringing term does not change
342
The Static Electric Field
Chap.4
much with the virtual displacement, so we can simplify dC / dx to
dC = LE _ LEo dx d d
+
d (Cmnging) ~ L (E _ Eo) dx d
(4.96)
Substituting (4.96) into (4.95) gives
Hence the direction of the force is such as to draw the dielectric slab further into the region between the capacitor plates. It is interesting to calculate the work W required (i.e., the energy that needs to be supplied by the batteries) to place the dielectric slab completely inside the capacitor. The total work to move the slab from x = 0 to x = w can be rewritten as w
Fx dx = iEi(E  Eo)Lwd = i fv EE· Edv  i fv EoE · Edv
W = 0
=
i f v (EoE · E + P · E) dv  i f v EoE · E dv = i f v P · E dv
which is the additional energy needed to polarize the inserted dielectric, as was discussed in Section 4.12.
The physical origin of electrostatic forces on dielectrics. Although we were able to calculate the force on the dielectric slab using the principle of virtual work, the physical origin of the force on a dielectric is interesting in its own right. This question relates to our most basic everyday experiences with electrostatics, which were discussed at the start of this chapter. Why does a charged (e.g., by rubbing) object pick up small dielectric objects? Why did Thales of Miletus observe that pieces of amber rubbed in silk attracted small pieces of straw? At first thought, one might conclude that the amber had one kind of charge and that the straws had the opposite charge. But the straws were not rubbed onto anything and thus were electrically neutral. Although they do not have any net charge, they were nevertheless attracted to the charged amber. What was the physical reason for this attraction? The answer lies in the fact that the electric field produced by the charged amber (or any charged object of finite size) is nonunifor1n. As shown in Figure 4.63, a dielectric placed in an electric field is polarized. There are polarization charges of both signs, which experience attraction and repulsion forces due to the electric field. However, there is net attraction, because the electric field nearer to the source (at the top in Figure 4.63) is stronger than that farther away. If the electric field were uniform (constant intensity everywhere), there would be no net attraction. Indeed, if pieces of straw are placed between the plates of a large parallelplate capacitor, they are not attracted to either of the plates, regardless of the intensity of the electric field. The spatial variation (or the nonunifor1nity) of the electric field is the fundamental reason why dielectrics experience a force in the presence of electric fields. A dielectric always tends to move towards regions of stronger electric fields, as illustrated in Figure 4.63.
Microelectromechanical Systems (MEMS)
Sec. 4.14
343
E
F

E
Figure 4.63 A dielectric object in an electric field. The force experienced by the object is proportional to the gradient of the square of the magnitude of the electric field. The dielectric tends to move toward regions of higher electric field.
As we have seen in Example 4.36, the force experienced by the dielectric depends on the square of the field. In fact, it can be shown that for small objects the force on a dielectric is proportional to the gradient of the square of the electric field,  VE 2 . This is because the induced polarization charges are proportional to the electric fields, and for given distributions of charges the electrostatic force is proportional to the electric field. As a result, the object experiences a net force only if the square of the field varies in space. The constant of proportionality that determines the magnitude of the force depends on the dielectric constant of the object as well as its size and shape. The calculation of the electrostatic force on a dielectric object can in general be a quite difficult problem to solve. The case of the dielectric slab between parallel plates considered in Example 4.36 is thus a very special case, where the calculation of the force is facilitated by the principle of virtual work. In actual fact, the force on the dielectric slab is produced by the nonuniform fringing fields (see Figure 4.62), whose intensity decreases with distance away from the plates. The direct calculation of the force would require the accurate determination of these fringing fields, evaluation of the gradient of the square of magnitude of the electric field, and its integration over the entire body of the dielectric. It is clear that the principle of virtual work comes in rather handy in this case.
4.14 MICROELECTROMECHANICAL SYSTEMS (MEMS) Microelectromechanical systems (MEMS 125 ) are miniature devices that integrate electronic circuits with mechanical machines that interact with the surrounding physical environment. The field of MEMS has its roots in the integrated circuit (IC) industry. One of the primary advantages of MEMS technology is the ability to integrate mechanical systems with electronic circuits a process called monolithic integration. This ability 125 One
of the most widely referenced early papers on MEMS was published by K.E. Petersen: ''Silicon as a Mechanical Material," Proceedings of the IEEE, 1982, 70(5), p. 420457. For a modem reference on many of the broad applications covered by MEMS technology at an appropriate level, see C. Liu Foundations of MEMS, 2nd ed., Pearson, 2012.
344
The Static Electric Field
Chap.4
reduces the overall manufacturing costs and allows for greater uniformity among the constituent electrical and mechanical components. Since the transistor was invented in 1947 at the AT&T Bell Laboratories, the blistering pace of innovation and advancements in integrated circuits has steadily pushed forward the capabilities of microfabrication and miniaturization. Against enor1nous technical challenges, Moore's Law an observation by Gordon Moore, one of the cofounders of Intel, that the density of integrated transistors doubles every 1218 months remains valid even today. This continuously advancing fabrication technology forms the backdrop that enabled the functional integration of mechanical devices with analog and digital electronics. Today, threedimensional mechanical structures can be machined out of bulk or thin film silicon, often on the same silicon die. A large portion of commercial applications of MEMS technology is centered around sensors and actuators. In general, sensors provide a way to monitor some physical environmental variable, such as pressure, temperature, or incident radiation. Actuators, on the other hand, take an input control signal, such as a voltage or current and produce a force or torque in order to generate mechanical movement. Sensors and actuators are both examples of transducers, which transfor1n power from one energy domain (such as mechanical or ther1nal) to another (such as electrical). Our study of forces resulting from charge separation on a capacitor provides the theoretical basis for a common class of MEMS transducers: electrostatic sensors and actuators. Electrostatic transducers convert power between electrical and mechanical domains. As a basic example, consider again the parallelplate capacitor. If the distance between the two plates changes as a result of an external force, the overall capacitance also changes. This principle forms the basis for electrostatic sensing of position. Conversely, if more or less charge were deposited on the capacitor, perhaps by varying the voltage or by using a current source, the attractive force on the plates would change. If there was some mechanical means of movement, the surface would also change position, thus forming a simple actuator. The length scales of MEMS devices generally range from 1 µm to 1 cm. Due to their small size, the gravitational force is usually negligible compared with the electrostatic force, and can be ignored. Electrostatic sensors and actuators are found in a wide array of modern commercial products. We review two such products, one involving electrostatic sensors and the other electrostatic actuators, in order to illustrate some of the possibilities enabled by MEMS technology: automobile airbags and the Digital Light Processing (DLP) chip developed by Texas Instruments for video projection systems. Airbag systems in automobiles are designed to detect rapid changes in acceleration (such as the rapid deceleration during a collision) in order to deploy the bag. This industry was one of the first to use accelerometers designed using MEMS technology. In the typical configuration, two electrodes consisting of interdigitated fingers fabricated on a silicon substrate are used (see Figure 4.69 in Example 4.38). This pair of electrodes has an associated capacitance, determined by the size of the fingers and the distance between them. One electrode is held fixed, while the other is suspended in equilibrium and is connected to a mechanical spring with spring constant k. If the mass of the suspended electrode is m and the device is accelerating at a rate a, there is a force F = ma on the suspended electrode. This force pushes the movable electrode toward the fixed electrode,
Sec. 4.14
Microelectromechanical Systems (MEMS)
345
changing the distance between them, thereby changing the overall capacitance. This small change in capacitance is then detected by electronics that are fabricated on the same piece of silicon, which then relays a signal to deploy the air bag. Integrating the mechanical structure (the suspended electrode) with the sensing electronics helps to reduce the cost of this important safety component by eliminating the need for manual integration of the electrical and mechanical components. The DLP projection system, 126 which was invented by Dr. Larry Hornbeck of Texas Instruments in 1987, uses an array of torsion hingemounted microscopic mirrors to reflect a light source onto a projection screen. Each mirror, which measures approximately 16x 16 µm 2 , is responsible for one pixel of the projected image. The monolithic integration of the mirrors and logic circuitry enables individual control over a large, dense array: up to 2 million mirrors can be micromachined on a single semiconductor chip. The mirror itself is mounted on a plate, called a yoke, that is free to rotate around a torsion hinge by a deflection angle of ± 10°. The deflection angle is kept precise by using mechanical stops. Electrodes are placed on either side of the torsion hinge. When the yoke and the electrodes are uncharged, the plate is stable at zero deflection, held in place by the mechanical restoring force of the torsion hinge. When the mirror and yoke are charged up to a bias potential, then applying a voltage differential across the electrodes causes an electrostatic attraction toward the electrode with a larger net opposite charge. To change the orientation of the mirror, the mirror bias voltage is removed, allowing the restoring force of the torsion rod to bring the mirror flat. The bias voltage is then applied again, causing the mirror to rotate to the new state of the electrodes. While each mirror has only two states, ''on'' (reflecting light toward the screen) or ''off' (reflecting light away from the screen), fast switching of the state allows shades of grey by controlling the percent of time the mirror is in the ''on'' state. Each mirror can oscillate in the kHz range to create an illusion of a grey scale. In the following sections, we revisit the attractive force between the two plates in a capacitor and show how this force may be used to design an electrostatic actuator. We consider both a chargecontrolled and a voltagecontrolled actuator.
4.14.1 ParallelPlate Electrostatic Actuators Consider the model of a parallelplate electrostatic actuator shown in Figure 4.64. The capacitor plates each have an area A and are separated by a gap distance d. The bottom plate is anchored, and the top plate moves with a mechanical restoring force dictated by the spring constant k. The origin is chosen at the position of the top plate when there is no charge on the capacitor and no force from the spring, with y increasing as the top plate moves toward the bottom plate. We define do to be the gap distance with y = 0, so d (y) = do  y. The capacitance is
AE
AE
C=  d doy 126 For
(4.97)
a modem overview reference, see J. Hornbeck, Combining digital optical MEMS, CMOS and algorithms for unique display solutions. Electron Devices Meeting, 2007. IEDM 2007. IEEE International., 2007.
346
The Static Electric Field
+Q
k
Chap.4
k A
A
+
Q (a)
(b)
Figure 4.64 Model of a parallelplate electrostatic actuator. The bottom plate of a parallelplate actuator is held fixed, while the top plate is suspended by a spring with spring constant k and is free to move in the vertical direction. The two plates have area A and are separated by distanced. Two control mechanisms are shown: (a) Charge control, where the current source controls the charge Q on each plate and (b) Voltage control, where the voltage source controls the potential difference across the capacitor plates.
In this coordinate system, Fspring = kyy. We assume that the restoring force of the spring remains linear throughout the range of motion. As before, we assume an ideal parallel plate capacitor, ignoring fringing fields at the edge of the capacitor plates and assuming the electric field is constant between the plates. This assumption introduces little error when the linear dimensions of the plates are much larger than the gap length d. We also consider the general case where the per1nittivity of the material between the two plates is uniform and equal to E. From (4.88), and noting the coordinate system in Figure 4.64, we can write the electrostatic force pulling the two plates together as (4.98a) (4.98b) (4.98c) For (4.98b) we used the fact that127 =Ed, as shown in Example 4.23. Equation (4.98c) can be derived from (4.98b) using the capacitance of the parallel plate capacitor, C = Q/ = EA/d. There are two ways to control the electrostatic force between the two plates: by controlling the charge Q, through a current source, or by controlling the potential difference , through a voltage source. These two configurations are shown in Figure 4.64a and b. When the force is expressed in ter1ns of the charge Q, it is independent of the 127 In
this section, we use instead of Vo for the potential difference between the capacitor electrodes since this value is not fixed in a transducer. The time dependence of (t) is implicit: it changes either in response to an external agent changing the capacitance (sensor) or due to an internal agent controlling the charge or voltage (actuator).
Sec. 4.14
Microelectromechanical Systems (MEMS)
347
distance d between the two plates (equation (4.98c)), whereas the force depends on the gap distance when expressed in ter1ns of the potential difference (equation (4.98b) ). We consider these two configurations separately.
Charge control. We first consider the case of moving the actuator by controlling the amount of charge on the capacitor plates. Since the electrostatic force is independent of the gap distance d when expressed in terms of the charge Q, the equations dictating the gap distance as a function of the charge are easy to derive. This independence of gap distance leads to some desirable properties in ter1ns of the mechanical considerations of the design. The total force acting on the upper plate is the vector sum of the electrostatic and spring forces: F + Fspring· At equilibrium, the total force must be zero. Therefore, at equilibrium we have Q2 2AE = ky
)
Q2 Yeq = 2kAE
(4.99)
The gap between the two plates is thus Q2
d = do  Yeq = do 
kA
2
E
(4.100)
The gap of the chargecontrolled actuator monotonically decreases as the charge increases and is therefore stable over the range of valid movement. Eventually, assuming the full range of motion is possible, the gap becomes zero when enough charge is placed on the capacitor. The total amount of charge Qo needed to close the gap entirely is >
Qo = Jdo2kEA
(4.101)
The stable deflection as a function of the deposited charge is a valuable attribute of the chargecontrolled actuator, since it enables a full range of motion as a function of the charge Q. However, the above f or1nulation ignored the parallel parasitic capacitance Cp between the mobile electrode and ground. Since capacitances involved in such applications are typically small, on the order of femtoFarads, parasitic capacitances, which alter the amount of charge that must be deposited by the current source, provide a practical limitation to a chargecontrol design approach. A parasitic capacitance that is at least half as large as the actuator capacitance at equilibrium can cause a socalled charge pullin effect, where a positive feedback loop occurs after a critical amount of charge is deposited on the electrodes and the top plate snaps down. 128 Example 4.37: Potential difference across a chargecontrolled actuator. Find the maximum potential difference across a chargecontrolled actuator. What is the corresponding gap distance? 128 For
a detailed discussion of this effect, see J. Seeger and B. Boser. Charge control of parallelplate, electrostatic actuators and the tipin instability, Journal of Microelectromechanical Systems 12(5), pp. 656671, 2003.
348
The Static Electric Field
Chap.4
Solution: Since the electric field is constant, the potential difference across the plates is given by (using (4.100)) 2
= Ed = Q d = Q
EA
EA
do 
Q
(4.102)
2kAE
To find the maximum potential difference, we differentiate (4.102) with respect to Q, set the result to 0, and first solve for the maximum charge: 2
d = do _ 3Q = 0 dQ EA 2kA2E2
>
where Qo is given by (4.101). Substituting
d _ Q~ax O 2kAE
_ Qmax
EA
max 
do2kAE 3
Qmax
Qo
v'3
(4.103)
into (4.102), we have
..J'do2kAE/3
EA
d 0

d0 2kAE/3 2kAE (4.104)
The deflection distance at this charge level is (using (4.99))
Q~ax
do2kAE/3 2kEA
Ymax = 2kAE
do 3
(4.105)
and so the gap width at the maximum voltage point is dmax = do  do/3 = 2do/3. The normalized voltage and deflection distance are plotted as a function of the deposited charge in Figure 4.65.
1.0
I I I
I
0.8
I I I
0.6
I
0.4
y!d0 , ,
,
0.2 0.0
Figure 4.65 Plot of nonnalized deflection and potential difference. The poten
I
, ,
I I
, , ,
... ... ...
""='1.1...'~
0.0
0.2
0.4
0.6
Q!Qo
0.8
1.0
tial difference across the chargecontrolled capacitor and the deflection distance y are plotted against the normalized charge Q/ Qo. The maximum potential difference occurs at QI Qo = 1I .J3 ~ 0.577. At this point, the separation between the upper and lower plate is d = 2do/3.
Voltage control. Unlike the charge control configuration, the voltage control actuator design is not susceptible to parasitic capacitances (since a parallel capacitance Cp does not alter the potential difference across the actuator electrodes) and is therefore a popular choice in practice. However, this ease of implementation comes at the cost of a reduced range of motion. For a voltagecontrolled actuator, we begin with (4.98b), which expresses the electrostatic force on the plates in ter1ns of the potential difference. As before, we seek to
349
Microelectromechanical Systems (MEMS)
Sec. 4.14
find the equilibrium position of the upper plate by setting the total force to zero. This equilibrium is achieved when )
2k (do  Yeq) 2
(4.106)
Because the electrostatic force is inversely proportional to the square of the gap separation d and therefore the deflection distance y, the solution for the equilibrium position cannot be derived as easily as with the chargecontrolled configuration. The resulting equation for Yeq is a cubic function, and so there may be more than one solution in the allowable range of motion. The characteristics of the solutions can be more readily seen by solving for the potential difference in (4.106) as a function of displacement:
=
Yeq2k AE (do  Yeq)
(4.107)
The potential difference, nor1nalized by the maximum value (4.104) found in the chargecontrolled example, is plotted in Figure 4.66. As seen, for each potential difference, there are two equilibrium values y that balance out the electrostatic and spring forces. While the forces cancel at each solution satisfying (4.106), the solution also needs to be stable. Let Ftot be the net force in the y direction. At a solution point that satisfies (4.106), F tot = 0. At a stable solution, the incremental net force ~F resulting from a small downward deflection ~y would bring the top plate back up to the equilibrium position. For small deflections,
~F = 3Ftot ~y
ay
(4.108)
Thus, the derivative of the total force at a stable equilibrium point must satisfy
ay
=
1 :Rp(r')dv' 4nEo v' Ir  r'l 2 1
Eo V 2 V
r
p(r')dv'
E · dl
4n Eo v' Ir  r'I
00
t p represents the total volume charge density, which in general may
include the volume bound polarization charge density  V · P.
In dielectric materials where we can write D = EE, we can express the free charge density directly in terms of the Efield and electrostatic potential, as summarized in Table 4.3. The free charge density p excludes the bound polarization charge density  V · P, so that V · D = p. TABLE 4.3
FREE CHARGE DENSITY
E
v. EE t p represents the free volume charge density, which
excludes the volume bound polarization charge density V · P.
PROBLEMS 4.1 4.2
4.3
1\vo point charges. Two identical point charges 1 m apart from each other in free space are experiencing a repulsion force of 1 N each. What is the magnitude of each charge? 1\vo point charges. Two small identical spheres have charges of + 20 nC and 5 nC, respectively. (a) What is the force between them if they are apart by 10 cm? (b) The two spheres are brought into contact and then separated again by 10 cm. What is the force between them now? 1\vo suspended charges. Two small, identical, electrically charged conducting spheres of mass 2.5 g and charge + 150 nC each are suspended by weightless strings of length 12 cm each, as shown in Figure 4.70. Calculate the deflection angle ().
Chap.4
359
Problems
l
F
+ ______ J. ______ +
e1 Q mg
4.4
Q ', mg
F
Figure 4.70 Two Problem 4.3.
suspended
charges.
Zero force. (a) Three point charges of Qi = +40 nC, Q2 = 20 nC, and Q3 = + 10 nC are all situated on the x axis in such a way that the net force on charge Q2 due to Qi and Q3 is equal to zero. If Qi and Q2 are located at points (2, 0, 0) and (0, 0, 0), respectively, what is the location of charge Q3? (b) Q3 is moved to a different position on the x axis such that the force on itself due to Qi and Q2 is equal to zero. What is the new position of Q3? 4.5 Three charges. Three identical charges of charge Q are located at the vertices of an equilateral triangle of side length a. Determine the force on one of the charges due to the other two. 4.6 Three point charges. Three point charges of values + 150 nC, + 100 nC, and +200 nC are located at points (0, 0, 0), (1, 0, 0), and (0, 1, 0), respectively. (a) Find the force on each charge due to the other charges. Which charge experiences the largest force? (b) Repeat part (a) for 100 nC as the second charge. 4.7 1\vo point charges. Two point charges of +Q and Q are located at (0, 0, 0) and at (0, 4a, 0) respectively. Find the electric field at Pi (0, 0, 3a) and at P2(0, 4a, 3a ). Sketch the orientations of the fields. 4.8 Zero field from three charges. Two point charges of +Q each are located at (10, 0) cm and ( 10, 0) cm, while a third one of charge 2Q is at (0, 10) cm. Find the coordinates of the point where the electric field is zero. 4.9 Three point charges. Two point charges of+ 10 nC each are located at points (0.46, 0, 0) and (0.46, 0, 0), respectively. (a) Where should a third point charge of + 15 nC be placed such that E = 0 at point (0, 1, O)? (b) Repeat part (a) for a third charge of 15 nC. (c) With the 15 nC charge located as you determined in part (b), is there another point at which the electric field E = O? If so, specify this point. 4.10 Four point charges on a square. Four point charges of +50 nC each are located at the comers of a square of side length 10 cm located on the xy plane and centered at the origin. (a) Find and sketch the electric potential on the z axis. (b) Find and sketch the electric field on the z axis. 4.11 Six point charges on a hexagon. Six identical point charges of + 25 nC each are situated in space at the corners of a regular hexagon whose sides are each of length 6 cm. (a) Find the electric potential at the center of the hexagon. (b) Determine the energy required to move a point charge of 25 nC from infinity to the center of the hexagon. 4.12 1\vo straightline charges. Consider two uniformly charged wires, each of length 1 m and a total charge +100 nC, with their ends separated by 1 m, as shown in Figure 4.71. (a) Find the electric potential ct> at the point P, midway between the two wires. (b) Find the electric field E at point P.
360
The Static Electric Field
Chap.4
y
!Q + + + + + + +
lm
lm
!'T't.....1+ P(0.5, 0) ++++++++++ Q
lm
1
= +lOOnC
x
Figure 4.71 Two straightline charges. Problem 4.12.
+ 10 nC
4.13
Seven point charges on a cube. Seven identical point charges of
4.14
1\vo line charges. A uniform line charge of P1i = 4n x 8.85 pcmi is located between the points ( 5, 0) and ( 2, 0) m and another such line of positive charge (i.e., P12 = 4n x
each occupy seven of the eight comers of a cube 3 cm on each side. Find the electric potential at the unoccupied comer.
8.85 pcmi) between (5, 0) and (2, 0) m, as shown in Figure 4.72. (a) Find the electric potential at (1, 0) m. (b) Find the electric field E at the same point. At which points, if any, is the electric field E zero? At which points, if any, is the electric potential zero? y
6 5
21
1 2
5 6
x
Figure 4.72 Two line charges. Problem 4.14.
4.15
Circular ring of charge. A total charge of Qi is distributed uniformly along a halfcircular ring as shown in Figure 4.73. Two point charges, each of magnitude Q2, are situated as shown. The surrounding medium is free space. (a) Find Q2 in terms of Qi so that the potential at the center of the ring is zero. (b) Find Q2 in terms of Qi so that the electric field E at the center of the ring is zero. y
R
2R
4.16
Figure 4.73
Circular ring of charge.
Problem 4.15 .
Semicircular line charge. A thin line charge of density Pl is in the form of a semicircle of radius a lying on the xy plane with its center located at the origin, as shown in Figure 4.74. Find the electric field at the origin for the cases in which (a) the line charge density Pl = Po is a constant and (b) the line charge density varies along the semicircular ring as p1 = Po sin¢.
Chap.4
Problems
361
z
/ / / / / / / /
/
,,.
/
/
/
a 
'¢1 I
Pz
4.17
y
Figure 4.74 Semicircular line charge. Problem 4.16.
Charge on a hemisphere. The curved surface of a hemisphere of radius a centered at the origin carries a total charge of Q uniformly distributed over its curved surface, as shown in Figure 4.75. (a) Find the electric potential on the z axis. (b) Find the electric field on the z axis. (c) Repeat parts (a) and (b) if the charge Q is uniformly distributed throughout the volume of the hemisphere.
z a
Figure 4.75 Charge on a hemisphere. Problem 4.17.
4.18
Sheet of charge with hole. An infinite sheet of uniform charge density Ps is situated coincident with the xy plane at z = 0. The sheet has a hole of radius a centered at the origin. Find (a) the electric potential and (b) the electric field E at points along the • z axis. 4.19 Spherical charge distribution. A charge density of p(r) = Kebr
4.20
where K and b are constants, exists in a spherical region of space defined by 0 < r < a. (a) Find the total charge in the spherical region. (b) Find the electric field at all points in space. (c) Find the electric potential at all points in space. (d) Show that the potential found in part (c) satisfies the equation V 2 =  p (r) / Eo for both r < a and r > a. The electron charge density in a hydrogen atom. According to quantum mechanics, the electron charge of a hydrogen atom in its ground state is distributed like a cloud surrounding its nucleus, extending in all directions with steadily decreasing density such that the total charge in this cloud is equal to qe (i.e., the charge of an electron). This electron charge distribution is given by ( ) = pr
qe
na 3
e
2r/a
where a is the Bohr radius, a :::'. 0.529 x 10 10 m. (a) Find the electric potential and the electric field due to the electron cloud only. (b) Find the total electric potential and the electric field in the atom, assuming that the nucleus (proton) is localized at the origin.
362
4.21
The Static Electric Field
Chap.4
Spherical shell of charge. The space between two concentric spheres of radii a and b (a < b) in free space is charged to a volume charge density given by
K p(r) = 2r ;
a < r < b
where K is a constant. (a) Find the total charge in the shell. (b) Find the electric field at all points in space. (c) Find the electric potential at all points in space. (d) What happens if b >a?
4.22
Space charge between parallel plates. The space between two perfectly conducting parallel plates is filled with space charge (i.e., free charge) of density given as: p = Po sin ( :z ) where d is the separation distance of the plates. Other than the space charge, the medium between the plates is air, with permittivity Eo. The upper and lower plates, located at z = 0 and z = d are kept at potentials of ct> = 0 and ct> = Vo, respectively. Determine expressions for the potential ct> (z) and the electric field E(z) between the plates.
4.23
Spherical charge distribution. A spherical charge distribution exists in free space in the region 0 < r < a given by r2 p(r) =Po 1 a2 (a) Find the total charge. (b) Determine E everywhere. (c) Determine ct> everywhere. (d) Sketch both IEI and ct> as a function of r.
4.24
4.25
4.26
Spherical charge with a cavity. A spherical region of radius b in free space is uniformly charged with a charge density of p = K, where K is a constant. The sphere contains an uncharged spherical cavity of radius a. The centers of the two spheres are separated by a distance d such that d + a < b. Find the electric field inside the cavity. Charge on a hollow metal sphere. A hollow metal sphere of 20 cm diameter is given a total charge of 1 µC. Find the electric field and the electric potential at the center of the sphere. Electric field. An electric field in empty space is given as: E(x, y, t) =
x sinx cosy sin wt+ y cosx siny sin wt
V m 1
Determine the energy required to move a positive unit test charge in the presence of this electric field from the point (0, 0) to (1, 1) along the shortest straight line in the xy plane. Determine the energy required to move the same test charge from point (1, 1) back to (0, 0) but via a different path, by first going fro (1, 1) to (1, 0) and then from (1, 0) to (0, 0).
4.27
4.28
A 1farad capacitor. To get an idea about the physical size of a 1F capacitor, consider a parallelplate capacitor with the two metal plates separated by 1 mm thickness of air. Calculate the area of the metal plates needed so that the capacitance is 1 F. Gate oxide capacitance of a MOS transistor. A basic MOS transistor consists of a gate conductor and a semiconductor (which is the other conductor), separated by a gate dielectric. Consider a MOS transistor using silicon dioxide (Si02) (Er = 3.9) as the gate oxide. The gate oxide capacitance can be approximated as a parallelplate capacitor. The gate oxide capacitance per unit area is given by _ Eox Cox tox
Chap.4
363
Problems
where Eox and fox are the pennittivity and the thickness of the gate dielectric. (a) If the 6 thickness of the Si02 layer is 2 x 1o cm, find the gate oxide capacitance per unit area. (b) If the length and the width of the gate region are L = 5 x 104 cm and W = 2 x 103 cm, respectively, find the total gate capacitance.
4.29
RG 6 coaxial cable. A coaxial cable (RG 6) designed for interior use, such as connecting a TV set to a VCR, has a perunitlength capacitance listed as 17.5 pF/ft. If the relative dielectric constant of the insulator material in the cable is Er ~ 1.64, find the ratio of the inner and outer radii of the insulator. 4.30 Radius of a highvoltage conductor sphere. Consider an isolated charged metallic conductor sphere in a dielectric medium at an electric potential of 500 kV. Calculate the minimum radius of the sphere such that dielectric breakdown will not occur if the surrounding dielectric is (a) air (EBR = 3 MVm 1), (b) a gaseous dielectric such as sulfur hexafluoride (SF6) (Er~ 1 and EBR = 7.5 MVm 1), (c) a liquid dielectric such as oil (Er = 2.3 and EBR = 15 MVm 1), and (d) a solid dielectric such as mica (Er = 5.4 and EBR = 200 Mvm 1). 4.31 Parallelplate capacitor. A parallelplate capacitor is constructed from two aluminum foils of 1 cm2 area each placed on both sides of rubber (Er = 2.5 and EBR = 25 MVm 1) of thickness 2.5 mm. Find the voltage rating of the capacitor using a safety factor of 10. 4.32 Energy in a capacitor. A 9V battery is connected across a parallelplate airfilled capacitor. The battery is subsequently removed, and a block of solid dielectric (E = 2Eo) is inserted between the plates. (a) What is the voltage across the capacitor after the introduction of the dielectric? (b) Compare the total electrostatic energy stored in this capacitor before and after the introduction of the dielectric. Comment and explain any differences. Neglect all fringing effects. 4.33 Coaxial capacitor. Consider a coaxial capacitor as shown in Figure 4.76. Given a = 5 mm, l = 3 cm, and the voltage rating of the capacitor to be 2 kV with a safety factor of 10, what is the maximum capacitance that can be designed using (a) oil (Er = 2.3 and EBR = 15 MVm1) and (b) mica (Er= 5.4 and EBR = 200 MVm 1).
Figure 4.76 Coaxial capacitor. Problem 4.33.
4.34
Coaxial capacitor with two dielectrics. A coaxial capacitor consists of two conducting coaxial surfaces of radii a and b (a < b ). The space between is filled with two different dielectric materials with relative dielectric constants Elr and E2r, as shown in Figure 4.77. (a) Find the capacitance of this configuration. (b) Assuming that l = 5 cm, b = 3a = 1.5 cm, and oil and mica are used, calculate the capacitance. (c) Redo part (b) assuming that only oil is used throughout.
364
The Static Electric Field
Chap.4
b
a
Figure 4.77 Coaxial capacitor with two dielectrics. Problem 4.34.
4.35
Capacitor with spacers. The crosssectional view of an airfilled coaxial capacitor with spacers made out of material with permittivity E is shown in Figure 4.78. (a) Find the capacitance of this coaxial line in terms of E, a, b, and¢. (b) If the spacers are to be made out of mica (E = 6Eo), determine the angle¢ such that only 10% of the total energy stored by the capacitor is stored in the spacers. (c) Consider the capacitor without the spacers (i.e., ¢ = 0). For a given potential difference Vo between the inner and outer conductors and for a given fixed value of b, determine the inner radius a for which the largest value of the electric field is a minimum.
Figure 4.78 Coaxial spacers. Problem 4.35.
capacitor
with
4.36 Earth capacitor. Consider the earth to be a large conducting sphere. (a) Find its capacitance (the earth's radius is '""'6.371 x 106 m). (b) Find the total charge and energy stored on the earth (take the electric field on the surface of the earth to be 100 Vm 1). (c) Find the maximum charge and energy that can be stored on the earth. 4.37 Coaxial capacitor with variable f. A coaxial capacitor of inner radius a and outer radius b is filled with a dielectric material whose relative permittivity varies as Er = 10 a /r over the region from r = a to r = b. Find the capacitance per unit length and compare with the capacitance of the same coaxial cable when filled with air. 4.38 Planar charge. A surface charge distribution Ps (x, z) exists on the xz plane, with no charge anywhere else (i.e., p = 0 for IYI > 0). Which of the following potential functions are valid solutions for the electrostatic potential in the halfspace y > 0, and what is the corresponding charge distribution Ps(x,z) on the xz plane? ¢>1 =
ey coshx
¢>4 =
sinx siny sinz
¢>2 =
ey cosx
¢>3 =
e,./iy cosx sinx
Chap.4
Problems
365
Parallel power lines. An isolated pair of parallel power lines a distance of d 1 apart have a potential difference of VAB and are located a distance h above a pair of telephone wires, as shown in Figure 4.79. The parameter values are di = 1 m, a= 2 cm, VAB = 440 V, h = 60 cm, and d2 = 15 cm. (a) Find the direction and magnitude of the electric field at points 1 and 2. Take the midpoint between the power lines as the origin of your coordinate system. (b) Determine the potential difference ct> 12 between points 1 and 2.
4.39
0
y
1        di      1
x
 0  0~A
B
h
1 0
0
2
  dz Figure 4.79 Parallel power lines above telephone lines. Problem 4.39.
Field under highvoltage line. Many 60 Hz highvoltage transmission lines operate at an rms alternating voltage of 765 kV. (a) What is the peak electric field at ground level under such a line if the wire is 12 m above the ground? (b) What is the peak potential difference between the head and feet of a 6ft tall person? (c) Is the field sufficient to ignite a standard (110 V) fluorescent lamp of 2 ft length? Capacitance of the twowire line with dielectric sleeve. A twowire line consists of two metallic conductors of radius a enclosed by dielectric (permittivity E) sleeves of radii b separated by a distance d as shown in Figure 4.80, with d >> a. The space surrounding the dielectric sleeves is air, with permittivity Eo. (a) Determine the capacitance per unit length of this twowire line. (b) The values of wire radius and separation distance are given to be a = 1 cm and d = 20 cm, respectively, while the radius of the dielectric sleeve is b = 5 cm. If the dielectric sleeve is made of mica (E = 5.4E 0 ,E~ca = 200 MVm 1), determine the maximum operating voltage of the twowire line capacitor. Note that the breakdown field 1 for air is E~1~ = 3 Mvm .
4.40
4.41
.. ,2a ~ 1
I
I
I I I I
I I I I
I~
/
,, b
~
 
      
       
