Branislav M. Notaros Electrostatic Field in Free Space Page 1 Dielectrics, Capacitance, and Electric Energy Page 61 Steady Electric Currents Page 124 Magnetostatic Field in Free Space Page 173 Magnetostatic Field in Material Media Page 221 Slowly Time-Varying Electromagnetic Field Page 263 Inductance and Magnetic Energy Page 311 8 Rapidly Time- Varying Electromagnetic Field Page 351 10 6 3 1 0 10 9 10 12 s r adio microwaves waves i / \ 15 10 18 1C > # 9 j 0 /[Hz] o infrared i 6 i i 10 3 Page 408 x-rays ultraviolet cosmic rays 7 -rays K > Uniform Plane Electromagnetic Waves 2l A[m] i i 10 3 1C -* -9 io io~ 12 icr 15 10 Reflection and Transmission of Plane Waves Page 471 integrated circuits 11 Field Analysis of Transmission Lines Page 533 12 Circuit Analysis of Transmission Lines Page 576 13 Waveguides and Cavity Resonators short circuit Page 662 14 Antennas and Wireless Communication Systems Page 713 ELECTROMAGNETICS Branislav M. Notaros Redeem access to Pearson eText, hundreds of conceptual questions and MATLAB exercises at: www.pearsonhighered.com/notaros Use a coin to scratch Do not use a knife off the coating and reveal your student access code. it may damage the code. or other sharp object as IMPORTANT: The access code on this page can only be used once to establish a the access code has already been scratched off, it may no longer be valid. If this is the case, visit the website above and select “Get Access” to purchase a new subscription. subscription. Technical Support is If available at www.247pearsoned.com. PEARSON Pearson Education One Lake I Street, Upper Saddle River, NJ 07458 Electromagnetics Electromagnetics Branislav M. Notaros Department of Electrical and Computer Engineering Colorado State University Prentice Hall Upper Saddle River Boston Columbus San Francisco New York London Toronto Sydney Singapore Tokyo Montreal Madrid Hong Kong Mexico City Munich Paris Amsterdam Cape Town Indianapolis Dubai Vice President and Editorial Director, Engineering and Computer Science: Marcia Senior Editor: Andrew J. Horton Gilfillan Editorial Assistant: William Opaluch Vice President, Production: Vince O'Brien Senior Managing Editor: Scott Disanno Production Editor: Pavithra Jayapaul, TexTech International Senior Operations Supervisor: Alan Fischer Operations Specialist: Lisa McDowell Executive Marketing Manager: Tim Galligan Art Director: Kenny Beck Cover Designer: LCI Designs Art Editor: Greg Dulles Media Editor: Dan Sandin Composition/Full-Service Project Management: TexTech International Copyright (c) 2011 by Pearson Education, Inc., Upper Saddle River, New reserved. Manufactured in the United States of America. This publication Jersey 07458. All rights protected by Copyright and permissions 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 materials from this work, please submit a written request to Pearson Education, Inc., Permissions Department, 1 Lake Street, Upper Saddle River, NJ is 07458. 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 effective- 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, ness. performance, or use of these programs. Library of Congress Cataloging-in-Publication Data Notaros, Branislav M. Electromagnetics / Branislav M. Notaros, cm. ISBN 0-13-243384-2 p. Electromagnetism 1. — Textbooks. I. Title. QC760.N68 2010 537-dc22 2010002214 Prentice Hall an imprint of is PEARSON www.pearsonhighered.com ISBN-13: 978-0-13-243384-6 0-13-243384-2 ISBN-10: To the pioneering giants of electromagnetics Michael Faraday, James Clerk Maxwell, and others (please see the inside back cover) for providing the foundation of this book. To my professors and colleagues Branko Popovic (late), for making To Antonije Djordjevic, and others me nearly all understand and fully love my students for teaching in all my classes me to over this stuff all these years teach. To Olivera, Jelena, and Milica for everything else. Contents Preface xi 2.2 Polarization Vector 2.3 Bound Volume and 64 Densities 1 Evaluation of the Electric Field and 2.4 Electrostatic Field in Free Coulomb’s Law 1.1 Space 63 Surface Charge 1 Potential due to Polarized Dielectrics 2 Generalized Gauss’ 2.5 Law 68 70 1.2 Definition of the Electric Field Intensity 2.6 Characterization of Dielectric Materials 2.7 Maxwell’s Equations for the Electrostatic 1.3 Vector 7 Continuous Charge Distributions 1.4 On 1.5 Electric Field Intensity Vector 1.6 Definition of the Electric Scalar 1.7 Electric Potential due to the Volume and Surface Charge Distributions Distributions Differential Relationship Gradient 1.11 3-D and 2-D 1.14 1.15 1.16 1.17 1.19 22 Poisson’s and Laplace’s Equations Finite-Difference Charged Conductors 2.13 Analysis of Capacitors with Capacitor Homogeneous 88 95 2.16 Energy of an Electrostatic System 102 Electric Energy Density 104 2.17 Dielectric Breakdown Systems 108 2.15 in Electrostatic 43 46 3 48 1.20 Method 1.21 Charged Metallic Bodies Image Theory 51 Moments for Numerical of Analysis 3.1 49 Electric 3.2 3.3 2 and Electric 61 Polarization of Dielectrics 62 Currents 124 Current Density Vector and Current Intensity 125 Conductivity and Form 2.1 86 Dielectrics Steady Energy for Analysis of Capacitors with Inhomogeneous 2.14 Charge Distribution on Metallic Bodies of Dielectrics, Capacitance, 82 2.12 Arbitrary Shapes of Method Numerical Solution of Laplace’s Equation 84 Definition of the Capacitance of a Dielectrics Electrostatic Shielding Boundary 79 2.11 23 Electric Dipoles 26 Formulation and Proof of Gauss’ Law 28 Applications of Gauss’ Law 31 Differential Form of Gauss’ Law 35 Divergence 36 Conductors in the Electrostatic Field 39 Evaluation of the Electric Field and and 75 2.10 between the Field in Electrostatics Potential due to 1.18 Conditions 18 1.9 1.10 Dielectric-Dielectric 2.9 21 and Potential Homogeneous Media due to Given Given Charge Voltage 75 Electrostatic Field in Linear, Isotropic, 2.8 10 1.8 1.13 9 16 Potential 1.12 Field 8 Integration 71 Ohm’s Law in Local 128 Losses in Conductors and Joule’s Local Form 132 3.4 Continuity Equation 3.5 Boundary Conditions Currents Law in 133 for Steady 137 vii Contents viii 3.6 Distribution of Charge in a Steady Current 3.7 Relaxation Time 3.8 Resistance, Law 3.9 3.10 1 Maxwell’s Equations for the Time-Invariant Electromagnetic Field 258 6 Duality between Conductance and Slowly Time-Varyinq Electromaqnetic Capacitance Field 146 External Electric Energy Volume Sources 6.1 149 6.2 Analysis of Capacitors with Imperfect Dielectrics 152 6.3 Analysis of Lossy Transmission Lines with Steady Currents 3.13 1 Joule’s 140 Inhomogeneous 3.12 1 139 Ohm's Law, and and Generators 3.1 5. 138 Field Induced Electric Field Intensity Vector 264 Slowly Time-Varying Electric and Magnetic Fields 269 Faraday’s Law of Electromagnetic Induction 156 Grounding Electrodes 263 Maxwell’s Equations for the Slowly Time-Varying Electromagnetic Field 6.5 Computation of Transformer 6.6 Electromagnetic Induction due to Motion 283 6.7 Total Electromagnetic Induction 6.8 Eddy Currents 162 4 Induction Magnetostatic Field 4.1 in Free Space 173 Magnetic Force and Magnetic Flux Density Vector 174 Law 4.2 Biot-Savart 4.3 4.4 Magnetic Flux Density Vector due to Given Current Distributions 179 Formulation of Ampere’s Law 185 4.5 Applications of Ampere’s 4.6 Differential Form of Law Ampere’s Law 4.9 Magnetic Vector Potential 201 Proof of Ampere’s Law 204 Magnetic Dipole 206 The Lorentz Force and Hall Effect 209 Evaluation of Magnetic Forces 211 7.1 195 of Conservation of Magnetic Flux Magnetostatic Field in Material Media 7.3 198 7.4 7.5 7.6 5 221 5.2 5.3 Materials 223 Magnetization Volume and Surface Current 5.5 5.6 5.7 5.8 5.9 Field 227 Generalized Ampere’s Law 234 Permeability of Magnetic Materials 236 Maxwell’s Equations and Boundary Conditions for the Magnetostatic Field 239 Image Theory for the Magnetic Field 241 Magnetization Curves and Hysteresis 243 Magnetic Circuits - Basic Assumptions for the Analysis 5.10 Kirchhoff’s 247 Laws for Magnetic Circuits 312 318 Analysis of Magnetically Coupled Circuits 324 Magnetic Energy of Current-Carrying Conductors 331 Magnetic Energy Density 334 Internal and External Inductance in Terms of Magnetic Energy 342 Self-Inductance Mutual Inductance 8 8.1 Displacement Current 8.2 Maxwell’s Equations for the Rapidly 250 352 Time- Varying Electromagnetic Field