 
()
I
I I 1 ~1
d
Figure 4.80 Twowire transmission line. Problem 4.41.
366
The Static Electric Field
Chap.4
4.42 Thundercloud fields. A typical thundercloud can be modeled as a capacitor with horizontal plates with 10 km2 area separated by a vertical distance of 5 km. Just before a large lightning discharge, the upper plate may have a total positive charge of up to 300 C, with the lower plate having an equal amount of negative charge. (a) Find the electrostatic energy stored in the cloud just before a discharge. (b) What is the potential difference between the top and bottom plates? (c) What is the average electric field within the cloud? How does this value compare to the dielectric breakdown field of dry air (3 MVm 1)? 4.43 1\vo conducting spheres. Consider a pair of small conducting spheres with radii a, b, small compared with the separation distanced between their centers (i.e., a, b 1020 SmI have been measured), although at slightly higher temperatures, the normal temperaturedependent resistance is observed. Most superconductors are metallic elements, compounds, or alloys and exhibit transitions into superconducting states at critical temperatures approaching absolute zero (0 K or 273°C). Some examples of metallic elements that exhibit superconductivity are aluminum (Tc = 1.2 K), lead (Tc = 7.2 K), and niobium (Tc = 9.2 K). Interestingly, some metals do not become superconducting; for example, copper is not superconducting even at 0.05 K. Until recently, the Nb3AlNb3Ge (Tc = 21 K) alloy was believed to have the highest critical temperature, and alloys and compounds of niobium (e.g., niobium with a tin coating) are widely used as superconducting wire and tape in magnets operating at liquidhelium temperatures. A practical use for superconductivity has been in generating intense magnetic fields for research projects. IO Niobiumtin compounds can support current densities of over 109 Am 2. In 1986, it was discovered that some oxides become superconducting at temperatures above the boiling temperature of liquid nitrogen (77 K). For example, yttriumbariumcopper oxide, or YBa2Cu307 , has Tc = 80 K, so its superconductivity can be utilized by cooling with liquid nitrogen. Since then, materials with Tc values as high as 125 K have been discovered. II
5.2 CURRENT FLOW, OHM'S LAW, AND RESISTANCE
Consider a piece of current carrying material such as the conductor shown in Figure 5.3. If at any time current is directed toward a boundary between the conductor and the
Figure 5.3 An arbitraryshaped conductor. The surfaces A 1 and A 2 are maintained at potentials 1 and 2, respectively and are assumed to be equipotential surfaces (i.e., they are coated by material with conductivity much larger than that of the conductor).
9 See
R. de Bruyn Outober, Heike Kamerlingh Onnes's discovery of superconductivity, Scientific American, pp. 98103, March 1997. 1°For more information on the practical uses of superconductivity, see the August 1989 issue of the Proceedings of the IEEE. 11V. Z. Kresin and S. A. Wolf, Fundamentals of Superconductivity, Plenum, New York, 1990.
Sec. 5.2
375
Current Flow, Ohm's Law, and Resistance
surrounding dielectric, charge accumulates at the boundary and produces a steadily increasing electric field. To maintain a constant electric field and a steady current flow, we must thus have J and E parallel to the conductor boundaries. Consider any two surfaces A and A', both of which are perpendicular to J, with their circumferences defined by the conductor boundaries. The currents passing through these surfaces, namely I = JA J · ds and I'= JA' J · ds, respectively, must be equal, since otherwise charge accumulates indefinitely in the region between the two surfaces. Because the two surfaces were chosen arbitrarily, the current must be the same through any other surface that cuts across the wire. For homogeneous media (i.e., where a is not a function of position), the distribution of current flow lines is the same as that of the electric field, and the charge density p 2 inside the conductor is zeroi so that the current lines do not begin or end anywhere in the conductor. The electric field within the conductor is presumed to be set up and maintained by means of a potential difference applied between the crosssectional surfaces at the ends of the conductor, namely Ai and A2 in Figure 5.3. For simplicity, we take these end surfaces to be equipotential surfaces. i 3 Note that no current can flow outside the conductor (a = 0) and that charge cannot indefinitely accumulate at the boundary between the conductor and the surrounding dielectric. Thus, the electric field lines, and thus the lines of current flow, are always parallel to the side edges of the wire, regardless of how curved or twisted the conductor may be. Since the conservative property of the electric field requires (see Section 4.11) that the tangential component of Eis continuous across any interface, the electric field just outside the wire must have the same tangential component as that just inside. In general, the electric field outside the wire also has a component perpendicular to the side surface of the wires, which is supported by surface charge induced on the surface of the wire. i 4 Further discussion of the structure of the electric field at the boundary between a conductor and an insulator is found in Section 5 .6. The flow of current (i.e., J) at each point in the material is in the direction of the electric field E. The potential difference i2 between the two equipotential surfaces Ai and A2 at potentials
2  i
i2
z2  zi
l
Eo=    
a discussion of Ohm's law and contributions of G. S. Ohm, see M. S. Gupta, Georg Simon Ohm and Ohm's Law, IEEE Transactions on Education, 23(3), pp. 156162, August 1980. 16 Note that we have assumed (on the basis of symmetry) the electric field to be constant everywhere inside the rectangular conductor.
Sec. 5.2
377
Current Flow, Ohm's Law, and Resistance l I I
I
Ai ~1
//
/
..z
A
~
E
~2
u
Figure 5.4 A rectangular conductor of unifonn cross section. The end surfaces are
/
/
coated with highly conducting material and are thus equipotential surfaces.
z=z 2
so that the current is given by aE. ds = a12A I= A l
)
12
R=1
2 
1
I
l aA
(5.5)
The preceding expression is not specific to a rectangular block and is in fact valid for the resistance of a conductor of length l having a uniform crosssectional area A of any shape. Example 5.1: Copper wire. Consider a copper wire 1 km long and 1 mm in radius. (a) Find the resistance R. (b) If the wire carries a current of 1 A, determine the duration of time in which the electrons drift across the length of the wire.
Solution: (a) For copper, we have a = 5.8 x 107 Sm 1 . For l = 103 m and a 1mm radius, which gives A= n x 106 m 2, using (5.5) we have R =
103 m 5.8 x 107 Sm 1 x n x 106 m2
~
5.49Q
(b) A current of 1 A and a crosssectional area A = n x 1o 6 m 2 implies a current density of J ~ 3.18 x 105 Am2. Since J = ncqevd and nc = 8.45 x 1028 elm 3 , we have lvd I ~ 2.35 x 105 mis. In other words, the electrons can drift across a distance of 1 m in about rvl2 hours. A particular tagged electron can traverse the entire 1km length of this copper wire in approximately rv492 days! On the other hand, we know from transmissionline analysis that an applied signal (e.g., a change in current through or voltage across a line) travels to the other end of the line at nearly the speed of light in air (i.e., approximately rv3.33 f.,lS for a 1km wire). The transmissionline signal propagates outside the conductors (or wires) as an electromagnetic wave, not via the drifting motion of electrons within the conductors that constitute the line.
Example 5.2: A microstrip trace. Consider a printed circuit board microstrip trace, as shown in Figure 5.5. If the metal trace is made of copper with thickness t = 34.3 f.Lm, find the trace resistance per centimeter for a trace width w of (a) 0.25 mm and (b) 0.5 mm.
378
Steady Electric Currents Copper trace
~...
w
.. 't
Chap.5
1
Insulator t
Figure 5.5 A microstrip trace. The crosssectional view of a microstrip trace.
Solution: Using (5.5) where l = 1 cm = 0.01 m, a = 5.8 x 107 Sm 1, and A = wt = 3.43 x 105 w m 2 : (a) For W1 = 2.5 x 104 m, we have Rl ~ 20.1 mncm 1, and (b) for w2 = 2w1 = 5 x 104 m, we have R2 = Rl/2 ~ 10.05 mQcm 1.
Example 5.3: Curved bar. The resistance of a rectangular block of metal was discussed in this section and found to be given by R = l/(aA), where a is the conductivity, l is the length, and A is the crosssectional area. As a more complicated geometry, consider the resistor in the form of a curved bar, which is made of the rectangular block bent to form the arc of a circle, with the two end surfaces remaining flat, as shown in Figure 5.6a. The edges are coated with perfectly conducting material so that they constitute electrodes at uniform potential. Find the resistance R between the electrodes. Solution: This problem has a cylindrical symmetry with respect to the 00' axis, as shown. The equipotential surfaces are the intersections of the curved rectangular bar and the planes
2
Flat washer b
I I I I I I I I
.... _ a
I I
wt
I I I I I;
I V I I /I a I I I I"' I I lf'O I I I 1 r,1 II
'i ~

....
(c)
,.....Yo' (b)
0
a
 
(a) (d) Figure 5.6 A curvedbar resistor. (a) A rectangular bar bent into a circular arc of rectangular cross section. (b) A differential angular element along the arc. (c) A flat washer is equivalent to a curved bar of arc length ¢ 0 ~ 2n. (d) The flat washer can be viewed as a parallel combination of elemental resistances of radial width !::..r, as shown.
Sec. 5.2
379
Current Flow, Ohm's Law, and Resistance
passing through the 00' axis. The current flow lines J are circular arcs centered on the 00' axis (i.e., J is in the " direction). We may suppose from symmetry that the E vector is the same at all points inside the bar that are equidistant from the 00' axis. However, the electric field must vary with radial distance from the axis (r) across the curved bar, since the inward portions of the end faces are closer to each other than the outward portions " are (i.e., E must be a function of r, or E = J/a = E (r) ). To determine how the electric field varies with radial distance, we can use the fact that fc E · dl = 0 (equation (4.20)). Consider a closed contour ABCD, as shown in Figure 5.6b. We have
c
E · dl =
where the first term is the contribution from path segment AB and the second term is that from path segment CD. We thus have
Ee/> !:l.r
+
dE r !:l.r dr
+
dE (!:l.r ) 2 = 0 dr
where the last term drops out as !:l.r
dr
>
r
> 0. Therefore
K Ee/>= r
>
where Ki and K are constants. The functional form of Ee/> is thus determined. Note that we could alternatively have found ct>(r) using Laplace's equation, namely V 2 ct> = 0, and subsequently determined E from E =  V ct>. The voltage between the end faces is thus o
o
K
Kl
 r d
V·E=E
Combining the two preceding equations we find ap a +p=O at E
Rewriting and integrating, we have ap a  at= p E
> >
a E 0
t
pa~
at= 
Po ~
>
a
 t =In E
p Po
p ( x,y,z,t ) =Po ( x,y,z ) e (a/E)t
where po(x,y,z) is the initial value of the free charge density at t = 0. The initial charge distribution throughout the conductor decays exponentially with time at every point, completely independent of any applied electric fields. If the charge density is initially zero, it remains zero at all times thereafter. The time Tr = E /a is is referred to as the relaxation time and is the time required for the charge at any point to decay to 1/ e of its original value. The relaxation time is extremely short for good conductors and relatively large for insulators or dielectrics. In fact, whether a material is considered a conductor or an insulator is decided on the basis of the relaxation time. When rr is extremely short compared with measurable times or times of interest in a given application, the material is considered to be a conductor; whereas when rr is very long, the material behaves like an insulator. For most metals, rr is indeed far too short to measure or observe; for example, for copper we have a = 5.8 x 107 Sm 1 and E = Eo,23 so that ircopper r v 10 19 s. In fact, ir is very small for all but the poorest of conductors. Even for distilled water, rrH2 o '.: : :'. 105 s. On the 23 The
dielectric constant E for a metallic conductor is not easily measurable, since any polarization effect is completely overshadowed by conduction. Nevertheless, based on measurements of reflectivity of metals and the fact that atomic resonances for metals lie in the ultraviolet and xray ranges, metallic conductors can be treated as if their dielectric constant is Eo at frequencies up to and including the visible range (i.e., rv 10 15 Hz).
Sec. 5.6
Boundary Conditions for Steady Current Flow
389
other hand, for a good dielectric Tr is very large; for example, Tramber ~ 4 x 103 s, Trmica ~ 1020 hours, and Trquartz ~ 50 days. Consider a thought experiment in which we suppose that at t = 0, charge is concentrated within a small spherical region located near the center of a very large conducting sphere. In every other region of the conductor, the charge density is initially zero. Starting at t = 0, the charge within the small spherical region begins to fade away exponentially, but since charge anywhere in the conductor can only decrease in time (because of the expression p = p0 e(a/E)t derived previously), no charge can appear anywhere within the conductor. Where then does the charge in the small spherical region go? Because charge is conserved, the exponentially vanishing charge near the center must begin to appear at the surface of the conducting sphere, no matter how great the radius of the conducting sphere may be. However, the surface charge must make its appearance at the exact instant that the interior charge begins to decay, because the total charge in the system is constant. 24 Although we are concerned in this chapter with electrostatic fields, the concept of relaxation time is also used to deter1nine the electrical nature (i.e., conductor versus insulator) of materials for timevarying fields. At any given frequency of operation f, a material is considered a good conductor if Tr is much shorter than the period T = 1/ f, that is, if Tr > T. We can now see that some materials that are considered to be good conductors at certain frequencies tend to become insulators at sufficiently higher frequencies. For example, seawater (a = 4 Sm 1) is considered to be a good conductor at frequencies up to r v 100 MHz but is an insulator for frequencies above rv 10 GHz. Further discussion of conducting or dielectric properties of materials at different frequencies is provided in Sections 8.3 and 11.2.1.
5.6 BOUNDARY CONDITIONS FOR STEADY CURRENT FLOW Most applications of steady current flow involve considerations of the interfaces between currentcarrying conductors and dielectrics (or insulators) or between two conducting materials of different conductivity. The manner in which the current density J and the electric field E behave across such interfaces is governed by the boundary conditions, which are for1nulated and discussed in this section. We separately consider conductordielectric and conductorconductor interfaces.
24This
interesting observation at first appears to indicate that this phenomenon may be used to transmit signals with infinite velocity. However, in order to convey such a signal at a given moment (t = 0), we must, until this moment, prevent charge on the small sphere from being dispersed. This can be done by means of an insulating envelope (e.g., a thin membrane) that might be withdrawn at the given moment to initiate the signal. Before this could happen, however, an induced charge equal and opposite to the charge on the small sphere would appear on the external surface of the envelope, and at the same time an induced charge equal to the original charge in the small sphere would be produced on the surface of the large sphere. The withdrawal of the insulating envelope would merely cause the charges on its two sides to unite (M. Abraham and R. Becker, Electricity and Magnetism, Blackie & Son Limited, Glasgow, pp. 260, 272, 1944).
390
Steady Electric Currents
Chap.5
5.6.1 Current and Electric Field at Conductor  Dielectric Interfaces In Section 4.7, we showed that the surfaces of metallic conductors were equipotentials and that, therefore, the tangential component of electric field on a metallic surface must necessarily be zero (i.e., Et = 0). On the other hand, when a conductor with finite conductivity carries a current, an electric field given by E = J/a exists within the medium. At the boundary of any conductordielectric interface, the current must flow tangentially to the boundary surface, as shown in Figure 5.12a, since current cannot flow across the boundary into the dielectric region with zero conductivity. Thus, on the conductor side of the interface, we have Et = lt/a. By the continuity of the tangential electric field at a boundary, the tangential field on the dielectric side must also be Et. If a potential difference is applied between the two ends of a good (but not perfect) conductor, 25 current flows, and the body of the conductor is no longer an equipotential. Equipotentials
,,
, ' Dielectric ( E, u = 0)
..
Et, It .. " +Vo
"
1=0
i I I I I ___
Battery
Conductor ( u)
,
I,
,
'"
'"
,
\
Perfect!y conducting wire
I
I
(a)
1=0 ++ + + +
+++++Load RL
E
    
I (b)

(c)
,
r.
I U
I ,
Conductor ( u)
1 




......,_____,
(d)
Figure 5.12 Conductordielectric interface. (a) Current must be tangential to the surface since it cannot flow into the dielectric region. (b) Battery connected to the two ends of a conducting bar. (c) Two metal strips separated by air or another dielectric are maintained at a potential difference of V0 . (d) The connection of a load across the metal strips leads to current flow, which in turn means that the electric fields on the surfaces of the metal strips are not entirely vertical. 25 In
practice this can be achieved by connecting the wire across a battery, which is rather unwise in most cases since the resistance of a piece of conducting wire is typically extremely small (much smaller than the internal resistance of the battery), and the battery is shorted.
Sec. 5.6
Boundary Conditions for Steady Current Flow
391
Assuming that the two ends of the conducting bar are coated with perfectly conducting material so that they are equipotentials, the potential varies uniformly along such a currentcarrying wire, as shown in Figure 5.12b. Assuming that the electric field E is uniform, the potential difference d over any differential length dl of the wire is E dl. The integral f E dl carried out over the full length l of the wire is equal to the applied potential difference Vo and is also given by Vo = IR, where R is the resistance of the wire and I is the total current. The field is the same just inside and just outside of the wire and is entirely tangential (i.e., parallel to the axis of the wire). Note, however, that in practice a conducting wire is very rarely used in the configuration implied in Figure 5.12b, since for typical applied potentials (e.g., Vo = 1 V from a battery) the current is rather large, and the wire shorts out the battery. As an example, consider a silver wire of radius 1 mm and length 10 cm. The resistance of such a wire is approximately rv5 x 104 Q, which means that a current of I rv 2000 amperes flows when this single wire is connected across a 1V battery! In reality, however, any battery has an internal resistance much larger than the resistance of the wire, so that most of the electromotive force of the battery would appear across its internal resistance, resulting in all of the battery power being dissipated internally. This would lead to rapid generation of internal heat, which could cause the battery housing to melt down or burst. In practice, a common use of conducting materials involves two or more conducting objects that are separated by dielectrics (or free space) and that are at different potentials.26 In such cases, a surface charge distribution Ps exists on the conductor surfaces, and a component of electric field nor1nal to the conductordielectric boundary exists, given by En = Ps / E. The amount of charge on any small portion of the surface of a conductor is equal to the average potential of that portion multiplied by its capacitance to ground or to the appropriate nearby conductor. An example of such a situation is shown in Figure 5. l 2c, where two metal strips are maintained at potential difference of Vo by a battery connection. Just inside the conductors, we must have En = 0 since the the current density J (and hence E) must be tangential to the boundary, as argued in Section 5 .2. If no path is available between the metal strips for current flow (e.g., Figure 5.12c), then the current J = 0, and hence we also have Et = 0, so that the total field inside both conductors is zero. If a load is now connected between the two conducting strips so that current can flow, the electric field lines no longer terminate on the conductors at right angles but are tilted, as shown in Figure 5.12d. Typically, the tangential electric field Et inside the conductors is much smaller than the nor1nal electric field En due to the external circuit connections. This circumstance is implied by the size of the arrows representing Et and En in Figure 5. l 2d; however, the size of Et and thus the slant of the electric field lines are nevertheless greatly exaggerated. In reality, the tilt angle e of the field lines is much smaller than a degree, since Et 12 = Vo = 1 V, in this case the equivalent line charge density on the conductors is p1 ~ 1.2 x 10 11 Cm 1 . Again from Chapter 4, we know that the electric field at a distance r from a line charge density of Pl is Er = Pl/ (2n Eor). Using the value of Pl just found, we find that the electric field midway between the wires (i.e., at r = 0.5 cm) due to one of the conductors is of order En ~ 43 Vm 1. It is thus obvious that En >>Et. In this case, we have e = tan 1 (Et I En) r v 0. 007° ' 27 so that the tilt of the electric field lines due to the current flow is indeed extremely small. The charges that reside on the surfaces of wires carrying steady currents, which are induced via the potential differences between the wires and other conductors, thus play a fundamental role in the production of steady currents. The distributed capacitances (also called stray capacitances) by which these charges are induced are the sources of electric fields much larger than the field that supports the electron drift that constitutes the current. The surface charges must of course be in motion as are those inside the wire; however, the amount of charge (i.e., charge density p) at any place must remain constant, being maintained by steady currents.
5.6.2 Bending of Current Flow at Interfaces between Two Conductors The boundary conditions for current flow across an interface between two conducting materials can be derived in a manner entirely analogous to the derivation of the electrostatic boundary conditions that was undertaken in Section 4.11. Figure 5.13 shows the interface between two media of conductivities a 1 and a1. To determine the conditions for the component of the steady currents nor1nal to the interface, we consider a cylindrical Gaussian surface, as shown in Figure 5. l 3a, and use the fact that
v
27 Note
(V · J) dv =
J · ds = 0
that the electric field due to the other conductor is in the same direction and essentially doubles the electric field. Also note that for the case of the twowire configuration, the electric field varies with distance between the conductors, being much larger near the wires. Thus, the tilt of the lines is in fact substantially smaller than the rv0.007° calculated here.
Sec. 5.6
393
Boundary Conditions for Steady Current Flow l1n
lit
fl.s
b
... a
(Tl
fl.h
n
d
(Tl
b
fl.h
.... ,
, __
\I'
A
\ .... /
c
oz a2
lzt
J2n
(b)
(a) (}l
I I I I I I I
(c) Figure 5.13 The boundary between two different conductors. (a) A differential pillboxshaped closed surface. (b) A differential closed contour. (c) The bending of current flow at the boundary for 02 > 01.
Assuming the height ~h of the cylinder to approach zero (~h > 0), the only nonzero contribution to the surface integral comes from the top and bottom surfaces. In other words, (5.9)
)
which implies that
If medium 2 is a perfect dielectric (a2 = 0), we must have current can flow perpendicular to the surface; that is,
l2n
= 0, meaning that no
)
Thus, the current flow at the surface of a conductor with a perfect insulator must be parallel to the surface, as mentioned earlier.
394
Steady Electric Currents
Chap.5
For the tangential components of the current, we consider the tangential component of the electric field around a closed differential contour, as shown in Figure 5.13b. The fundamental equation we use here is the same as that used in Section 4.11, namely, that the electrostatic field is conservative, or in other words, :fc E · dl = 0. Naturally, the condition we find when we let ~h > 0 is then also the same as in Section 4.11, namely, that E It = E2t. In terms of the tangential components of the current, we then have

(5.10)
Together, (5.9) and (5.10) indicate that upon crossing a boundary, the current line bends by an amount proportional to the ratio of the conductivities. Noting the definition of 81 and 82 from Figure 5.13c, we can write lit tan81 = l 1n
or
Note that if medium 1 is a good conductor and medium 2 is a lowloss dielectric (i.e., a 1 >> a1), the current enters medium 2 at a right angle to the boundary for practically all angles of incidence from medium 1 (i.e., 12 ~ l2n). This corresponds to the requirement (noted in Section 4.7) that the electric field is normally incident to the surface of a good conductor. In general, the condition (5.9) requires that the normal component of the electric field be discontinuous across the interface. Using the boundary condition (4.70), we also have indicating that the discontinuity of the normal component of electric flux density necessitates a surface charge layer Ps at the boundary. Using a1E1n = a1E2n, we can write
Note that the surface charge Ps will vanish only for the special case of E2/E1 = a1/a1. Alternatively, we can write Ps as
Ps =In where rr1 and rr2 are the charge rearrangement (or relaxation) times in media 1 and 2, respectively. If both media are metallic conductors, we have E1 ~ E2 ~ Eo, so that we have 1 1 Ps = Eoln
Sec. 5.7
Duality of J and D: The ResistanceCapacitance Analogy
395
5.7 DUALITY OF J AND D: THE RESISTANCECAPACITANCE ANALOGY In most materials, both the current density J and the electric flux density D are linearly proportional to the electric field. As a result, there exists a dual relationship between J and D in regions where nonconservative fields are not present, that is, in regions outside the interior of the batteries and in the absence of timevarying fields. In this section, we analyze some of these dual relationships. If current enters and leaves a conducting medium via two ''perfect'' conductors (also referred to as electrodes), the equivalent resistance is given in ter1ns of the potential difference 12 = 2  1 between the perfect conductors as R = 12/I, where I is the total current leaving the positively charged electrode. Taking an arbitrary surface completely enclosing the positive electrode, this current is given by I=
s
E · ds = a
J · ds =a
s
E
D · ds = aQ S
E
where Q is the total charge induced on the positive electrode due to the capacitance between it and the other (negative) electrode. Noting that this capacitance is given by C = Q / 12, we thus have 12
R= 
I

E12
aQ

E
aC
Note that the definition of C depends on the existence of static charge on the electrodes. This charge is proportional to 12 and is independent of whether a current also exists. Note, however, that all of the electric flux, and thus current, that leaves one electrode enters the other. Consider now a pair of electrodes that can be placed either in a dielectric (or free space) or in a conducting medium in which there are no boundaries that would disturb the field and the current patterns. The electric field pattern must be the same in both cases because charge is distributed in the same way on the conductors in each case, since V · E = 0 holds in the intervening space. Thus, if we can find the capacitance for the case when the electrodes are immersed in a dielectric, we can then deter1nine the resistance when the conducting medium is present simply by using the for1nula R = E/(aC). Note that since C is proportional to the pertnittivity E (see, e.g., the general definition of capacitance given by (4.52)), it is evident that the resistance R does not depend on the value of E. Sitnilarly, since R is inversely proportional to a (see (5.3)), the capacitance C does not depend on a. This duality relationship can be easily verified by considering the case of a slab of lowconductivity material of thickness d sandwiched between and in contact with parallel planar electrodes of area A, as shown in Figure 5. l 4b. The resistance of this arrangement is R = d/(aA), and the capacitance of the same configuration when the conducting slab is replaced by a dielectric (Figure 5.14a) is C = EA/d, consistent with RC= E/a. The duality relationships between linear homogeneous dielectric and conducting media are summarized in Table 5.2 and in Figure 5.14.
396
Steady Electric Currents A
1
d
d
(b)
(a)
E
Chap.5
or a
= 0
= Vo
JorD 2
1
s
(c) Figure 5.14 Duality between J and D. The fact that the configuration of the electric field lines is determined only by f c E · dl = 0 brings about the duality between resistance and capacitance between two arbitrary, perfectly conducting bodies (electrodes). (a) A dielectric medium sandwiched between two electrodes. (b) A conducting medium between two electrodes. (c) Two electrodes of arbitrary shape surrounded by a homogeneous medium, either a dielectric or a conductor.
In homogeneous, linear, isotropic, and timeinvariant materials, an important consequence of the duality is E
RC= a
=Tr
(5.11)
which is consistent with our previous discussions of Tr as the relaxation time constant of 28 a material. The relation (5.11) can be very useful in deriving expressions for resistance of electrode configurations for which we already have the capacitance or vice versa. that rr = E /a is akin to the RC time constant of a firstorder capacitive circuit, indicating that the behavior of a lossy dielectric medium under different applied voltage/current conditions can be modeled by such an RC equivalent circuit. 28 Note
Sec. 5.7
Duality of J and D: The ResistanceCapacitance Analogy
397
TABLE 5.2 DUALITY BETWEEN DIELECTRIC AND CONDUCTING MEDIA
Conducting Media
c
Dielectric Media
E · dl = 0
c
D=EE
J=aE
v ·J =
E · dl = 0
V ·D =0
0
J = aE = a\7
D =EE= EV
l1n = l2n
Din= D2n
11 al It =
R=
iv E1 lt
11 a2 2t
=
iv E2 2t
__ J:L _ E ·_ dl fs aE · ds
For example, we had derived in Example 4.28 that the capacitance of two concentric spheres is 4rrEab C=ba Using (5.11), we can immediately deter1nine that, if the region between two concentric spheres is filled with a homogeneous conducting material, the resistance between the two spheres is ba R=aC 4rraab E
Note that for b > oo, this resistance becomes one half of the resistance of the buried hemispherical electrode found in Example 5 .4. This result is to be expected, since the geometry of the electric field (and thus that of the current) is exactly the same. But a full spherical electrode at the same potential Vo has twice as much current flow. In other words, since half of the space in Example 5 .4 is free space, half as much current flows for the same potential difference Vo, thus resulting in twice as much resistance. Example 5.6: Resistance of a coaxial shell. A cross section of a coaxial line consists of an inner conductor of radius a and an outer shell of radius b separated with a dielectric with E, as shown in Figure 5 .15. When the coaxial line is used as a transmission line, the current flows along the inner conductor and returns in the outer shell (or vice versa). The capacitance (per unit length) of this configuration was found in Example 4.27 to be Cu= (2nE)/ln(b/a). When the space between radii a and b is filled with imperfect dielectric with conductivity a, a radial leakage current tends to flow between the inner conductor and the outer shell. Find the resistance that determines this leakage current.
398
Steady Electric Currents
Chap.5
l /
/
/
a,
I
E
I
I
I / ...... . . . . '\ I I I I I I I I
I I I I I \
\
\
I
\
I
\
\
\ I
I I I I I \
I
          '\   2"" / \ \ '\
' ......
Figure 5.15 Resistance of a coaxial shell. When the space between the inner and outer conductors of a coaxial line is filled with lossy material, leakage current J flows between the two conductors, as the arrows show .
Solution: Since the electric field line configuration is radial in both cases, we can use the duality of D and J to find the resistance R over a length l. Using (5 .11), 1 E 1 E R=   = Ca Cul a
1 b ln2nal a
Note that the leakage resistance of the coaxial shell is identical to that of the thin circular semiconductor contact discussed in Example 5 .5 (with h = l). This result is to be expected, since the contact in Example 5 .5 is simply a coaxial shell that has very small height h >Np). Find the conductivity of thentype Si sample at 300K. For the mobility of majority charge carriers, use µe = 1194 cm2 (Vs) 1 . (b) Repeat part (a) if boron acceptor atoms with a concentration of Na = 10 16 cm 3 are added to a pure sample of Si (which is called ptype Si since the holes are now the majority carriers, that is, Np ~Na >>Ne). For the mobility of majority charge carriers, use µP = 444 cm2 (Vs) 1 . (c) Compare your results in parts (a) and (b) with the conductivity of the intrinsic Si and comment on the differences. 5.11 A silicon resistor. A silicon bar 1 mm long and 0.01 mm2 in crosssectional area is doped with Nd = 1017 cm 3 arsenic (As) atoms. Find the resistance of the bar and compare with the resistance of the same bar made of pure silicon (see Problem 5.9). For the mobility of the free arsenic electrons, use µe = 731 cm2 (Vs) 1 . 5.12 A silicon resistor. A sample of ptype silicon is 8 µm long and has a crosssectional area of 2.5 µm x 2.5 µm. If the measurements show that the average hole concentration and the resistance of the sample are Nd = 10 16 cm 3 and R = 18 kQ, respectively, find the hole mobility µP. 5.13 Sheet resistance Rsq· Consider the resistance of a uniformly doped ntype Si layer of length l, width w, and thickness t, as shown in Figure 5.19. This resistor can be divided into square sheets of dimension w on each side, as shown. The resistance of any one of these square sheets is called the sheet resistance, denoted by the symbol R8q, in units of Q(sq) 1 . (a) Show that the sheet resistance is given by
(b) Show that the total resistance of the Si layer is given by l R = Rsq w
(c) Calculate the total resistance of a Si layer that has a length of 50 µm, width of 5 µm and a sheet resistance of 150Q(sq) 1 .
l w
l_ t
tI
I
w
T w
Figure 5.19 Sheet resistance. Problem 5.13.
5.14 Integratedcircuit (IC) resistor. Consider an integratedcircuit (IC) resistor of length l, width w, and thickness t that is made of Si doped with 1016 cm 3 phosphorus atoms (i.e., Ne ~ 2.5 x 1016 cm 3 ). Given t = 2 µm, find the aspect ratio, w /l, such that the resistance of the IC resistor is 10 kQ at 300K. Take µe = 1000 cm2 (Vs) 1 .
Chap.5
409
Problems
5.15
An ionimplanted IC resistor. An ionimplanted ntype resistor layer with an average doping concentration Nd = 4 x 1017 cm 3 is designed with 1 /Lm thickness and 2 /Lm width to provide a resistance of 3 kQ for an IC chip. Find the sheet resistance and the required length of the resistor. Assume the electron mobility as /Le = 450 cm2 (Vs) 1 . 5.16 Diffused IC resistor. An IC resistor is frequently fabricated by diffusing a thin layer of ptype impurity into an ntype isolation island, as shown in Figure 5.20. If contacts are made near the two ends of the ptype region and a voltage is applied, a current will fl.ow parallel to the surface in this region. It is not possible to use R = l/(aA), however, to calculate the resistance of this region because the impurity concentration in it is not uniform. The impurity concentration resulting from the diffusion process is maximum near the surface (x = 0) and decreases as one moves in the x direction. (a) Show that the sheet resistance of the ptype layer is given by 1
t
a(x) dx
Rsq = 0
(b) Assuming the conductivity of the ptype layer decreases linearly from ao at the surface (x = 0) to al >a above it. If the conductivity of the water is a, find the resistance (per unit length) between the wire and the planar bottom of the lake. Tuo conducting spheres. Two metallic conducting spheres of radii ai and a2 are buried deep in poorly conducting ground of conductivity a and permittivity E. The distance b between the spheres is much larger than both ai and a2. Determine the resistance between the two spheres.
This page intentionally left blank
The __. tatic
In the preceding chapters, we discussed various aspects of electrostatics or the configurations of static electric fields and their relationship to stationary charges and conducting or dielectric boundaries. Our discussions in Chapter 4 involved charges that were assumed to be at rest, and the various effects studied were consequences of Coulomb's law the only experimental fact that we needed to introduce. In Chapter 5 we considered motion of charges under the influence of the Coulomb force resulting from an applied static electric field, constituting steady electric current. We now focus our attention on a series of new phenomena caused by the steady motion of charges. These phenomena constitute the subject of magnetostatics and can be understood as the consequences of a new experimental fact: namely, that currentcarrying wires exert forces on one another. In a certain sense, magnetostatics is an unfortunate name for the class of phenomena discussed in this chapter. First, the phenomena are caused by charges in motion, making the ''static'' part of the name a misnomer. Secondly, the ''magnet'' part of the name immediately suggests phenomena involving actual magnets. In reality, the physical phenomena involving natural magnets are highly complex, and it is very difficult, if not impossible, to construct a theory of magnetostatics based on experiments with magnets. It is much easier to for1nulate the theory of magnetostatics based on experiments involving moving charges or electric currents, a task which we undertake in this chapter. As we noted in Chapter 4, our earliest experiences with electricity date back to ancient times and involve the attraction of objects by other objects. Interestingly enough, the subject that we now call magnetism also began with the observation that certain natural minerals, readily found near the ancient city of Magnesia in Anatolia, which is now part of Turkey, could attract other materials. 1 For centuries, magnetism was thought 1That
lodestone attracts iron was first noted in print by Roman poet and philosopher Lucretius [99?55? B.C.] in his philosophical and scientific poem titled De rerum natura (On the Nature of Things). Lodestone is a form of iron oxide called magnetite, found in the shape of elongated fragments. The name lodestone comes from Middle English lode, ''course," because a compass needle made of this material can be used for navigation. In the same manner, the lodestar is the pole star, which marks north in the sky. Also see G. L. Verschuur, Hidden Attraction, The Mystery and History of Magnetism, Oxford University Press, New York, 1993.
415
416
The Static Magnetic Field
Chap.6
to be independent of electricity. Phenomena involving magnets were studied extensively because lodestone was readily available, and performing experiments with magnets was relatively simple. 2 However, construction of a theory of magnetostatics based upon results of such experiments requires that we introduce highly artificial concepts and fictions such as a ''magnetic pole." After many years of experimentation, and the accidental discovery in 1820 (see the next paragraph) of the experimental fact that an electric current can exert a force on a compass needle, it was realized that the simple understanding of magnetostatic phenomena lay in the relatively difficult experiments involving moving charges rather than in the relatively simple experiments involving magnets. We adopt this modern viewpoint of magnetostatics, founded on Ampere's experimentally based law of force between currentcarrying wires. That wires carrying electric currents produce magnetic fields was first discovered3 apparently by accident by H. C. Oersted in 1820. Within a few weeks after hearing of Oersted's findings, AndreMarie Ampere announced4 that currentcarrying wires exert forces on one another, and J.B. Biot and F. Savart repeated Oersted's experiments and put forth5 a compact law of static magnetic fields generated by current elements in a circuit, which is known as the BiotSavart law. The most thorough treatment of the subject was undertaken by Ampere in the following three years, culminating in an extensive memoir6 in 1825. This work established Ampere's law, which describes the law of force between two current elements and is analogous to Coulomb's force law in electrostatics. Ampere's studies also led him to postulate that magnetism itself was due to circulating currents on an atomic scale, thus closing the gap between the magnetic fields produced by currents and those produced by natural magnets. This exciting period of development of the experimentally based underpinnings of magnetostatics was one of the most interesting in the history of science and was not without controversy. 7 After the development of the special theory of relativity, it was 2 An
example is the work of Pierre de Maricourt, a native of Picardy [his Epistola Petri Peregrini de Maricourt de magnete was written in 1629], who experimented with lodestone and needles and introduced the concept of magnetic poles. Also notable was the work of William Gilbert [Gulielmi Gilberti, de Magnete, Magneticisque corporibus, et de magno magnete tellure, London, 1600], who was the personal physician of Queen Elizabeth. Gilbert studied the properties of magnets and realized that magnets set themselves in definite orientations with respect to the earth because the earth is itself a giant spherical magnet. He was thus the first to discover the earth's magnetic field. Gilbert was rather carried away with this new discovery, however, and conjectured that magnetic forces also accounted for the earth's gravity and the motions of the planets. 3 H. C. Oersted, Experiments on the Effect of Current of Electricity on the Magnetic Needle, privately distributed pamphlet dated July 21, 1820. English translation in Ann. Philos., 16, p. 273, 1820. 4 A.M. Ampere, Memoir on the mutual action of two electric currents, Ann. Chimie et Phys., 15, p. 59, 1820. 5 J.B. Biot and F. Savart, Ann. Chimie et Phys., 15, p. 222, 1820. 6 A.M. Ampere, On the mathematical theory of electrodynamic phenomena uniquely deduced from experiment, Mem. Acad., pp. 175388, 1825. About 50 years later, J. C. Maxwell described Ampere's work as being ''one of the most brilliant achievements in science." Also see L. P. Williams, AndreMarie Ampere, Scientific American, pp. 9097, January 1989. 7 For an excellent account of the history and a thorough treatment of underlying fundamentals see R. S. Elliott, Electromagnetics, IEEE Press, Piscataway, New Jersey, 1993.
Sec. 6.1
Ampere's Law of Force
417 8 law.
realized that magnetostatic theory may be derived from Coulomb's Nevertheless, in this text we adopt an experimentally based formulation of Maxwell's equations, so Ampere's law of magnetic force and the BiotSavart law of magnetic induction are considered quantitative statements of experimental facts. The study of magnetostatics constitutes the second major step in our quest to understand the foundations of the laws of electromagnetics. In magnetostatics we deal with magnetic fields produced by steady currents, which are themselves constant in time and therefore do not allow inductive coupling between circuits or the coupling between electric and magnetic fields. Yet, mastering the behavior of static magnetic fields and the techniques of solution of magnetostatic problems is essential to the understanding of the more complicated electromagnetic phenomena. Further1nore, many natural phenomena and the principles of some important industrial and technological applications are based in magnetostatics. In this connection, it suffices to say that magnetic recording was a more than $20 billion industry 9 in 1984. Today, annual sales of just hard disk drives, which record data by magnetizing a thin film of ferromagnetic material (see Section 6.8.3) on a disk, are still over $30 billion worldwide. Our coverage of magnetostatics in this chapter also brings us one step closer to a full understanding of the underlying physical basis of the transmission line behavior discussed in Chapters 2 and 3. An important physical property of a transmission line is its distributed inductance, which comes about because of the magnetic fields generated by the transmission line currents. In Chapters 2 and 3, we took it for granted that any twoconductor system has some inductance and we relied on for1nulas (in Table 2.2) to determine the distributed inductances of a few common transmission line structures. In this chapter, we define the physical basis of the concept of inductance and discuss how the inductance of different types of currentcarrying conductors can be determined using fundamental laws of magnetostatics.
6.1 AMPERE'S LAW OF FORCE Before we for1nally write down the magnetic force law, it is useful to review the new experimental facts in simple ter1ns. Figure 6.1 illustrates the direction of the force between currentcarrying wires as first experienced by Oersted, its dependence on the orientation of the wires, and the direction of the current flowing in them. The experimental facts indicate that two parallel wires carrying like (i.e., in the same direction) current attract 8 L.
Page, A derivation of the fundamental relations of electrodynamics from those of electrostatics, Am. J. Sci., 34, p. 57, 1912; see Section 4.2 of R. S. Elliott, Electromagnetics, IEEE Press, Piscataway, New Jersey, 1993, for a clear treatment. It should be realized, however, that such derivations implicitly make other assumptions in addition to using special relativity; see Section 12.2 of J. D. Jackson, Classical Electrodynamics, 2nd ed., Wiley, 1975. 9 See, for example, R. M. White (Ed.), Introduction to Magnetic Recording, IEEE Press, Piscataway, New Jersey, 1984.
418
The Static Magnetic Field
z. A
U1
z.
/
I I I I
/ / A
u 1 xR A
,x
/
. . . . . . . .Uz ... .....
/ / /
/
/
I I I I
/
/
12
/
ii2 x (u1 x R)
/
/
(a)
x
r
u2 x (u1 x R) = 0 /
/ A
/
/
Ul
11
/ /
/ / /
ii1 x ,R
A
U1
y
~
z•
Chap.6
~~
~~'\.,,,../ ~1
. A
/
A
Uz    ........y
U1XR
/
R , , ., ,,,..I
,x
,,,..,,,..
/
(b)
.A
/

z, Uz
.........
"""'
y
(c)
Figure 6.1 Magnetic force between straight currentcarrying wires. (a) Two infinitely long straight parallel wires carrying current in opposite directions repel one another. (b) Two straight parallel wires carrying current in the same direction attract. (c) When a small wire element carrying a current /2 is oriented perpendicular to the wire carrying current / 1, it feels no magnetic force, regardless of the direction of its current.
one another while those carrying opposite current repel, and that when a small 1° currentcarrying wire element is oriented perpendicular to another currentcarrying wire (see Figure 6. lc) it feels no magnetic force. This set of experimental facts is represented by expressing the force between two currentcarrying wires as a double cross product, 11 namely
where F 12 is the force exerted on wire 2 by wire 1, R is the vector from wire 1 to wire 2, fi. 1 and fi.2 are unit vectors along wires 1 and 2 in the direction of currents 11 and 12, respectively, and k is a proportionality constant. Note from Figure 6.la that the vector (11fi. 1 x R) is in the z direction, but the cross product of 02 x (fi. 1 x R) is in the y direction, indicating repulsion of the wires carrying oppositely directed current. Note that we could have just as well written an expression for the force F21 due to wire 2 at the location of wire 1, which points in the y direction. The magnitude of the magnetic force is experimentally determined to be inversely proportional to the square of the distance between wires. 10
The reason we consider a small element in Figure 6.lc is that a long wire oriented in the z direction and carrying current /2 would in fact experience a torque (see Section 6.10 on magnetic forces and torques). More precisely, the portion of the long wire above the xy plane would feel a force in the +x direction, while that below the x y plane would experience a force in the  x direction. 11The cross product of two vectors A and B is a vector, denoted by (A x B), with its magnitude equal to the product of the magnitudes of the two vectors times the sine of the angle '!frAB between them and its direction following that of the thumb of the right hand, when the fingers rotate from A to B through the angle '!frAB· Namely, A x B  filAI IB I sin lfrAB where fi is normal to both A and B and its direction is in the direction of advance of a righthanded screw as A is turned toward B. In rectangular coordinates, and noting that xx y = y x = x, and xx= y, we can use the distributive property of the cross product to write
z,
+ yAy + ZAz) x (XBx + yBy + zBz) i(AyBz AzBy) + y(AzBx AxBz) + z(AxBy
z
z
Ax B = (XAx Ax B =
AyBx)
The cross product is sometimes referred to as the vector product, since the result is a vector quantity.
Sec. 6.1
Ampere's Law of Force
419
Figure 6.2 Ampere's law of force. Two separate circuits C 1 and C2 , carrying currents of /1 and /2, respectively, exert a force on one another.
The simplest form of Coulomb's law (Section 4.2) dealt with forces between individual point charges. By analogy, we might expect that we should similarly consider forces between elementallength currentcarrying wires. In practice, however, steady currents must necessarily flow in complete circuits, so our fundamental experimental law must describe the total force between two complete circuits. Consider two idealized 12 complete circuits C1 and C2, consisting of two very thin conducting loops (wires) carrying filamentary currents of /1 and /2, respectively. With respect to an arbitrary origin, as shown in Figure 6.2, the position vectors describing points on the two loops are taken to be rl and r2 as indicated. We examine the force exerted on the circuit C2 by the circuit C 1, neglecting for the purposes of this discussion the forces between current elements within the same loop. The vector distance from an elemental length dl1 along C1 to another dl2 on C2 is thus (r2  rl) = R =RR, R is the unit vector 13 directed from dl1 to dl2, and R = IRI = lr2  rl I is the distance between the two current elements. With dl1 at rl = X.x1 + YY1 + zz1 and dl2 at r2 = ix2 + YY2 + zz2, we have R = [(x2  x1) 2 + (y2  y1) 2 + (z2  z1) 2]112, and R = (r2  rl)/ lr2  rl I = R/R. In his extensive experiments, Ampere found that the total vector force F 12 exerted on C2 by C 1 (both of which are located in free space) due to the mutual interaction of the currents /1 and /2 can be expressed 14 as A
µo F12= 4Jr
A
(6.1)
Equation (6.1) is referred to as Ampere's law of force and constitutes the foundation of magnetostatics. In MKS units, F 12 is measured in newtons, the currents / 1 and /2 in amperes, and the lengths dl 1, dl2, and R in meters. The proportionality constant is µo/ (4n) because of our use of MKS units and includes the 4n ''rationalization'' factor so 12The
circuits are idealized in the sense that the batteries that would have to be sustaining the currents are not shown. It is assumed that such batteries are some distance away and that the conducting leads from them are twisted closely together; one of Ampere's first experiments showed that two oppositely directed currents close together produced no effect on another current. 13 That is, from the ''source'' point to the ''field'' point or ''observation'' point, since we are aiming to write an expression for the force experienced by the circuit C2 due to the presence of C 1. A comparison between Figure 6.2 and Figure 4.3 illustrates the similarity between the fundamental force laws of magnetostatics and electrostatics. 141t may appear incredible that a formula with the generality implied in (6.1) could have been established from a few experiments on circuits of special and simple shapes as was done by Ampere. Indeed, (6.1) represents a generalization from results found in special arrangements of current loops; however, it should be noted that (6.1) continues to be valid (for steady currents), providing consistent results for every experiment involving a certain arrangement of loops that has been carried out since the time of Ampere.
420
The Static Magnetic Field
Chap.6
that a 4n factor does not appear in Maxwell's equations. In the MKS system of units, µo is defined to have precisely the value µ 0 = 4n x 107 henrys per meter~ 1.26 µHm 1 . This constant is called the permeability offree space; for practical purposes µo is also the per1neability of air. 15 From (6.1), we see that the dimensions of µ 0 are force(current) 2 , 2 2 2 or NA . Thus, we have 1 H = 1 NmA = 1 JA . Since the newton and the meter are determined independently, the above choice of the value of µo constitutes a definition of the unit of electric current or ampere (and hence also the unit electric charge or coulomb). Example 6.1: An extrahighvoltage de transmission line. An extrahighvoltage directcurrent (de) overhead transmission line consists of two very long parallel wires a = 10 m apart and each located h = 35 m above ground as shown in Figure 6.3. If the wires carry an equal current of I = 3000 A each flowing in opposite directions, find the repulsion force per unit length on each line. Neglect the effects of the ground plane. a
lz
y
I I
    Zz =
I I
a
@  ®
11
+ 1/2
x
z ~~ Fi2
lz
    Zz = 1/2
h Ground
(b)
(a)
Figure 6.3 Extrahighvoltage de transmission line. (a) Crosssectional view. (b) Top view.
Solution: The wires are assumed to be in the z direction, with /1 dl1 = z/ dz1 and /2 dl2 = z/ dz2 as shown. Using Ampere's law of force we can calculate the net force exerted by one of the wires on a unit length section of the other wire as 2
Fi2 =
µo/
4n
1/2 z2=1/2
00
,..
(dz2)z x (dz1z x R)
Zl =00
where R = J (z2  z 1) 2 + a 2 is the magnitude of R shown in Figure 6.3b and R = [z(z2 z 1) + ia]/R and z1 and z2 are the integration variables over wires 1 and 2, respectively. Note that F 12 is the force exerted on a 1m long segment of wire 2 (thus the limits of the z2 integral are from to + by the entire wire 1 (thus the limits of ±oo for the z1 integral).
i
15 The
i)
physical meaning of permeability and its dimensions will become clearer in Section 6.8, when we discuss magnetic materials and inductance. At this point we may note that permeability has the same significance for magnetostatics as permittivity has for electrostatics.
Sec. 6.1
Ampere's Law of Force
421
,..
Substituting for R and R, we have µo/ 2 F12 =  
4n
where we have used we find
112
00
(  dz2)z x
y
z2 =1 / 2
zx R=
a dz1
z1 =oo [(z2  z1)2
+ a2]3/ 2
ya[(z2  z1) 2 + a 2] 112. Performing the integration16 over z1 00
1/ 2
(  dz2)z x
y
z2=1 / 2
112
Zl =00
dz2
z2=1 / 2
"µo/2 =X  
2na
Using the numerical values given as I = 3000 A and a = 10 m, we find the repulsion force to be _1 ,. . (4n x 107 Hm 1)(3000 A) 2 F12 = x = 0.18 Nm 2n(l0 m) This lateral repulsion force can be compared with the weight of a line segment of 1 m length. Assuming the wires to have outside diameters of a few centimeters, their mass would be a few kg, and their weight a few tens of newtons. Thus, the perunitlength magnetostatic repulsion force is quite small compared with the perunitlength weight of the wires. However, it should be noted that the force between the two wires is proportional to the square of the current, so it can reach substantial values when much larger currents (tens to hundreds of kA) may flow due to accidental shorts (see Problem 6.3).
Ampere's force law is another example of a force law describing ''action at a distance," analogous in this sense to Coulomb's law for electrostatics. Just as it was useful to divide Coulomb's law into two parts by using the concept of an electric field as an inter1nediary to describe the interaction between charges, we can use Ampere's force law to define an appropriate field that may be regarded as the means by which currents exert forces on one another. The socalled 17 magnetostatic induction field or magnetic flux density B can be defined by rewriting (6.1) as follows: F12
(6.2)
= I,,_a
B12 16 This
is a rather common integral already encountered in Example 4.5. Using integral tables, or simply a change of variables, it can be shown that d~
17 For
~
J + const. b2 ~2 + b2
historical reasons, the term ''magnetic field'' is generally used for a different vector, which will be defined and discussed in Section 6.8. In retrospect, it would have been more proper to refer to B as the magnetic field. The terms ''magnetostatic induction field'' or ''magnetic flux density'' are in fact quite inappropriate descriptions of the nature of the B field (also see Section 6.2.2). For this reason, and whenever possible, we shall try to refer to B simply as the ''B field."
422
The Static Magnetic Field
Chap.6
where A
µo
/1 dl1 x R
4n
R2
B12= 
(6.3)
where B 12 is the B field at point r2 (i.e., at the location of dl2 at a distance R from dl 1) due to the current 11 in circuit C 1 . Equation (6.2) evaluates the force on circuit C2 in ter1ns of the interaction of its current 12 with the field B 12, which is set up by the current /1 in circuit C 1 . The currentfield interaction takes place over the entire circuit C2, while the field B12 depends only on the current and configuration of the circuit C1, which sets up the field. In the MKS system of units, the B field has units of18 tesla (T) or weber 2 per meter . Since one tesla is relatively large as a practical value, B is often also given in the CGS units of gauss (G), where 1 T = 104 G. For reference purposes, the earth's magnetic field on the surface is rv0.5 G, the B fields of per1nanent magnets range from a few to thousands of gauss, while the B field at the surface of neutron stars is believed to be 1012 Gauss. The most powerful permanent magnets, 19 such as samariumcobalt or neodymiumironboron magnets, have fields of 30004000 gauss, and several of them could easily lift an entire refrigerator. Worldrecord magnetic fields of nearly a million gauss have been achieved with pulsed electromagnets of novel designs. 2 Fields of such intensity represent a burst of energy comparable to an exploding stick of dynamite; the resultant forces imposed on the currentcarrying wire surpass the tensile strength of copper and the resistive losses generate enough heat to melt the copper wire. Table 6.1 provides additional examples of B field values in different applications, ranging from hightechnology applications such as superconducting quantum interference devices21 to relatively lowtechnology (but no less important) applications such as cow magnets. 22
°
18 Nikola
Tesla (18561943) was a brilliant electrical engineer who invented the induction motor and many other useful electromagnetic devices and made alternating current practical. See M. Cheney, Tesla: Man Out of Time, PrenticeHall, Englewood Cliffs, New Jersey, 1981. Wilhelm E. Weber was a professor of physics at Gottingen and a close colleague of J. K. F. Gauss. For most of his professional life, Weber worked in collaboration with Gauss on the study of magnetic phenomena. 19 P. Campbell, Permanent Magnet Materials and Their Design, Cambridge University Press, 1994. 20 G. Boebinger, A. Passner, and J. Bevk, Building worldrecord magnets, Scientific American, pp. 5866, June 1995. 21 Superconducting quantum interference devices (SQUIDs) are the basis for scanning SQUID microscopes, which can image magnetic fields at the surface of samples under study with an unprecedented sensitivity. SQUID imaging of fields as low as 10 11 gauss in a 1 cm2 area was reported by J. Clarke, SQUIDs, Scientific American, pp. 4653, August 1994. At the level of sensitivity of these devices, nearly everything is magnetic. Applications of SQUIDs range from diagnosis of brain tumors to tests of relativity. For further information, see J. Kirtley, Imaging magnetic fields, IEEE Spectr., pp. 40 48, December 1996. 22 The cow magnet is an alnico (AlNiCoFe) cylinder three inches long and a halfinch in diameter; it has rounded ends and is usually coated with plastic for protection against corrosion and breakage. Grazing cows are encouraged to swallow the cow magnet as a protection against the socalled ''hardware disease," which occurs when a grazing cow indiscriminately swallows sharp steel objects, such as bits of wire used to bale hay, which can cause damage to the walls of her intestines. The cow magnet remains in the cow's stomach, attracting to it any steel objects she later swallows, and can be retrieved and reused when the cow is slaughtered. For more on cow magnets and other interesting applications, see J. D. Livingston, Driving Force: The Natural Magic of Magnets, Harvard University Press, Cambridge, Massachusetts, 1996.
Sec. 6.1
Ampere's Law of Force TABLE 6.1
423
TYPICAL B FIELD VALUES IN SELECTED APPLICATIONS
B Field (Gauss)
Application
1011 108 106 104 10 3 0.5 10800 50100 100200 10002000 2000 30004000 10310 5 105 10510 6 108 1012
Sensitivity of a scanning SQUID microscope* Human brain Intergalactic and interstellar magnetic fields Human heart Earth's magnetic field Refrigerator memo magnets Electron beam of CRT (computer or TV) Magnetic read switch 1horsepower electric motor Cow magnets* Powerful permanent magnets* Magnetic resonance imaging (MRI) Highenergy particle accelerators Pulsed electromagnets* White dwarf stars Neutron stars *See references in text.
The vector field B can be calculated at any field point r using (6.3) even if there is no current element there to experience a magneto static force. When a currentcarrying wire is placed in a region per1neated by a B field, it experiences a magnetostatic force given by F = f Idl x B, as is evident from (6.2), which can be calculated as illustrated in Example 6.2. As noted in Section 4.3, the notion of a ''field," however abstract, not only is useful and convenient in thinking and working with actionatadistance phenomena but may also be regarded as an actual physical entity in its own right. 23 For our purposes in the rest of this chapter, it is more useful24 to continue our discussions of the implications of (6.1) using the socalled Biot Savart law, which quantifies the B field produced at a point by currents in its vicinity. Example 6.2: A semicircular loop in a B field. A semicircular loop of wire of 1 m diameter carrying a current of I = 10 A lies in a uniform B field of magnitude Bo = 1.5 T, as shown in Figure 6.4. Determine the total magnetostatic force experienced by the loop. Solution: loop is
Using (6.2), the total magnetostatic force on the straight portion of the current a
a
(/ dx)x x zBo = y/Bo
Fstr = a
dx a
= y2/Boa = y2(10 A)(l.5 T)(0.5 m) = y15 N 23 Which
it certainly was for M. Faraday; see Footnote 42 in Section 4.3. 24 Direct use of (6.1) to determine the force on a circuit is feasible only in relatively simple geometries. It is usually easier to determine the force on a particular circuit ( C2) through the concept of the B field produced by the current of another circuit (C 1).
424
The Static Magnetic Field
Chap.6
tY @
@
@ B
I
Fare @ Fstr
@
@
..x
I
@
@
@
Figure 6.4 A semicircular current loop in a B field. The B field is directed out of the page, as represented with a circle having a central dot.
Similarly, the total force experienced by the semicircular arc portion of the current loop is 7r'
7r'
" (la d µolr
4n ,..
+a z'=a [r2
dz'
=
if> µolr
+ (z')2]3/2
4n
z' r2Jr2 + (z')2
+a
z'=a
µala
=2 2nrJr2 + a For an infinitely long conductor, or at small distances from a finitelength conductor (i.e., r oo) in addition to being infinitely long (i.e., surface current flowing on the whole y = 0 plane), then the y components of the B field would cancel because of symmetry. Thus, the total B field in the x direction is • given as
y>O y> a), the B field
at the
,..µoNI zl ,..µoNI Z 21
Example 6.9: A solenoidtype magnetic filter. A magnetic filter involves a solenoidtype magnet used to separate very fine metallic particles from food products such as flours and com starch. The solenoid has a 50 cm diameter and 1 m length. If the solenoid has 1000 turns of uniformly wound wire with a current of 100 A flowing through it, calculate the magnetic field (a) at its center and (b) at the end of its axis. Assume an aircore solenoid.
Solution:
Using a= 25 cm and l = 1 m, we have
(a)
"(4n x 107 Hm 1 )(1000 tums)(lOO A) = z '.:::'. 0.112 T = 1120 gauss (1 m)v/ 1 + (0.5)2
Bctr
and (b)
Bend
_,.. (4n x 107 Hm 1)(1000 tums)(lOO A) ,....., O T_  z  .061  610 gauss (2 m)Jl + (0.25)2
That Bend is approximately onehalf of Bctr can be seen by considering a very long solenoid of length 21 and with 2N turns. The field at the center of this solenoid is µo(2N)I / (21), half of which is equal to Bend (by superposition, we expect the field to be halved when the contributions of the elemental loops from one half of the solenoid are removed). Thus, for a long solenoid, the B field at the ends of its axis is onehalf that of the field at its center. The variation of the B field along the axis of the solenoid between its two ends is shown in Figure 6.14, for two different values of the ratio of the solenoid radius to its length. We see that for a long solenoid (a /l = 0.1), the magnetic field is effectively confined to the region inside the solenoid and is nearly unifor1n except close to the edges. However, for a shorter solenoid (a/ l = 0.5), there is more leakage of the B field outside the solenoid, and also the field is not as uniform within the solenoid. It is interesting to note, however, that the B field at the end (i.e., at z = l /2) of the solenoid is approximately equal for both values of a/ l. In the foregoing discussion, we confined ourselves to the evaluation of the B field at points along the axis of the solenoid. For a long and slim solenoid (i.e., l >> a), the field is nearly unifor1n within the solenoid even at points off its axis, as we shall show in the next section using Ampere's law.
6.2.2 The Duality of the B and E Fields The B field is in many ways analogous to the electric field E that we worked with in Chapter 4, rather than to the electric flux density D. In this section we comment on
The BiotSavart Law and Its Applications
Sec. 6.2
1
437
a= O.ll
0.5 a= O.Sl
l
l/2
l
l/2
z
' ,,
'
2a B ,
'I
/
,
I I
Figure 6.14 The variation of the B field along the axis of a solenoid. Normalized magnitude of Bz (i.e., B (z) / B~ where B~ = µ.,oNI / l) is plotted versus z for two different ratios of radius versus length, namely a/l = 0.1 and 0.5.
this duality. The basic experimental fact of magnetostatics, as elucidated31 by Ampere, is that currentcarrying wires exert forces on one another. The BiotSavart law is one statement of this experimental fact, as discussed at the beginning of Section 6.2. In terms of Ampere's law of force, the magnetic force Fm felt by a current element Idl in the presence of a magnetic field B is Fm= I dl x B
Noting that current is charge in motion, a charge q moving at a velocity v is equivalent to an element of current I dl = qv, and hence, in the presence of a magnetic field B, would experience a force Fm given by (6.8) This force is called the Lorentz magnetic force, and (6.8) is sometimes 32 taken as the defining equation for B. In many ways, the quantity B is analogous to the electric field E, as can be seen by comparing the electric and magnetic force expressions given below:
Fe= qE = q V' 31 A.M.
p(r')R.dv' 2 4JTEoR
Ampere, On the mathematical theory of electrodynamic phenomena uniquely deduced from experiment, Mem. Acad., pp. 175388, 1825. 32The Lorentz force acting on a particle with charge q is the vector sum of the magnetic and electric Lorentz forces; that is, F = qE + qv x B. Also see Section 6.10.1.
438
The Static Magnetic Field
Chap.6
A
J(r') x Rdv'
Fm= qv x B = qv x V'
1 2 R
4nµ 0
We can see that both E and B act on a charge q to produce force, both are related to their respective sources (p for E and J for B) in a similar manner, and both are mediumdependent (through E01 and µ 0 ) quantities. In view of this similarity, it is unfortunate that B is usually referred to as the magnetic flux density, which conveys the incorrect impression 33 of a similarity between B and the electric flux density D. In this text, we may succumb to this convention and on occasion refer to B as the magnetic flux density, although in most cases we simply refer to it as the ''B field." '
6.3 AMPERE'S CIRCUITAL LAW In electrostatics, problems involving symmetries of one form or another can often be solved more easily using Gauss's law than by direct application of Coulomb's law. For static magnetic fields, the BiotSavart law and Ampere's force law are the point relationships that are analogous to Coulomb's law, whereas Ampere's circuital law serves the same purpose as Gauss's law. Ampere's circuital law, hereafter referred to simply as Ampere's law, is a mathematical consequence34 of the BiotSavart law, much as Gauss's law was that of Coulomb's law. Ampere's law can simply be stated as ,.
,.
0), then B reduces to the form obtained in Example 6.10.
Example 6.14: The infinitely long solenoid. In Section 6.2.1 we considered the magnetic field inside a tightly wound solenoid of finite length and derived an expression for the B field along its axis using the BiotSavart law. In this example, we consider a very long solenoid and find the B field inside it using Ampere's law. Solution: Consider the cross section of a closely wound solenoid as shown in Figure 6.19. For an infinitely long solenoid, the B field everywhere inside must be parallel to the axis
The Static Magnetic Field
444
Chap.6
B ~
B
c
b
l
(N turns)
e
,
,,
I I I I I I
I I I I I
~
tI tI
I I I I I I
I I I I I
~
~ =~
I I I I I
I
:;:::

I
~
,.,.  
~
~
'
I
I I
d
I
1~
I .. I ~ I I I __ _J __
...J
a

e
I
·~
I I I I I
c
d
f
f
(b)
(a) Figure 6.19
Cross section of a solenoid. 36
(i.e., B = zBz, with Br = 0 and B = 0) of the solenoid, and the field outside the solenoid must be zero. 37 Applying Ampere's law to a closed contour indicated in Figure 6.19 as abed, selected to be of length l and containing N turns of the solenoid coil, gives
+
B·dl= ab
+ be
+ cd
da
Note that the integral along cd is zero because B = 0 outside the solenoid, and the integrations along be and da are zero since B = zBz is perpendicular to di. Thus, we have
B ·di= µoNI ab
>
Bl= µoNI
>
B = µoNI l
which is identical to the result found in Section 6.2.1 for Bctr for the case of a very long solenoid. This result indicates also that the B field is constant across the cross section of the solenoid; note that we did not in any way restrict the location of segment ab in Figure 6.19 to be on the axis of the solenoid. In the context of the above formulation, it is easy to show that the z component of the B field at either end of a long solenoid would be half of that at its center. To see this, 36 The
reason for this is the fact that the integral of the B field over any closed surface is identically zero, as will be shown later in Section 6. 7. Considering an axial cylinder of unit length inside the solenoid as shown in Figure 6.19b, the integral :f B · ds over its surface is simply 2rrrBr, since the integrals over the two end faces cancel. Thus, Br must be zero in order for :f B · ds to be zero. Note that B = 0 because all of the current flow is in the cl"> direction and we know from the BiotSavart law that B must be perpendicular to the direction of current flow. 37 If the field outside was not zero, then the line integral of B around the contour cdfe would not be zero (as it should be, since it does not enclose any current) unless the field were the same along cd and ef (no contribution along the ce and df paths since Br = 0), which would in turn imply that a constant field exists everywhere outside the solenoid, contrary to experimental observation.
Ampere's Circuital Law
Sec. 6.3
445
we can cut a very long solenoid into two. If the same current I is maintained in the two parts, the magnetic flux density in the newly created ends must drop to half of its original values; otherwise the field would not combine to its original value when the two ends were reconnected. Note, however, that the radial component Br is not zero at the ends of a finite length solenoid.
Example 6.15: The toroid. If the long solenoid of Example 6.14 is bent into a circle and closed on itself, we obtain a toroidal (doughnutshaped) coil. A toroidal coil with a rectangular cross section is shown in Figure 6.20a. Find the B field inside the toroid.
b
I .. 1
I
(b)
(a)
Figure 6.20 A toroid. (a) Threedimensional view showing the windings. (b) Crosssectional view showing two Amperian contours C 1 and C2.
Solution: Let the toroid consist of N turns of wire carrying a current /, uniformly and tightly distributed around its circumference as shown in Figure 6.20a. Because of cylindrical symmetry, the magnetic field B has only a¢ component. We apply Ampere's law to any circular contour C1 of radius r such that a < r < b lying inside the toroid, as shown. By symmetry the field B must be the same everywhere along such a contour. Noting that the total current crossing through the area enclosed by the contour C1 is NI, we have (Bcf>~) · (~r d
B"' _ 1_lo_N_I .,,,  2nr
For any circular path C2 that lies outside the toroid (r > b ), the total net current crossing through the enclosed area of C2 is N (/  /) = 0, so that fc B · di = 0, yielding B = 0. Note that if we consider a contour with radius r 0) closed path lying in the yz plane. For the B field, we know from Ampere's law that the circulation of B along any closed contour C (i.e., line integral of B along C) is equal to the total current passing
Sec. 6.4
Curl of the Magnetic Field: Differential Form of Ampere's Law
447
through the surface area S enclosed by C times µo. For a differential surface element, we can assume that the current density is constant at all points over the surface, so the total current is IJI~s, if the surface is chosen to be orthogonal to the direction of the current flow so that the net circulation is maximized. Accordingly, we have curl B
= max
6.4.1 Curl in the Rectangular Coordinate System
We now utilize the definition of curl to obtain a convenient expression in ter1ns of partial derivatives. For this, we simply need to evaluate curl B at a general point in a given coordinate system. With reference to Figure 6.21, we can evaluate the component of curl B in the x direction by conducting a line integral along the path abcda, which, as shown, lies entirely in the yz plane. Note that this is an infinitesimally small path with dimensions ~y and ~z, so the magnitude of B at different sides can be simply related to the value of the field at the center point xo,yo,zo. From Ampere's law, we have B · dl = B1y~Y
+ B2z~Z 
B3y~Y
 B4z~Z = µolx~Y~Z
abcda
(B1y  B3y)
~y
+ (B2z I,,..
 B4z)
~z = µolx~Y~Z
....,,...,
(8Bz I ay) D..y
(8By /8z)D..z
z
By
Bx Liy
3
d
4
~
c 2
Liz
Lis= LiyL\z
/ /
a/ /
/
/
b
1
/
(curl B)x
Figure 6.21 Evaluation of the x component of curl B.
448
The Static Magnetic Field
Chap.6
from which (curl B)x =
.11m i abcda B · dl
~s+0
~s
aBz = ay
where the partial derivatives are to be evaluated at the center point (xo, yo, zo). By taking line integrals along other contours lying in the xz and xy planes, we can show that curl B =
x
+y
+z
in rectangular coordinates. Using the del operator we can express the curl of B as
I curl B = V x B = µoJ
(6.11)
J
which is the differential for1n of Ampere's law. A convenient method for remembering the expression for V x B in a rectangular coordinate system is to use the determinant for1n
VxB=
x"
y"
z"
a
a
a
ax
ay
az
Bx
By
Bz
Example 6.16: Circulation in a fluid. Further physical insight into the nature of the curl operation can be gained by considering a simple example. Consider the vector field representing the velocity on a water surface. Suppose this velocity is given by
v = y(vox) as shown in Figure 6.22. Find the curl of this vector field.
Solution:
Consider the closed contour marked abcda in Figure 6.22. We can easily see that there is a net circulation fabcda v · dl around this path, since the velocity field is stronger on side ab than it is on side cd, and there is no contribution to the line integral from the sides be or ad. The line integral of the velocity vector around the contour abcda is thus given by V ·
dl
abcda
=VO
X
+
~x
2
~y
vo x  2
since the velocity v does not vary with y. Since curl v =
~x
"s = zvo "
~s
~y
=
vo~x~y
= ~x ~y, and
or
s = z, we have
Vxv=zvo
Sec. 6.4
Curl of the Magnetic Field: Differential Form of Ampere's Law
y
t t t t t
t t t t
t t t t t t t t t t t t t c I.t t~ ~ _,I bf tLiyl: t t t t :I t I t t t T IP(x!y, z)j t j t t :t t t t : t t d '11:X rrI a t t t t t t t t t t t t t t t t
t t t t t t t t t
t t t t
449
v = y(vox)
x
Figure 6.22 Velocity field on a water surface.
which is the same result that one would obtain by using the differential expression, namely,
,..
Vxv=z
Bvy
ax
= z(vo
+ 0) =
zvo
We note that the velocity fl.ow itself is entirely in the y direction and does not bend or curve around in any way. However, if we placed an infinitesimal wooden chip with its length in the x direction at the point on the surface where we have evaluated the curl, the chip would tend to rotate, since the speed of the fl.ow is greater on side ab than on side ed. As viewed from above, the wooden chip rotates in the counterclockwise direction, which, according to our definitions is perpendicular to the plane of circulation and is in the direction defined by the fingers of the right hand when the thumb points in the direction of the curl vector (i.e., the righthand rule).
6.4.2 Curl in Other Coordinate Systems Although we use the notation V x A to indicate the curl of a vector A, expression of curl as a cross product with a vector del operator (having the form V = X.(a/ax) + y(a/ay) + z(a/az) for rectangular coordinates) is useful only in the rectangular coordinate system. In other coordinate systems, we still denote the curl of A by V x A, but note that the specific derivative expressions for each component of curl will need to be derived from the physical definition of curl as given in (6.10), using a differential surface element appropriate for that particular coordinate system. To derive the differential expression for curl in the spherical coordinate system, we consider the cuboid volume element shown in Figure 6.23b. To determine the r component of the curl, we consider the curvilinear contour ABCDA. Note that this contour is perpendicular to r, and the sense of rotation is related to r by the righthand rule if we go around the contour in the A > B > C > D > A direction. The lengths
The Static Magnetic Field
450
Chap.6
z z
.::::...... I I ;I I ._.',..I'...~/ I I A, ' 'l.. ,...i ... ) / \.f/ _..,I ,,, _.. ....... ,,,
       
_..