Densities 5.4 311 Rapidly Time-Varying Electromagnetic 351 Magnetization Vector 222 Behavior and Classification of Magnetic 5.1 Inductance and Magnetic Energy 7.2 Law 4.13 289 294 7 193 Curl 4.12 277 187 4.8 4.11 276 177 4.7 4.10 271 6.4 8.3 Electromagnetic Waves 8.4 Boundary Conditions 8.5 8.6 8.7 357 361 for the Rapidly Time- Varying Electromagnetic Field 363 Different Forms of the Continuity Equation 364 for Rapidly Time-Varying Currents Time-Harmonic Electromagnetics 366 Complex Representatives of Time-Harmonic Field and Circuit Quantities 369 Contents 8.8 Maxwell’s Equations in Complex Domain 8.9 8.10 8.11 8.12 10.9 Lorenz Electromagnetic Potentials 376 Computation of High-Frequency Potentials and Fields in Complex Domain 381 Theorem 389 Complex Poynting Vector Poynting’s Wave Propagation Media 373 11 533 Transmission Lines TEM Waves in Lossless Transmission Lines with 11.2 9 in Multilayer 520 Field Analysis of 11.1 397 Homogeneous Electrostatic 534 Dielectrics and Magnetostatic Field 538 Distributions in Transversal Planes Uniform Plane Electromagnetic Waves 408 ix 11.3 Currents and Charges of Line Conductors 539 9.1 Wave Equations 9.2 Uniform-Plane-Wave Approximation 411 Time-Domain Analysis of Uniform Plane 11.4 Analysis of Two-Conductor Transmission Waves 412 Time-Harmonic Uniform Plane Waves and Complex-Domain Analysis 416 The Electromagnetic Spectrum 425 Arbitrarily Directed Uniform TEM Waves 427 Theory of Time-Harmonic Waves in Lossy Media 429 11.5 Transmission Lines with Small Losses 11.6 Attenuation Coefficients for Line 11.7 Conductors and Dielectric 550 High-Frequency Internal Inductance of 9.3 9.4 9.5 9.6 9.7 409 9.8 Explicit Expressions for Basic Propagation 9.9 9.10 Wave Propagation Wave Propagation Conductors 439 9.11 Skin Effect 441 Parameters in Good Good Dielectrics 436 Polarization of Electromagnetic Circuit Parameters of Transmission Lines 11.9 11.10 Waves 12.1 458 10.3 10.4 10.5 10.6 10.7 10.8 567 576 Telegrapher’s Equations and Their Solution in 12.3 and Transmission of Plane Waves 471 10.2 563 Multilayer Printed Circuit Board Complex Domain 12.4 472 Normal Incidence on a Penetrable Planar Interface 483 Surface Resistance of Good Conductors 492 Perturbation Method for Evaluation of Small Losses 497 Oblique Incidence on a Perfect Conductor 499 Concept of a Rectangular Waveguide 505 Oblique Incidence on a Dielectric Boundary 507 Total Internal Reflection and Brewster Angle 513 581 Circuit Analysis of Lines a Perfectly Conducting 577 Circuit Analysis of Lossless Transmission Lines 10 Plane 557 Transmission Lines with Inhomogeneous Circuit Analysis of Transmission Lines 12.2 Normal Incidence on 556 12 9.14 10.1 547 Evaluation of Primary and Secondary Dielectrics in 9.13 Reflection 540 Transmission Lines 11.8 433 Wave Propagation in Plasmas 447 Dispersion and Group Velocity 452 9.12 Lines Low-Loss Transmission 581 Reflection Coefficient for Transmission Lines 583 12.5 Power Computations of Transmission 12.6 Transmission-Line Impedance 12.7 Complete Solution Lines 589 Current 12.8 and 597 Short-Circuited, Open-Circuited, and Matched Transmission Lines 12.9 592 for Line Voltage Transmission-Line Resonators 601 608 12.10 Quality Factor of Resonators with Small 12.11 The Smith Chart - Construction and Basic Losses 610 Properties 12.12 614 Circuit Analysis of Transmission Lines Using the Smith Chart 618 x Contents 12.13 Transient Analysis of Transmission 14.1 628 Thevenin Equivalent Generator Pair and Reflection Coefficients for Line Transients 630 Step Response of Transmission Lines with Purely Resistive Terminations 634 Analysis of Transmission Lines with Pulse Excitations 640 Bounce Diagrams 646 Transient Response for Reactive or Nonlinear Terminations 649 Lines 12.14 12. 15 12.16 12.17 12.18 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 13 14.10 Waveguides and Cavity Resonators 13.1 662 Analysis of Rectangular Waveguides Based on Multiple Reflections of Plane 13.2 663 Propagating and Evanescent Waves 13.4 Dominant Waveguide Mode General TE Modal Analysis 13.5 TM Modes in a Rectangular 13.3 Waveguides Waveguide 666 of Rectangular 13.7 13.8 Waves 680 Power Flow along of TE and TM a Waveguide 681 Waveguides with Small Losses 684 Waveguide Dispersion and Wave 688 Waveguide Couplers 692 Rectangular Cavity Resonators 696 Electromagnetic Energy Stored in a Cavity Resonator 700 Quality Factor of Rectangular Cavities with Small Losses 703 Velocities 13.12 13.13 13.14 14.14 14.15 APPENDICES 676 Modes 677 Wave Impedances 13.11 14.12 715 720 Steps in Far-Fieid Evaluation of an Arbitrary Antenna 722 Radiated Power, Radiation Resistance, Antenna Losses, and Input Impedance 730 Antenna Characteristic Radiation Function and Radiation Patterns 736 Antenna Directivity and Gain 740 Antenna Polarization 745 Wire Dipole Antennas 745 Image Theory for Antennas above a Perfectly Conducting Ground Plane 751 Monopole Antennas 754 Magnetic Dipole (Small Loop) Antenna 758 Theory of Receiving Antennas 760 Antenna Effective Aperture 766 Friis Transmission Formula for a Wireless Link 768 Antenna Arrays 772 Far Field and Near Field 671 Cutoff Frequencies of Arbitrary Waveguide 13.9 1 668 13.6 13.10 14.1 14.13 Waves Electromagnetic Potentials and Field Vectors of a Hertzian Dipole 1 Quantities, Symbols, Units, Constants and 791 2 Mathematical Facts and Identities 3 Vector Algebra and Calculus Index 4 Answers to Selected Problems 14 Antennas and Wireless Communication Systems 713 796 Bibliography Index 809 806 802 801 ) Preface E lectromagnetic theory is a fundamental under- pinning of technical education, but, at the same one of the most time, difficult subjects for students to master. In order to help address this difficulty contribute to overcoming on electromagnetic fields it, here is and another textbook and waves for undergrad- uates, entitled, simply, Electromagnetics. This text provides engineering and physics students and other users with a comprehensive knowledge and firm and wave computation and most importantly, outstanding (by the judgment of students so far) workedfor electromagnetic field problem solving, and, out examples, and tions, homework problems, conceptual ques- MATLAB exercises. The goal is to sig- improve students’ understanding of electromagnetics and their attitude toward it. Overall, the book is meant as an “ultimate resource ” for undergraduate electromagnetics. nificantly grasp of electromagnetic fundamentals by emphasizing both mathematical rigor and physical under- standing of electromagnetic theory, aimed toward is distinguishing features of the 371 practical engineering applications. The book The designed primarily (but by no means realistic book examples with very detailed and instrucoupled to the theory, includ- ctive solutions, tightly exclusively) for junior-level undergraduate university ing strategies for problem solving and college students in electrical and computer engineering, physics, and similar departments, for both fully supported two-semester (or two-quarter) course sequences and one-semester (one-quarter) courses. It includes 14 chapters on electrostatic fields, steady electric currents, magnetostatic fields, slowly time-varying (low-frequency) electromagnetic fields, rapidly time- varying (high-frequency) electromagnetic fields, all of them. It also introduces many new pedagogical features not present in any of the existing texts. This text provides cally and new style many nonstandard practically important sections 650 realistic theoreti- and chapters, and approaches to presenting challenging topics and abstract electromagnetic phenomena, innovative strategies and pedagogical guides end-of-chapter problems, strongly and by solved examples (there example for every homework problem Clear, rigorous, complete, of material, balance and with is a demo logical presentation of breadth and depth, balance of static (one third) and dynamic ( two uni- form plane electromagnetic waves, transmission lines, waveguides and cavity resonators, and antennas and wireless communication systems. Apparently, there are an extremely large number of quite different books for undergraduate electromagnetics available (perhaps more than for any other discipline in science and engineering), which are all very good and important. This book, however, aims to combine the best