y
..::::
..:::
. . . . zd     \,... __ \ ... ...
.......
y
\
\
\
\

... ...... ...... \
x
(a)
(b)
Figure 6.23 Derivation of curl in other coordinate systems. (a) Cylindrical coordinates. (b) Spherical coordinates.
of the sides are AB= rd(), CD= rd(), BC= r sin(()+ d()) d, and DA= r sin() d. The line integral is given by A· dl = Aor d() +
3 A + A d()
3()
ABCDA
= 
Ao + 3Ao
r sin(() + d()) d
3Ao
d rd()  Ar sin() d 3 3 .
rd() d + () (A s1nB)r d() d 3 3
where we have retained small quantities up to second order. The area of the contour ABCDA is ~s = r 2 sin() d() d, so [V
X A]r
=
A· dl
J:
YABCDA ~S
1
= .r sin()
3
() (Ao sin()) 3
3A
3
The other two components can be obtained by considering the other sides of the cuboid in Figure 6.23b, namely, the contour ABFEA for the component and the contour BCGFB for the() component. The complete expression for the curl in spherical coordinates is 1 curl A = V x A = r  r sin() A
3
() (A sin()) 3
1 3Ar sin() 3
3
3Ao
3
 (rA)
3r
3 ( ) 3Ar  rAo  3r
3()
Sec. 6.4
Curl of the Magnetic Field: Differential Form of Ampere's Law
451
Similar reasoning in a cylindrical coordinate system using a volume element such as that shown in Figure 6.23a leads to the curl expression in cylindrical coordinates: curl A = V x A =
+
aA
r
az
Example 6.17: Curl of the B field due to an infinitely long cylindrical conductor. In Example 6.10, the B field of an infinitely long straight cylindrical currentcarrying conductor of radius a was found to be if> µolr r µol 2nr
r>a
Find V x B.
Solution: Since Bis expressed in cylindrical coordinates, we use the cylindrical coordinate expression for curl. Note that the only nonzero component of B is B, which varies only with r. Thus, the curl of B is 1 B
1
zr Br (rB) = zr
V x B=
B
+
BB"' r  o/ Br
so that for r < a, we have V x B=
1
zr
µolr µol  + r2na2 2na2
where J is the current density vector in the conductor given by J = for r > a, we have 1 µol µol V x B = zr =0 2nr2 r 2nr
z/ /(na 2 ).
Similarly,
as expected, since J = 0 for r > a.
Example 6.18: Curl of the electrostatic dipole field. As an example of a field that has both r and () components, consider the electric field of the electric dipole as derived in Section 4.4.3. For a dipole consisting of two opposite charges ±Q separated by a distance d, the electric field at faraway points (r >> d) was shown to be (equation (4.28)) E=
Qd
4nEor
" [r2cos8 + 0 sin()] 3
Find the curl of the electric field at faraway points.
Solution:
The curl of E in spherical coordinates is given by "1
V x E = r
B BEr Br (rEe)  B()
=
"1

r
2 sin () r3
2 sin ()
+
r3
=0
452
The Static Magnetic Field
Chap.6
Example 6.19: Curlfree fields. Show that the following vector fields, shown in Figure 6.24, are curlfree: (a) Ai = X.x + yY. (b) A2 = x sinx ycosy. (c) A3 = x sinx cosy+ ycosx siny. 2 2 (d) A4 = lr = x(x I + y 2) + y(y I + y 2 ). (e) As = r2r. (f) A6 = cl>r 1.
Solution: Since these vector fields are all two dimensional, they can have curl only in the z direction, given by BAy BAx [V x A]z =
ax
ay
in rectangular coordinates, and 1 [V x A]z = r
B(rAcf>)
BAr
Br
8 P(r, ¢, z = 0)
I
z' =a I I I I
Figure 6.30 Vector potential A around a wire of length 2a.
Solution: Taking the current I to flow in the z direction and noting that f 8 , J ds' = /, where S' is the crosssectional area of the wire, the vector potential A on the z = 0 plane is given by (6.15):
A= µo 4n 47 For
V'
µ0 J  dv' = R 4n
a
zl dz'
a
R
example, if a wire is looped around an infinitely long solenoid, Faraday's law (to be discussed in Chapter 7) requires that an electromotive force be induced on the loop if the current in the solenoid varies with time. We might then ask: How can a B field that never touches a wire have an effect on the charges in it? The apparent conceptual difficulty is resolved when we realize that the induced emf can be directly determined using A, which is nonzero.
Sec. 6.5
Vector Magnetic Potential
463
From the geometry of the problem (Figure 6.30), and since r' = zz', r = rr, we have R = + (z') 2 , so that the A field at a distance from the wire on the z = 0 plane • • 1s given as
Jr 2
r
A= ZAz =µal 4n
a
dz'
a
Jr2 + (z')2
,. . µal =Z 1n 2n
a
+ r
"'µal
=Z  
2n
ln
z'
r +
z' r
a
2
+1 0
a2
r
2+1
where the integration is carried out either by change of variables or by using integral tables. By using B = V x A in cylindrical coordinates, and noting that there is no dependence on ¢, we have
so that on the bisecting plane, where z = 0, we have
B =_µal a ln
2n Br
a
r +
a2
r
2+1
µal
a/r
2nr
Jl + (a/r)2
µal
2nr Jr2
a
+ a2
Note that this result is identical to that obtained in Example 6.3. We can observe that for r a, on the other hand, we have B varying as 1/ r 2 , consistent with the field of a very short current element, as is expected from the BiotSavart law.
Example 6.23: Field and potential at the center of a spherical shell. Consider a spherical shell "' of inner and outer radius a and b, respectively, and conducting a uniform current density J = lo (where lo is a constant), which is always tangent to circles of constant latitude, as shown in Figure 6.31. Find the vector potential A and the B field at the center of the sphere (i.e., at z = 0). Solution:
From (6.15), we have /J,Q
A(r = 0) = 
4n
J(r') dv' = µo v' R 4n
J(r') dv'
v' Ir' I
where V' is the volume of the shell, and the origin of the coordinate system is located at the center of the sphere that is, at the field point (in other words, r = 0). Note that the magnitude of J is uniform throughout the volume V' of the spherical shell. Each current density element J dv' at a given point on the shell produces a differential vector magnetic potential dA at the center of the sphere, which is in the same direction as the current element. However, in view of the spherical symmetry, for each such element there exists another one
464
The Static Magnetic Field
z
Chap.6
z
• a
J rI
// / / /
'x
•
    ...
y
// I
b
I I I I
(b)
(a)
Figure 6.31 A spherical currentconducting shell. (a) The density J is everywhere tangent to circles of constant latitude. The polar axis of the sphere is along the z axis. (b) Vertical crosssectional view along the polar axis.
located diametrically opposite and having oppositely directed current flow. Thus, the vector potential at the center of the sphere is A(r = 0) = 0. On the other hand, B(r = 0) does not vanish. From (6.6), we have
µo B(r = 0) = 4n
J_(_r'_)_x_R _" dv' R2
V'
"
,..
where J(r') = lo is the current flowing between r =a and r = b, and R is the unit vector pointing from a source point r' in the spherical shell to the origin (i.e., the observation point in this case) and is the inward,..normal to the spherical surface. From symmetry considerations, only the component of J x R that is parallel to the polar axis of the sphere contributes to the B field at r = 0. Noting that this component is proportional to sin()', where ()' is the spherical coordinate of the source point, we have
µo B(r = 0) = lo 4n
. ()' ,.. sin d , z v V'
R2
where R 2 = (r') 2 . Note also that because the spherical coordinate volume element is given as dv' = (r') 2 sin()' d()' d a, with I fixed, we approach an infinitely thin shell conducting a finite total current /, and the B field at the center is
µol B(r = 0) = z4a
Sec. 6.5
465
Vector Magnetic Potential
The spherical shell carrying a uniform current provides an interesting example of a configuration for which A= 0 but Bi= 0. This circumstance is not at all surprising when we realize that curl is essentially a vector differentiation; thus, the case in hand is analogous to a onedimensional function that crosses zero with a nonzero slope.
Example 6.24: Saddleshaped gradient coils. Magnetic resonance imaging (MRI) scanners use gradient coils to produce transverse gradients (nonuniformities) in the main field of the scanner in order for imaging to be possible. The magnetic fields of the gradient coils strengthen the main field (typically a constant field of a required strength and uniformity produced by superconducting or other types of coils or permanent magnets) in one region while weakening it in another, resulting in a total field whose strength varies in a continuous fashion in a desired direction. Two types48 of gradient head coils are shown in Figures 6.32a and b. Typically, the gradient fields are much weaker than the main field, so much less powerful electromagnets can be used. Saddleshaped coils are used in MRI scanners to produce transverse gradients. The optimum geometry49 of saddle coils consists of two semicircular loops of wire connected by straight wires. The semicircular portions are active in creating the gradient; the straight portions serve only to connect the loops and do not contribute. A pair of identical saddleshaped coils wound on the surface of a cylinder with a circular cross section oriented in the z direction is shown in Figure 6.32c. Find the B field at the center of the imaging system (i.e., at the origin of the coordinate system in Figure 6.32c).
Solution:
First we consider the B field due to the arcs of the two coils. The vector magnetic potential at a point along the axis due to an element of current on the circular arc is given by the integrand of (6.20):
dA = µol dl' 4n R " where di' = a d  ') integral is zero, so evaluating the second integral yields
=
sin'. Note that the first
2
A= if>µola sin8 4r 2
We can now find B from (6.14). Using spherical coordinates, since Ar = Ae = 0 and aA/3 = 0, it follows that ,... ,... B = rBr + 9Be where
51 See
1 a . µoia 2 cose Br = . (A sine) = r sine ae 2r 3
Section 4.4.3 for the electric dipole, where a similar expansion was shown in detail.
470
The Static Magnetic Field
Chap.6
and 1 Bo =  
a (rA ) =
r ar
2
µola sin() 
4r3
It is interesting to compare these expressions with those obtained for the electric field due to an electric dipole. For the electric dipole (from Section 4.4.3), we have "p cos() E =r 2rr Eor 3
+
sin() 4rr Eor 3
a" p
where p = Qd is the electric dipole moment. For the magnetic dipole we have "µola 2 cos() B =r 3
2r
,.. µola 2 sin() +9 3
4r
As these expressions are identical in for1n, we can associate the magnetic dipole with a dipole moment m such that m = z lm l = ZI (rra 2 ), where rra 2 is the total area of the circular loop. For a magnetic dipole with N turns, the magnetic dipole moment is 2 m = zNirra . The B field for the magnetic dipole can be rewritten as µo Im I cos () B =r 2rrr 3 A
+
a" µo Im I sin () 4rrr 3
(6.22)
Although (6.22) was derived for a circular loop, it is also valid for small, currentcarrying loops of other symmetric shapes (e.g., a square loop), with m = ZNIA, where A is the area of the loop. The similarity of the fields for the magnetic and electric dipoles is illustrated in Figure 6.34. The top panels show the nor1nalized electric and magnetic dipole field lines as given, by (4.28) and (6.22), respectively, which of course have identical shapes but are valid only at large distances from the dipoles. The bottom panels show the closeup views, illustrating that the electric field lines terminate on the two charges, whereas the magnetic field lines close on themselves. Note that magnetic dipole moment mis a vector, which has a magnitude lm l and a direction along the axis of the magnetic dipole loop determined by the direction of the current flow using the righthand rule. The vector A due to a magnetic dipole can be written in terms of its dipole moment m as A= µom x r = µo lm l sine 4rrr 2 4rrr 2 Noting that written as
1 vr
=
2 r/r
(see Footnote 40 in Section 6.5), this result can also be
A= µo 4rr
r" mxr2
µo 4rr
Vx m
r
Sec. 6.6
471
The Magnetic Dipole
~ /
 
 
'\
+QEB E
10
B
®I
Qe \
/
/
'
'
 


(b)
(a)
B
E
(d)
(c)
Figure 6.34 The field lines of electric and magnetic dipoles. (a) Field lines as defined by (4.28) for the E field at large distances from an electric dipole. (b) B field lines at large distances as given by (6.22) for a magnetic dipole. (c) Closeup view showing the termination of the E field on the charges. (d) Closeup view showing the selfclosure of the B field lines.
Using the above, we can also write an expression directly relating B and m. Using (6.14), we have B=
vx
A= µ
0
4n
v
x
v
x m
r
which can be simplified52 to the f or1n 1 r
52 Using
the fact that V2 (1/r) = 0 for r
=/=
0 and the identity
V x V x G  V(V · G)  V 2 G
(6.23)
472
The Static Magnetic Field
Chap.6
Example 6.25: Field of a long uniform solenoid. As an application of the magnetic dipole results derived above, consider the B field of a long uniform solenoid as shown in Figure 6.35, having N turns per meter and carrying a current I. The length of the solenoid is l, and its radius is a. Determine the B field at point P(x,y,z) such that r >>a, Ri >>a, and R2 >>a.
z

.........
(}
'
Ri
'
~
'
'\
//
///
' r
// /
/
/
//
/ /
//
l
 7,
/ R/ /
/
/
/
/
y
/
//
P(x,y,z)
i dz'/// R2 / /
/ /
2a
Figure 6.35 B field of a solenoid.
Solution:
We can consider the solenoid as consisting of many magnetic dipoles stacked on top of one another, and since r >>a and Ri, R2 >>a, we can use the magnetic dipole expressions derived above for distant observation points. For a segment of length dz', we have Ndz' turns, each with current/, so that the magnetic dipole moment of that differential segment is m = ZN/dz' n a 2 . Although we can first find the vector potential A and then use B = V x A to find the B field, we can also use the direct relationship between B and m as given above. Using the expression for the B field of a magnetic dipole shown in (6.23), the differential contribution to the field at P due to the differential segment of the solenoid is dB=
/J,Ov
4n
m·V
1
=
R
/LO
4n
V N/dz'na
2
z· V
1 
R
where R is the distance between the elemental magnetic dipole and point P as shown in Figure 6.35. Note that V(R) 1 = R/R 2 = R/R 3 , as derived in Footnote 40 in Section 6.5. To find the total B field at P we simply need to integrate over the length of the solenoid. In other words, l/2
v
l/2
Noting that R
1 R
· ZN/na 2 dz'
= Ir  r'I =ix+ yy + z(z  z'), we have B=
Nla 2µo 4 Nla 2µo 4
l/2 V
v
l/2 l/2 l/2
A
R I · zdz R2 (z  z') dz' [x2 + y2 + (z _ z')2]3/2
Sec. 6.7
Divergence of B, Magnetic Flux, and Inductance
=
N/a2
µo V[x2
+ Y2 + (z
_ z')2]t/2
4
473
112 l/2
2
Nla µo
1
112
R
_ 112
V
4
2 Nla µo V
4
1
1
Rt
R2
where Rt and R2 are, respectively, the distances to the observation point from the top and bottom of the solenoid. The evaluation of the gradient in the above expression for B is difficult in the general case, because of the complicated dependence of Rt and R2 on the spatial coordinates. However, when the point P is very far away from the solenoid (i.e., r >> l), it can be shown that R}t ~ rt[l  (l/2r) cos8] and R:;t ~ rt[l + (l/2r) cos8], so we have
B=
Nla 2µo
4
V
l cos8 r2
= µo (Nlna 2l)
4n
I,,_
V"
... 2 cos 8 r 3 r
il sin 8 3
+u
r
(6.24)
lms1ndl
We note that the above expression for B is identical to (6.22), if we replace lml with the dipole moment of the solenoid, namely Imslnd I = NI n a 2 1. Thus, as far as the field at large distances is concerned, the solenoid produces the same type of field structure as the elementary magnetic dipole, as if it had a total dipole magnetic moment of Nlna 2l.
6.7 DIVERGENCE OF B, MAGNETIC FLUX, AND INDUCTANCE As we discussed in Section 4.5, it is often advantageous to think of a vector field as representing the flow of something. Faraday's experiments on electric displacement were the basis for the concepts of electric flux and electric flux density D, which was shown (Section 4.6) to originate from or ter1ninate at sources (or electric charges), as described by the relation V · D = p. In this section, we first discuss the concept of magnetic flux and the divergence of the B field. In fact, in engineering practice we more often think in ter1ns of magnetic flux than electric flux. Much of our present view of magnetic flux has its origins in Faraday's visualizations of the magnetic force as stretching out in all directions from the electric current that produces it, filling all space as a magnetic field. Through this magnetic flux, which per1neates all space, seemingly isolated circuits are inductively coupled; hence, we next consider the important concept of inductance.
6. 7.1 Divergence of B and Magnetic Flux We saw in Section 6.5 that the magnetic field B can be derived from the curl of an auxiliary vector potential function A. This result leads at once to an important physical property for the B field. The divergence of the curl of any vector is identically equal to zero, and hence V · V x A = 0, from which it follows that the divergence of B is identically zero; that is, (6.25)
The Static Magnetic Field
474
Chap.6
This property of B is a mathematical consequence of the formulation of Ampere's experimentally based force law or its alternative statement in the BiotSavart law. In Section 6.8, we show that the effects of material bodies on the distribution of the magnetic field can be accounted for by their equivalent volume and surface magnetization currents, which also produce magnetic fields, as given by (6.6). Thus, even in the presence of material media it is possible to derive B from the curl of a vector potential A, and V · B = 0 is valid in general. As was discussed in Chapter 5 in connection with steady currents for which V · J = 0, the fact that V · B = 0 requires that the flux lines of B are always continuous and form closed loops. Thus, (6.25) implies that there are no sources of magnetic fields. (In this connection, ''source'' is used in a mathematical sense; literally, electric current is the source of magnetic fields.) In a mathematical sense, (6.25) means that there exist no magnetic charges (or free magnetic poles), corresponding to electric charges, from which B field lines can emerge or onto which they can terminate. If the divergence of B were not zero, magnetic field lines could originate at sources (at which V · B :j:. 0), just as electric field lines originate at charges (V · D = p ). However, at the present time, there is no experimental evidence whatsoever for the existence of free magnetic poles 53 or ''magnetic charge.'' On the contrary, all available experimental data can be interpreted on the basis of V · B = 0. All magnets have both a North and a South pole, and the field B is continuous through the magnet. For this reason the magnetostatic field B is fundamentally different from the electrostatic field E. Our discussion of the concepts of flux, flux tubes, and flux lines, given in Section 4.5 for electric fields, also applies to the B field. In thinking about the B field, it is often useful to consider it as representing the flow of something. It is in this context that the B field is often thought of as representing a magnetic flux density, although the implied analogy between Band the electric flux density Dis somewhat misleading, as discussed in Section 6.2.2. The magnetic flux through a differential area ds is given by the dot product of B with ds, namely B · ds. The dot product selects the nor1nal component of B through the surface ds, as shown in Figure 6.36a, so the total magnetic flux \II passing through the surface S is given by \II
53 The
=
s
B · ds
(6.26)
lack of magnetic poles or charges is the one outstanding asymmetry between electrostatics and magnetostatics. If magnetic poles did exist, say with a volume density Pm, the divergence of the B field would then be V · B = JLoPm· If the magnetic charges moved with a velocity v, the quantity Pm v would constitute a magnetic conduction current density Jm, which in turn would produce an electric field, just as electric conduction current density J produces a B field. P. A. M. Dirac [19021984] carried out an extensive theoretical investigation of magnetic poles and found no fundamental reason for isolated magnetic charges (magnetic monopoles) not to exist. However, he found that a quantum of magnetic charge would be so great that if pairs of opposite sign combined together, it would take highly energetic cosmic rays to separate them. See P. A. M. Dirac, The theory of magnetic poles, Phys. Rev., 74, p. 817, 1948. For a brief discussion of experimental attempts to detect magnetic monopoles, see pp. 136137 of M. Schwartz, Principles of Electrodynamics, Dover, 1972. Also see Sections 6.12 and 6.13 of J. D. Jackson, Classical Electrodynamics, Wiley, 2nd ed., New York, 1975.
Divergence of B, Magnetic Flux, and Inductance
Sec. 6.7
475
B B
c
'
'
Figure 6.36 Magnetic flux through a surface S. (a) An open surface S0 enclosed by a contour C and (b) a closed surface Sc enclosing a volume V .
(b)
(a)
in units of webers (Wb). The magnetic flux passing through the surface S bounded by the contour C is said to link the contour C and is commonly referred to as the flux linkage. For a closed surface S, however, as in Figure 6.36b, just as much flux leaves the surface as enters because of the continuous nature of the flux lines. Hence the integral of B · d s over a closed surface is always equal to zero. Mathematically, this result follows from (6.25) using the divergence theorem. Since V · B = 0, we have
v
V · B dv =
s
B · ds = 0
(6.27)
where the volume V is enclosed by the surface S. We have thus established the fact that the integral of the B field on any closed surface is zero, a fact that was utilized in Example 6.14 to deduce that the B field everywhere inside an infinitely long solenoid is directed along the axis of the solenoid. The magnetic flux that links a contour C may also be expressed in terms of the vector potential A. Since B = V x A, we have \II =
s
B · ds =
s
V x A· ds
The latter integral may be transfor1ned to a contour integral by using Stokes's theorem: \II =
s
V x A· ds =
c
A· dl
This latter integral is sometimes more convenient to evaluate than f s B · d s when determining the magnetic flux linked by a contour C. 6. 7 .2 Inductance
In Chapter 4 we pointed out that electrostatic problems are often difficult because the distribution of the electric charges is not known and is deter1nined by the configuration of metallic conductors, thus requiring the solution of Poisson's equation. The concept
The Static Magnetic Field
476
Chap.6
of capacitance was introduced as a measure of the distribution of the electric field in the vicinity of conductor configurations. In practice, calculation of magnetostatic fields is relatively simpler than that of electrostatic fields, because the magnetostatic problem usually involves a known distribution of currents, from which the B field can be found using the BiotSavart law. However, practical problems are nevertheless complicated by the magnetic properties of the surrounding medium. Fortunately, a detailed description of the B field is rarely desirable; more often than not, some overall measure of the field is sufficient. Inductance is this single measure of the distribution of the magnetic field near a currentcarrying conductor. Capacitance was introduced as a measure of the ability of a conductor configuration to hold charge per unit applied voltage, or store electrical energy, and was shown to be a property of the physical arrangement of the conductors. In a similar vein, inductance is another property of the physical layout of conductors and is a measure of the ability of a conductor configuration to link magnetic flux, or store magnetic energy. For our purposes, we define flux linkage as the integral of the magnetic field over the area enclosed by a closed circuit. Before defining inductance, we must introduce the concept of flux linkage. Consider two neighboring closed loops C1 and C2 as shown in Figure 6.37. If a current /1 flows around the closed loop C 1, a magnetic field B 1 is produced, and some of this magnetic field links C2 (i.e., will pass through the area S2 enclosed by C2). 54 This magnetic flux produced by the current I 1 flowing around C 1 but linked by the area S2 enclosed by C2 can be designated as
using the same subscript indexing as in previous cases, where the first subscript indicates the source of the quantity while the second indicates the location at which it is observed. In practice, inductances typically consist of coils with multiple turns. If C 1 consists of
B
Figure 6.37 Two magnetically coupled circuits. 54 We
choose to not give a precise definition of the term ''links," because intuition gives a better feel for this particular concept than any formal definition. Linking of one circuit by another can be thought of as the magnetic flux produced by one circuit ''passing through'' or connecting with another circuit.
Sec. 6.7
Divergence of B, Magnetic Flux, and Inductance
477
multiple turns Ni, then the total flux produced is N i times larger (see Section 6.2.1); namely, Ai2 =Ni \lli2. The mutual inductance Li2 between the two coils is thus defined as
Noting that in general C2 has N1 turns, the magnetic flux produced by C i is thus linked N1 times by C2. Some of the magnetic flux produced by Ii links the area Si enclosed by the closed contour Ci, so that we can also define the selfinductance of loop Ci as
where Aii =Ni \llii is the the total flux linked by a single turn of Ci, and \llii is the magnetic flux produced by a single turn of Ci and linked by a single turn of Ci. Note that if a current /2 flows around the closed loop C2, it generates a magnetic flux that links Ci, so that the mutual inductance between the two loops is given by NiN2 \112i
L2i =   
12
where \112i is the flux generated by the current /2 flowing around a single turn of C2 and linking the area enclosed by a single turn of Ci. To show that L i2 = L2i , we can rely on the representation of the B field by a vector potential. Consider the expression for the mutual inductance Li2:
where Bi is the total field produced by C i (with Ni turns) at the surface S2 enclosed by C2 (with N1 turns). Noting that Bi = V x A i we can rewrite Li2 as
where we used Stokes's theorem. The vector potential Ai is related to its source current Ii through (6.15), so we have A i = µoNili
4n
dli
c1 R
where R = Ir  r' I, with rand r' being the positions of the observation and source points, respectively. We can thus write (6.28)
478
The Static Magnetic Field
Chap.6
which indicates that L12 = L2 1, since the dot product is commutative and the line integrals can be interchanged. Equation (6.28) is known as the Neumann formula for mutual inductance. It underscores the fact that the mutual inductance is only a function of the geometrical arrangement of the conductors. We shall see in Section 6.8 that if the medium the circuits are located in is a magnetic material, the constant µo needs to be replaced by another constant to be defined as magnetic per1neability µ. Thus, inductance also depends on the magnetic properties of the medium that the circuits are located in, as represented by its per1neability µ. In general, carrying out the double line integral is quite involved. In most cases, we take advantage of the symmetries inherent in the problem to determine the flux linkage or stored magnetic energy so that we can find the mutual inductance without resorting to (6.28). In some cases, however, it actually is easier to use (6.28); we shall use this for1nula in Example 6.31 to deter1nine the selfinductance of a currentcarrying loop. For any given conductor configuration, the selfinductance of the closed loop C can be evaluated in a manner very similar to the procedure used to find the capacitance in electrostatics. We can first assume a current I to flow in the closed loop, from which we can determine B using Ampere's law or the BiotSavart law or the vector potential A. This magnetic field is proportional to the current I. We can then find the flux linked by the circuit by conducting an integral, namely \II=
s
B · ds
where S is the area enclosed by the closed loop C. Given that the number of turns in the loop is N, and that all the turns produce and link the same flux, the selfinductance is then given by L = (N 2 \IJ) /I. Note that I will cancel out so that L depends only on the geometrical arrangement (i.e., shape and dimensions) of the circuit C and the magnetic properties of the medium as mentioned in the preceding paragraph. Example 6.26: Selfinductance of a long solenoid. Find the self inductance of a solenoidal coil of length l, radius a, and total number of turns equal to N. Solution: In Section 6.2.1, it was shown that for a long solenoid the intensity of the B field at the end points of the axis is half that at the center because of the flux leakage near the ends. However, this leakage is mainly confined to the ends of the solenoid and can be neglected. So, a good approximation is to assume that the B field is constant over the entire interior of the solenoid, being equal to its value at the center given by the result derived in Example 6.14, namely
B
~
Bctr
~
,..µoNI z l
As a result, the total flux linkage of every individual turn of the solenoidal coil is
Divergence of B, Magnetic Flux, and Inductance
Sec. 6.7
479
where A is the crosssectional area of the solenoid (equal to A = n a 2 for a solenoid with a circular cross section of radius a). Since there are N turns, the total flux linkage of all N turns is A = N\11 = µ0N2/A l Thus, the selfinductance of a long solenoid is
2 L= A= µoN A I
l
To get a feel about the orders of magnitude involved, let us calculate the selfinductance of a solenoid of 100 turns wound uniformly over a cylindrical wooden core(µ= µo) of length 10 cm and diameter 1 cm. Substituting these values yields
L =
(4n x 107 Hm 1) x (100) 2 x n(0.5 x 102) 2 m 2 10 x
102
~
m
9.87 µH
Using a linear ferromagnetic core material (Section 6.8) with µ = 1000µ,o instead of the wooden core increases the selfinductance of the 100turn solenoid by a factor of 1000, yielding L ~ 9.87 mH.
Example 6.27: Inductance of a toroid. Find the inductance of a toroidal coil having N turns, similar to that shown in Figure 6.20 but having a circular cross section. For a toroidal coil, we assume55 that the mean radius of the toroid is much greater than the diameter of the coil; that is, rm = (a + b) /2 >> b  a. With this assumption, the magnetic field B is approximately uniform throughout the inner part of the toroid (also referred to as the core) and is given by
Solution:
a 0.8a is 59 L
rv
10µ on:a 2N 2
9a
+ 101
Inductance of a single loop of wire. The extreme case of a short solenoid is a singleconductor loop of arbitrary shape. In general, in evaluating the selfinductance of such a loop, we have to account for the internal and external inductances of the wire separately. For most circuits the total magnetic flux generated by a current can be partitioned into the portion lying outside the conductor plus the flux that lies wholly inside the conductor. The storage of magnetic energy and flux linkages associated with the internal flux lead to an internal inductance, while those associated with the fluxes outside the conductor are represented by the external inductance. Up to now, we have implicitly assumed circuits consisting of filamentary currents, thus neglecting internal inductance. For most highfrequency applications, internal inductance is negligible, since the magnetic fields (and thus magnetic fluxes) are confined to a very thin region on the surface of the currentcarrying conductors. Determination of internal inductance using the linking of magnetic flux requires the introduction of the concept of partial flux linkages and unnecessarily complicated analysis. 60 It is generally much easier to evaluate the internal inductance by considering the magnetic energy stored inside the conductor, as will be done in Example 6.32, after we briefly introduce the concept of magnetic energy. We show in Example 6.32 that the de internal inductance per unit length of an infinitely long wire (or the inner conductor of a coaxial line) is µ / (8n:), where µ is the magnetic permeability (defined in Section 6.8) of the material the wire is made of. It is shown in Section 6.8 thatµ r v µo for most materials, including highly conducting metals such as copper and aluminum, and that µ is substantially different from µo only for magnetic materials such as iron. Thus, the internal selfinductance per unit length of a wire is in most cases simply equal to µo/(8n:). For a thin wire of total length l bent into an arbitrary loop as shown in Figure 6.42, the magnetic field near the surface is very nearly the same as that for an infinitely long
59 H.
A. Wheeler, Simple inductance formulas for radio coils, Proc. I. R. E., 16, pp. 13981400, 1928. 6°For examples of partial flux linkage analyses, see Section 511 of C. T. A. Johnk, Engineering Electromagnetic Fields and Waves, John Wiley & Sons, New York, 1988, and Section 8.5 of R. Plonsey and R. E. Collin, Principles and Applications of Electromagnetic Fields, McGrawHill, New York, 1961.
486
The Static Magnetic Field
, ' .... ,,,/
;
~
... ........
Chap.6
' ',
dl2
I
\
I
\
I I
'
'11 I
C2
\ 1 1
Ri2
\
''
\
II
}
, ,' __ ,
' , ____ '
/
Figure 6.42 A conductor of finite radius d bent into a closed loop.
wire, provided the radius of curvature of the wire is much greater than the radius of the wire at all points. In other words, we may treat the wire locally as though it were part of an infinitely long wire. Thus, the internal selfinductance of any loop of mean length l 61 is µol/(8n). Thus, the internal selfinductance of a thin wire (d
The magnetic dipole moment of this unperturbed orbit is the current times the area. The current is the charge per unit time that passes any point on the orbit, or simply the charge 65 For
further discussion, see R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics, AddisonWesley, Reading, Massachusetts, 1964. 66 See Section 3435 of The Feynman Lectures on Physics, AddisonWesley, Reading, Massachusetts, 1964.
Sec. 6.8
493
Magnetic Fields in Material Media B0
z•
z.
Bo
I I
(_J_) w electron
v electron ~r..;...;;:;;;.n
qe
Figure 6.44 Perturbation of electron orbit by applied B field. The nuclei are positively charged by an amount qe '.'.: : '. 1.6 x 10 19 C. (a) No external B field. (b) The
I I
qe,At ,,..,,,..,, /
•
presence of the external B field produces a magnetic force Fm.
(b)
(a)
qe divided by the time taken for one revolution, or 2n /wo. The unperturbed magnetic dipole moment of the electron orbit is thus ,...
2
,...
mu = ZI (na ) = zqe
WO
2n
2
,... q e Woa
2
na = z   
2
19
Since the charge of the electron qe = 1.6 x 10 C, the current flow is in the direction opposite to that of the electron, so the dipole magnetic moment is oriented in the z direction. In the presence of the external field B = zBo, the electron motion is influenced by an additional force given by Fm = q e v x B, where v is the linear velocity of the electron, whose magnitude is given by lvl = wr, where w is the new angular velocity and r is the new orbital radius, both of which may be different from their values in the absence of the external field. The force balance equation that is now required for the electron to orbit is
Assuming 67 that the orbital radius remains constant, so that we haver~ a, we have
Since we expect the applied field B = ZBo to produce only a small perturbation (i.e., (w  wo) > l) points. While (6.24) is based on the dipole approximation (i.e., a distant observation point, or r >>a and r >> l), it is possible to use the magnetization current sheet model of the permanent magnet to evaluate the B field everywhere, both inside and outside the magnet. Methods for such evaluations are described elsewhere. 70 However, an approximate sketch of the magnetic field lines around a cylindrical bar magnet is shown in Figure 6.46b. Note the similarity between the B field distribution for a bar magnet and that for a finite length solenoid, shown in Figure 6.41a.
Just as we differentiated between the free charge density and the bound polarization charge density Pp in Chapter 4, we need to separately account for a free current density (due to the net transport of charge carriers across a given crosssectional area) and the bound magnetization current density given by (6.37). In Chapter 5, we considered one type of free current density: the conduction current density, which is sustained by an electric field in a conducting material through J = a E. However, conduction current is 7°For
a comparative discussion of the B fields for a cylindrical rod magnet and a finitelength solenoid, see Section 9.4 of M. A. Plonus, Applied Electromagnetics, McGrawHill, 1978. For a discussion of the formal procedure used to determine the fields around a permanent magnet, see Section 6.6 of J. Van Bladel, Electromagnetic Fields, Revised Publishing, Hemisphere Publishing Corp., New York, 1985.
Sec. 6.8
499
Magnetic Fields in Material Media
not the only mechanism through which we can have free current. For example, one could imagine a cloud of suspended electrons blowing in the wind. This current may not be driven in accordance with Ohm's law, but it nevertheless represents a net movement of charges, and so can be defined a free current density. We now consider the B field inside (i.e., Jsm = 0) a magnetic material body in which a free current distribution J exists. This current produces a macroscopic B field, which magnetizes the material, resulting in the magnetization current density Jm given by (6.37). Since the total effective current density is then the vector sum of J and Jm, we can write the differential for1n of Ampere's law as V x B = µo(J + Jm)
Noting that we have Jm = V x M, we can rewrite this as Vx
8 µo
M
=J
Therefore, the vector [ (B / µo)  M] has as its source only the free current J. Thus, to eliminate the necessity of dealing with the magnetization vector M, we introduce a new vector H, defined as (6.39) The vector H is called the magnetic field intensity vector, and it is directly analogous to the electric flux density vector D in electrostatics in that both D and H are mediumindependent and are directly related to their sources. The dimensions of H are amperes per meter. We can also express Ampere's law in ter1ns of Has follows:
c
H · dl =
s
J · ds =I
or
VxH=J
(6.40)
where I is the total net free current passing through the surface S enclosed by C. Note that the mediumindependent nature of His once again apparent. Taking the divergence of (6.39), we obtain V·H=
1 µo
V·BV·M=V·M
since V · B = 0. Note that V · H =  V · M is in general nonzero and can be thought to correspond mathematically to an equivalent magnetic charge. For most materials other than those which are ferromagnetic, the magnetization M is directly proportional to the applied external field B, so that, based on (6.39), M is also directly proportional to H. It is thus customary to write M=xmH
(6.41)
500
The Static Magnetic Field
Chap.6
where Xm is the dimensionless proportionality constant called the magnetic susceptibility of the material that we introduced in (6.35). The value of Xm provides a measure of how susceptible a material is to becoming magnetized by an applied magnetic field. Xm is negative for diamagnetic materials and positive for paramagnetic materials. For ferromagnetic materials, Xm in general depends on the applied magnetic field and the past history of magnetization of the material. Substituting (6.41) into (6.39), we find (6.42) where the quantity µ = µo(l + Xm) = µoµr is called the magnetic permeability and is entirely analogous to the electric per1nittivity E in electrostatics. Note that µr is the relative per1neability, analogous to Er in electrostatics. Just like electric permittivity E, the per1neability µ is a macroscopic parameter71 that is used to account for the microscopic effects of material bodies in the presence of external magnetic fields. In practice, the per1neability µfor a given material can be determined experimentally. When its value is known, µ can be used in (6.42) to relate B and H directly, thus eliminating the necessity of taking the magnetization M into account explicitly. It is important to note that (6.42) only holds for a class of materials and under certain conditions. In particular, since (6.42) originates in (6.41), it depends primarily on the linear relationship between M and H. Magnetic materials, especially ferromagnetic materials discussed in the next section, often exhibit per1nanent magnetization and highly nonlinear behavior, in which case (6.41) (and thus (6.42)) is not valid. Equation (6.42) also requires that the material be isotropic that is, that Xm be independent of the direction of B. In anisotropic materials, such as ferrites (see Section 6.8.3), a B field in one direction can produce magnetization (and thus M) in another direction. Accordingly, the relation between B and H must be expressed72 as a type of matrix, more commonly called a tensor. We recall from Chapter 4 that the introduction of the electric flux density vector D enabled us to simplify Gauss's law by focusing on the free charge density. Similarly, introduction of the quantity H, together with the per1neability µ, allows us to simplify Ampere's law by using only the free (e.g., conduction) current density. The BiotSavart law (6.6) and Ampere's law as stated in (6.9) both depend on the total current density, which in general consists of the free current density J and the magnetization current density Jm. However, in calculating the H field through (6.40), we only need to account for the free current, which in practice is usually externally driven (such as a current flowing through a wire). While the magnetic field intensity vector H is analogous to electric flux density vector D, in practice H is used much more often then D. In a laboratory, we typically 71 We
should note here that in the interior of any material the values of M and B may vary rapidly with both time and space. However, in any practical problem that involves the measurement of fields, the field is sampled over regions of space much larger than atomic dimensions. Thus, the quantities B, M, and H in the above expressions are all macroscopic quantities. 72 See Section 13.12 of S. Ramo, J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics, 3rd ed., John Wiley & Sons, Inc., New York, 1994.
Sec. 6.8
Magnetic Fields in Material Media
501
control conduction currents through wires and voltages on metal conductors. Thus, by reading our equipment dials we can directly calculate the line integrals of the electric field E using (4.19) and the H field using (6.40). If instead of controlling voltages, there existed an easier means to place a known amount of charge on each conductor, then D instead of E would be the more widely used quantity in electrostatics. In retrospect, it is quite unfortunate that the duality of E and B (see Section 6.2.2) and that of D and H were not appreciated during the early development of the theory of electromagnetics, since the use of the reciprocal of µo in relating H to B would have been more appropriate. Also, the common references to H as the ''magnetic field strength'' and to B as the ''magnetic flux density'' are rather misleading. 73 For paramagnetic and diamagnetic materials the value of µ differs only insignificantly from µo. For one of the strongest diamagnetic materials, bismuth, we have µ = 0.999833µ0, whereas for the paramagnetic tungsten µ = 1.000078µ0. The per1neability of air is µ = 1.00000037µ0. For copper and water, we have µ = 0.999991µ0. Thus, it is very common to assume the relative permeability µr = 1 for diamagnetic and paramagnetic materials. However, for ferromagnetic materials, discussed in the next section, µr is in general much larger than unity and in some special cases can be as large as 106 ! The relative per1neability (µr) values for selected materials are given in Table 6.2.
6.8.3 Ferromagnetic Materials The value of µ is generally much larger than µo for ferromagnetic materials, which exhibit large magnetizations even in the presence of very weak magnetic fields. Only the three elements iron, nickel, and cobalt are ferromagnetic at room temperature and above. Almost all ferromagnetic alloys and compounds contain one or more of these three elements or manganese, which belongs to the same group of transition elements in the periodic table. The ease of magnetization of these materials results from quantum mechanical effects that tend to align neighboring atomic spins parallel to each other even in the absence of an applied magnetic field. In ferromagnetic materials the diamagnetic and paramagnetic effects also occur, but the contributions of these effects to the total magnetization is negligible even in relatively large fields. The per1nanent magnetization of cobalt, 73 The
inappropriateness of the term ''magnetic field strength'' for H is discussed in Section 2 of A. Sommerfeld, Electrodynamics, Academic Press, 1952, reproduced here with permission of Academic Press, Inc. In Sommerfeld' s words (p. 11): ''We may indicate finally a subdivision of physical entities into entities of intensity and entities of quantity. E and B belong to the first class, D and H, to the second. The entities of the first class are answers to the question 'how strong,' those of the second class, to the question 'how much.' In the theory of elasticity, for example, the stress is an entity of intensity, the corresponding strain, one of quantity; in the theory of gases, pressure and volume form a corresponding pair of entities. In D the quantity character is clearly evident as the quantity of electricity (i.e., flux) that has passed through; in H the situation is slightly obscured by the fact there are no isolated magnetic poles. We are in general inclined to regard the entities of intensity as cause, the corresponding entities of quantity as their effect." On p. 45 of his text, Sommerfeld further states, ''The unhappy term 'magnetic field' for H should be avoided as far as possible. It seems to us that this term has led into error none less than Maxwell himself." For an interesting discussion of the question of the relative physical merits of B versus Hin connection with permanent magnets, see pp. 136139 of M. Abraham and R. Becker, Electricity and Magnetism, Blackie & Son Limited, Glasgow, 1944.
502
The Static Magnetic Field TABLE 6.2
Chap.6
RELATIVE PERMEABILITY OF SELECTED MATERIALS
Material
Relative Permeability (µr)
Air Aluminum Bismuth Cobalt Copper Iron (Purified: 99.96% Fe) Iron (Motor grade: 99.6%) Lead Manganese Manganesezinc ferrite Mercury Nickel Nickelzinc ferrite Oxygen Palladium Permalloy: 78.5% Ni, 21.5% Fe Platinum Silver Supermalloy: 79% Ni, 15% Fe, 5% Mo, 0.5% Mn Tungsten Water
1.00000037 1.000021 0.999833 250. 0.9999906 280,000. 5,000. 0.9999831 1.001 1,200. 0.999968 600. 650. 1.000002 1.0008 70,000. 1.0003 0.9999736 1,000,000. 1.00008 0.9999912
nickel, and iron leads to the magnetic permeabilities of µc 0 = 250µ0, µNi = 600µ0, and µFe = 6000µ0. For purified iron,µ ranges from 10,000µo to 200,000µo; for supermalloy (79% Ni, 15% Fe, 5% Mo, 0.5% Mn)µ can be up to rvl06 µo. Typically, the value ofµ for a ferromagnetic material is not unique, because of strong nonlinearities. In practice, the relation between B and H for ferromagnetic materials is represented graphically in the for1n of a curve known as the hysteresis curve. The magnetic properties of ferromagnetic materials are strongly temperaturedependent. In the absence of an applied field, they exhibit spontaneous magnetization below a temperature Tc, known as the Curie temperature, and are strongly paramagnetic above that temperature, susceptibilities (Xm) decreasing with increasing temperature. The Curie temperatures for iron, nickel, and cobalt are 1043 K, 631 K, and 1393 K, respectively. Some elements that are neighbors of the ferromagnetic elements on the periodic table, such as chromium and manganese, also have strong quantum mechanical coupling forces between the atomic dipole moments. However, for these elements the coupling leads to antiparallel alignment of electron spins between adjacent atoms, so the net magnetic moment is zero. Elements with this property are called antiferromagnetic. Antiferromagnetism is also strongly temperaturedependent. At temperatures above the Curie temperature, the spin directions suddenly become random, and the material exhibits paramagnetic properties.
Sec. 6.8
Magnetic Fields in Material Media
503
A number of oxides containing iron, nickel, or cobalt exhibit a magnetic behavior between ferromagnetism and antiferromagnetism and are called ferrimagnetic materials. The magnetic moments of these molecules alternate from atom to atom in an unequal manner, resulting in a net magnetic moment, but one that is much smaller than those of ferromagnetic materials. The most common ferrimagnetic materials are Fe304 and the family of ferrites described by the chemical formula KO· Fe203, where K is any divalent metal such as Fe, Co, Ni, Mn, Mg, Cu, Zn, Cd, or a mixture of these. These are ceramiclike compounds with very low conductivities (for ferrites a = 104 to 1 Sm 1 as opposed to 107 Sm 1 for ferromagnetic Fe) and thus exhibit very low eddy current losses at high frequencies. Thus, ferrites are commonly used in highfrequency and microwave applications as cores for FM antennas, transfor1ners, and phase shifters. An important property of ferrites is that they are anisotropic in the presence of magnetic fields, meaning that the H and B vectors are in different directions. Microwave devices that utilize the anisotropy of ferrites include isolators, gyrators, and some directional couplers. Further discussion of ferromagnetic materials is beyond the scope of this book. 74
6.8.4 The Meissner Effect in Superconductors
It was mentioned in Section 5 .1 that some materials become superconducting below a critical temperature, with their conductivity becoming nearly infinite (> 1020 ). This phenomenon is accompanied by a type of repulsion or exclusion of the B field from the interior of the material. In effect, the material becomes perfectly diamagnetic, with B = 0 within the metal independent of the existence and nature of current carriers in the neighborhood of the metal, as if its per1neability µ = 0. This phenomenon was discovered in 1933 by W. Meissner and R. Ochsenfeld and is known as the Meissner effect. The description of the Meissner effect as a complete repulsion of the B field is an oversimplification. In fact, for a metal below the critical temperature the B field penetrates into the surfaces to optical distances, 50200 nm. This result is true for de fields as well as timevarying fields and may be understood by modeling the superconductor as a dense collisionless plasma. 75 The Meissner effect completes our understanding of a ''perfect'' conductor as one in which no static or timevarying electric or magnetic fields exist. A perfect conductor is usually understood to be one in which no electric field (at least none that varies more slowly than the relaxation time rr, which is typically of order 19 s for metallic conductors) can exist, since otherwise the abundantly available rv 10free electrons would be in perpetual motion. Maxwell's equations imply that there must then be no timevarying magnetic fields inside a perfect conductor either. However, a
74 For
an excellent and relatively concise discussion at an appropriate level, see Chapters 9 and 10 of M.A. Plonus, Applied Electromagnetics, McGrawHill, New York, 1978. 75 T. Van Duzer and C. W. Turner, Principles of Superconductive Devices and Circuits, Elsevier, New York, 1981. Plasma is the fourth state of matter, consisting of an ionized collection of positive and negatively charged particles. Plasmas are inherently diamagnetic, since motions of both positively charged and negatively charged particles tend to reduce the magnetic field, as illustrated for a single electron in Figure 6.44.
The Static Magnetic Field
504
Chap.6
truly static magnetic field should in principle be completely uncoupled from the electric field and should not be affected by the conductivity of the material, even if a = oo. To the degree that superconductors are the physical materials that can be viewed most nearly as ''perfect'' conductors, the Meissner effect ensures that the interiors of perfect conductors are completely free of electric or magnetic fields. 76
6.9 BOUNDARY CONDITIONS FOR MAGNETOSTATIC FIELDS In some magnetostatic problems we deal with interfaces between two or more different types of materials. The manner in which B or H behaves across such interfaces is described by the boundary conditions. The boundary conditions are derived from fundamental laws of magnetostatics, which we reiterate below. Starting with the experimental fact that electric currents create magnetic fields, as expressed by the BiotSavart law, we have derived the following fundamental differential and integral laws of magnetostatics:
VxH=J
c s
H · dl =
s
J · ds
B · ds = 0
Ampere's law No magnetic charges
Note that Ampere's law was first derived in ter1ns of B and used as such during most of this chapter. However, on the basis of our discussions in Section 6.8, we now understand that the more general version of Ampere's law relates the H field directly to the free current sources (i.e., J), independent of the material media. The magnetic properties of different physical media are accounted for via the permeability µ of the material. For linear (magnetization M proportional to B) and isotropic (magnetization Min the same direction as B) media, we have or
B=µH
(6.43)
Note that (6.43) is valid even for inhomogeneous materials as long as we allowµ to be a function of position, that is, µ(x,y,z). A very common type of inhomogeneity encountered in practice occurs at the interface between two magnetically different materials. To establish a basis for solving such problems, we now study the behavior of H and B in crossing the boundary between two different materials and derive the conditions that B and H must satisfy at such interfaces. These conditions are referred to as the boundary conditions; they are similar in nature to the boundary conditions for electrostatic fields studied in Section 4.11. 76 For
further reading on this interesting subject, see Section 13.4 of S. Ramo, J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics, 3rd ed., John Wiley & Sons, Inc., New York, 1994; also see Section 9.10 of W. B. Cheston, Elementary Theory of Electric and Magnetic Fields, John Wiley & Sons, Inc., New York, 1964.
Sec. 6.9
505
Boundary Conditions for Magnetostatic Fields
The boundary conditions must necessarily be derived using the integral for1ns of the fundamental magnetostatic laws, since the differential forms apply only at a point. The methodology we use for derivation is identical to that used in Section 4.11 for electrostatics. For the normal component of the magnetic field, we consider the surface of a differential pillbox, as shown in Figure 6.47 a. Since the contributions from the side surfaces can be made infinitesimally small by taking fl.h > 0, using (6.27) we have
s
(6.44)
B · d s = B 1 · fi  B2 · fi = 0
Hence the nor1nal components of the magnetic field B are always continuous across an interface between two different materials. Since B = µH, the magnetic field intensity H is not continuous across a boundary:
For the parallel components of the magnetic field, we consider the line integral of H around a closed contour abcda such as that shown in Figure 6.47b. In general there may be a free current in a layer of vanishing thickness flowing along the boundary between two materials, as occurs with timevarying fields and when one of the materials
a d
A
n
ilh
.... \
" __ ,
\/
/
(a)
(b)
B1 µl µz> µl
01
I I I I I I I I I I I
02
B2
I
(c) Figure 6.47 The boundary between two magnetic media. (a) A differential pillboxshaped closed surface. (b) A differential closed contour. (c) B vectors at the boundary for the case JL2 > µ, 1.
The Static Magnetic Field
506
Chap.6
is an excellent conductor. From Figure 6.47b, since the contributions from the sides can be made infinitesimally small by taking fl.h > 0, applying (6.40), we have a
~w
c
H1 · dl d
~h/2
H2 · dl = b
l dh dw 0
~h/2
H 1tfl.w  H2tfl.w = l fl.w fl.h H lt  H 2t = l fl.h = ls
Since the relationship between the surface current density J s and H must satisfy the righthand rule, this boundary condition can be compactly expressed as (6.45) where fi is the outward unit normal pointing from medium 2 at the interface and Js is normal to the page pointing inward. In the absence of any free surface currents at the interface (i.e., Js = 0), the tangential component of the H field is continuous across the interface between two materials: (6.46) The tangential component of H is also continuous across the interface when the conductivities of both media are finite, since in such cases currents are defined by volume current densities instead of surface currents, so l fl.h > 0 as fl.h > 0 and so ls= 0. When one of the media (say, medium 2) is a perfect conductor, the magnetic field in its interior is zero (see the discussion in Section 6.8.4), so we have H2 = 0 and thus from (6.45),
I fi
x H1 =
Js I (Perfect conductor)
(6.47)
as the basic boundary condition on the surface of a perfect conductor. Note that another consequence of the fact that B = 0 inside a perfect conductor is that B2n = 0, which implies (based on (6.44)) that B i n = 0. In other words, magnetic fields cannot terminate nor1nally on a perfect conductor. The magnetic field immediately outside a perfect conductor must always be tangential to the surface. Even in the absence of free surface currents, since B = µH, the tangential component of B is not continuous across a boundary between two magnetically different materials. Instead, from (6.46) we have (6.48) The discontinuity of the tangential B field can be understood in ter1ns of the bound magnetization current. Integrating (6.37) over the differential surface element bounded
Sec. 6.9
507
Boundary Conditions for Magnetostatic Fields
by the closed contour abcda shown in Figure 6.47b, and noting that fl.h
s
V x M · ds =
c
M · dl = (Mir  M2r)fl.w =
s
> 0 we have
Jm · ds
Since fl.h > 0, the surface integral of Jm selects only the surface magnetization current density at the boundary:
s
Jm · ds = lmfl.hfl.w = lsmfl.W
Combining the two previous expressions gives (6.49) To write (6.49) in terms of the B field, we first note from (6.39) that
so that
=0 from (6.46)
Substituting this result into (6.49) gives
Noting that the magnetic field and surface current density vectors are related via the righthand rule, we can write this result more compactly as
where fi is the outward unit normal pointing from medium 2 at the interface. At the surface between two magnetically different media, the tangential component of the B field is discontinuous by the amount µolsm, as if there exists a surface current density at the boundary. This surface current arises due to unequal magnetization surface current densities at the interface between the two different media (see Figure 6.45). This result is analogous to the discontinuity of the nor1nal electric field due to effective surface charge density at the interface between two dielectrics, as discussed in Section 4.11. A consequence of (6.44) and (6.48) is the bending of magnetic field lines across material interfaces, as illustrated in Figure 6.47c. Consider a magnetic field B that
508
The Static Magnetic Field
Chap.6
is oriented at an angle 81 from the nor1nal in medium 1. From (6.44) and (6.46) we have
B 1cos 81 = B2 cos 82 µ1B1 sin 81 = µ1B2 sin 82 where B 1,2 are the magnitudes of B 1,2. From these results we can find
µ2 tan 82 = tan 81 µ1 and µ
2
µ1
2
sin 81
+ cos2 81
The above relationships indicate that magnetic flux lines are bent further away from the normal in the medium with the higher magnetic permeability. Most materials have magnetic permeabilities of µ ~ 1, except for ferromagnetic materials, for which µ can be very large. If medium 2 is a ferromagnetic material and medium 1 is vacuum or air (or any other nonmagnetic material), µ1 >> µ 1, so tan 81 oo, B approaches that found in Example 6.6. Square loop of current. A current I flows in a square loop of side length a as shown in Figure 6.56. Find the B field at the point P(a/4, a/4). y I
a
.p 
a
6.14
x
Figure 6.56 Square loop of current. Problem 6.13.
A wire with two circular arcs. Consider a loop of wire consisting of two circular and two straight segments carrying a current /, as shown in Figure 6.57. Find B at the center P of the circular arcs.
524
The Static Magnetic Field
Chap.6
b / / / /
p
.11}!_0 _.____1.,.__
.
a
___.
...
Figure 6.57 Wire with two circular arcs. Problem 6.14.
Wire with four 90° bends. An infiniteIy long wire carrying current I = 1 A has four sharp 90° bends 1 m apart as shown in Figure 6.58. Find the numerical value (in Wbm 2 ) and direction (i.e., the vector expression) of the B field at point P(l, 0, 0).
6.15
z

/
/ /
I= 1A
/
1m
/
lm
/ /
/ / /
/ /
I
/ /
..... .....
..... .....
..... .....
..... ..... / /
P(l, 0, 01,,,,"'
y
/
x
/ /
Figure 6.58 Wire with four 90° bends. Problem 6.15.
Infinitely long copper cylindrical wire carrying uniform current. An infinitely long solid copper cylindrical wire 2 cm in diameter carries a total current of 1000 A, distributed uniformly throughout its cross section. Find B at points inside and outside the wire. 6.17 Helmholtz coils. Two thin circular coaxial coils each of radius a, having N turns, carrying current/, and separated by a distanced, as shown in Figure 6.59, are referred to as Helmholtz coils for the case when d = a. This setup is well known for producing an approximately uniform magnetic field in the vicinity of its center of symmetry. (a) Find B on the axis of symmetry the z axis of the Helmholtz coils. (b) Show that dB z /dz = 0 at the point P midway between the two coils. (c) Show that both d 2 Bz/dz 2 = 0 and d 3Bz/dz 3 = 0 at the midway when d =a. (Note that d is called the Helmholtz spacing, which corresponds to 6.16
d
N
a /
/
/
__ .L__

z
N I
Figure 6.59 Helmholtz coils. Problem 6.17.
Chap.6
525
Problems
the coil separation for which the second derivative of Bz vanishes at the center.) (d) Show that Bz at the midpoint P between the Helmholtz coils is
Bz
6.18 6.19
~
0.8992 x 10
_ 6 NI
a
T
(e) Find Bz at the center of each loop and compare it with the value at the midpoint between the coils. Helmholtz coils. Design a pair of Helmholtz coils, separated by 0.5 m, to produce a magnetic field of 100 µT midway between the two coils. Take I= 1 A. Helmholtz coils. Consider the Helmholtz coils discussed in Problem 6.17. Calculate and sketch Bz along the axis of symmetry between the two coils if one of the coils is connected backward (i.e., its current is switched).
6.20
Square Helmholtz coils. Consider a pair of square coils, similar to Helmholtz coils, each of same side a, having N turns, and carrying current /, separated by a distance d. Show that the Helmholtz spacing d (i.e., the coil separation for which d 2Bz/dz 2 = 0 at the center) is equal81 to d ~ 0.5445a. (Note that the Helmholtz spacing for circular coils is d =a, where a is the radius of each coil.) To simplify your solution, take each side to be of unity length and use a simple iterative procedure. 6.21 A circular and a square coil. Consider a pair of coils, similar to Helmholtz coils, separated by a distanced, where one of the coils is circular in shape with radius a and the other square with side 2a, each having N turns and carrying current /, and d =a. For a = 25 cm, I = 1 A, and N = 20, calculate and plot the magnitude of the B field along the axis of symmetry between the two coils. 6.22 Magnetic field of a solenoid. For an aircore solenoid of N = 350 turns, length l = 40 cm, radius a = 2 cm, and having a current of 1 A, find the B field (a) at the center and (b) at the ends of the solenoid. 6.23 Magnetic field of a surface current distribution. A circular disk of radius a centered at the origin with its axis along the z axis carries a surface current flowing in a circular direction around its axis given by
Js = b  a and compare it with the result of part (b).
6.39 Inductance of a rectangular toroid. Consider an aircore rectangular toroid as shown in Figure 6.20 with the values of its dimensions given by a = 3.8 mm, b = 6.4 mm, and t = 4.8 mm, respectively. (a) Find the total number of turns to be wound on this core such that its total inductance is around 1 mH. (b) Repeat part (a) for a powdered nickeliron core (assume /Lr = 200) having the same geometric dimensions. 6.40 Inductance of a rectangular toroid. A 50mH toroid inductor is to be designed using a molypermalloy powder core with /Lr = 125, a = 7 .37 mm, b = 13.5 mm, and t = 11.2 mm. Find the approximate number of turns N required. 6.41 Inductance of a circular toroid. Consider a toroid of circular cross section of radius a and mean radius rm as shown in Figure 6.65. Show that the inductance of this coil is given by
6.42
where /Lr is the relative permeability of the core material. Inductance of a twowire line. Determine the inductance per unit length of a twowire transmission line in air as shown in Figure 6.39, designed for an amateur radio transmitter, with conductor radius a = 1 mm and spacing d = 6 cm. (b) Repeat part (a) if the conductor spacing is doubled (i.e., d = 12 cm).
Chap.6
• • • • • • • • •
•
_ • __ •

529
Problems
___.
2a
/ •
•
•
•
rm
•
rm
_ /
• •
•

•
\ .......
• •



•
µ •
6.43
6.44
6.45

• • • • •
•
,,, .......
•
\
I
Figure 6.65 Inductance of a circular toroid. Problem 6.41.
Inductance of a thin circular loop of wire. (a) Find the inductance of a single circular loop of wire with d = 2 mm and a= 5 cm. (b) Repeat part (a) with d = 0.5 mm and a= 10 cm. Refer to Figure 6.43. Mutual inductance between two solenoidal coils. An aircore solenoid 25 cm long and 2.5 cm diameter is wound of 1000 turns of closely spaced, insulated wire. A separate coil of 100 turns, 3 cm long, with about the same diameter is located at the center of the longer coil (where Bis approximately constant). Find the mutual inductance between the two coils. Mutual inductance between a wire and a circular loop. Find the mutual inductance between an infinitely long, straight wire and a circular wire loop, as shown in Figure 6.66.
a
d
6.46
Figure 6.66 Wire and circular loop. Problem 6.45.
Long wire and loop. An infinitely long wire carrying current I passes just under a circular loop of radius a, also carrying the same current I as shown in Figure 6.67. (a) Find the
z
I
d>>a
I
y Figure 6.67 Long wire and loop. Problem 6.46.
530
6.47
The Static Magnetic Field
Chap.6
magnitude and direction of the B field at point P1. Note that d >> a. (b) Repeat (a) for point P2. (c) What is the mutual inductance between the wire and the loop? Inductance and energy of a solenoid. An aircore solenoid 30 cm long and 1 cm diameter is wound of 1000 turns of closely spaced, insulated wire. (a) Find the inductance of the solenoid. (b) Find the energy stored in the solenoid if it is carrying a current of 10 A.
6.48 1\vo square coils. Two square coils each of side length a are both located on the xy plane such that the distance between their centers is d. (a) Find a simple expression for the mutual inductance between the two loops for the case d >> a. (b) Find the mutual inductance between the two coils for the case d = 4a. 6.49 Fixedlength piece of wire. You have a fixed length of copper wire of 1 mm diameter and 20 cm length. You can bend this wire in any shape or form to obtain as large an external selfinductance as you can. (a) What is the value of the maximum inductance, and what arrangement would work best? You cannot use any magnetic materials. (b) If you had the tools necessary to melt the copper wire and form it into a longer wire of diameter 0.1 mm instead, how would your answer to (a) change? Also compare and comment on the resistance of the two different wires. 6.50 Magnetic energy. A conductor consists of a cylinder with radius b with a hole of radius a (a < b) drilled coaxially through its axis. The current density is uniform and corresponds to a total current of I. Find the magnetic energy stored inside per unit length of the wire. 6.51 Explosion in a power transformer. A current transformer used for 500 kV transmission lines has a single primary coil which is connected to the high voltage line via two parallel wires that are 20 cm apart and carrying the same current in opposite directions. A fault occurs on the line and causes the current of each wire of the transformer to reach a peak value of 100 kA which results in an explosion in the oil/paper insulation structure of the wires. Calculate the perunitforce on each wire and explain what happened. 6.52 Force between a long wire and a loop. A long wire extending along the z axis is situated near a rigid loop as shown in Figure 6.68. Find an expression for the force and torque (about the origin) experienced by the loop.
b /
/ / / /
/
//
11
¢0
h
0   L...t~1 a
Figure 6.68 Force between a long wire and a loop. Problem 6.52.
an
Oersted's accidental discovery in 1820 of the close connection between electricity and magnetism led to almost immediate advances in technology. The fact that wires carrying current experience forces and torques when in the presence of a magnetic field was put to good use in directcurrent electric motors (Section 6.10) and in the design of galvanometers, sensitive instruments for measuring electrical current. These instruments were crucial in facilitating scientific experimentation. However, a much broader range of industrial and technological applications of electricity and magnetism was yet to come, with Faraday's discovery in 1831 of electromagnetic induction the fact that magnetic fields that change with time produce electric fields. This experimentally based fact constitutes the last of our three experimental pillars of electromagnetics 1 and is the subject of this chapter. Electrostatics (Chapter 4) deals with the effects of stationary charges, the spatial distributions of which are deter1nined by the presence of conducting bodies and applied potentials. The governing laws of electrostatics, derived from Coulomb's force law, can be summarized as follows:
VxE=O V ·D= p
c s
E · dl = 0 D · ds =
v
p dv
where p is the free volume charge density, and the fields E and D are related via D = EE, with E being a scalar constant for linear, isotropic, homogeneous, and timeinvariant media. 1The
first pillar of electromagnetics is the fact that electric charges attract or repel one another in a manner inversely proportional to the distance between them (Coulomb's law). The second pillar is the fact that currentcarrying wires create magnetic fields and exert forces on one another (Ampere's law of force or alternatively the BiotSavart law).
531
532
TimeVarying Fields and Maxwell's Equations
Chap. 7
Magnetostatics (Chapter 6) deals with the physical effects produced by charges in motion (i.e., steady currents). The governing laws of magnetostatics, derived from Ampere's force law (expressed also in the BiotSavart Law), can be summarized as follows: VxH=J
H · dl =
s
s
J · ds
B · ds = 0
where J is the free volume current density, and the fields B and H are related via 1 H = µ B where µ is a constant scalar for linear, isotropic, homogeneous, and timeinvariant media. Up to now, we have exclusively considered static cases and paid little attention to electric and magnetic effects produced by charges, currents, or circuits whose positions or 2 intensities vary with time. Under the static assumption, the electric and magnetic fields are unrelated, and the field vectors E, D and B, H form independent pairs. 3 However, following Oersted's findings that electric currents produce magnetic fields, M. Faraday and many others started thinking about the possibility that magnetic fields may in turn produce electric fields, and that the two sets of preceding equations may be coupled. After many years of various trials,4 Faraday carried out the classic experiment5 of induction on August 29, 1831, and showed that E and B were indeed related, but only for the case in which the fields change in time, that is, under nonstatic conditions. At about the same time, J. Henry of Albany Academy in New York studied similar effects and arrived at similar conclusions. However, Faraday's results were published earlier and had the greatest impact in the scientific world. 6 As a result, the law of electromagnetic induction is commonly referred to as Faraday's law. Faraday's law of electromagnetic induction, namely that magnetic fields that change with time induce7 electromotive force, is the third and final experimental 2 0ccasionally,
it was necessary to apply Gauss's law to timedependent phenomena with little or no justification, for example, in the derivation of the continuity equation (Section 5.5). 3 In a conducting medium (a =I= 0), a static electric field causes a steady current to flow (J = a E), which in tum gives rise to a static magnetic field. However, at steady state, the electric field can be completely determined from the static charges or potential distributions. The magnetic field is simply a consequence of the current and does not affect the electric field. 4 For a brief historical account, see R. Elliott, Electromagnetics, IEEE Press, New Jersey, 1993. Another excellent review of the history and development of electromagnetics, with Chapter VI devoted to Faraday, is given by E. Whittaker, History of the Theories of the Aether and Electricity, T. Nelson and Sons, Ltd., London, 1951. It is interesting, for example, that a number of scientists, including Ampere, only narrowly missed discovering the law of induction. 5 M. Faraday, Experimental Researches in Electricity, Taylor, London, Vol. I., 1839, pp. 1109. 6 Henry' s lack of promptness in announcing his results, combined with the fact that the New World was remote from European centers, caused his results to be viewed as merely confirmation of Faraday's findings. 7 The word ''induce'' is used quite literally in this context; to induce means to cause the formation of. The word ''induction," on the other hand, has several meanings in nontechnical parlance, including the initiation of a person into military service. In our context, the word induction refers to the process by which the electromotive force is brought into existence by the timevarying magnetic field.
Chap. 7
TimeVarying Fields and Maxwell's Equations
533
8 fact
that for1ns the basis for Maxwell's equations and is arguably the most important of our three experimental pillars of electromagnetics. Without it, electromagnetic waves would not propagate through free space, and wireless communication would not be possible. Faraday's discoveries also for1n the basic underpinnings of modern electrical technology; without electromagnetic induction, the workhorses of our industrial world electric motors and generators would not be possible. Our coverage of Faraday's law in this chapter completes our understanding of the underlying physical basis of the transmission line behavior discussed in Chapters 2 and 3. The fact that temporal variation of a voltage applied between the two conductors of a transmission line induces electromotive force, which in turn affects the current flow, is the very basis of the distributed inductance of the line. Completing our formulation of Maxwell's equations later in the chapter also allows us to exhibit the similarity between the fundamental equations that relate electric and magnetic fields for electromagnetic waves and the transmission line equations that relate line voltages and currents. Natural solutions of both the electric/magnetic field equations and the voltage/current equations are in the form of waves that propagate with a finite speed, leading to the distributed circuit effects studied in Chapters 2 and 3, and the propagation, reflection, and guiding of electromagnetic waves covered in Chapters 810. As we discuss timevarying magnetic fields and Faraday's law in the first three sections of this chapter, we are concerned with variations that are slow enough so that radiation effects are negligible. This is the socalled quasistatic approximation, which implies that the system of conductors carrying currents has dimensions much smaller than a wavelength. A wide range of problems in laboratory physics and engineering lie within the domain of the quasistatic approximation. The criteria for the validity of the quasistatic approximation are the same as those discussed in Chapter 1 in connection with determining the applicability of the lumped analysis and the necessity of a distributed circuit treatment. Basically, the quasistatic approximation restricts us to considering circuits of such sizes and rates of change of current that the electromagnetic disturbance propagates over much of the useful parts of the circuit before the current has changed significantly. Under such conditions, we implicitly assume that at any given instant, the magnetic fields everywhere in the circuit are strictly proportional to the current at that instant. In general, the quasistatic approximation amounts to calculating all fields as if 8 We
note once again that the complete set of Maxwell's equations can be derived via transformation of Coulomb's law, using special theory of relativity. This was first shown in 1912 [L. Page, A derivation of the fundamental relations of electrodynamics from those of electrostatics, Am. J. Sci., 34, pp. 5768, 1912]; a clear treatment is given in Chapter 5 of R. S. Elliott, Electromagnetics, IEEE Press, New Jersey, 1993. However, such derivations must necessarily make other implicit assumptions; see Section 12.2 of J. D. Jackson, Classical Electrodynamics, Wiley, New York, 2nd ed., 1975. Even before the introduction of the special theory of relativity, and about ten years after Faraday's discovery, H. von Helmholtz and Lord Kelvin recognized and suggested [Helmholtz, Uber die Erhaltung der Kraft, 1847; Kelvin, Trans. Brit. Assoc., 1848 and Phil. Mag., December 1851] that it is possible to deduce the existence of induced currents from energy considerations. For a brief discussion, see Article 481 of G. H. Livens, The Theory of Electricity, Cambridge University Press, Cambridge, UK, 1918. Nevertheless, and in view of the profound importance of the law of electromagnetic induction in forming the basis of Maxwell's equations, in this book we consider Faraday's law to be based on experimental fact.
TimeVarying Fields and Maxwell's Equations
534
Chap. 7
they were stationary, without having to account for the travel time of fields from one point of the circuit to another. Formally, the quasistatic approximation is valid so far asap/at is negligible compared with V · J. When freecharge density varies rapidly enough that ap/at is comparable to V · J, time variations of electric fields produce magnetic fields, leading to propagation of electromagnetic waves, as is discussed in Section 7 .4.
7.1 FARADAY'S LAW Michael Faraday was completely innocent of all mathematics, but at the same time he was one of the most imaginative and visual scientific thinkers of all time. Since Oersted demonstrated that an electric current could deflect a compass needle (i.e., generate a magnetic field), Faraday was convinced that a magnetic field or a magnet should be 9 capable of producing a current. In 1831, Faraday set up an apparatus consisting of an iron ring in the shape of a toroid on which were wound two coils of wire as sketched in Figure 7 .1. The primary coil was connected through a switch to a voltaic cell; the ends of the secondary coil were joined with a wire that ran above a compass. Any current induced in the secondary coil would thus deflect the compass needle to the left or right of its nor1nal position, depending on its flow direction. Upon closing the switch, current flowed in the primary coil, producing a magnetic field in the iron ring that passed through the secondary coil. Faraday observed a momentary deflection of the compass needle detecting a brief surge of current induced in the secondary coil. However, the compass needle quickly settled back to zero, indicating that the induced current existed only during the initial transient. In other words, no current flowed in the secondary coil once the current in the primary coil became steady and the field in the toroid reached
Iron ring (toroid)
Wire
Battery Switch
.___, Primary coil Secondary ,__, coil
Compass beneath the wire
Figure 7.1 Faraday's ironring experiment. A sketch showing the elements of the apparatus used by Faraday that led to the discovery of the law of electromagnetic induction (also called Faraday's law). The iron ring provides a path for the magnetic flux coupling the two circuits.
9 See
H. Kondo, Michael Faraday, Scientific American, IO, pp. 9198, 1953 and Chapter 6 of G. L. Verschuur, Hidden Attraction, The Mystery and History of Magnetism, Oxford University Press, New York, 1993.
Sec. 7.1
535
Faraday's Law
its final value. When the switch was opened, terminating the current in the primary coil and thus eliminating the magnetic field in the toroid, another momentary deflection was observed, opposite in polarity with respect to the one observed when the switch was closed. Based on these extraordinary observations, Faraday deduced that a magnetic field could be generated by a steady current, but that a current could be induced only by a changing magnetic field. 10 This meant that the mathematical description of this phenomenon, which came to be known as electromagnetic induction, or simply induction, had to depend on the time variation of the magnetic flux. Later, Faraday investigated the possibility of generating a steady current in the secondary coil and realized that a continuously moving magnetic field would be necessary. For this purpose he made a copper disk rotate with its edge between the poles of a magnet and found that current flowed from the center toward the edge of the disk (or vice versa). This was the world's first dynamo or electric generator. Faraday also used the motion of a magnet in and out of a helical coil to create a current, as sketched in Figure 7 .2. He also determined that a more efficient way to produce currents was to move coils of wire in a magnetic field, the principle on which electric generators are built. These experiments enabled Faraday to for1nulate his famous law of electromagnetic induction, or Faraday's law, which states that when the magnetic flux enclosed by a loop
Wire coil
Permanent magnet I I
I I
I I
Moved in/out
~,,,,
Compass beneath the wire
Figure 7.2 A magnet moving in a coil. A sketch of one of Faraday's experiments used to induce a current in a helical coil of wire by moving a magnet into and out of the coilthe principle of the electric generator. The induced current is detected by the deflection of the compass needle located under the wire combining the ends of the coil.
10 The
transitory nature of the electromotive force produced, due to its dependence on the change in magnetic flux, contributed to the elusive nature of the effect. An outstanding example of this was the experiments of Jean Daniel Colladon in Geneva in 1825. Colladon made a large helix of insulated copper wire 810 cm long and 45 cm in diameter, and he connected the ends to a sensitive galvanometer. He then brought a permanent magnet up to one end of the helix, expecting to produce a permanent current flow to be registered by the galvanometer. However, being the careful scientist he was, he wanted to place the galvanometer far from the magnet so that it would not be affected by the magnetic field of the magnet itself. His solution was to connect the galvanometer to the helix with 50 m of wire and to place it under a glass jar in another room well away from the magnet. There is little doubt that Colladon' s galvanometer registered the temporary current caused by his movement of the magnet; however, by the time he walked to the other room, the galvanometer needle had returned to its zero reading and Colladon missed his chance of discovering electromagnetic induction. (From W. A. Atherton, The history of electromagnetic induction, Am. J. Phys., 48(9), pp. 781782, September 1980.)
TimeVarying Fields and Maxwell's Equations
536
Chap. 7
of wire changes with time, a current is produced in the loop, indicating that a voltage, or an electromotive force (emf), is induced. The variation of the magnetic flux can result from a timevarying magnetic field linking a stationary and fixed loop of wire or from a loop of wire moving in a static magnetic field or from both, that is, a loop of wire moving in a timevarying magnetic field. The mathematical form of Faraday's law is d\J! °Vind=  
dt
(7.1)
where °Vind is the induced voltage across the terminals of the loop C, and \J! is the total magnetic flux linking 11 the closed loop C, as shown in Figure 7.3. Faraday's law states that the induced voltage °Vind around a closed loop C is equal to the negative of the time rate of change of the magnetic flux linking C. Note that \J! is the total magnetic flux linking the contour C, given by (6.26): \J! =
s
B · ds
where S is the area of the surface enclosed by contour C, as shown in Figure 7.3. Having an induced voltage °Vind appear across the terminals of the loop means that current will flow if a resistance is connected across the terminals. In this sense, this induced voltage is similar to the electromotive force as defined and discussed in Chapter 5. The electromotive force is the agent that ''pushes'' the electrons that constitute the current. Faraday thus discovered that changing magnetic fields produce an electromotive force or emf, which acts just like the emf generated in other voltage sources, such as a chemical battery, a piezoelectric crystal (emf produced in response to mechanical pressure), or a
B
di
s
c ds