features and advantages of are: thirds) fields, no missing steps emphaand ordering the material in a course or courses, Flexibility for different options in coverage, sis, including the transmission-lines-first approach Many nonstandard derivations, topics and subtopics and new explanations, proofs, examples, pedagogical style, and interpretations, visualizations 500 multiple-choice conceptual questions (on the Companion Website), checking conceptual under- standing of the book material 400 MATLAB computer exercises and projects the Companion tions (tutorials) Website), and many (on with detailed solu- MATLAB codes (m files) www.pearsonhighered.com/notaros The following sections explain these and other tures in more detail. fea- XI Preface xii WORKED EXAMPLES AND HOMEWORK PROBLEMS a strong appreciation for both mentals and its theoretical funda- its practical applications. “Physical” nontrivial examples are good also - and - as The most important feature of the book is an extremely large number of realistic examples, with for instructors detailed and pedagogically instructive solutions, and and discussion in the class than the “plug-and-chug” or purely “mathematical” end-of-chapter (homework) problems, strongly and fully supported by solved examples. There are a worked examples, all tightly coupled for lectures much more they are interesting recitations and suitable for presentation logical examples. total of 371 to the theory, strongly reinforcing theoretical con- cepts and smoothly and systematically developing the problem-solving skills of students, and a total of CLARITY, RIGOR, AND COMPLETENESS 650 end-of-chapter problems, which are essentially offered and meant as end-of-section problems (indications appear at the ends of sections as to which problems correspond to that section). Along with the number and type of examples and problems (and questions and exercises), the most homework problem attention to clarity, completeness, and pedagogical always an example or whose detailed solution soundness of presentation of the material throughout the entire text, aiming for an optimal balance of breadth and depth. Electromagnetics, as a fundamen- Most importantly, for each or set of problems, there set of examples is in the text provides the students and other readers with all nec- characteristic feature of the book is its consistent essary instruction and guidance to be able to solve tal problem on their own, and to complete all homework assignments and practice for tests and exams. The abundance and quality of examples and problems are enormously important for the success of the course and class: students always ask for more and more solved examples, which must be relevant for the many problems that follow (for homework and exam preparation) - and this is exactly what this book attempts to offer. Examples and problems in the book emphasize physical conceptual reasoning and mathematical synthesis of solutions, and not pure formulaic (plugand-chug) solving. They also do not carry dry and too complicated pure mathematical formalisms. The primary goal is to teach the readers to reason through different (more or less challenging) situations and to help them gain confidence and really understand and like the material. Many examples and problems have plete physical explanations for (almost) everything the a strong practical engineering context. show and explain every ample discussions of approaches, strategies, and alternatives. Very often, solutions are presented in more than one way to aid understanding and development of true electromagnetic problemSolutions to examples step, with By acquiring such skills, which are browsing through the book pages in a quest for a suitable “black-box” formula or set of formulas nor a skillful use of pocket calculators to plug-and-chug, the reader also acquires true confidence and pride in electromagnetics, and solving skills. definitely not limited to a skillful science and engineering discipline, provides within com- scope and rigorous mathematical models its for everything it covers. Thus, besides a couple of Coulomb’s law) model to building the most impres- experimental fundamental laws (like that have to be taken for granted for the build on, all other steps in and exciting structure called the electromagnetic theory can be readily presented to the reader in a consistent and meaningful manner and with enough detail to be understandable and appreciable. This is exactly what this book attempts to do. Simply speaking, literally everything is derived, proved, and explained (except for a couple of expersive imental facts), with many new derivations, expla- and visualizations. and important concepts and derivations are regularly presented in more than one way to help students understand and master the subject at hand. Maximum effort has been devoted to a continuous logical flow of topics, concepts, equations, and ideas, with practically no “intentionally skipped” steps and parts. This, however, is done in a structural and modular manner, so that the reader who feels that some steps, derivations, and proofs can be bypassed at the time (with an opportunity of redoing it later) can do nations, proofs, interpretations, Difficult so, but this discretion is left to the reader’s discretion (or to the and advice of the course instructor), not the author’s. Overall, (or all my approach is to provide all possible necessary) explanations, guidance, and detail Preface transmission lines, waveguides, and antennas). In whereas students’ actual understanding of the mate- addition, the book features a favorable balance of “on their own feet,” and ability independent work are tested and challenged to do through numerous and relevant end-of-chapter problems and conceptual questions, and not through filling the missing gaps in the text. On the other hand, I am fully aware that static in the theoretical parts rial, and examples xiii in the text, their thinking brevity may seem attractive to students at first glance (one third) and dynamic (two thirds) fields. or a sequence of courses using cover the book material, would completely text Ideally, a course this with a likelihood that some portions would be given to students as a reading assignment only. book allows the a lot of flexibility and However, many dif- ferent options in actually covering the material, or means fewer pages for readHowever, most students will readily acknowledge that it is indeed much easier and faster to read, grasp, and use several pages of thoroughly explained and presented material as opposed to a single page of condensed material with too many missing parts. During my dealings with students over so many years, I have been constantly told that they in fact prefer having everything derived and explained, and host of sample problems solved, to a lower page count and too many important parts, steps, and explanations missing, and too few detailed solutions, and this was the principal motivation for parts of my writing this book. able for different areas of emphasis and specialized because it typically ing assignments. This approach, in my opinion, is also good for have a self-contained, ready-touse continuous “story” for each of their lectures, instead of a set of discrete formulas and sample facts with little or no explanations and detail. On the other hand, the instructor may choose to present only main facts for a given topic in class and rely on students for the rest, as they will be able to quickly and readily understand all reading assignments from the book. Indeed, I expect that every instructor using this text will have different “favorite” topics presented in class with all details and in great instructors, as they depth, including a number of examples, while “giv- away” some other topics to students to cover on their own, with more or less depth, including worked ing examples. it, and ordering the topics in a course (or courses). One 1-7, do scenario is to quickly go through Chapters just basic concepts and equations, and a couple of examples in each chapter, quickly reach Chapter 8 (general Maxwell’s equations, etc.), and then do everything else as applications of general Maxwell’s equations, including selected topics from Chapters 1-7 and more or less complete coverage of all other chapters. This scenario would essenreflect the inverse tially (nonchronological) order of topics in teaching/learning electromagnetics. In fact, there may be many different scenarios suit- outcomes of the course and the available time, of them advancing in chronological order, all through Chapters 1-14 of the book, just with different speeds and different levels of coverage of individual chapters. To help the instructors create a plan for using the book material in their courses and students and other readers prioritize the book contents in accordance with their learning objectives and needs, Tables 1 and 2 provide classifications of all book chapters and sections, respectively, in two levels, indicating which chapters and sections within chapters are suggested as more likely candidates to be skipped or skimmed (covered lightly). This is just a guideline, and I expect that there will be numerous extremely creative, effective, and diverse combinations of book and subtopics constituting course outlines and meet the preferences, interests, and needs of instructors, students, and other book users. Most importantly, if chapters and sections are topics learning/training plans, customized to best OPTIONS IN COVERAGE OF THE MATERIAL This book promotes and implements the direct or chronological and not inverse order of topics in teaching/learning electromagnetics, which can briefly and then dynamic topics, or first fields (static, quasistatic, and rapidly time-varying) and then waves (uniform plane waves. be characterized as: first static skipped or skimmed in the class, they are not skipped nor skimmed in the book, and the student will always be able to quickly find and apprehend additional information and fill any missing gaps using pieces of the book material from chapters and sections that are not planned to be covered in detail. XIV Preface Table would Classification of 1. book chapters in two groups, where “mandatory” chapters are those that be covered in most courses, while some of the “elective” chapters could be skipped (or skimmed) based on specific areas of emphasis and desired outcomes of the course or sequence of courses and the available time. In selecting the material for the course(s), this classification at the chapter level could be combined with the classification at the section level given in Table 2. likely “Mandatory” Chapters: 1, 3, 4, 6, 8, 9, “Elective” Chapters: 12 14 2, 5, 7, 10, 11, 13, 1. Electrostatic Field in Free Space 2. Dielectrics, Capacitance, 3. Steady Electric Currents 5. Magnetostatic Field in Material Media 4. Magnetostatic Field 7. Inductance and Magnetic Energy 6. Slowly Time-Varying Electromagnetic Field 10. 8. Rapidly Time-Varying Electromagnetic Field 11. Field 9. Uniform Plane Electromagnetic Waves 13. Waveguides and Cavity Resonators 14. Antennas and Wireless Communication Systems 12. Circuit in Free Space Analysis of Transmission Lines and Electric Energy Reflection and Transmission of Plane Waves Analysis of Transmission Lines 2. Classification of book sections in two “tiers” in terms of the suggested priority for coverage; if one or more sections in any of the chapters are to be skipped (or skimmed) given the areas of emphasis and specialized outcomes of the course or courses and the available time, then it is suggested that they be selected from the “tier two” sections, which certainly does not rule out possible omission (or lighter coverage) of some of the “tier one” sections as well. Table Chapter 1. “Tier Electrostatic Field in Free Space 2. Dielectrics, 1. Capacitance, and Electric Energy One” Sections 1-1.4, 1.6, 1.8-1.10, 1.13-1.16 2.1,2.6,2.7,2.9,2.10,2.12, “Tier Two” Sections 1.5, 1.7, 1.11, 1.12, 1.17-1.21 2.2-2.5,2.8,2.11,2.14, 2.17 2.13,2.15,2.16 3. Steady Electric Currents 3.1-3.4,3.8,3.10,3.12 3.5-3.7,3.9,3.11,3.13 4. Magnetostatic Field in Free Space 4.1, 4.2, 4.4-4.7, 4.9 4.3, 4.8, 5. Magnetostatic Field 5.1,5.5,5.6,5.8,5.11 5.2-5.4, 5.7, 5.9, 5.10 6. Slowly Time-Varying Electromagnetic Field 6.2-6.5 6.1, 6.6-6.8 7. Inductance and Magnetic Energy 7.1, 7.4, 7.5 7.2, 7.3, 7.6 8. Rapidly Time-Varying Electromagnetic Field 8.2, 8.4, 8.6-8. 8, 9. Uniform Plane Electromagnetic Waves 9. 3-9.7, in Material Media 10. Reflection and Transmission of Plane 1 1 Field Analysis of Transmission Lines . 12. Circuit Waves Analysis of Transmission Lines 8.11,8.12 9.11,9.14 4.10-4.13 8.1,8.3,8.5,8.9,8.10 9.1,9.2,9.8-9.10, 9.12,9.13 10.1, 10.2, 10.4-10.7 10.3, 10.8, 10.9 11.4-11.6, 11.8 11.1-11.3, 11.7, 11.9, 11.10 12.1-12.6, 12.11, 12.12, 12.15 12.7-12.10, 12.13, 12.14, 12.16-12.18 13. 14. Waveguides and Cavity Resonators 13.1-13.3, 13.6, 13.8, 13.9, 13.4, 13.5, 13.7, 13.10, 13.11, 13.12 13.13, 13.14 Antennas and Wireless Communication 14.1, 14.2, 14.4-14.6, 14.8, 14.3, 14.7, 14.9-14.13 Systems 14.14,14.15 Preface Table xv Ordering the book material for the transmission-lines-first approach; Chapter 12 (Circuit 3. is written using only pure circuit-theory concepts (all field-theory aspects of transmission lines are placed in Chapter 11 - Field Analysis of Transmission Lines), so it can be taken at the very Analysis of Transmission Lines) beginning of the course (or at any other time in the course). Note that two sections introducing (or reviewing) complex representatives of time-harmonic voltages and currents (Sections 8.6 and 8.7) must be done before Chapter 12. Section 8.6: Time-Harmonic Electromagnetics Section 8.7: Complex Representatives of Time-Harmonic Chapter 12: Circuit Analysis of Transmission Lines (or a selection of sections from Chapter 12 - see Table 2) Chapters 1-11, 13, 14 or a selection of chapters (see Table TRANSMISSION-LINES-FIRST APPROACH One possible Field and Circuit Quantities 1) and sections (see Table 2) an invaluable resource. They are also ideal for and discussions (so-called active teaching and learning) to be combined with traditional lecturing - if so desired. in-class questions exception from the chronological sequence of chapters (topics) in using this text implies a different placement of Chapter 12 (Circuit Analysis of Transmission Lines), which is written in such a manner that it can be taken at any time, even at the very beginning of the course, hence constituting the transmission-lines-first approach to the course and learning the material. Namely, the field and circuit analyses of transmission lines are completely decoupled in the book, so that any field-theory aspects are placed in Chapter 11 (Field Analysis of Transmission Lines) and only pure circuit-theory concepts are used in Chapter 12 with per-unit-length characteristics (distributed parameters) of the lines being taken for granted (are assumed to be known) from the field analysis if the circuit analysis is done first. Table 3 shows the transmission-lines-first scenario using this book. teaching In addition, conceptual questions are perfectly suited for class assessments, namely, to assess stu- and evaluate the effectiveness between the course “pretest” and “posttest” scores, and especially in light of ABET and similar accreditation criteria (the key word in these criteria is “assessment”). Selected conceptual questions from the large collection provided in the book can readily be used by instructors as partial and final assessment dents’ performance of instruction, usually as the “gain” instruments for individual topics at different points in the course on the Companion Website, and comprehensive collection of MATLAB computer exercises, strongly coupled to the book material, both the theory and the worked examples, and designed to help students develop a stronger intuition and a deeper understanding of electromagnetics, and find it more attractive and lika The book provides, on the Companion Website, a 500 conceptual questions. These are multiplechoice questions that focus on the core concepts of the material, requiring conceptual reasoning and understanding rather than calculations. They serve as checkpoints for readers following the theoretical parts and worked examples (like homework problems, conceptual questions are referred to at the ends of sections). Generally, conceptual questions may appear simple, but students often find them harder than the standard problems. Pedagogically, they are total of for the entire class. MATLAB EXERCISES, TUTORIALS, AND PROJECTS The