11 With
Oii'
v ind
+
Figure 7.3 Illustration of Faraday's law. Voltage °Vind is induced between the terminals of loop C due to a timevarying magnetic field B that links the area S enclosed by C, where the direction of the differential length dl on the contour and the direction of the differential area ds on the surface S are related by the righthand rule.
loops of N turns, where each turn links the same amount of flux, we can also write A = N \If, where A is the total flux linkage of N turns and \If is the flux linked by each turn. We then have °Vind= N d\lf /dt = dl\./dt.
Sec. 7.1
537
Faraday's Law
ther1nocouple (emf resulting from a temperature gradient). The term ''electromotive force'' is somewhat misleading because an emf is actually not a force but rather a line integral of a force per unit charge (i.e., an electric field). More precisely, emf is defined as the tangential force per unit charge along the wire integrated over its length, around the complete circuit. Having a nonzero emf across the ter1ninals of the loop in turn means that the line integral of the electric field around the loop is not zero, since the force that moves the current carriers is ultimately due to an electric field. Note that this electric field is certainly not an electrostatic field, because otherwise its integral around the closed circuit must be zero. Nor can the source of the force be a magnetostatic field, since the stationary charges cannot experience magnetic forces. The field that provides the force to drive the current is an entirely new kind of electric field produced by the new physical effect embodied in Faraday's law. 12 In words, Faraday's law states that a changing magnetic field induces an electric field. The induced voltage °Vind can thus be expressed as °Vind=
c
E · dl
(7.2)
where E is the induced electric field. Combining (7 .1) and (7 .2), the mathematical statement of Faraday's law is
c
E · dl = 
d dt s
B · ds
(7.3)
where S is the surface enclosed by the contour C. According to (7 .3), an electric field induced by a changing magnetic field exists in space regardless of whether there are conducting wires present. If conducting wires are present, an induced current flows through these wires. Faraday's law is the basic principle on which electric generators operate; mechanical energy is supplied to change the magnetic flux \II that links the coil C (e.g., by rotating coil C and thus the surface S) and thus to produce induced voltage across its ter1ninals through (7 .3). Faraday himself invented the first directcurrent generator in 1831 and the prototype of the modern electric generator in 1851. The minus sign in (7 .3) indicates that the induced emf is in a direction that opposes the change in the magnetic flux that caused the emf in the first place. In other words, the induced voltage leads to current flow 13 in a direction that produces an opposing magnetic
12This
new kind of physical force that drives electric current is also called an electric field to recognize the fact that it moves charges in the same manner as an electrostatic field. Keep in mind, however, that this new field is quite different from an electrostatic field, whereas the line integral of an electrostatic field around a closed path is identically zero, that of an induced field is obviously not zero. 13 In typical circuits, which have nonzero resistance, the induced currents may be tiny fractions of the original currents that produce the magnetic fluxes linking the circuit. In other words, the magnetic fields produced by the induced currents are usually negligible, so that although the circuit opposes the changing flux by establishing a current flow, the resulting currents typically fall far short of fully compensating for the flux change.
538
TimeVarying Fields and Maxwell's Equations
Chap. 7
14
flux \II. This statement of experimental fact is known as Lenz' s law. In terms of the correct polarities of the two integrals in (7 .3), it is important to note that the orientation of the two integrals cannot be independently chosen. In other words, if di is chosen such that one finds the first integral by going counterclockwise around the contour C, then the direction of d s in the second integral must be outward from the page assuming that the area S enclosed by loop C lies in the same plane as the loop. The righthand rule applies here; with the fingers of the right hand pointing in the direction of di along C, the thumb points in the direction of d s. The relationships between the polarity of the induced voltage °Vind, the relative orientations of the induced electric field E, and the other quantities (e.g., di) in (7.3) are depicted in Figure 7.4. We take the contour C to be enclosing the surface S, as shown in Figure 7 .4a, and we take the orientation of the surface element d s to be in the z direction (out of the page), in which case di must be counterclockwise, as shown in Figure 7 .4a. For the purposes of this discussion, we take the magnetic field B to be pointing out of the page, unifor1nly present everywhere, and steadily increasing in time (i.e., aB /at > 0).
,.  .... .........
,,   .... ' ' ,. ' ' ' I I ,..... ' \ I I ' \ \ I I B(t) ', \ \ /
/
c~
B(t), ds
1
1 (
@
\ \
\ \
s
@
I,,
\
I
I ,
/
I I
I
\ \' ' . . _.... E/ I ' ' ' ' ..... .... ___  ,, ' .... .... ___ .... ....""' ,,,
I
/
(a)
(b) Wire
Wire
I(t)
y
2
B(t) x
@
E
B(t)
@ I(t) 1
(c)
(d)
Figure 7.4 The polarity of the induced voltage and the electric field. The case illustrated is for 8B /8t > 0. 14 We
note here that although Lenz's law is taken to be an experimental fact, it is in fact a consequence of the conservation of energy. In other words, if the polarity of the induced current were opposite to that dictated in Lenz's law, magnetic energy would not be conserved, as is discussed later in Section 7.3. We expect this result on a purely heuristic basis; if the polarity of the induced currents were such as to enhance the effect that produced the currents, the magnetic flux would increase without limit.
Sec. 7.1
539
Faraday's Law
The orientation of the induced electric field E is shown in Figure 7.4b. Note that this field is induced regardless of whether conducting wires are present. Since aB /at > 0 and B and ds are both out of the page, the righthand side of (7.3) is negative. The fact that d s is out of the page requires that di must be in the counterclockwise direction, so the induced electric field E must encircle the B field in the clockwise direction in order for the lefthand side of (7 .3) to also be negative. Now consider what happens when a conducting wire loop is present, with a small gap across which the induced voltage appears as shown in Figure 7 .4c. Note that the thickness of the wire is exaggerated for convenience. At the instant this wire loop is introduced into the system, the induced electric field causes the free electrons inside the conductor to move to one end and leave the other end positively charged, as shown in Figure 7 .4c. Assuming that the wire is a good conductor such as copper, this rearrangement of charge within the wire occurs in rvl0 19 sand produces an electrostatic field Ees due to the separation of charge, which cancels the induced field, so that the net electric field inside the conductor is zero. The potential difference between the ends 2 and 1 is 2
12
E · di
= 2  1 = 1
This relation is consistent with the definition of electric potential in Section 4.4, where 12 is the work done in moving a unit positive test charge from point 1 to point 2 (e.g., see E ·di. Note that if we integrate (4.19)) where [W /q]a~b = ab = (b)  (a) = the total electric field around the contour C in the counterclockwise direction, the only contribution to the integral comes from across the gap (since the total electric field is zero within the wire), where E ·di is negative, which in turn results in a positive value of 12. If a load resistance R is connected across the gap (Figure 7 .4d), current flows in the clockwise direction, tending to produce magnetic flux opposing the increase in B; this is consistent with the orientation of the induced electric field, which drives the current carriers. If the loop did not have a gap, then current would still flow in the clockwise direction, in accordance with the wire resistance and selfinductance of the loop. In effect, the external resistor R can then be thought of as the internal resistance of the wire. Turning now to the choice of the polarity of °Vind, we note from (7.3) and the definition of °Vind= :f E ·di that for aB /at > 0, and for ds chosen to be in the same direction as B as is the case in Figure 7 .4, °Vind as determined by (7 .3) is negative. Thus, the polarity of °Vind in Figure 7.4 must be chosen to be positive at end 1 and negative at end 2, so that we have °Vind = 12, that is, a negative value, consistent with the result obtained from the equation (7.3). More generally, in applying (7.3) to any loop with a gap, the polarity of °Vind must be defined to be positive on the terminal at which the di element, the polarity of which is determined by the right hand rule once ds is chosen, points outward from that terminal. For further elaboration of Lenz' s law, consider the circuit shown in Figure 7 .5a, which consists of a magnetic core material 15 having two separately wound coils of wire
J:
15 The
coils.
magnetic material enhances the mutual inductance, and thus the inductive coupling, between the two
TimeVarying Fields and Maxwell's Equations
540
v~ >• : Rz~ •
Lz Sl Battery

~
t
t= 0 11 ~
L 12
'
I
~~
S2 .
T
' 
~
Chap. 7
•
J2(t)
: L1
T

t
(a)
(b)
Figure 7.5 Illustration of Lenz's law. (a) A solenoidal core with two separately wound coils of wire. A battery is connected to or disconnected from the lower coil through switches Sl and S2. (b) The variation of the currents / 1 and /2 as a function of time for the case when the switch Sl is closed at t = 0 and S2 is closed at t = t 1.
(similar to Faraday's experiment shown in Figure 7.1, except the circuit in Figure 7.5 is a solenoid instead of a toroid). The lower coil is connected to a battery through the switch Sl. When the switch Sl is closed (say at time t = 0), a current 11 starts to flow in the lower coil. As a result of the increase in the magnetic flux linking the upper coil, a current 12 is induced in the direction opposite to 11, opposing the change in the magnetic flux linking the upper coil. Note that, in general, the current 11 does not begin to flow immediately after switch S 1 is closed but is instead governed by a time constant deter1nined by the inductance and resistance of the lower coil, as well as on the mutual inductance between the two coupled coils. Similarly, the establishment of the current 12 is governed by a time constant dependent on R1, L2, as well as on the mutual inductance between the coils. If the wire were a perfect conductor, the induced current 12 would flow in the upper coil as long as current 11 flows in the lower coil. However, in a practical coil, the resistive losses of the wire cause the current 12 to decay exponentially with a time constant determined by the selfinductance and resistance (L1, R1, L 2 , and R1) of each coil, as well as the mutual inductance L12 between the two coupled coils. Thus, at steady state, the current 12 becomes zero and a constant magnetic flux passes through the upper coil due to the current 11 flowing in the lower coil. Later, when the switch S2 is closed (say at time t = t 1), the upper coil tries to keep the flux constant by inducing a current 12 in the upper coil that is in the same direction as the current 11 in the lower coil in order to maintain constant flux, until 11 gradually decreases to zero. Both 11 and the induced current 12 die off with time (in a manner defined by the solution of the differential equation describing the coupled circuit consisting of L 1, £i, R 1, R2, and L 12, that is, not necessarily a simple exponential decay) due to the resistive losses in their respective loops. Figure 7.5b shows the variation of the two currents 11 and 12 as a function of time as just discussed.
541
Faraday's Law
Sec. 7.1
Example 7.1: Emf induced in a coil. Consider a singletum coil of wire of radius a, as shown in Figure 7.6a. The region the coil lies in is permeated by a magnetic field B(t) = zBz (t) = zBo sin(wt). Find the induced voltage between the two open end terminals of the coil. I(t) ~
z
I(t) ~
x
y
@
RL
B(t)
Bz(t) Bo
B(t) y ds
Z•
°Vind(t)
/
x
@ B(t)
di

/
,...,. /
+
, ,. ....
/
''
/
''
wt
Bo
(b)
(a)
l(t)RL, °Vind(t) .... .... .... 27r '
(c)
Figure 7.6 Emf induced in a coil. (a) A stationary coil located in a timevarying magnetic field. (b) The best way to think about the direction of the induced current and the polarity of the induced voltage °Vind is to view the coil as a voltage source. (c) The variation of the magnetic field Bz, the induced emf °Vind = IRL as a function of time. The direction of induced current flow is determined by Lenz's law.
Solution:
Using (7.1), the induced voltage is given by
d
d
[ZB0 sin(wt)] · (z ds) °Vind (t) =  B · ds =  dt s dt s = _ _:£[Bo sin(wt)](rra 2 ) = wna 2Bo cos(wt)
dt
To appreciate the polarity of °Vind, it is best to plot the variation of °Vind(t) over one cycle together with Bz (t). In fact, it is more illuminating to think in terms of the current I (t) that would fl.ow in the coil if an external load were connected to the loop, as shown in Figure 7.6b. The plots of Bz (t), I (t), and °Vind (t) in Figure 7.6b clearly illustrate Lenz' s law in effect; when the magnetic field Bz (t) exhibits its largest rate of increase (at t = 0), the current I (t) is maximally negative (i.e., flows clockwise so that its associated magnetic field is in the z direction and opposes the increase in Bz (t)). Negative (i.e., clockwise) current flows throughout the interval 0 < wt < n /2, but with the magnitude of the current decreasing as the rate of change of Bz (t) decreases, with I (t) eventually equal to zero at wt = n /2 when Bz (t) has reached its peak value Bo, so that BBz/Bt = 0. Similar manifestations of Lenz's law can be observed during the rest of the cycle. Note that, in this context, the coil is best viewed as a voltage source (or battery) for a load RL connected to it. If the coil consists of N turns, the surface S is enclosed N times, where each turn induces a voltage. These voltages all add in series, so that the total voltage induced across
TimeVarying Fields and Maxwell's Equations
542
Chap. 7
the terminals of the coil is N times greater and is given by °Vind(t) = N wna 2 Bo cos(wt)
The induced voltage is thus proportional to the rate of change of the B field (w ), the number of turns (N), and the intensity of the field (Bo) linking each tum with the area of the coil.
Example 7.2: Pair of lines and a rectangular loop. Consider a rectangular loop in the vicinity of a pair of infinitely long wires, as shown in Figure 7. 7. The two wires carry a current I flowing in opposite directions, the magnitude of which increases at a rate di/ dt > 0. Find the electromotive force induced in the loop (i.e., the °Vind that would be measured across the small opening if the loop were broken) for the two different placements of the rectangular loop, as shown in Figure 7.7.
I
w
I
w
I
I
/ind
h
h
a
t r
b
(a)
(b)
b I I I I I
I I
r+b
Figure 7.7 Pair of parallel wires and a loop. (a) The loop outside the region between the wires. (b) The loop in between the two wires equidistant from each wire.
Solution:
We know from Example 6.3 that a current I in an infinitely long straight wire " produces a field B = B at a distance r of
µol
B"'.,.,  2nr
(a) Since the B field due to both infinitely long straight wires is given by the same expression, the total flux linking the loop can be written as a+w
\II= _µo_I
2n
a
h dr
r
b+a+w b+a
h dr
r
(b + a)(a + w) µoh/ ln 2n a(b+a+w)
543
Faraday's Law
Sec. 7.1
The induced emf in the rectangular loop is then given by Ojf' .
_
(b + a)(a + w) 1n 2n a (b + a + w)
µoh
_
v ind 
di _ µoh a(b +a+ w) 1n dt 2n (b + a) (a + w)
di dt
Note that the polarity of the emf is such that the induced current /ind in the loop flows in the counterclockwise direction when di/ dt is positive. (b) When the loop is between the wires (Figure 7.7b), placed equidistant from both straight wires, the magnetic fluxes linking the loop due to the current flowing in each wire are equal in magnitude and in the same direction. Thus, using (6.26), the total flux linking the rectangular loop is (b+w)/2
µolh \JI =2 
2n
(bw)/2
dr
r
µolh b +w   ln n bw
The induced emf is °Vind
d\11 µ 0h b +w = = ln dt n b  w
di dt
Note that this time, the polarity of the induced voltage is such that the induced current /ind in the loop flows in the clockwise direction when di/ dt is positive.
Example 7.3: Jumping ring. A conducting ring of radius a is placed on top of a solenoidal coil wound around an iron core, as shown in Figure 7.8. Describe what happens to the ring when the switch is closed.
I I

~
/
Vo
_ I 
~
~
I I
.._
~
•
•
I
I
I I I I I I I I I I I I
• •
4 4 4
1
> >
'
>
• • • •
' '
1
> > I
I
I
I

' '
' '
1
"> )
I
1
I
>
' _,. ....   1\  .... __ __ .... ' \
'\ B
~
• •
I
I I I I I I I I I I
,
JI
~
I
1
\
..
,
(a)
z
A
z
A
r
I I /
(b)
Figure 7.8 Jumping ring. (a) A conducting ring placed on top of a solenoidal coil. (b) A cylindrical coordinate system.
y
544
TimeVarying Fields and Maxwell's Equations
Chap. 7
Solution: Before the switch is closed, the conducting ring has zero flux linkage. When the current is first turned on in the solenoid by closing the switch, the flux produced by the solenoid links the ring in the upward direction, inducing a current in the ring opposite in direction to that in the coil (i.e., Lenz's law). Since currents in opposite directions repel, the repulsion force lifts the conducting ring off the top of the solenoid. Using (6.52), this force is given by 2rr
F=
c
"
I dl x B =
lcf>(a d> a, b, which is the case in hand with d = 0.1 m. Using this expression for L12, we find
~ (4rr x 10 )rr(l0)(10)(0.02) (0.01) ~ 0.00 7
L
2
2(0.1)3
12
2
8
H
µ
Thus, we see that the mutual inductance between the two coils is much less than their respective selfinductances. The main reason for this is that the coils are relatively far apart, with d = 10 cm. Note that L 12 is inversely proportional to d 3 , so that it is larger by approximately a factor of 1000 if the coils are 1 cm (instead of 10 cm) apart. However, care should be exercised here since the expression for L12 derived in Example 6.30 is only valid for the cased>> a, b. The sign of the mutual inductance L 12 depends on the definition of the polarity of the currents / 1 and /2 in the two circuits. If these are defined such that positive values of /1 and / 2 produce magnetic flux in the same direction, then L 12 is positive. The total magnetic energy stored in the system of two parallel coaxial coils is thus
2 Wm= iL1111 ± ~
+ i£i2lf
i (12 x 106 H)(103 A) 2 ± (8 x 109 H)(lo3 A)(lo3 A) + ~(5 x
106 H)(l0 3 A) 2
2
~
L12/1/2
(8.5 ± 0.008)
pJ
Note that the mutual coupling, numerically small in this case, can either slightly decrease or slightly increase the stored energy, depending on the relative directions of the two currents /1 and /2.
Example 7 .9: A solenoid with a secondary winding. Find the total magnetic energy stored by the solenoid shown in Figure 7.16, which has a secondary winding wound concentrically over the main winding. Neglecting leakage flux from the ends, determine the total magnetic energy stored in this system if currents of /1 = /2 = 1 mA are flowing through the two coils and other parameters are Ni = 1000, N2 = 100, 11 = 10 cm, 12 = 5 cm, and a = 1 cm. Assume a nonmagnetic core (µ = µo).
l1
..
Ni • • • •
µ
l2
.. \
\
I I I
•
• •
•
I I I
2a
Figure 7.16 A solenoid with a secondary winding. The primary winding (N1 turns) in fact covers the full length (l 1) of the core, but the middle portion is not shown to avoid clutter in the diagram.
TimeVarying Fields and Maxwell 's Equations
562
Chap. 7
Solution: From Example 6.26, we know that the selfinductance of a solenoid of length l and having N turns with a crosssectional area of A is L = µ 0N 2A/l. Thus, we have L
2
11
= µoN 1 na
2
7
2
2
= (4n x 10 )(1000) [n(0.01) ]
11
~ .
0.10
3 95
mH
and similarly, we find ~2 ~ 0.079 mH. The mutual inductance between the two coils is given by L12 = N2A12//1. Once again invoking the result of Example 6.26, the flux produced by the primary coil is
µ0N1I1na 2
A12=    
11
Since this flux is linked N2 times by the secondary coil, we have
Upon substitution of the parameter values, this yields L12 ~ 0.395 mH. Thus, we see that the mutual inductance between the two coils is larger than the selfinductance of the secondary coil. The total magnetic energy stored in this system is given by 2
Wm= ~L1111 ± L12/1/2 ~ ![(3.95 x ~
+ ~~2/f 103)(10 3) 2 + (0.079 x
(2.01 ± 0.395)
103)(10 3) 2 ] ± (0.395 x 103)(10 3 )(10 3 )
nJ
If the currents / 1 and /2 are positive with directions as shown in Figure 7 .16, they reinforce one another so that L12 is positive and Wm ~ 2.41 nJ. Otherwise, the magnetic flux produced by /1 is reduced as a result of that produced by /2, so that the total energy stored is Wm~ 1.62 nJ.
7.3.3 Magnetic Energy in Terms of the Fields As we did for electrostatics in Section 4.12, it is possible to express the stored magnetic energy of a configuration of currentcarrying loops solely in ter1ns of the B and H fields. To this end, we start with the general expression derived in Section 7.3.2 for the energy of N currentcarrying loops, namely
From Section 6.7, we recall the Neumann for1nula for the mutual inductance between two circuits Ci and Cj as
Sec. 7.3
563
Energy in a Magnetic Field
where Rij is the distance between the differential lengths dli and dlj. Using this, we have
N
µo '"°'I· 4Jr ~ J j=l
Recalling from (6.20) in Section 6.5 the definition of the vector potential A, we recognize that the expression in the square brackets is the vector potential A, at a point on the loop identified by subscript i, due to (or produced by) currents flowing in all the loops. Therefore, we can write
In ter1ns of the more general free volume current density J, and noting that :fc I dl is equivalent to f v J dv, the preceding expression can be further generalized as (7.11) where the volume V is the entire region of space within which J · A ::j:. 0. Equation (7.11) is analogous to (4.81) associated with electrostatics. This expression identifies the quantity ~J ·A as the volume density of magnetostatic energy. Using the relationships between A, B and J, H, we can write Wm entirely in ter1ns of the B or H fields. To do this, we proceed in a manner identical to that used in electrostatics. If we assume Jin (7.11) to be the free volume current density (thus excluding the bound magnetization current V x M), we can substitute J = V x H (Ampere's law). Using the vector identity (see Appendix A) V · (H x A)
=(V x H) · A 
(V x A) · H
equation (7.11) can be rewritten as
Wm = = =
i f v J · A dv = i f v (V x H) · A dv i f v (V x A) · H dv + i f v V · (H x A) dv 1fv B · H dv + ~ :f (H x A)· ds 8
where S is the surface that encloses the volume V and where we have used Stokes's theorem to rewrite the second ter1n as a surface integral. Note that the surface S and the volume V must be large enough to enclose all currents in the region of interest. Since there are no other restrictions on the choice of the surface S kern1.5pt, we can extend it to infinity. In this case, the surface integral reduces to zero because dependencies of the various quantities at large distances from the currents (i.e., at points from which the
TimeVarying Fields and Maxwell's Equations
564
Chap. 7
entire distribution of currents looks just like an elemental current) are
IAI
~
1
, r
IHI ~
1 r2
,
and
ldsl ~
r
2
so that lim (H x A) · d s
r+oo
In other words, the magneto static energy Wm reduces to simply the volume integral ter1n, or (7.12) Note that since we let S extend to infinity, we must have V > oo, meaning that the integral in (7 .12) has to be carried out over all space in which the magnetic field is nonzero. Based on (7 .12), we can define the volume energy density for the magnetostatic field as (7.13) For linear26 and isotropic materials, where the per1neability µ is a simple constant, we have H · B = µH 2 , so that the magnetic energy density is Wm= ~µH 2 in units of Jm3. Alternatively, magnetic energy density can be expressed in terms of the B field 2 as Wm = (B / µ). The relationship between magnetic energy density and the magnetic field H is entirely analogous to the corresponding expression for the electrostatic field, 2 namely We= , found in Section 4.12. As in the electrostatic case, we note that the two alternative expressions for magnetic energy density, namely J · A and H · B, are quite different. The f or1ner implies that magnetic energy exists at places where currents exist and is zero where J = 0. However, the latter indicates that magnetic energy exists wherever the fields exist. As before, both points of view have merit, and it is neither necessary nor possible to determine which one is more ''correct." It is not possible to localize energy or to decide whether it is associated with the current or the field. Thus, the quantities J · A and H · B represent magnetic energy density only to the extent that their volume integral over space is the total stored magnetic energy.
!
iEE
!
!
!
!
26 Even
equation (7 .12) is not the most general form of magnetic energy applicable for nonlinear materials. The work done by external sources in establishing a magnetic field of B in material media, which by definition is the magnetic energy stored in that field, is given by B
Wm=
v
H·dB dv 0
where the integral is to be carried out over all space. For further discussion, see Section 2.15 of J. A. Stratton, Electromagnetic Theory, McGrawHill, New York, 1941. On the other hand, equations (7.12) and (7.13) apply to both linear and anisotropic materials, in which the permeability is a tensor, such as in the case of ferrites. See Section 5 of A. Sommerfeld, Electrodynamics, Academic Press, New York, 1952.
Sec. 7.3
565
Energy in a Magnetic Field
Example 7.10: Magnetic energy near long, straight, currentcarrying wires. Compare the magnetic energy stored per unit length by two separate infinitely long, straight wires of different radii a and b such that b > a; each wire has a circular cross section and carries a uniformly distributed current I. Solution: The B field at any position r in the vicinity of a wire carrying uniformly distributed current and having a circular cross section of radius a was determined in Example 6.10 to be
if> µair
B=
r µol
r>a
2nr
To determine the magnetic energy stored per unit length, we can integrate cylindrical volume of 1m axial length. In other words,
i (BJ/ µo) over a
1 Wm=2µo 1
µ512
2µo (2na2)2 o µ0/2
r4
4na 4
4
r=a r=O
II
+ ,.,.,o
[2
4n
II
[ln r] r
oo =
ra
,.,.,o
[2
16n
+ oo
which indicates that the energy residing in the magnetic field surrounding a 1m length of such a wire is infinite. This result is physically not surprising, since the very premise of sustaining a current I in an infinitely long wire is nonphysical. In reality, any such current carried in one direction by a long straight wire has to return via another wire or set of wires. This return current would generate B fields in the opposite direction, thus reducing the net total B field so that the stored magnetic energy is not divergent. An interesting aspect of the preceding expression for Wm is the fact that the first term, which represents the magnetic energy stored within the wire in the region r < a, is independent of the radius a of the wire. This internal magnetic energy given by (Wm)int = µ 01 2 /(16n) is the quantity represented by the internal inductance of a wire discussed in Section 6.7. In other words, using (6.32) we have L· _ 2(Wm)int _ /J,Q mt /2  8n
Although the total magnetic energy stored per unit length for a straight wire is divergent, the difference between the energy stored by two different wires of radii a and b is finite. Since the internal stored energy is the same for both radii, the difference between the total magnetic energies stored per unit length for the two different wires is given by 1 µ5/2 W W =     2 m m 2µo (2n) a
b
b
a
1 µ0/2 b  2 2nrdr = ln r 4n a
TimeVarying Fields and Maxwell 's Equations
566
Chap. 7
7.3.4 Magnetic Forces and Torques in Terms of Stored Magnetic Energy In Section 4.13, we saw that electrostatic forces can be expressed in ter1ns of spatial derivatives of stored electrostatic energy. The underlying principle used was called the principle of virtual displacement, which involves consideration of how the energy of a system changes for a small virtual change in geometry. In an analogous manner, magnetic forces and torques can also be derived from stored magnetic energy using the principle of virtual displacement. In this manner, one can usually calculate the lifting force of magnets. 27 In this section, we briefly discuss this topic with analogy to Section 4.13. In Section 4.13, we noted that the relationship between electrostatic force and stored energy depended on the premise under which we allowed the virtual displacement to occur. Specifically, it was possible to consider virtual displacement under the constraint of constant charge or constant voltage. For magnetostatics, we can consider a system of currentcarrying loops with constant currents or constant magnetic flux linkages. In the following, we consider only the case of constant current. Consider a solenoid with a movable, permeable core, where µ > µo, as shown in Figure 7 .17. A constantcurrent generator is attached to the coil, and the core of the solenoid is allowed to move a distance ~x, as shown. Assume that the movement of the core by an amount ~x changes the magnetic flux linked by the coil from \111 to \112. Note 2 that the magnetic energy stored in this system is given by Wm= ~LI , or alternatively Wm = ~I \II, since L = \II/I. Thus, the change in flux corresponds to a change in stored magnetic energy of
This additional energy must be provided by the generator in order to maintain the current I constant. The generator can do this by canceling out the induced emf by supplying a voltage of  °Vind, in which case the amount of work done by the generator is
x
Solenoid
~x
I I
I ~t
.. 
I I
Current source
Figure 7.17 Solenoid with a movable core. 27 For
a discussion at an appropriate level, see Chapter 10 of M. A. Plonus, Applied Electromagnetics, McGrawHill, New York, 1978.
Sec. 7.3
567
Energy in a Magnetic Field
where !l.t is the time duration over which the virtual displacement is assumed to occur. In order for energy to be conserved, we must have >
fl.Wm Fx =  !l.x
where Fx !l.x is the virtual work done to move the core a distance !l.x. In other words, F x is the magnetic force exerted on the core by the coil, and F x must be overcome in order to move the core. We have thus arrived at a result analogous to that found for electrostatic forces in Section 4.13. For the case shown in Figure 7.17, under the constantcurrent constraint, the H field everywhere inside the coil remains constant. With the magnetic energy density being 4µH 2 , and for µ > µo (i.e., paramagnetic or ferromagnetic material), the total stored magnetic energy decreases when the core moves to the right, so that less of it is within the coil. Thus, F x = dWml dx is negative, and the core experiences a force that tends to pull it within the coil, in a manner analogous to the case of the dielectric slab between parallel plates analyzed in Example 4.36. Note that if the material were diamagnetic, that is, ifµ < µo, then the direction of the force would be reversed, tending to push the core out of the coil. In practice, however, µ is significantly different from µo only for ferromagnetic materials, so that magnetic forces are significant only when the core is made out of such materials; in this case, the magnetic force tends to pull the core in, under constantcurrent conditions. More generally, it can be shown that, under the constraint of constant currents, the magnetic force acting on a circuit is given by [F]1 =const. = V Wm
If the circuit is constrained to rotate about the acting on it is given by
z axis, then the z component of the torque
Magnetic forces in terms of mutual inductance. The principle of virtual displacement can also be used to deter1nine magnetic forces and torques between currentcarrying circuits. Consider, for example, two parallel coaxial loops carrying currents 11 and 12 and with the z axis passing through their centers, as shown in Figure 6.40. Based on what we found in Section 7.3.2, and also discussed in Example 7.8, the magnetic energy stored in this system is given by
If the currents 11 and 12 are maintained (i.e., kept constant) in both circuits (presumably by driving them with current sources), then the mutual inductance between the two coils is the only quantity that varies as a result of any relative motion of the two coils. Thus, an infinitesimal displacement by an amount !l.z of either circuit with respect to the other
568
TimeVarying Fields and Maxwell's Equations
Chap. 7
results in a change of energy given by ~Wm
= 1112
~L12 ~z
~z
= 1112 ~L12
Note that ~Wm is a positive quantity when 11 and 12 have the same sign and when ~L12 is positive. Thus, the force between the two circuits, given by the gradient of Wm, tends to pull the circuits in the direction in which L12 increases most rapidly. In other words, the two circuits attract each other whenever their currents are in the same direction, since their mutual inductance obviously increases when they move closer. When 11 and 12 are in opposite directions, ~Wm is negative and the circuits repel one another. In general, it can be shown that under constantcurrent conditions, the magnetic force between two circuits is proportional to the gradient of mutual inductance or
and if the circuit is constrained to rotate about the torque acting on it is given by
z axis, then the z component of the
7.4 DISPLACEMENT CURRENT AND MAXWELL'S EQUATIONS
In a series of brilliant papers between 1856 and 1865, culminating in his classic paper, 28 James Clerk Maxwell formulated the complete classical theory of electromagnetics. 29 He provided a mathematical framework for Faraday's results, clearly elucidated the different behavior of conductors and insulators under the influence of fields, imagined and introduced the concept of displacement current, and inferred the electromagnetic nature of light. His theoretical framework predicted the existence of electromagnetic waves in the absence of any experimental evidence. His hypotheses were to be confirmed 23 years later (in 1887) in the experiments of Heinrich Hertz. 30 The crucial step in Maxwell's development of the theory of electromagnetics was his introduction of the notion of a displacement current. In the following section, we discuss this concept. 7.4.1 Displacement Current and the General Form of Ampere's Law
Before Maxwell developed his theory of electromagnetics, Ampere's circuital law was known as the relationship between a static magnetic field Hand the conductioncurrent 28 For
an excellent account with passages quoted from Maxwell's papers, see Chapter 5 of R. S. Elliott, Electromagnetics, IEEE Press, New Jersey, 1993. 29 J. C. Maxwell, A dynamical theory of the electromagnetic field, Phil. Trans. Royal Soc. (London), 155, p. 450, 1865. 30 H. Hertz, On the finite velocity of propagation of electromagnetic actions, Sitzb. d. Berl. Akad. d. Wiss., February 2, 1888; for a collected English translation of this and other papers by H. Hertz, see H. Hertz, Electric Waves, MacMillan, London, 1893.
Sec. 7.4
density
31
569
Displacement Current and Maxwell's Equations
Jc. From Chapter 6, the differential form of this relationship is V x H =Jc
Another fundamental relation concerning conductioncurrent density was derived in Chapter 5 on the basis of conservation of charge. This relation, known as the continuity equation and given in (5.7), is repeated here: ap
V ·Jc+
=0
at
(7.14)
In magnetostatics, we simply have V ·Jc= 0, since ap/at = 0 by definition. We can then relate Ampere's law to the continuity equation by taking the divergence of the for1ner, V · (V x H) = V · Jc
0 = V ·Jc
+)
where we have observed (see Appendix A) that the divergence of the curl of any vector is identically zero. We see that Ampere's law is indeed consistent with the principle of conservation of charge under static conditions. However, in the more general case when the field and source quantities are allowed to vary with time, Ampere's law, in its for1n used in Chapter 6, is clearly inconsistent with the fundamental principle of conservation of charge. Faced with this inconsistency, Maxwell developed the timevarying for1n of Ampere's circuital law by postulating an additional current term Ja such that V x H = Jc +
an at
= Jc + Ja
(7 .15)
where Jc is the free conduction current density and (7.16)
is the displacementcurrent density. Note that the divergence of (7 .15) gives V · (V x H) = 0 = V · Jc + V ·
an at
= V · Jc +
a at
(V · n)
Substituting for V · n = p from Gauss' law, we have ap
0 = V ·Jc+ at
Thus, we see that (7.15) is consistent with the continuity equation (7.14) and hence with the principle of conservation of charge. 31 The
current density term in V x H = J includes all forms of charge transport except for the bound magnetization current density V x M. In this section, we use the label Jc to distinguish this free current density (which in practice often is conduction current) from the displacement current term Jd.
TimeVarying Fields and Maxwell's Equations
570
Chap. 7
Equation (7.15) states that the net circulation (or curl) of H is the sum of the conductioncurrent density Jc plus the displacementcurrent density Ja = an I at. Note that this new term has the dimensions of a current density and indicates that V x H may be generated (or sustained) at a point in space (even in the absence of Jc) when a timevarying electric field is present. This relationship between n and H is analogous to Faraday's law (7. 7), which states that V x E may be generated at a point in space when a timevarying magnetic field is present. The combined action of the V x H and V x E equations is precisely what leads to the propagation of electromagnetic waves, as discussed in Section 7 .4.2 in connection with Figure 7 .20. Without the displacement current, electromagnetic waves do not exist. It is not surprising that this ter1n was not discovered earlier (e.g., in Ampere's experiments); in currentcarrying wires, and especially at low frequencies, the magnitude of Ja is much smaller than that of Jc (see the discussion that follows). On the other hand, for electromagnetic waves propagating in free space, where there cannot be any conduction current, the displacement current density Ja = an;at is the only current ter1n. The displacement current density J a consists of two terms corresponding to the time derivative of both the electric field and polarization vectors of n in (4.64):
Ja =
an = at
Eo
aE + aP at at
(7.17)
The first term states that a timevarying electric field generates a magnetic field. It is not surprising that the time rate of change of electric field E behaves like a current. In any given configuration of conductors, any increase (decrease) in E implies a buildup (decrease) of charge, which requires current flow. Consider a parallelplate capacitor having plates separated by free space, as shown in Figure 7. l 8a. A current le charges up the lefthand plate to a surface charge density Ps, which means that current moves
.. E
(a)
S1
Ic
Figure 7.18 Displacement current concept. (a) Flow of current le charges up the
c
(b)
lefthand plate, thus increasing the electric field. (b) Two different surfaces S1 and S2, enclosed by the same contour C. Note that S2 is shaped like a sack, with C being the circumference of its opening.
Sec. 7.4
Displacement Current and Maxwell's Equations
571
away from the righthand plate to create an opposing charge density  Ps. The resulting electric field between the capacitor plates is E = Ps / Eo. If the current flow is continuous, Ps (and thus E) increases with time. Thus, current flow into the capacitor is associated with a time rate of change of E (i.e., aE/at) between the capacitor plates. In effect, the displacement current, given by EoaE/at, ''takes over'' the conduction current in the freespace region between the capacitor plates. The physical nature of the second tertn, aPI at is quite different. This term is called the polarization current density: aP (7.18) Jp= at As P changes with time, the electrons in a dielectric are displaced with respect to their parent nuclei and move to and fro in accordance with aE/at. Although the charge never leaves the parent molecule, the backandforth motion constitutes a true alternating current. Indeed, we can check if this current satisfies the continuity equation (5.7) for bound charges: aP a apP (7.19) v · Jp = v · at = at cv · P) =  at where Pp is the bound polarization volume charge density. Hence in timevarying fields, the polarization current accounts for the creation of the bound charge in regions where v. p # 0. To further illustrate the need for a displacement current, consider an arbitrary contour C 1 encircling both the wire and two different surfaces S 1 and S2, as shown in Figure 7.18b. Considering surface S1, we have from Ampere's law
Jc · ds =le
H · dl = C1
S1
where le is the conduction current in the wire, which passes through S 1· When we consider surface S2, however, we have a problem. Without the displacement current term, we have no conduction current through the surface S2 so that we would have to conclude :fc H · dl = 0, which contradicts what we found for surface S 1 enclosed by 1 the same contour C1 ! This difficulty is resolved by including the displacementcurrent term. Let A be the area of the parallelplate capacitor and a be the separation between its plates. The capacitance is then given by C =EA/a. When a displacement current Id flows through the capacitor, the potential across its plates is deter1nined by the wellknown voltagecurrent relationship for a capacitor, namely d ld=Cdt
assuming that the material between the capacitor plates is a perfect dielectric with a = 0 (i.e., le = 0). But the potential difference across the plates is =Ea, where E is the electric field between the plates. Recalling from Example 4.26 that C =EA/a, we have
dE Ca d(EAE) ld=Ca = dt EA dt
d(EEA) d dt dt
572
TimeVarying Fields and Maxwell's Equations
Chap. 7
Note that if we neglect any fringing effects so that the electric field between the capacitor plates is unifor1n over the area A and zero elsewhere, the integral of D · d s over any arbitrary surface is simply equal to EEA. Therefore, we see that the choice of either surface S1 or S2 gives the same result for :fc H · dl, illustrating that le = /d and that consistent 1 results are obtained by including the displacementcurrent ter1n in Ampere's law. While we can construct a strong plausibility case for its existence, the real physical bases of the displacement current are the three experimental pillars of electromagnetics, namely, Coulomb's law (equation (4.1)), the BiotSavart law (equation (6.5)), and Faraday's law (equation (7.1)). After all, Maxwell introduced the 3D/3t term in order to reconcile these physical experimental facts with the principle of conservation of charge (i.e., the continuity equation (5.7)). The conclusive experimental proof of the presence of displacement current was provided by Hertz's experiments in 1887, when he demonstrated that electromagnetic waves (whose existence, as we discuss in the next section, depend on the displacement current) can indeed travel through free space. Example 7.11: A parallelplate capacitor. Consider a parallelplate capacitor consisting of two metal plates of 50cm2 area each, separated by a porcelain layer of thickness a = 1 cm (for porcelain, Er= 5.5 and a= 1014 Sm 1). If a voltage (t) = 110J2cos(l20nt) Vis applied across the capacitor plates, find (a) le, (b) lct, and (c) the total current I through the capacitor. Solution: E(t) =
Using (4.48), the electric field in the porcelain layer is (t)
a
110J2cos(l20nt) V ~ 4 1 = = (l.lv2) x 10 cos(l20nt) Vm0.01 m
The conduction (le) and the displacementcurrent (Jct) densities are then (a)
le= aE(t)
= (10 14
Sm 1)[(1.1J2)
4
x 10 cos(l20nt) Vm 1 ]
= l.1J2 x 10 10 cos(l20nt) Am2
(b) lct =
dD dt
= E
dE(t) dt
dE(t)
= ErEo  
dt
= (5.5 x 8.85 x 10 12 Fm 1)[l.lh x 104 x 120n sin(l20nt) V(ms) 1]
~ (2.86 x 104 ) sin(l20nt) Am 2
Note that the conduction current is in phase with the electric field, whereas the displacement current is 90° out of phase. Also note that the amplitude of the conduction current is ""5 x 105 times smaller than the displacement current, as expected since porcelain is an excellent insulator. (c) The total current I can be found by integrating the total current density over the crosssectional area of the capacitor plates as
(Jc +Jct) · ds = (le + lct)A
I = A
573
Displacement Current and Maxwell's Equations
Sec. 7.4
since Jc and Jd are both approximately uniform and perpendicular to the plates throughout the dielectric region. Note also that IJclmax
~lt = ~2t
(7.27)
where fi is the unit vector perpendicular to the interface and outward from medium 2 (shown as fi in Figure 7.22b). 40 That
the continuity equation can be derived from equations (7.2lb) and (7.21c) indicates that Maxwell's equations (7.2l b) and (7.21c) are not entirely independent, if we accept conservation of electric charge as a fact.
Sec. 7.5
583
Review of Maxwell's Equations
/\
n
~w
(b)
(a)
Figure 7.22 Interfaces between two different materials. The boundary conditions for the electromagnetic fields are derived by applying the surface integrals to the cylindrical surface as shown in (a) and the line integrals to the rectangular contour in (b).
(2) It can be shown by applying (7.21c) to the same contour C in Figure 7.22b that the tangential component of the magnetic field 'Je is continuous across any interface: )
'/1f, 1t = '/1f, 2t
(7.28)
except where surface currents (Js) may exist, such as at the surface of a perfect conductor (i.e., a = oo ):
fi x 'Je1 =
:J s
Noting that the field 'Je2 inside the perfect conductor is zero, as discussed in Section 6.8.4. (3) It can be shown by applying (7.2lb) to the surface of the cylinder shown in Figure 7.22a that the nor1nal component of electric flux density ~ is continuous across interfaces, except where surface charge (Ps) may exist, such as at the surface of a metallic conductor or at the interface between two lossy dielectrics (a
# 0): ~In
)