book MULTIPLE-CHOICE CONCEPTUAL QUESTIONS and very able. provides, large MATLAB is chosen principally because it is a generally accepted standard in science and engineering education worldwide. There are a total of 400 MATLAB exercises, which are referred to regularly within all book chapters, at the ends of sections, to supplement problems and conceptual questions. Each section of this collection starts with a comparatively very large num- ber of tutorial exercises with detailed completely XVI Preface worked out files). solutions, as well as MATLAB codes (m This resource provides abundant opportunities and homework for instructors for assigning in-class projects - if fascinating biographies of famous scientists and pio- neers in the field of electricity and magnetism. There my are a total of 40 biographies, which are, in view, not only very interesting historically and informa- so desired. terms of providing the factual chronological review of the development of one of the most imprestive in VECTOR ALGEBRA AND CALCULUS and complete theories of the entire and technological world - the electromagnetic theory - but they also often provide additional technical facts and explanations that complement the sive, consistent, scientific Elements of vector algebra and vector calculus are presented and used gradually across the book sections with an emphasis on physical insight and immediate links to electromagnetic field theory concepts, instead of having a purely mathematical review in a separate chapter. They are fully integrated with the development of the electromagnetic theory, where they actually belong and really come to life. The mathematical concepts of gradient, divergence, curl, and Laplacian, as well as line (circulation), surface (flux), and volume integrals, are literally derived from physics (electromagnetics), where they naturally emanate as integral parts of electromagnetic equations and laws and where their physical meaning is almost obvious and can readbe made very visual. Furthermore, the text is written in such a way that even a reader with litily tle background in vector algebra and vector calculus indeed be able to learn or refresh vector analy- will sis concepts directly through the Appendix ters (please see 3 first several chap- - Vector Algebra and Calculus Index). LINKS I also feel that some basic knowledge about the discoverers - who made such epochal scientific achievements and far-reaching contributions to humanity - like Faraday, Maxwell, Henry, material in the text. Hertz, Coulomb, Tesla, Heaviside, Oersted, Ampere, Ohm, Weber, and throughout all of explanations for chapters. its all It elements of made an irre- engineering and physics students. SUPPLEMENTS The book (for accompanied by the Solutions Manual is instructors) with detailed solutions to all end-of-chapter problems (written in the same manin the examples in the book), answers to all conceptual questions, and MATLAB codes (m files) for all MATLAB computer exercises and projects, as well as by PowerPoint slides with all illustrations from the text and by other supplements. Pearson eText of the book is also available. ner as the solutions www.pearsonhighered.com/notaros TO CIRCUIT THEORY The book provides detailed discussions of the links between electromagnetic theory and circuit theory others should be placeable part of a sort of “general education” of our ACKNOWLEDGMENTS contains physical circuit theory, for both dc and ac regimes. All basic circuit-theory equations (circuit laws, element laws, etc.) are derived from electromagnetic theory. The goal is for the reader to develop both an appreciation of electromagnetic theory as a foundation of circuit theory and electrical engineering as a whole, as well as an understanding of limitations of circuit theory as an approximation of field theory. is based on my electromagnetics teachand research over more than 20 years at This text ing the University of Belgrade, Yugoslavia (Serbia), Colorado at Boulder, University of Massachusetts Dartmouth, and Colorado State University of University, in students advice, at Fort my acknowledge these ideas, Collins, U.S.A. I gratefully colleagues and/or former Ph.D. institutions enthusiasm, whose initiatives, discussions, co-teaching, and co-authorships have shaped my knowledge, teaching style, pedagogy, and writing in electro- HISTORICAL ASIDES magnetics, including: Prof. Branko Popovic (late). Throughout almost Prof. Milan Ilic, Prof. Miroslav Djordjevic, Prof. Antonije Djordjevic, Prof. Zoya Popovic, Gradimir Bozilovic, Prof. Momcilo Dragovic (late), Prof. all chapters of the book, dozens of Historical Asides appear with quite detailed and Preface Branko Kolundzija, Prof. Vladimir Petrovic, and Jovan Surutka (late). All I know in electromagnetics and about its teaching I learned from them or with them or because of them, and I am enormously Prof. thankful for that. am I over ing all grateful to all these years for my all students in the joy them electromagnetics and I all my have had classes in teach- for teaching me to teach better. I Nada Ana Sekeljic, my current Ph.D. students Manic, and Sanja Manic for MATLAB computer and codes, checking the derivaand examples in the book, and solving selected their invaluable help in writing exercises, tutorials, tions end-of-chapter problems. gratitude to Prof. Milan with the manager Scott Disanno, for expertly leading the book production, Marcia Horton, Vice President and Editorial Director with Prentice Hall, for great conversations and support in the initial phases of the project, and Tom Robbins, former Publisher at Prentice Hall, for the first encouragements. I hope they enjoyed our dealings and discussions as extensively as my Ilic, initial I owe a particular debt of colleague and former Ph.D. student for his outstanding work and help electronic artwork in the book. colleagues and former students Andjelija Ilic My and Prof. Miroslav Djordjevic, as well as Olivera Notaros, also contributed very significantly to the artwork, for which I am sincerely indebted. I would like to express my gratitude to the reviewers of the manuscript for their extremely detailed, useful, positive, and competent comments that I feel helped me to significantly improve the quality of the book, including: Professors Indira Chatterjee, Robert J. Coleman, Cindy Harnett, Jianming Jin, Leo Kempel, Edward F. Kuester, Yifei Li, Krzysztof A. Michalski, Michael A. Parker, Andrew F. Peterson, Costas D. Sarris, and Fernando L. Teixeira. Special thanks to all members of the Pearson Prentice Hall team, who all have been excellent, and particularly to my editor Andrew Gilfillan, who has been extremely helpful and supportive, and whose input was essential at many stages in the development of the manuscript and book, my production I did. my thank I especially thank xvii wife ECE Olivera Notaros, Department who also Colorado State University, not only for her great and constant support and understanding but also for her direct involvement and absolutely phenomenal ideas, advice, and help in all phases of writing the manuscript and production of the book. Without her, this book would not be possible or would, at least, be very different. I also acknowledge extraordinary support by my wonderful daughters Jelena and Milica, and I hope that I will be able to keep my promise to them that I will now take a long break from writing. I am very sad that the writing of this book took me so long that my beloved parents Smilja and Mile did not live to receive the first dedicated copy of the book from me, as had been the case with my previous teaches in the at books. Finally, on a very personal note as well, I really hope convey at least a portion of my admiration and enthusiasm to the readers and help more and more students start liking and appreciatlove electromagnetics and teaching that this book it, and I will ing this fascinating discipline with endless impacts. I am proud of being able to do that in my classes, and am now excited and eager to try to spread that mesmuch larger audience using this text. Please your comments, suggestions, questions, and corrections (I hope there will not be many of these) sage to a send me regarding the book to notaros@colostate.edu. Branislav M. Notaros Fort Collins, Colorado “I believe but cannot explain that the author’s confidence is somehow student as a trust that the text they are reading and learning from —Anonymous reviewer of the book manuscript is transferred to the worth their time.” ' Electrostatic Field in Free Space [ Introduction: E lectrostatics ics is the branch of electromagnet- that deals with phenomena associated with which are essentially the consequence of a simple experimental fact - that charges exert forces on one another. These forces are called electric forces, and the special state in space due to one charge in which the other charge is situated and which causes the force on it is called static electricity, the electric field. Any charge distribution in space with any time variation tric field. The charges at rest due to time-invariant (charges that do not change in time and do not move) or electrostatic a source of the elec- is electric field is field. called the static electric field This is the general electromagnetic the simplest form of field, and its physics and mathematics represent the foundation of the entire electromagnetic theory. On the other hand, a clear understanding of electrostatics for many and forces and systems. electric fields, charges, electronic devices We is essential practical applications that involve static shall begin in electrical our study of electrostatics by investigating the electrostatic field in a air (free space), and which will vacuum or then be extended to the analysis of electrostatic structures composed of charged conductors in free space (also in this chapter). In the next chapter, we shall evaluate the electrostatic field in the presence of dielectric rials, and include such materials in mateour discussion of general electrostatic systems. 1 2 Chapter 1 Electrostatic Field in Free Space HISTORICAL ASIDE The first dates city with B.C., of Miletus PXeKxpov an (624 B.C.-546 B.C.), (Mayvrjala), wrote that amber rubbed in wool attracts pieces of straw or feathers - which we now Charles Augustin Coulomb was a and in between like and unlike charges (i.e., between two charges of the same or opposite polarity) using his genuine the torsion balance apparatus, in a course of experi- ments originally compass. experimentalist in electricity word “magnet” Greece named Magnesia which ancient Greeks first noticed origin of the basic law for the electrostatic force Engineering Corps of the French Army and a brilliant The 800 B.C.) that pieces of the black rock they were standing on, now known as the iron mineral magnetite (Fe 304 ), attracted one another. de in Our experiences (ca. (1736-1806) colonel elektron ) for amber. relates to the region in time, all ( to ancient times. pher and mathematician, one of the greatest minds of fric- “electron” for the with magnetism, on the other side, also trace back Greek philoso- ancient name subatomic particle carrying the smallest amount of (negative) charge comes from the Greek word back to the when Thales a manifestation of electrification by is tion. In relation to this, the electri- century sixth know record of our ex- periences magne- aimed He measured tion or repulsion that He at improving the mariner’s the electric force of attrac- two charged small pith balls graduated in 1761 from the School of the Engineering Corps ( Ecole du Genie), and exerted on one another by the amount of twist pro- was in charge of building the Fort Bourbon on Martinique, in the West Indies, where he showed his engineering and organizational skills. each of the balls and inversely proportional to tism. In 1772, Coulomb returned to France duced on the torsion balance, and demonstrated an inverse square law for such forces - the force is the square of the distance between their centers. came out to be an underpinning of the whole area of science and engineering now known as electromagnetics, and of all of its applications. Upon the outbreak of the French Revolution in This result with impaired health, and began his research in applied mechanics. In 1777, he invented a torsion balance measure small forces, and as a result of his 1781 memoir on friction, he was elected to the French Academy ( Academie des Sciences ). Between 1785 and 1791, he wrote a series of seven papers on electricity and magnetism, out of which by far the most important and famous is his work on the theory of attraction and repulsion between charged bodies. Namely, Coulomb formulated in 1785 the to 1.1 The COULOMB S 1789, to Coulomb work in retired to a small estate near Blois, peace on his scientific memoirs. His last post was that of the inspector general of public instruction, under Napoleon, from 1802 to 1806. The law of electric forces on charges now bears his name - Coulomb’s law - and his name is further immortalized by the use of coulomb (C) as the unit of charge. LAW basis of electrostatics that the electric force proportional to the product of the charges of Fe i is 2 an experimental result called Coulomb’s law. It states to a point charge Q\ in a on a point charge Qi due F Section 1.1 vacuum (or air) is given by 1 (Fig. Coulomb's Law 3 1 1) . Fel2 1 Q1Q2 Attsq R2 = R 12 (1 - 1 ) . Coulomb's law With R12 denoting the position vector of Q2 relative to Qi, R = IR12I is the distance between the two charges, R12 = R12/R is the unit vector 2 of the vector R12, and £0 is the permittivity of a vacuum (free space), £0 = (p 10“ 12 and F = pF/m 8.8542 farad, the unit for capacitance, is (1 which be studied will 2) . permittivity of a vacuum in the By point charges we mean charged bodies of arbitrary shapes whose dimensions are much smaller than the distance between them. The SI (International System of Units 3 ) unit for charge is the coulomb (abbreviated C), named in honor next chapter). of Charles Coulomb. This which is is a very large unit of charge. c in magnitude ((^electron QiQ 2 /( 47teor2 ) sion The charge of an electron, negative, turns out to be 1.602 x 1 (T 19 C (1.3) Eq. ( 1 1) . Fe i2 represents the algebraic intensity (can be of with respect to the unit vector R12. sign or polarity (like charges), this intensity Q2 are of the the same orientation as R12, and the force between charges the electric force between unlike charges (Q1Q2 < 0 ) is is is If positive, Q\ and Fe i2 has repulsive. Conversely, attractive. and noting that R21 = — R12, we obtain that F e 2i e i2; he., the force on Q\ due to Q2 is equal in magnitude and opposite in direction to the force on Q2 due to Q\. This result is essentially an expression of Newton’s third law - to every action (force) in nature, there is an By reversing the indices 1 and 2 in Eq. ( 1 1) . =— opposed equal If sition, charge of electron, magnitude = — e). The unit for force (F) is the newton (N). The expres- in arbitrary sign) of the vector same = Figured Notation Coulomb's law, given in by Eq- (1-1). reaction. we have more than two which also particular charge is point charges, we can use the principle of superpo- on a by each a result of experiments, to determine the resultant force - by adding up vectorially the partial forces exerted on it of the remaining charges individually. is carried out component by component an arbitrary number of vectors), most frequently in the Cartesian coordinate system. Cartesian (or rectangular) coordinates, x, y, and z, and coordinate unit vectors, x, y, and z (unit vectors along the x, y, and z directions), are shown in Fig. 1 2 The unit vectors are mutually perpendicular, and an arbitrary vector a in Cartesian coordinates can be represented as In the general case, vector addition (for . a Tn typewritten work, vectors are = ax x + a y y + a z z. e.g., F, whereas in handwritten work, they are denoted by placing a right-handed arrow over the symbol, as F. 2 All unit vectors in this text will be represented using the “hat” notation, so the unit vector in the x direc- tion (in the rectangular coordinate system), for example, widely used notations for unit vectors would represent 3 SI is the modernized version of the metric system. International d’Unites. is this given as x (note that vector as a v The abbreviation , is i*, and some y, z) unit vectors . (1.4) commonly represented by boldface symbols, Figure 1.2 Point M(x, and coordinate of the alternative u*, respectively). from the French name Systeme in the Cartesian coordinate system. Cartesian vector components 6 . 