~2n = Ps
(7.29)
(4) A consequence of applying (7.2ld) to the surface of the cylinder in Figure 7.22a is that the nor1nal component of the magnetic field
:J + at
v ·~ =0 In addition, a complete specification of the field quantities involves the constitutive relations, which in simple media (i.e., linear, isotropic, homogeneous media) are ~ = µ 1 ~ and= E~. PROBLEMS 7.1
Stationary rectangular loop. Consider a fixed singletum rectangular loop of area A with its plane perpendicular to a uniform magnetic field. Find the voltage induced across the terminals of the loop if the magnetic flux density is given by (a) B (t) =Bateat and (b) B(t) =Boeat sin(wt).
7.2
Stationary circular loop. Consider a 20tum circular loop of wire of 15 cm diameter with its plane perpendicular to a uniform magnetic field, as shown in Figure 7.6. If the magnetic flux density B is given by
B = 10 cos(l20nt)
G
find the rms value of the induced current through the 1on resistor at the following time instants: (a) t = 0, (b) t = 10 ms, (c) t = 100 ms, and (d) t = 1 s.
7.3
1\vo circular coils. Two circular coils of 5 cm radius each have the same axis of symmetry and are 1 m apart from each other. Coil 1 has 10 turns and coil 2 has 100 turns. (a) If an alternating current of 100 A amplitude and 1 kHz frequency is passed through coil 1, find the amplitude of the induced voltage across the open terminals of coil 2. (b) Repeat part (a) for a frequency of 10 kHz.
7.4
1\vo concentric coils. Consider the two concentric coils shown in Figure 7.9. The magnetic flux density produced at the center of the two coils due to the current I flowing in the larger coil is given by B = z2.5 x 1051 T
(a) If the larger coil has 30 turns, find its radius. (b) If the smaller coil has 75 turns and its radius is 1 cm, find the total flux linking the smaller coil. (c) If the current I in the larger coil is given by I (t) = 10 cos(l20nt) A, find the induced current through the resistor R assuming R = ion.
7.5
Triangular loop and long wire. An equilateral triangular loop is situated near a long currentcarrying wire, as shown in Figure 7.23. The wire is part of a power line carrying
Chap. 7
587
Problems
60Hz sinusoidal current. An ac ammeter inserted in the loop reads a current of amplitude 1 mA. Assume the total resistive impedance of the loop to be O.OlQ. Find the amplitude of the current I in the long wire.
a= 0.1 m
d = 0.2 m
I
7.6
Figure 7.23 Triangular loop and long wire. Problem 7 .5.
Faraday's law. Consider a circular loop (radius a =5 cm) of wire lying in the xy plane with its center at the origin, in the presence of a zdirected magnetic field B(r, t) = zBo [1  (10 m 1)r] cos(2nf t)
7.7
where r is the polar coordinate and Bo= 10 mT (milliwebers/m2 ). The loop wire is made up of copper (a =5.8x107 S/m) and has a crosssectional area of 1 mm2 • It is known that the wire would melt if the total current I flowing through it exceeds 20 A (/ > 20 A). Stating all assumptions, calculate the maximum frequency f for which this wire can be used in the presence of this magnetic field. Toroidal coil around a long, straight wire. A long, straight wire carrying an alternating current of I (t) = 100 cos(377t) A coincides with the principal axis of symmetry of a 200turn coil wrapped uniformly around a rectangular, toroidalshaped iron core of inner and outer radii a= 6 cm and b = 8 cm, thickness t = 3 cm, and relative permeability µr = 1000, as shown in Figure 7.24. Calculate the induced voltage °Vind between the terminals of the coil. (This type of coil, called a current transformer, is used to measure the current in a conductor wire that passes through it.)
200turn coil
~==J:::i
Iron core
t µr= 1000
I
7.8
Wire
Figure 7.24 Toroidal coil around long wire. Problem 7.7.
Current transformer. A current transformer is used to measure the current in a highvoltage transmission line, as shown in Figure 7 .25. The circular toroidal core has a mean diameter of 6 cm, circular cross section of 1 cm diameter, and relative permeability of µr = 200. The winding consists of N = 300 turns. If the 60Hz current of amplitude 1000 A flows through the highvoltage line, find therms value of the open circuit voltage induced across the terminals of the toroid.
TimeVarying Fields and Maxwell's Equations
588
Chap. 7
6cm
~,J...
100 Mm
Sec. 8.2
TimeHarmonic Uniform Plane Waves in a Lossless Medium
613
wavelength ranges of the electromagnetic spectrum and selected applications for each range. Maxwell's equations and the results derived from them thus encompass a truly amazing range of physical phenomena and applications that affect nearly every aspect of human life and our physical environment. In this section, we briefly comment on a few of the many applications listed in Table 8.1. At frequencies in the ultraviolet range and higher, physicists are more accustomed to thinking in ter1ns of the associated energy level of the photon (a quantum of radiation), which is given by hf, where h ~ 6.63 x 1034 Js is Planck's constant. Cosmic rays, consisting typically of photons at energies 10 MeV or greater, are constantly present in our universe; they ionize the highest reaches of Earth's atmosphere and help maintain the ionosphere at night, in the absence of solar radiation. Shortduration bursts of yrays, which bathe our solar system (and our galaxy) about three times a day, are believed to be produced in the most powerful explosions in the universe, releasing (within a few seconds or minutes) energies of 1051 ergs more energy than our sun will produce in its entire ten billion years of existence. 8 Brief flashes of yrays have been observed to be originating from Earth, 9 and represent the highest natural energy emission within Earth's atmosphere. These bursts of yrays are generated from electrons accelerated by strong electric fields associated with thunderstorms. The exact generation mechanism is still an active area of research; recent theoretical work has shown that sufficient electron acceleration for yray production may result from strong electric fields within thunderclouds, 10 from intense electromagnetic impulses generated by individual lightning flashes, 11 or from strong electric fields above a thundercloud after a large lightning discharge to ground. 12 It is also interesting to note that only a very narrow portion of the electromagnetic spectrum is perceptible to human vision, namely the visible range. We can also ''feel'' infrared as heat, and our bodies can be damaged by excessive amounts of microwave radiation, Xrays, and yrays. Applications of Xrays, ultraviolet, visible, and infrared light are far too numerous to be commented on here and include vision, lasers, optical fiber communications, and astronomy. The part of the electromagnetic spectrum between the lowfrequency end of infrared and the highfrequency end of the millimeter range is called terahertz radiation. The frequency in this region is measured in terahertz (THz), or 10 12 Hz, corresponding to wavelengths in the submillimeter range. This frequency band is used by astronomers to study the chemical composition of celestial bodies. In addition, artificial generation of terahertz waves is an active area of research. Typically, optical lasers are modified to generate 8 G.
J. Fishman and D. H. Hartmann, Gammaray bursts, Scientific American, pp. 4651, July 1997. 9 G. J. Fishman, P. N. Bhat, R. Mallozzi, J. M. Horack, T. Koshut, C. Kouveliotou, G. N. Pendleton, C. A. Meegan, R. B. Wilson, W. S. Paciesas, S. J. Goodman, and H. J. Christian, Discovery of intense gammaray flashes of atmospheric origin, Science, 264, pp. 13131316, May 1994. 10 J. R. Dwyer and D. M. Smith, A comparison between Monte Carlo simulations of runaway breakdown and terrestrial gammaray flash observations Geophys. Res. Lett., 32(22), 2005. 11 U. S. Inan and N. G. Lehtinen, Production of terrestrial gammaray flashes by an electromagnetic pulse from a lightning return stroke, Geophys. Res. Lett. 32(19), 2005. 12 N. G. Lehtinen, T. F. Bell, and U. S. Inan, Monte Carlo simulation of runaway MeV electron breakdown with application to red sprites and terrestrial gamma ray flashes, J. Geophys. Res. Space Phys., 104(Al 1), 1999.
614
Waves in an Unbounded Medium
Chap.8
radiation at the lower submillimeter frequency range, such as by mixing two signals or using a pulsed laser to dislodge electrons from a semiconductor that are then accelerated by an applied field. 13 Technological advancements in THz radiation sources have enabled a growing number of practical applications that leverage this frequency band. Unlike visible or infrared light, some frequencies of terahertz radiation can penetrate lowmoisture biological tissue, clothing, and some polymers, enabling novel imaging technologies for a wide variety of medical, security, and quality control applications. The submillimeter band is also a candidate for use in high data rate wireless transmission applications. Each decade of the electromagnetic spectrum below the millimeter range of frequencies is divided 14 into designated ranges, with the acronyms indicated in Table 8.1: extremely high frequency (EHF), super high frequency (SHF), ultra high frequency (UHF), very high frequency (VHF), high frequency (HF), medium frequency (MF), low frequency (LF), very low frequency (VLF), ultra low frequency (ULF), super low frequency (SLF), extremely low frequency (ELF). The microwave band of frequencies is vaguely defined as the range from 300 MHz up to 1 THz, including the millimeter range. It is extensively utilized for radar, remote sensing, and a host of other applications too numerous to cite here. 15 In radar work, the microwave band is further subdivided into bands with alphabetical designations, which are listed in Table 8.1. The VLF range of frequencies is generally used for global navigation and naval communications, using transmitters that utilize huge radiating structures. 16 Although VLF transmissions are used for global communications with surface ships and submarines near the water surface, even lower frequencies are required for communication with deeply submerged submarines. The Sanguine system operated by the U.S. Navy utilizes two sets of large (22.5 km length) orthogonal horizontal antennas, one located in Wisconsin and the other in Michigan at a distance of 240 km, and operates 17 at 72 to 80 Hz. The lowest frequencies of the experimentally investigated electromagnetic spectrum are commonly used to observe socalled micropulsations, which are electromagnetic waves at frequencies 0.00110 Hz generated as a result of largescale currents flowing in the earth's auroral regions and by the interaction between the earth's magnetic field 18 and the energetic particles that stream out of the sun in the form of the solar wind.
13 For
a summary of some modem terahertz generation technologies, see Nagatsuma, T. Generating Millimeter and Terahertz Wave, IEEE Microwave Magazine, 2009. 14 There is a certain arbitrariness in the designations of these frequency ranges. In geophysics and solar terrestrial physics, the designation ELF is used for the range 3 Hz3 kHz, while ULF is used to describe frequencies typically below 3 Hz. 15 See J. Thuery, Microwaves: Industrial, Scientific and Medical Applications, Artech House, BostonNorwood, Massachusetts, 1992. Also see 0. P. Gandhi, editor, Biological Effects and Medical Applications of Electromagnetic Energy, PrenticeHall, 1990. 16 See A. D. Watt, VLF Radio Engineering, Pergamon Press, New York, 1967; J.C. Kim and E. I. Muehldorf, Naval Shipboard Communications Systems, PrenticeHall, Englewood Cliffs, New Jersey, 1995. 17 C. H. Richard, Sub vs. Sub, Orion Books, New York, 1988; M. F. Genge and R. D. Carlson, Project ELF Electromagnetic Compatibility Assurance Program, IEEE J. Oceanic Eng., 9(3), pp. 143153, July 1984. 18 J. A. Jacobs, Geomagnetic Micropulsations, SpringerVerlag, New York, Heidelberg, and Berlin, 1970; J. K. Hargreaves, The SolarTerrestrial Environment, Cambridge University Press, 1992.
Sec. 8.3
Plane Waves in Lossy Media
615
These natural signals are also often used for geophysical prospecting and magnetotelluric studies. 19 Another potentially very important practical use of this band may yet emerge, based on experimental evidence of detectable electromagnetic precursors produced many hours prior to major earthquakes. 20
8.3 PLANE WAVES IN LOSSY MEDIA Many of the more interesting electromagnetic applications involve the interactions between electric and magnetic fields and matter. The manner in which waves interact with matter is introduced in Chapter 11. For now, we represent the microscopic interactions of electromagnetic waves with matter in ter1ns of the macroscopic parameters E, µ, and a. In general, most dielectric media exhibit small but nonzero conductivity or complex permittivity and can absorb electromagnetic energy, resulting in the attenuation of an electromagnetic wave as it propagates through the medium. The performance of most practical transmission lines, and of other devices that convey electromagnetic energy from one point to another, is limited by small losses in conductors or dielectrics. The inherently lossy nature of some media (e.g., seawater, animal tissue) determines the range of important applications (e.g., submarine communications and medical diagnostic implants). All media exhibit losses in some frequency ranges; for example, although air is largely transparent (lossless) over the radio and microwave ranges, it is a highly lossy medium at optical frequencies. The fact that upper atmospheric ozone is lossy at ultraviolet frequencies (i.e., it absorbs ultraviolet light) protects life on Earth from this harmful radiation. When a material exhibits a nonzero conductivity a, the electric field of a propagating wave causes a conduction current of Jc = a E to flow. This current, which is in phase with the wave electric field, leads to dissipation of some of the wave energy as heat within the material, with the power dissipated per unit volume being given by E · J, as discussed in Section 5.8. This dissipation requires that the waves electric and magnetic fields attenuate with distance as they propagate in the lossy material, much like the attenuation of voltage and current waves propagating on a lossy transmission line as discussed in Section 3.7. To deter1nine the characteristics of uniform plane waves in lossy media, we start with Maxwell's equations and follow a procedure quite similar to that used in the previous section. In a conducting (or lossy) medium the J term in (7.23c) is nonzero even in the absence of external sources, since a conduction current of Jc = a E flows in response to the wave' s electric field. We then have V x H = aE + jwEE 19 M.
(8.15)
N. Nabighian, (Ed.), Electromagnetic Methods in Applied Geophysics, Vol. 2: Application, Parts A and B, Society of Exploration Geophysics, 1991. 20 A. C. FraserSmith, A. Bernardi, P. R. McGill, M. E. Ladd, R. A. Helliwell, and 0. G. Villard, Jr., Lowfrequency magnetic field measurements near the epicenter of the Ms 7 .1 Loma Prieta earthquake, Geophys. Res. Lett., 17, pp. 14651468, 1990; also see B. Holmes, Radio hum may herald quakes, New Scientist, p. 15, 23/30 December 1995.
616
Waves in an Unbounded Medium
Chap.8
and, taking the curl of (7 .23a), we have Vx Vx E
=V(V · E) 
2
V E = jwV x B
Using the fact that V · E = 0 for a simple, sourcefree medium, and substituting from (7.23c), we find 2
V E  j wµ(a
+ jwE)E =
0
or (8.16) where y = .Jjwµ(a + jwE) is known as the propagation constant (or wave number) and is in general a complex quantity, which can be expressed in terms of its real and imaginary parts as y =a+ j {3. Note that the complex propagation constant y for uniform plane waves in an unbounded lossy medium is analogous to the complex propagation constant y given in (3.60) and derived in Section 3.7 for propagation of voltage and current waves on a lossy transmission line. Realizing that (8.16) represents three scalar equations similar to (8.7), we limit our attention once again to unifor1n plane waves propagating in the z direction (i.e., all field quantities varying only in z ), with the electric field having only an x component, for which (8.16) becomes (8.17) which, when we replace Ex(z) with V (z), is identical to equation (3.59) in Section 3.7 that describes the variation of the voltage phasor V (z) on a lossy transmission line. The general solution of (8 .17) is Ex(Z) = Cieyz
+ C2e+yz
= C1eazejf3z
+ C2e+aze+jf3z "...
(8.18)
For a,{3 > 0, the two terms Ex+(z) and Ex(z) represent waves propagating in the +z and z directions, respectively. Note that the constants C 1 and C2 are in general complex numbers. The instantaneous electric field can be found from (8.18) in the same way21 as described in (8.12). We find ~x (z, t) = C1eaz cos(wt  {3z)
'...,,
+ C2e+az cos(wt + f3z)
~: (z,t)
(8.19)
~; (z,t)
assuming C 1 and C2 to be real. The nature of the waves described by the two terms of (8.19) is shown in Figure 8.6. We see, for example, that the wave propagating in the +z 21 ffie{ C
constant.
1 eaz ej /3z ejwt} = C1 eazme{ej(wt/3z)} = C1 eaz
cos(wt  {Jz), where
C1
is assumed to be a real
Sec. 8.3
617
Plane Waves in Lossy Media ~~ (z, t)
~x (z,
.. v
VP
p
  
t)
•

,/
,/
_. _. _.
Cz
1

 
z/A.

Cz

3
 
.._

z/A. .......
.......
Figure 8.6 Snapshots of waves in a lossy medium. The two terms of equation (8.19) plotted as a function of z /A. at wt = 0.
direction has a decreasing amplitude with increasing distance at a fixed instant of time. In other words, the wave is attenuated as it propagates in the medium. The rate of this attenuation is given by the attenuation constant a, in units of npm 1 (see Section 3.7.1 for a discussion of the unit nepers per meter [npm 1] ). The quantity f3 (in radm 1) determines the phase velocity and wavelength of the wave and is referred to as the phase constant. We shall see below that the wavelength of a uniform plane wave in lossy materials can be substantially different from that in free space. We note once again that all aspects of the solution given in (8.19) are entirely analogous to that given in (3.66) for voltage and current waves on a lossy transmission line. In this connection, the similarity between Figure 8.6 and Figure 3.54a should also be noted. Note that the complex propagation constant y is given by y =a+ j f3 = Jjwµ(a
+ jwE) =
a
jwffe1 1j 
(8.20)
(J)E
Explicit expressions for a and f3 can be found 22 by squaring and equating the real and imaginary parts. We find I
µE 2
1

1+
a
 1/2 "2
1
WE
(8.21)

22 We have y 2 =(a+ jf3) 2 =jwµ(a
+ jwE)
a 2  {3 2 = w2µE
and
Real part
2af3
= wµa
Imaginary part
Squaring and adding the two equations and talcing the square root, we have
+ (a2
+ {32)2 =
w4µ2E2 1 + (;E )2 + a2
+ {32 =
w2µE
1 + (;E )2
Equation (8.21) follows by adding the expressions for (a 2 + {3 2) and (a 2  {3 2).
618
Waves in an Unbounded Medium
Chap.8
and µE
f3 = w
a
1+
2
2
WE
1/2
(8.22)
+1
Consider now only the wave propagating in the of (8.18):
+z
direction, the first term (8.23)
The wave magnetic field that accompanies this electric field can be found by substituting (8.23) into (7.23a), as we did for the lossless case, or by making an analogy with the losslessmedia solution obtained in the previous section. Following the latter approach, we can rewrite (8.15) as V x H = aE + jwEE =
jwEeffE
where E eff
=
j
E 
a/ w
(8.24)
It thus appears that the solutions we obtained in the previous section for a lossless medium can be used as long as we make the substitution E > Eeff. On this basis, the propagation constant may be written in a more compact form as
Y
•
=] W
(8.25)
µEeff
The intrinsic impedance of a conducting medium, which is also a complex quantity, can be found as µ 
rJ c =
1
rJ c Iei 11 =
µ 1
µ

Eeff
\
E
1
a •
Jw
ei (1/2) tan 1 [a /(wE)]
E
1+

/a WE
2 1/4
(8.26)

and using (8.23) the associated magnetic field H is
H = yH+(z) = Y
The instantaneous magnetic field
~ (z, t)
y
1 C1eaze1'{3 z 'fJc
can be found as
(8.27)
Sec. 8.3
619
Plane Waves in Lossy Media {3z = (27r/J..)z
5
5
0
0
1~
/   wt= 7r/2 wt=O
y
1 1 1
z l 1
x
1 0 ,_ __
0
1
5
5 0
0 0 1 1
wt= 7r
1
0
1
wt= 37r/2
1
Figure 8.7 Time snapshots of the electric and magnetic fields in a conducting medium. The curves shown are for C 1 = 1, > 1
Therefore, at 10 MHz, seawater behaves like a good conductor. The skin depth for seawater is given by (8.41): 1
8'.::::'.
;:::==================================== Jn x 107 rads 1 x 4n x 107 HIn 1 x 4 SInl
1
4n
Ill '.::::'.
8 cIll
The results clearly indicate that coilllllunication through seawater at 10 MHz is not feasible, since the skin depth is only 8 cin. However, 10 MHz electroinagnetic waves do penetrate lake water, Inaking coininunication possible at relatively shallow depths of tens of Ineters.
634
Waves in an Unbounded Medium
Chap.8
Example 8.12: VLF waves in the ocean. The electric field component of a uniform plane VLF electromagnetic field propagating vertically down in the z direction in the ocean (a = 4 Sm 1 , Er =
81,
/Lr =
1) is approximately given by ~(z, t) = XEoeaz cos(6n x 103 t  {3z)
where Eo is the electric field amplitude at z = o+ immediately below the airocean interface, at z = 0. Note that the frequency of the wave is 3 kHz. (a) Find the attenuation constant a and phase constant {3. (b) Find the wavelength A., phase velocity vp, skin depth 8, and intrinsic impedance rJc, and compare them to their values in air. (c) Write the instantaneous expression for the corresponding magnetic field, 'Je(z, t). (d) A submarine located at a depth of 100 m has a receiver antenna capable of measuring electric fields with amplitudes of 1 IL Vm 1 or greater. What is the minimum required electric field amplitude immediately below the ocean surface (i.e., Eo) in order to communicate with the submarine? What is the corresponding value for the amplitude of the magnetic field?
Solution: tan 8c =
At w = 6n x 103 rads 1 , using (8.28) the loss tangent of the ocean is given by a
~
WErEo
4 6n
x 103 rads 1 x
81
x
(8.85
x
10 12 Fm 1)
~
2.96 x 105
>> 1
Therefore, the ocean acts as a good conductor at f = 3 kHz. (a) Using the approximate expressions given by (8.39), we have
a=f3~
W/,la
(6n x 103 rads 1 )(4n x 107 Hm 1)(4 Sm 1)
2
2 ~ 0.218 npm 1 or radm 1
(b) The wavelength in the ocean is
2n
2n A. = f3 ,. . ., 0.218 ~ 28.9 m Since in air A. ~ 100 km at 3 kHz, the wavelength in the ocean is approximately 3464 times smaller than that in air. The phase velocity is vp =
f
A.~ 3
x
103
x
28.9 ~ 8.66
x
104 ms 1
Compared with c ~ 3 x 108 ms 1 in air, the phase velocity in the ocean at 3 kHz is approximately ,.....,3464 times smaller. The skin depth and the intrinsic impedance in the ocean are given by (8.41) and (8.40), respectively: 8 = 1/a
~
4.59 m
(4n x 10 Hm )(6~ x 10 rads ) eJ45° ~ . x 2eJ45° n 7 70 10 4 Sm7
r/c =
1
3
1
Compared with rJ ~ 377Q in air, the magnitude of r/c in the ocean is approximately 4900 times smaller. In addition, r/c in the ocean is a complex number with a phase angle of ,.....,45°.
Sec. 8.4
Electromagnetic Energy Flow and the Poynting Vector
635
(c) Using (8.27), the corresponding magnetic field is given by
'Ji(z, t) =
y Eo
l17cl
eaz cos(6n x 103 t

{Jz 
Emin = 1 µVm1
Therefore, the minimum value for Eo to establish communication with the submarine located at a depth of 100 m is Eo ~ 2.83 kVm 1 The corresponding value for Ho is
Ho =
Eo
lrJc I
~
13Eo
~
36.8 kAm
1
8.4 ELECTROMAGNETIC ENERGY FLOW AND THE POYNTING VECTOR Electromagnetic waves carry power through space, transferring energy and momentum from one set of charges and currents (i.e., the sources that generated them) to another (those at the receiving points). Our goal in this section is to derive (directly from Maxwell's equations) a simple relation between the rate of this energy transfer and the electric and magnetic fields ~ and 'Ji. We can in fact attribute definite amounts of energy and momentum to each elementary volume of space occupied by the electromagnetic fields. The energy and momentum exchange between waves and particles is described by the Lorentz force equation (6.51). This equation represents the forces q and velocity v) by electric and magnetic exerted on a charged particle (having a charge fields: ~elec = q~ and ~mag = qv x C1y...... .,,.,.  ....... .......
/ I
'
/
',
\
wt= 7r
'wt=O
I
I

~ \
', '
I I
.......
.......
wt= 37r/2
...... ......
/
y
'\\
I I I I I I \ \
~
'
\ \
I I I I

~ I \
\
/
\
I \
' '  wt= 37r/2
I
/ /
/
.......

(a)
Cix< C1y
'
/
\
I I \
.......
I
I
wt= 7r
/

(b)
Figure 8.20 Elliptical polarization. The loci of the tip of the total electric field vector shown at the origin for an elliptically polarized wave propagating in the +z direction represented by (8.67) with~ = n /2. (a) The major axis of the ellipse is along the x axis when C1x > C 1y · (b) The major axis of the ellipse is along they axis when C 1x < C1y · As in the case of circular polarization, it can be shown using the righthand rule that this wave is righthand elliptically polarized (RHEP).
wt=O
660
Waves in an Unbounded Medium
I I I
I '
1
',
1
',
I I
......
' ......
I
y
'
7r/2 < ( < 7f
I I I
I
;
I
(= 7r/2
I '
;
I I ......
I
I I
0 < ( < 7r/2
Chap.8
I ......
1' , ......
I
1
I I
'
7f < ( < 37r/2
37r/2 < ( < 27f
(= 37r/2
Figure 8.21 Elliptical polarization ellipses. The loci of the tip of the total electric field vector for an elliptically polarized wave as given by (8.67) for different ranges of values of~· It is assumed that C1x > C1y· Top panels are LHEP, whereas the bottom panels are RHEP waves. The straightline loci for the cases of~ = 0 and n, corresponding to linear polarization, are not shown.
at 45° with respect to the xaxis to elliptical (0 < ~ < re /2) to circular (~ = re /2) to elliptical (re /2 < ~ < re) and back to linear tilted at 45° (~ =re). We can show that the tip of the total electric field vector does indeed trace an ellipse by eliminating wt  f3z between the two equations of (8.67) and rewriting them in the for1n62 2
+
2
 2
~
~
x Y C1x C1y
cos ~ = sin ~ 2
(8.68)
This equation represents a tilted polarization ellipse, as shown in Figure 8.22. The major and minor axes of this ellipse do not coincide with the x and y axes of the rectangular coordinate system; rather, they coincide with the x' and y' axes obtained by rotating the x and y axes by a tilt angle l/f, as shown in Figure 8 .22, given by (8.69) where (8.70) The principal semiaxes Ax' and Ay' (in other words, the halflengths of the principal axes) can be written in ter1ns of C lx, Cly, ~, and l/f as
Ax' = Ay' =
Cfx cos l/f + Cfy sin l/f 2
2
+ C1xC1y sin(2l/f) cos~
Cfx sin l/f + Cfy cos l/f 2
2
C1xC1y
sin(2l/f) cos~
1/2
1/2
62 To eliminate wt  fJz, we can expand cos(wt  fJz +~),substitute cos(wt  /Jz) = ~x(z,t)/C1x, and use sin(·) = ./1  cos2(·). See Principles of Optics, M. Born and E. Wolf, 5th ed., Pergamon Press, Oxford, 1975, pp. 2430.
Sec. 8.5
661
Polarization of Electromagnetic Waves
x'
_______ ,..
Figure 8.22 The tilted polarization ellipse.
From power considerations, Ax', Ay', C ix, and
are related by
Cly
One can rewrite the equation of the polarization ellipse given by (8.68) with respect to the rotated axes x' and y' as 2
2
+
=1
where ~ x' and ~y' are the components of the wave electric field in the rotated coordinate system. Also of interest is the socalled axial ratio, AR, which is defined as the ratio of the major to the minor axes. If Ax' > Ay', then AR = Ax'/Ay', whereas if Ax' < Ay', then AR= (Ax' /Ay') 1, the value of which satisfies 1 ~AR~ oo. Example 8.22: Elliptical polarization. An electromagnetic wave travels in the x direction with its magnetic field given by 'K(x, t) = y12 cos(wt  fJx)
+ z8 cos(wt 
fJx
+ 70°)
in mAm 1 . Find the tilt angle, axial ratio, and rotation sense of the polarization ellipse.
Solution:
From geometry, we have C1y tana = C1z
8 12
2 3
662
Waves in an Unbounded Medium
Chap.8
from which 2 tan a = 2.4 1  tan2 a
tan(2a) =
So, the tilt angle measured from the z axis is given by
1/f =
i tan 1 [ tan(2a) cos ~] = i tan 1 [ (2.4) cos 70°] ~ 19.7°
The principal semiaxes Ax' and Ay' are found as
Ay' ~ (8 cos (19.7°) + 12 sin (19.7°) + 8 x 12 sin(2 x 19.7°) cos(70°))
112
~ 12.5
Ay' ~ (8 sin (19.7°) + 12 cos (19.7°)  8 x 12 sin(2 x 19.7°) cos(70°))
112
~ 7.23
2 2
2
2
2
2
2 2
So the axial ratio is 12.5 7.23
AR~
~
1.73
Figure 8.23 shows the polarization ellipse and the sense of rotation of the magnetic field vector as a function of time at the origin. Since the magnetic field vector rotates around the ellipse in the clockwise direction as shown, this is a LHEP wave.
z 8 \ \
\ \ \ \ \
19.7°
\ \
\ \ \ \
\ \ \ \ \ \
\
Figure 8.23 Polarization ellipse of Example 8.22.
Sometimes it is helpful to express an elliptically polarized wave as a sum of two oppositely rotating circularly polarized waves, as shown in Figure 8.24. To see this decomposition analytically, note that the total electric field phasor for a lefthand circularly polarized wave can be written as
Sec. 8.5
663
Polarization of Electromagnetic Waves
w ,,,.
/
A I
y
I /I
/I/ / //
I I
t ....
I  ..... , I I ..... I .................... I   .1 Cl2 ,. .......... © I .....,
,I
w
\

,~
I
t
I \ \
\ \
'
\
\
\ \ \ \
\
 + 
'
I I
\ \ \
I I
I '
''
''
~2 \     + \
''
'
''
'
..... 
/
/
I
I
I I I
w
/ /
'
..... ............... .....
___ __
_,,,..*""""'
""'
/
y
z 1• 
x
Figure 8.24 The decomposition of an elliptically polarized wave into two circularly polarized waves. Any elliptically polarized wave (~) can be decomposed into two counterrotating circularly polarized components ~1 and ~2· In the limiting case when~ is linearly polarized (i.e., { = 0 in (8.67)), 1~1 I = 1~2 1 . Otherwise, the sense of rotation of~ is in the direction of the larger of ~ 1 and ~ 2 .
whereas that for a RHCP wave is
the sum of the electric fields for these two waves is then
which is an elliptically polarized wave (since ~x and ~Y have different magnitudes and are 90° out of phase). Note that for elliptically polarized waves in which the major and minor axes are not aligned with the x (or y) and y (or x) axes, a similar type of decomposition can be used after rotating the coordinate system to align the polarization axes with the x (or y) and y (or x) axes.
664
Waves in an Unbounded Medium
Chap.8
Example 8.23: Two linearly polarized waves. Two linearly polarized waves propagating in the same direction at the same frequency are given by E1 (y) = iC1 ej f3y E2(y) = zC2ejf3y d 8 where C1, C2, and () are constants and f3 = wffe. Find the polarization of the sum of these waves for the following cases: (a) () = 0°, (b) () = n /2, (c) () = n /2 and C1 = C2 , and (d) () = n.
Solution:
The realtime expression for the sum of these two waves can be found as
= iC1 cos(wt  fJy)
+ zC2 cos(wt 
fJy
+ ())
(a) When () = 0°, the two components of the total electric field are in phase, therefore, resulting in a linearly polarized (LP) wave oscillating in the direction as shown in Figure 8.25a. (b) When () = n /2, the two components are 90° out of phase. Since the amplitudes are in general different, the total wave is elliptically polarized. Figure 8.25b shows the variation of the total electric field at y = 0 as a function of time (for C2 > C1) from which we can conclude that the wave is righthand elliptically polarized (RHEP). (c) When () = n /2 and C 1 = C2, the wave is circularly polarized (i.e., RHCP), the time variation of which is shown at y = 0 in Figure 8.25c. (d) When () = n, the two components are 180° out of phase. Rewriting the total electric field yields ~(y, t) = xC1 cos(wt  fJy)
+ zC2 cos(wt 
fJy
+ n)
= xC1 cos(wt  fJy)  zC2 cos(wt  fJy)
At any time instant t, the direction of the total electric field is given by the unit vector as " UE
=
=
~(y,t)
l~(y,t) I
"
C1
"
C2
x ;::::::==========  z ;::::::==========
cr+c:f
which does not change with time. Therefore, the wave is linearly polarized, as shown in Figure 8.25d.
Sec. 8.5
665
Polarization of Electromagnetic Waves
z
z LP
C2    
RHEP
wt = 0, 27r
I I
Y@
wt= 37rl2_
,
I
I
\
\
I
I I I I
'
' Ci ,
I
wt= 7r x
\
/
, wt= 0, 27r
I I
Ci'
I I
\
/\',
wt = 7r/2, 37r/2
C2
Ellipse
wt=7r
I

... /
wt= 7r/2
2
(a)
(b)
z
z RHCP
y®
y®
LP wt= 7r

/.;
wt= 7r
 
....... ~
\
\
I I
Ci'
I
\
/"Circle
,wt= 0, 27r I
.......