4 Chapter 1 Space Electrostatic Field in Free Here, a x ay and a z are the components of vector a , , system, and its magnitude a The is unit vector of a unity, = |a| |a| /a a Shown in Fig. 1.2 is is = a 1. Cartesian coordinate in the is = |a| (1.5) = a /a. Of course, the magnitude of a, and The sum of two vectors is given by +b= (a x + bx ) x + + by ) (a y + y (a z + bz ) of any unit vector, z. also the position vector r of an arbitrary point (1 .6) M(x, y, z ) in space, with respect to the coordinate origin (O), position vector of a point r where, using Eq. points O (1.5), r = = xx + yy + zi, = |r| yjx2 (1.7) OM + y2 + z 2 — is the distance 4 between and M. Example Three Equal Point Charges at Triangle Vertices 1.1 Three small charged bodies of charge Q are placed at three vertices of an equilateral triangle with sides a in air. The bodies can be considered as point charges. Find the direction and magnitude of the electric force on each of the charges. , Solution Even with no computation whatsoever, we can conclude from the symmetry of problem that the resultant forces on the charges, Fe i, Fe 2 and Fe 3 all have the same magnitude and are positioned in the plane of the triangle as indicated in Fig. 1.3(a). Let us compute the resultant force on the lower right charge - charge 3. Using the principle of superposition, this force represents the vector sum of partial forces due to charges 1 and 2, this respectively, that is [Fig. 1.3(b)], Fe3 From Coulomb's =Fet T F 23 3 law, Eq. (1.1), F e23 Fe3 and both forces are We of an equilateral triangle and (b) (1-8) Q2 = Fe 23 = (1.9) ' 4neoa 2 repulsive. note that the vector 2, i.e., , computation of the on Fe 3 = resultant electric force one of the charges; Example 1 .1 (vector superposition). Fe3 is positioned along the symmetry line between charges 1 between vectors Fe i 3 and Fe 23 and it makes the angle a = n/ with both vectors. The magnitude of the resultant vector is therefore twice the projection of any of the partial vectors on the symmetry line, which yields and Figure 1.3 (a) Three equal point charges at the vertices e magnitudes of the individual partial forces are given by Fel3 (b) , , 2 (Fel3 cos a) = 2 Fe 3 ^- = Fe uV?> = (1 i . 10 ) for 4 While dealing with a wide variety of vector quantities (draw) them as arrows tion. However, we shall in space, like the force always have in mind vector in Fe i electromagnetics, 2 in Fig. 1.1, we shall regularly visualize and computaand some other to aid the analysis that only position vectors, like r in Fig. 1.2, length vectors to be introduced later have this feature of their magnitude being the actual geometrical distance in space. Magnitudes of sizes (lengths) of arrows in nitudes of quantities of the which is all other vectors are measured in units different from meter, and the space that they are associated with can only be indicative of relative mag- same nature (with the same unit), like two forces acting on the same body, equally useful and will be utilized extensively in this text as well. . Section Q 2 Figure 1.4 Three point charges equal magnitude but with in different polarities at the an equilateral - computation of the resultant electric force on charge 3; for Example 1 .2. vertices of triangle Example Three Unequal Point Charges at Triangle Vertices 1.2 Determine the resultant force on the lower Assume that Q and a are given quantities. The only Solution Fe i 3 is now p is , = tt/3. Its 1.3 Point charges Q\ 1, i 3 Compute = 2Fe cos£) Three Point Charges = 1 Q 2 = —2 /xC, /zC, by Cartesian coordinates defined The and the angle magnitude is hence Fe3 = 2(Fe Example shown difference with respect to the configuration in Fig. 1.3 attractive, as indicated in Fig. 1.4. the line connecting charges 2 and and F e 23 right charge in the configuration (1 in i3 it is is .4. parallel to the partial forces, ^£ = Fel3 = - 1 that the force resultant force on charge 3 makes with any of in Fig. Fc ] 3 (1.11) Cartesian Coordinate System and Q 3 =2 /xC are situated in free space at points m, 0, 0), (0,1m, 0), and (0,0,1m), respectively. the resultant electric force on charge Q\. Solution From Coulomb’s law and Fig. 1.5(a), the magnitudes of the individual forces on the charge Q\ are -= fl=iS =9mNi 01 2) f where R is Q the distance of Fe3 t we decompose them , from 02 1 ( or Q3 )- In order to Cartesian coordinate system. Based on Figs. 1.5(b) and so the resultant force Fe2 i = — Fe2 F e3 i = cos a x i „ Fe3i COS /l X + Fe2 — i sin Fe3] Sin = Fe21 + Fe31 = 1 The Cartesian components ^eix its i /3 (c), a a y, „ Z, = (1 . 13 ) 7T £=4’ (1.14) is FC and add together vectors F e 2 and - into components in the into convenient components, in this case magnitude [Eq. — (1.5)] ^ei 0) of the vector Feiy — —Fe comes out = 74 x + ^e21 V2 „ "yfy ~ Fe amount i \7 „ Z). (1.15) to — Fe 2i~2~ ~ ^-26 mN, (1.16) to be 4 y + Fjlz = Fe21 = 9 mN. (1.17) 1 Coulomb's Law 5 Chapter 1 Electrostatic Field in Free Space Figure 1.5 Summation of electric forces in the Cartesian coordinate system: point charges in (a) three space, with partial force vectors Fe 21 and F e 3 i, (b) component decomposition of Fe 2 i, (c) decomposition of F e 3 i, and (d) alternative addition of forces using the head-to-tail and the cosine formula; Example 1 .3. rule for Note that Fig. 1.5(d ), 5 in Fe \ can alternatively be obtained using the head-to-tail combination with the cosine formula 6 which yields Pel Note is = portrayed in yj F\2 \ also that the vector + ^e 3 ~ 2 /re 2 lFe3 Fe i i is i cos y = F2 = \ 9 mN, y parallel to the line connecting charges = Q3 (1.18) J. and Q2 , and that it positioned at an angle of n/4 with respect to the plane xy. Four Charges at Tetrahedron Vertices Example 1.4 Four point charges Q are positioned in free space at four vertices of a regular (equilateral) tetrahedron with the side length 5 rule, as , By Find the electric force on one of the charges. a. the head-to-tail rule for vector addition, to obtain graphically the vector arrange the two vectors (usually translate b from (second vector) “connected" we draw is placed at the head of a to the tail of the second, (first its sum c = a + b, we way that the tail original position) in such a vector). In other words, the and hence the term “head-to-tail” head of the first first of b vector for this arrangement. is Then extending from the tail of a to the head of b, as in Fig. 1.5(d). add two vectors together is the parallelogram rule, where c = a + b corresponds to a diagonal of the parallelogram formed by a and b, which can also be seen in Fig. 1.5(d). To add more than two vectors, e.g., d = a + b + c, we simply apply the head-to-tail rule to add c to the already found a + b, and so on - the resultant vector extends from the tail of the first vector to the head of the last vector in the multiple head-to-tail chain, and a polygon is thus obtained, which is why this procedure is often referred to as the polygon rule. An ft In c (resultant vector) as a vector equivalent graphical method to an arbitrary triangle of side lengths a, /;, and c and angles a , fi , and y, the square of the length c of the 2 2 side opposite to the angle y equals c a + b 2 — lab cos y (and analogously for a and b and cos /l, respectively), and this is known as the cosine formula (rule) or law of cosines. = 2 2 using cos a Section Note Solution 1 7 Definition of the Electric Field Intensity Vector .2 that this configuration actually represents a spatial version of the planar configuration of Fig. 1.3. Referring to Fig. of the tetrahedron - charge 4. This force Fe4 = 1.6, let us find the force on the charge on the top given by is Fel4 + Fe24 + Fe34, (1-19) same magnitude, equal to Fe 14 = Q / (4n eqo 2 ) The horizontal components of the force vectors all lie in one plane and the angle between where all each two 2 the three partial forces are of the add up to 120°, so that they vectorially is . zero. Thus, the resultant vector component only, whose magnitude amounts component of each partial force, Fe 4 To determine coscr (as = 3 (Fe i 4 COS a) Fe 4 has to three times that of the vertical a vertical (1-20) . H/a) from the right-angled triangle A014 in Fig. 1.6, we first find 1 and point O) from the equilateral triangle A 123 (the base Figure 1.6 Four point charges at tetrahedron vertices; for Example definition of E electric field due 1 .4. the distance b (between charge of the tetrahedron), as 2/3 of the height of this triangle 2 in a Eq. (1.20) results 1. 