__
/
Ci
/
C2
I I I I I I I
wt= 37r/2 Ci /
CW rotation
1
"°c
x
x
Ci
Ci
/
I I I I I I
wt = 7r/2, 37r/2
Ci wt= 7r/2
C2
(c)
x

wt= 0, 27r
(d)
Figure 8.25 Two linearly polarized waves. The variation of the field vector at y = 0 is shown as a function of time at five different time instants, namely wt = 0, n /2, n, 3n /2 and 2n, for the cases when (a) e = 0°, (b) e = n /2, (c) e = n /2 and C1 = C2, and (d) e = n . The direction of propagation is they direction. Also note that when C1 =/= C2, it is assumed that C1 < C2.
8.5.4 Poynting Vector for Elliptically Polarized Waves
Consider the electric field phasor for an elliptically polarized wave given by
with its accompanying magnetic field, which follows as ,.. C1y j(f3zs) H(z ) = H 1 (z ) + H 2 (z ) = x e 11
+,..y C1x e jf3z 11
666
Waves in an Unbounded Medium
Chap.8
where E 1 = x(· ··)is associated with H 1 = y(· ··),whereas E 2 = y(· ··)is associated with H2 = x(· · ·). The timeaverage Poynting vector (8.58) for this wave can be shown to be equal to " 1 Sav = Z 217
Cfx + Cfy
(8.71)
It should be noted here that Sav is independent of ~. For a lossy medium the timeaverage Poynting vector of an elliptically polarized wave can be shown to be
Sav =
" 1
Z 2117c
I
c2
lx
+
c2
ly
~
2az
COS o/TJe
(8.72)
where a = ffie{y }. Example 8.24: Elliptical polarization. Calculate the total timeaverage power carried by the electromagnetic wave given in Example 8.22. Solution: The total timeaverage power per unit area is equal to the timeaverage Poynting vector for this wave given by 1 2 Sav = 11[C1y 2
+
1 2 C1z] ~ (377)[8 2 2
+
2
2 mwm(12) ] ~ 39.2
8.5.5 Polarization in Practice Polarization of electromagnetic waves and thus light can be observed in our everyday environment and is also put to very good use in practice. Direct sunlight is virtually unpolarized, but the light in the rainbow exhibits distinct polarization, as does the blue sky light itself. 63 Polarization studies were crucial to the early investigations of the nature of light64 and to the determination that X rays were electromagnetic in nature. 65 Much useful information about the structure of atoms and nuclei is gained from polarization studies of the emitted electromagnetic radiation. Polarized light also has many practical applications in industry and in engineering science. In the following, we describe some of the more common engineering applications at radio frequencies. 63 The
color of the sky is blue because of the scattering of sunlight from tiny molecules of air, a process which leads to polarized light. Sometimes, small clouds, hardly visible in air, can be seen more clearly in a reflection in water (e.g., a lake), because the light from the clouds is not polarized and is thus more efficiently reflected by the water than the blue sky light, which is polarized. Reflection and refraction of electromagnetic waves will be studied in Chapter 9. For more on the color of the sky and many other interesting phenomena, see M. Minnaert, The Nature of Light & Colour in the Open Air, Dover, 1954. 64 Polarization was discovered by E. L. Malus [Nouveau Bulletin des Sciences, par la Soc. Philomatique, p. 266, 1809]; also see C. Huygens, Traite de la Lumiere, 1690. 65 C. G. Barkla, Polarization in Secondary Rontgen Radiation, Roy. Soc. Proc. Ser. A, 77, pp. 247255, 1906.
Sec. 8.6
Arbitrari ly Directed Uniform Plane Waves
667
Linearly polarized waves are commonly utilized in radio and TV broadcast applications. AM broadcast stations operate at relatively lower frequencies and utilize large antenna towers and generate vertically polarized signals, with the wave electric field perpendicular to the ground. Although the primary polarization of the AM signal is vertical, there is typically also a component of the electric field in the direction of propagation, due to groundlosses. For maximum reception of an AM radio signal, the receiver antenna should be parallel to the electric field, or perpendicular to the ground (i.e., vertical). Most AM radio receivers use ferriterod antennas instead of wire antennas to detect the wave magnetic field, which is horizontal. For maximum reception, the ferriterod antenna should be parallel to the magnetic field. In North America, TV broadcast signals are typically horizontally polarized, which is why typical rooftop antennas are horizontal. Most FM radio broadcast stations in the United States utilize circularly polarized waves, so that the orientation of an FM receiving antenna is not critical as long as the antenna lies in a plane orthogonal to the direction from which the signal arrives. Polarization of an electromagnetic signal is initially determined by the source antenna that launches the signal. 66 However, as a signal propagates through some medium, its polarization may change. A good example of this occurs in propagation of waves through Earth's ionosphere, when a rotation of the plane of polarization occurs, known as Faraday rotation. Circularly polarized waves are utilized in some communication and radar systems, primarily to relax the requirement for the receiving antenna to be carefully aligned with the wave electric vector. Many international communication satellites also utilize circularly polarized signals. In radar applications, circular polarization is used because such a wave reflects from a metal target with the opposite sense (i.e., RHCP becomes LHCP), providing a method to distinguish metal targets (i.e., aircraft) from clouds and clutter. Use of polarized transmissions also provides a means of multiple simultaneous usage of the same frequency band. Most North American communication satellites use linear polarization, alternating between horizontal and vertical polarization between adjacent satellites. 67
8.6 ARBITRARILY DIRECTED UNIFORM PLANE WAVES In previous sections we used expressions for uniform plane wave electric and magnetic fields for which the direction of propagation was chosen to coincide with one of the coordinate axes, namely, the z axis. This choice obviously results in no loss of generality in our considerations of waves in an unbounded medium in this chapter. In Chapter 9, however, we consider electromagnetic wave propagation in the presence of planar boundaries, resulting in their reflection and refraction. When uniform plane waves 66 A
very interesting antenna that can selectively generate righthand or lefthand circularly polarized signals is the helical antenna, discovered by J. D. Kraus. See J. D. Kraus, The Helical Antenna, Proc. IRE, 37(3), pp. 263272, 1949. For detailed discussion of the helical antenna and other antennas see J. D. Kraus, Antennas, McGrawHill, New York, 1988. 67 W. Sinnema, Electronic Transmission Technology, PrenticeHall, 1988.
668
Waves in an Unbounded Medium
Chap.8
are nor1nally incident on an infinite planar boundary, the direction of propagation of the wave can be chosen to be along one of the principal axes without any loss of generality. However, the general case of reflection and refraction of waves at infinite planar boundaries involves waves that are incident on the boundaries at an arbitrary angle, in which case we shall need to work with unifor1n plane waves propagating in arbitrary directions. In this section, we derive general expressions for such waves and introduce the notation to be used. We also briefly discuss nonunifor1n plane waves, because such waves are encountered in later chapters.
8.6.1 Uniform Plane Waves in an Arbitrary Direction Before we proceed with deriving some general relationships concerning unifor1n plane waves, it is useful to note that the physical nature of the waves we consider here is still the same as those studied in Sections 8 .1 through 8 .5. In other words, the waves considered here are unifor1n plane waves in which the field vectors are confined to an infinite plane (which is the surface of constant phase) and exhibit spatial variations only in the direction perpendicular to that plane. Consider the vector wave equation (8.7) for timehar1nonic fields in a lossless medium, repeated here for convenience: (8.73) where f3 = wffe. We know that one solution of (8.73) is E(z) = Eoej,Bz = [xEox
+ yE0y]ej,Bz
A more general solution representing a wave propagating in an arbitrary direction r = XX + yy + ZZ IS A
A
A
•
E(x, y, z) = Eoej .Bxxj ,Byyj .Bzz = [XEox
+ yE0y + zEoz]ej .Bxxj ,Byyj .Bzz
(8.74)
It can be shown by direct substitution in (8.73) that (8.74) satisfies the wave equation as long as we have
We can rewrite (8.74) in a more compact form by using vector notation for the exponent: E(r) = Eoej ,Bk·r "
(8.75)
where k is the unit vector in the direction of propagation, as shown in Figure 8.26. Note " that k is given by " xf3x + y/3y + zf3z xf3x + y/3y + zf3z k [/3} + f3J + /3}]1/2 wffe
Sec. 8.6
669
Arbitrari ly Directed Uniform Plane Waves
Surface of constant phase
B
'' k ·r
Figure 8.26 Constantphase surface. The planar surface of constant phase, characteristic of a plane wave, and the orientation of the field vectors. "
The vector kfi is the wavenumber vector, commonly referred in other texts as k. For a uniform plane wave propagating in the z direction as considered in previous sections, the phase constant fi represents the space rate of change of wave phase (in units of radm 1) along the z axis. In the general case of a uniform plane wave propagating in an arbitrary direction, the component phase constants fix, /1y, and fiz represent the space rates of change of the wave phase as measured along the respective axes. These component phase constants and the corresponding component wavelengths are further discussed below in connection with Figure 8.27. " We note that k · r =constant represents the equation for the planes over which the wave phase is a constant, referred to as the planes of constant phase, or phase fronts (Figure 8.26). By substituting (8.75) into (7 .23b) for a sourcefree medium (i.e., V · E = 0), we find V · Eoej,Bk·r = 0 Eo · V(ej,Bk·r) = 0 Eo ·
,. . a ,. . a ,. . a x ax + y ay + z az
Eo. [j (X.fix
+ Yfiy + zfiz)ej(.Bxx+.ByY+.Bzz)] =
0
Eo · (j fikej,Bk·r) = 0  j fi (Eo · k )e j ,Bk· r = 0
670
Chap.8
Waves in an Unbounded Medium
',/
x '\ '\
'\
'\
\
'\
'\ '\
k
'\ '\
'\ '\ '\ \ '\ '\ \
'\
'\ '\ '\ '\ '\
'\ '\
\ \
'\ '\
'\ \
'\ '\
Az
\
'\
~
', A.z 
'\ '\
'\ '\
~
\ Az '\
'\ \
'\ '\
'\ '\ '\
\
'\ \
~
'
\
'\
,/
z
'\ '\ \
'\ '\ '\
'\ '\
'\
~
'\
'\
'\ '\
k ', '\
'
'
'\ \
'\ '\
'\ '\
\
'\
'\ '\
'\
'\
'\ '\
'\
'
',/
'
'
Figure 8.27 A uniform plane wave propagating at an angle 0 to the z axis. Although the wavelength in the propagation direction is A., the projections of the constant phase fronts on the z and x axes are separated by Az and Ax, respectively.
which yields "
k·Eo=O
(8.76)
Thus, Eo must be transverse to the propagation direction for (7 .23b) (V · E = 0) to be satisfied. (This result is the generalization of what we saw earlier for the simplest uniform plane wave propagating in the z direction, where we had to have Ez = 0 for the field to have zero divergence.) To find the magnetic field Hin terms of the electric field E, we use Faraday's law (equation (7.23a)): 1 H(r) = . V x E(r) 1wµ
We can further manipulate (8.77) as follows
H =
1
.
V x Eoej,Bk·r
1wµ 1 . Eo x Vej,Bk·r 1wµ 1    E o x (j fik)ej,Bk·r • 1wµ A
since Eo is a constant vector
A
see derivation of (8.76) above .
(8.77)
Sec. 8.6
Arbitrari ly Directed Uniform Plane Waves
=
f3
wµ
kx
671
Eoej f3k·r
1 " · Rk" 1 = k x Eoe ,_, ·r 'f}
'.._., E(r)
or 1" H(r) =  k x E(r)
(8.78)
'f}
The magnetic field vector H is thus perpendicular to both the electric field vector E and " k, as shown in Figure 8.26. Equations (8.75) and (8.78) constitute general expressions for transverse electro" magnetic (TEM) waves propagating in an arbitrary direction k. These expressions will be useful in the discussion of reflection and refraction of unifor1n plane waves obliquely incident on dielectric or conductor interfaces, which is the topic of the next chapter. As an example, we consider a uniform plane wave propagating in the xz plane (i.e., in a direction perpendicular to the y axis) but in a direction oriented at an angle e with respect to the z axis, as shown in Figure 8.27. Assuming that the wave electric field " is in the y direction, and noting that we have k = x sine + z cos e' the expression for the wave electric field phasor can be written as E(r) = yEoejf3k·r = yEoejf3(xsin8+zcos8)
whereas the wave magnetic field can be found using (8.78) as H(r) =
~k
x [yEoejf3(xsin8+zcos8) ]
'f}
=
~(xsinB + zcosB)
x [yEoejf3(xsinB+zcosB) ]
'f}
= Eo (xcose
+ zsinB)ejf3(xsinB+zcos8)
'f}
H(r) = Eo (xcose
+ zsinB)ej(f3xx+f3zz)
'f}
where f3x = f3 sine and f3z = f3 cos e. Note from Figure 8.27 that the expression for H(r) " could also have been written by inspection, since E x H must be in the direction of k. Further consideration of Figure 8.27 allows us to better understand the properties of a uniform plane wave. It was mentioned above that the component phase constants f3x,y,z represent the space rate of change of the phase of the wave along the principal axes. Instead of considering the rate of change of wave phase, we can consider the distances between successive equiphase surfaces on which the wave phase differs by exactly 2n. These distances are apparent wavelengths, because the phase shift is linearly proportional to distance in any direction. Along the propagation direction, we have the
672
Waves in an Unbounded Medium
Chap.8
conventional wavelength A over which the wave phase changes by 2n. Measured along any other direction, the distances between equiphase surfaces are greater, as is apparent from Figure 8 .27. The apparent wavelengths along the x and z axes are respectively Ax and Az, which are both greater than A. The apparent wavelengths Ax and Az are related to A by 1 A2 x
+
1 1 A2 = A2 z
•
since
Similarly, because a particular planar phase front moves uniformly (i.e., with same speed at all points on the plane) with time along the propagation direction at the phase velocity Vp = w / f3, its intercept along any other direction also moves uniformly in time but with a greater speed. In the time of one full cycle of the wave (i.e., Tp = 1/f ), a " phase front moves a distance along the k direction of A = 2n / f3 and a distance along the x direction of Ax =A/ sine = 2n /({3 sin B) = 2nI f3x. Thus, the various apparent phase velocities along the principal axes are in general given by
vx,y,
or z  x,y, or z
/3
and, in the case shown in Figure 8 .27, are related to one another as
Component phase velocities greater than the speed of propagation in the medium can be observed in water waves approaching at an angle to a breakwater or a shoreline. Visualize water waves striking a beach obliquely. The distance between successive crests (i.e., maxima), for example, is large along the beach, especially if the waves are only slightly off normal incidence. To keep up with a given crest, one has to run much faster along the beach than the speed with which the waves progress in their own directions of propagation.
8.6.2 Nonuniform Plane Waves Nonuniform plane waves are those electromagnetic waves for which the amplitudes of the electric and magnetic fields of the wave are not constant over the planes of constant phase. An example of a nonuniform plane wave is '(Ji(y,z,t) = xC 1 cos(ny) cos(wt  f3z) Ho(y)
which is a plane wave, because the constant phase surfaces are planes of z = constant (i.e., planes parallel to the xy plane), but it is nonuniform, because the amplitude of the
Sec. 8.7
Nonplanar Electromagnetic Waves
673
field is a function of y and hence varies as a function of position along the planes of constant phase. We shall encounter nonunifor1n plane waves when we discuss reflection of unifor1n plane waves from obliquely oriented boundaries in Chapter 9, and also when we study guiding of electromagnetic waves by metallic and dielectric boundaries in Chapter 10.
8.7 NONPLANAR ELECTROMAGNETIC WAVES We have seen that unifor1n plane waves are natural solutions of Maxwell's equations and the wave equations derived from them. The general character of these unifor1n plane waves is apparent in the functional form of any one of the electric or magnetic field components, namely
~ (r, t) = ffie Eoej .Bk·reiwt = ffie{Eo} cos(wt  f3k · r)  J>m{Eo} sin(wt  f3k · r) where Eo is, in general, a complex vector given by Eo = ffie{Eo} + j J>m{Eo}. As dis68 cussed before, such waves are plane waves because the surfaces of constant phase, that is, the surfaces over which ~ (r, t) is constant, are planes. These planar surfaces are " defined by k · r =constant. These waves are uniform because the field ~(r, t) does not vary as a function of position over the surfaces of constant phase; in other words, Eo is not a function of spatial coordinates. Examples of nonplanar waves are those electromagnetic fields for which the variation of the field quantities exhibits a more complicated dependence on spatial coordinates " than is implied by (wt  f3k · r). In general, the electric field vector of a nonplanar timehar1nonic wave can be expressed as ~(r,
t) = ~o cos[wt  g (r)]
where g ( ·) is some arbitrary function. The surfaces of constant phase are now given by g(r) =constant
which could be any arbitrary surface, depending on the function g ( ·). For example, if g(r) = g(x,y,z) = xy, then the surfaces of constant phase are hyperbolic cylinders, with infinite extent in z (i.e., no variation in z) and with the functional form in planes parallel to the xy plane given as xy =constant. Many of the properties of plane waves that we may take for granted may not be applicable when the wave fronts are nonplanar. For example, the speed of wave propagation, or the phase velocity Vp, may not be constant in space and may instead be a function of position. 69 68 Note
that the functional form of the quantity does not need to be a cosinusoid for it to be a uniform plane wave. In the general case, an electromagnetic field is considered to be a uniform plane wave if the field " quantities vary with space and time asf(wt  ,Bk· r), wheref(·) is any function. 69 See M. Born and E. Wolf, Principles of Optics, 5th ed., Section 1.3, Pergamon Press, Oxford, 1975.
674
Waves in an Unbounded Medium
Chap.8
8.8 SUMMARY This chapter discussed the fallowing topics:
•
Uniform plane waves. The characteristics of electromagnetic waves in sourcefree and simple media are governed by the wave equation derived from Maxwell's equations:

a2~
\7 2~  µE
at 2
= 0
Uniform plane waves are the simplest type of solution of the wave equation, with the electric and magnetic fields both lying in the direction transverse to the direction of propagation (i.e., z direction): ~x (z, ~Y (z,
t) =Pl (z  Vpt) 1
p1 (z
t) =
 vpt)
rJ
where Vp = 1/ ffe is the phase velocity, rJ = 5[E is the intrinsic impedance of the medium, and Pl is an arbitrary function.
•
Timeharmonic waves in a lossless medium. The electric and magnetic fields of a timehar1nonic unifor1n plane wave propagating in the z direction in a simple lossless medium are ~(z,
t) = iC1 cos(wt  {Jz)
and
~(z, t) = y C
1
cos(wt 
/3z)
rJ
where E and Hare the phasor quantities, and f3 = wffe is the propagation constant (C1 is assumed to be real).
•
Uniform plane waves in a lossy medium. The electric and magnetic fields of a unifor1n plane wave propagating in the
z
~(z,
direction in a lossy medium are
t) = iC1eaz cos(wt  {Jz)
and
H(z) = y A
C1 rJc
e
yz
~(z, t)
A
= y
C le az
lrJc l
cos(wt  {Jz  TJ)
where y = a + j f3 is the complex propagation constant and rJc = IrJc lejTJ is the complex intrinsic impedance of the medium. Tables 8.6, 8.7, and 8.8 summarize the equations for the (generally complex) propagation constants and intrinsic impedances considered in this chapter. Table 8.6
Sec. 8.8
675
Summary
TABLE 8.6 SUMMARY OF PROPAGATION CONSTANT AND INTRINSIC IMPEDANCE EQUATIONS
Category General
y=a.+j/3
1/c
j W J /LeffEeff
/Leff Eeff
Lossless:
µ
jwffe
Eeff = E
E
/Leff=µ
Lossy: Eeff = E
1
jw,J/i?Jl  j tan Oc
(1  j tan 8c)
E'
/Leff=µ
Lowloss: tan8c > 1
2
. +1w
1+
1
µ
tan8c
. tan Oc 1 +J 2
µ
1 2 (tan 8c) 8
µE'
Jl  j
E1
tan8c
µ ·45o el
2
E'
tan Oc
TABLE 8.7 SUMMARY OF LOSS TANGENTS FOR MAGNETICALLY LOSSLESS MEDIA
Category
tan~c
O' etT
O' etT
= 
W€
Ohmic (a)
+
Dielectric (E'')
Dielectric (E'') Ohmic (a)
a
+ WE WE
11
11
a
E
1
11
+ WE' E 1
E'' E' a WE
1
shows y and rJc for the general, lossless, and lossy cases, where the losses are strictly associated with a complex effective permittivity (µeff = µ is assumed to be real). Approximate expressions with tan De > 1 are provided for the lossy case. Table 8.7 summarizes the effective conductivity and loss tangent equations for each per1nutation of losses due to Joule heating and dielectric losses. Finally, Table 8.8 shows y and rJc for the lowloss dielectric and highloss conductor special cases considered in Section 8.3. These equations are readily derived by substituting the relevant loss tangent values from Table 8.7 into the last two rows of Table 8.6.
676
Chap.8
Waves in an Unbounded Medium
TABLE 8.8 COMMON PROPAGATION CONSTANT AND INTRINSIC IMPEDANCE EQUATIONS FOR MAGNETICALLY LOSSLESS MEDIA
Category
y=a.+j/3
1/c
Lowloss Dielectric: tan8c =
E'' E'
> 1
µwa
.
+1
_µ w _ ej 450
2
a
Another important parameter for good conductors is the skin depth 8, given by 1 1 8 =  rv ;::==== a  ,Jrcf µa
•
The skin depth for metallic conductors is typically extremely small, being rv3 .82 µm for copper at 300 MHz. Electromagnetic power flow and the Poynting vector. Poynting' s theorem states that electromagnetic power flow entering into a given volume through the surface enclosing it equals the sum of the time rates of increase of the stored electric and magnetic energies and the ohmic power dissipated within the volume. The instantaneous power density of the electromagnetic wave is identified as ~(z,
t) = ~(z, t) x ~(z, t)
although in most cases the quantity of interest is the timeaverage power density, which can be found either from ~(z, t) or directly from the phasor fields E and Has 1
Tp_
Sav(Z) = Tp o
~(z,
1 (il) * Sav(z) = ~Le{E x H }
t) dt
2
The timeaverage Poynting vector for a uniform plane wave propagating in the z direction in an unbounded medium is S (z) = av
•
z
c21
e 2az cos(~ )
2117c l
'PT/
Wave polarization. The polarization of an electromagnetic wave describes the behavior of its electric field vector as a function of time, at a fixed point in space. In the general case of an elliptically polarized wave, the two components of the electric field vector of a wave propagating in the z direction are given by ~x(z,t)
= C1x cos(wt  {3z)
~Y (z, t) = C1y
cos(wt  {3z
+ ~)
Chap.8
•
677
Problems
A wave is linearly polarized if~ = 0 or JT, and is circularly polarized if C1x = C1y and ~ = ±n/2, with the negative sign corresponding to a righthand circularly polarized wave. For any other value of ~, the wave is elliptically polarized, in the lefthand sense for 0 < ~ < JT, and righthand sense for JT < ~ < 2n. Uniform plane wave propagating in arbitrary direction. The electric and magnetic field phasors of a unifor1n plane wave propagating in an arbitrary direction " k in a simple lossless medium are given as E(r) = Eoej ,Bk·r
1" H(r) =  k x E(r) 11
PROBLEMS
8.1
Uniform plane wave. The electric field of a uniform plane wave in air is given by ~(z, t) = x3 cos(2n x l0 9 t  {Jz)
8.2
Vm 1
(a) Find the phase constant fJ and the wavelength A.. (b) Sketch ~x(Z, t) as a function oft at z = 0 and z = A./4. (c) Sketch ~x(z,t) as a function of z at t = 0 and t = n/w. Uniform plane wave. The electric field phasor of an 18 GHz uniform plane wave propagating in free space is given by E(y) = zl5ejf3y Vm 1
(a) Find the phase constant, phasor H(y).
fJ, and wavelength, A.. (b) Find the corresponding magnetic field
8.3
Uniform plane wave. A uniform plane wave is traveling in the x direction in air with its magnetic field oriented in the z direction. At the instant t = 0, the wave magnetic field has two adjacent zero values, observed at locations x = 2.5 cm and x = 7 .5  cm, with a maximum value of 70 mAm 1 at x = 5 cm. (a) Find the wave magnetic field 'lJe(x, t) and its phasor H(x). (b) Find the corresponding wave electric field ~(x,t) and its phasor E(x).
8.4
Broadcast signal. The magnetic field of a TV broadcast signal propagating in air is given as 'lJe(y, t) =
8.5
x0.3 sin(wt + 7 .2y) mAm 1
(a) Find the wave frequency f = w/(2n). (b) Find the corresponding ~(y,t). Uniform plane wave. A NASA spacecraft orbiting around Mars receives a radio signal transmitted by the UHF antenna of Curiosity Rover, with an electric field given by ~(z,t) = x75cos wt +
8nz
3
tLVm
1
(a) Determine the frequency Land the wavelength A. of this radio signal. (b) Find the corresponding magnetic field 'lJe(z, t).
678 8.6
Waves in an Unbounded Medium
Chap.8
Lossless nonmagnetic medium. The magnetic field component of a uniform plane wave propagating in a lossless simple nonmagnetic medium (µ = µo) is given by ~ (x, t) = z 0.5 sin[2n (10 t  0.5x  0.125)] µ T 8
(a) Find the frequency, wavelength, and the phase velocity. (b) Find the relative permittivity, Er, and the intrinsic impedance, rJ, of the medium. (c) Find the corresponding ~. (d) Find the timeaverage power density carried by this wave.
8. 7
A wireless communication signal. The electric field of a wireless communication signal traveling in air is given in phasor form as E(x) = l0d 50x CY  jz) Vm 1
(a) Find the frequency f and wavelength A.. (b) Find the corresponding phasorform magnetic field H(x). ( c) Find the total timeaverage power density carried by this wave.
8.8
Uniform plane wave. An 8 GHz uniform plane wave traveling in air is represented by a magnetic field vector given in phasor form as follows: H(y) = x 0.015ej f3y
+ z0.03ei 0 halfspace, as shown in Figure 9 .1. The interface between the two media is the entire xy plane
Sec. 9.1
Normal Incidence on a Perfect Conductor
691
x
Reflected ..
wave

E 1·
Incident wave
ff.1 1etk 1·
Medium2 (perfect conductor)
Medium 1 (El, µl , O"l = 0)
z =O
Figure 9.1 Uniform plane wave incident normally on a plane, perfectly conduct" " ing boundary. The unit vectors ki and kr represent the propagation directions of the incident and reflected waves, respectively.
(i.e., infinite in transverse extent). We arbitrarily 1 assume that the electric field of the incident wave is oriented in the x direction and that therefore its magnetic field is in the y direction. The phasor field expressions for the incident wave (presumably originating at z = oo) and the reflected wave are given as Incident Wave:
Reflected Wave:
where f31 = w µ1 E1 and 111 = ,Jµ1/E1 are respectively, the propagation constant and the intrinsic impedance for medium 1. The presence of the boundary requires that a reflected wave propagating in the z direction exist in medium 1, with phasor field expressions as given above. Note that the reflected wave propagates in the z direction, so its magnetic field is oriented in the y direction for an xdirected electric field. The boundary condition on the surface of the conductor requires the total tangential electric field to vanish for z = 0 for all x and y, since the electric field inside medium 2 (perfect conductor) must be zero. This condition requires that the electric field of the 1Note
that for normal incidence on a planar boundary of infinite extent in two dimensions, we lose no generality by talcing the direction in which the electric field of the incident wave vibrates to be along one of the principal axes.
692
Reflection, Transmission, and Refraction of Waves at Planar Interfaces
Chap.9
reflected wave is also confined to the x direction. The total electric and magnetic fields in medium 1 are E1 (z) = Ei(Z)
+ Er(Z) =
x[Eioe  jf3iz
+ Er0e+j/3iz]
Eio ·13 Er0 ·13 1 H1 (z) = Hi(Z) + Hr(Z) = y e  iz e +J iz ~1
~1
Since E1 (z) is tangential to the boundary, application of the boundary condition at z = 0, namely that E1 (z = 0) = 0, gives E1 (z = 0) = x[Eioe j/3iz
+ Er0e+j/3iz ]z=O =
x[Eio + Er0] = 0
from which
so that
______
E1 (z) = xEio .._,[e  jf3iz  e +j/3iz] = xEioj2 sin(fi1z)  j2 sin(/31z) With the constant Er0 determined in terms of Eio (i.e., Er0 = Eio), we can also write the total magnetic field in medium 1 as H1 (z) = Hi(Z)
+ Hr(z) =
y
1 ~1
. . 2Eio [Eioe  1 13 iz + Eioe+J131z] = y cos(fi1z) ~1
The corresponding spacetime functions ~ 1(z, t) and 'Je1(z, t) can be found from the phasors E1 (z) and H 1(z) as ~1 (z, t) = 9Jle{E1 (z)dwt} = 9Jle{x2Eio sin(fi1z)ejwt ejrr/2} "y_.I •
1
which gives ~1 (z, t) = x2Eio sin(fi1z) sin(wt)
'Je1 (z, t) = y2
E·o 1
cos(fi1z) cos(wt)
~1
2 assuming that Eio is a real number. 2
If, instead, Eio was a complex number, Eio = IEio l ej~ , the expressions for '&1 and ~1 would be 
'&1(z, t) = i21Eiol sin(,81z) sin(wt
~ 1 (z , t) =
+ ~)
y2 IEio I cos(,81z) cos(wt + ~) 171 amounting to a simple shift of the time origin.
(9.1)
Sec. 9.1
693
Normal Incidence on a Perfect Conductor
The fields described by (9 .1) are plotted in Figure 9 .2 as a function of distance z at different time instants. Note that the ~ and '/Jf, fields do not represent a propagating wave, because as time advances, the peaks (or nulls) always occur at the same points in space. Such a wave is termed to be a pure standing wave; it consists of a superposition of two waves traveling in opposite directions. The maxima and minima stand at the same location as time advances. This standing wave, which is established when a unifor1n plane wave is incident on a perfect conductor, is analogous in all respects to the one that occurs in the case of sinusoidal excitation of a lossless transmission line terminated in a short circuit (see Section 3.2). This analogy is illustrated in Figure 9.3. A pure standing wave such as the one just described does not carry electromagnetic energy, as expected on physical grounds, since the perfect conductor boundary reflects all of the incident energy. To verify this notion, we can consider the Poynting vector for the standing wave just described. Using the phasor forms of the fields and noting that
x wt = w/2 .. wt= w/6 wt=O,w~

~1
~lx
1
1
~
1
1
/
;
 '' .....
\
;~ ..... , ' ,
,,,
~;
.,___ ..,..k·
Perfect conductor
1
y.
~z
wt=  w/6 wt=  w/2~

~r ..

~'
wt= w/3 __,,
\
wt=w/2"

..... ' \ . \
' ' ..... wt=w__,,
wt=2w/3   ' wt=O__,  ,\
 3,\/4
 ,\/2
 ,\/4
z=0
Figure 9.2 Instantaneous electric and magnetic field wavefor1ns. Total electric and magnetic field waveforms in medium 1 are shown at selected instants of time. The ''standing'' nature of the waves is apparent as the peaks and nulls remain at the same point in space as time progresses. Note that medium 1 is a perfect dielectric, while medium 2 is a perfect conductor.
694
Reflection, Transmission, and Refraction of Waves at Planar Interfaces
Chap.9
171 Reflected~
Short circuit
Incident~
Zo II(z)I
IV(z)I
~
,_ /
'
\
I \
I
'
\
I
'
I \
I
z=O
,_/
'
/
I
\ \
I
,,\/2
,,\
' ~
I
\
171 IH1(z)I
IE1(z)I
/
I
\
Perfect conductor
I \
I
'
I
\
\