1-1.7; b2 (1.21) a in Fe 4 = 3Fe i4 : we have V a2 — COSO! 3 Problems so // b which substituted 7 , a/6 V6Q 2 T" 4nsoa 2 (1 Conceptual Questions (on Companion Website): and 1.1 .22) 1.2; MATLAB Exercises (on Companion Website). DEFINITION OF THE ELECTRIC FIELD INTENSITY 1.2 VECTOR The is a special physical state existing in a space around charged fundamental property is that there is a force (Coulomb force) acting on any stationary charge placed in the space. To quantitatively describe this field, electric field objects. Its we introduce a vector quantity called the electric field intensity vector, E. nition, it is equal to the electric force the electric field, divided by Qp , that e= The probe charge has Fe on a probe (test) point charge By defi- Q p placed in is, E (Gp - (1.23) o). (unit: V/m) be small enough in magnitude in order to practically not which are the sources of E. The unit for the electric field intensity we use is volt per meter (V/m). From the definition in Eq. (1.23) and Coulomb’s law, Eq. (1.1), we obtain the expression for the electric field intensity vector of a point charge Q at a distance R from the charge (Fig. 1.7) to affect the distribution of charges Q E= 4tt 8q 7 R2 R, Note that the orthocenter (point O in Fig. 1.6) of an equilateral 1, so into segments 2/z/3 and /i/3 long. Note also that h ratio 2 : being the side length of the triangle. (1.24) charge triangle partitions = C3a/2 (in its heights ( h ) in the an equilateral triangle), a in free to a point space — 8 Chapter Electrostatic Field in Free 1 R where Space is the unit vector along R directed from the center of the charge (source point) toward the point at which the field is (to be) determined observation (field or point). By superposition, the electric field Qn) at a point that is at (Q i, Q2, produced by N point charges R2 Rn, respectively, from intensity vector distances R\. the charges can be obtained as Figure 1.7 Electric intensity vector point charge field due to in free N a e = e + e2 + 1 space. where R /, / Problems 1 .3 : = 1,2, ... 1.8; ... + e^ = - 1 O 0-25) R<’ 4ne 0 *ri Rf ,N, are the corresponding unit vectors. MATLAB Exercises (on Companion Website). CONTINUOUS CHARGE DISTRIBUTIONS A point charge the simplest case of a charge distribution, which, mathematically, is corresponds to the space (three-dimensional) delta function. In the general case, however, charge can be distributed throughout a volume, on a surface, or along a line. Each of these three characteristic continuous charge distributions is described by a suitable charge density function. The volume charge density defined as volume charge density (in a volume v) is [Fig. 1.8(a)] (1.26) (unit: C/m 3 ) the surface charge density (on a surface S) is given by [Fig. 1.8(b)] surface charge density (unit: (1 .27) C/m 2 ) and the line charge density charge density (along a line line /) is [Fig. 1.8(c)] (1.28) (unit: C/m) symbol pv is sometimes used instead of p, a instead of ps and p\ In the above equations, dQ represents the elemental charge in a volume element dv, on a surface element dS, and along a line element dl, respec3 C/m 2 and C/m. tively, and the corresponding units for charge densities are C/m Note that the instead of Q' , . , , charge Q for the three characteristic charge distributions obtained as / d Q (adding charge elements d<2), which leads to The total Q = 'm v P dv, Qon JV d2 ( dv Figure 1.8 Three characteristic , / continuous charge distributions and charge elements: (a) volume charge, (b) surface charge, and (c) line charge. v (a) / \ J S and C?along / — JQ d/. in Fig. 1.8 is (1.29) r , Section 1 .4 On the Volume and Surface Integration 9 respectively. Special, but important, cases of continuous charge distributions are uniform volume, surface, and line charge distributions. A charge distribution is said to be uniform if the associated charge density is constant over the entire region (v, S, or /) with charges. The expressions in Eqs. (1.29) then become much simpler, Qinv = pv (/0 = const), Qalong Note that Q' (Q' length (p.u.l.) = const) is / = = PsS (p = const), Q = const). QonS s ' Q'l ( (1 also used to represent the so-called charge per unit of a long uniformly charged structure thin or thick cylinder), (e.g., defined as the charge on one meter (unit of length) of the structure divided by Q = „ Glp.u.i. y — Qalong = Qforlm ; / 1 l 1.5 A volume charge Nonuniform Volume Charge Distribution is where total r = Po - (0 < r < m, (|-31) charge per unit length, in in a distributed in free space inside a sphere of radius P(r) 1 length m and hence Q' numerically equals the charge on each meter of the Example .30) C/m structure. Sphere a. The charge density is (1.32) a). stands for the radial distance from the sphere center, and po is a constant. Find the charge of the sphere. we need to integrate in the first expreswe adopt dv in the form of a thin shown in Fig. 1.9. The volume of the shell is Solution Since the charge density depends on r only, sion in Eqs. (1.29) only with respect to that coordinate, and spherical shell of radius r and thickness dr, as dv = 4nr1 dr, 0-33) which can be visualized as the volume of a thin flat slab of the same thickness ( dr) and the same surface area (S = 4nr2 ), dv = S dr. The total charge of the sphere comes out to be Q= [ p dv Jv = [ Jr=0 po - 4n 2 dr = pqtt a 3 . 0-34) a Figure 1.9 Integration of nonuniform volume charge density in Example Problems 1 .4 : 1.9-1.12. ON THE VOLUME AND SURFACE INTEGRATION A few additional comments about the volume integration performed in Eq. (1.34), and similar multiple integrations, may be useful at this point. In general, our strategy for solving volume integrals, fv f dv, is to adopt as large volume elements dv as possible, the only restriction being the condition that / = const in dv. For instance, for the function p = p(r) in Eq. (1.32) and a sphere as the domain of integration, the largest volume element over which p = const is a thin spherical shell, with dv given in Eq. (1.33). This adoption enables us to perform the volume integration in Eq. (1.34) by integrating along r only, from 0 to a, whereas the adoption of the standard differential volume element (elementary curvilinear cuboid) in the spherical coordinate system (Fig. 1.10), dv' = dr(rd6) (rsin0d0) = 2 r sin 6 dr d6 d0 (1.35) 1 a sphere; for .5. 10 Chapter 1 Electrostatic Field in Free Space z r sin0 Figure 1.10 Standard differential ( volume element dv') in the spherical x coordinate system. would require two additional integrations, i.e., the integration in 6 from 0 to n and the integration in <p from 0 to 2 n. Note, however, that these two integrations are implicitly contained in the expression for dv in Eq. (1.33), as (1.36) Note also that the elementary cuboid dv' p(r 9 , , (p ). cylindrical A dv the for a charge den- all and Cartesian coordinate systems. example for our integration strategy trivial in cases [e.g., would be necessary three coordinates in the spherical coordinate system, p = Similar considerations apply to volume integrations associated with the depending on sity when / first = is the adoption of v in place of const in the entire integration domain v, yielding fv f dv — fv expression in Eqs. (1.30)]. The same principle is adopting surface elements dS for solving utilized for surface integrals. 1 .5 ELECTRIC FIELD INTENSITY VECTOR DUE TO GIVEN CHARGE DISTRIBUTIONS Using the superposition principle, the electric field intensity vector due to each of the (uniform or nonuniform) charge distributions p, p s and Q' can be regarded as the vector summation of the field intensities contributed by the numerous , equivalent point charges making up the charge distribution. Thus, by replacing in electric field Eq. (1.24) with charge element d Q = Q p dv, ps d S, or Q' d / and integrating, we get due to volume (1.37) due to surface (1.38) due to line charge electric field charge electric field charge (1.39) Section Note R that, in the general case, R and 1 Electric Field Intensity .5 Vector due to Given Charge Distributions vary as the integrals in Eqs. (1.37)-(1.39) We shall now apply these are evaluated, along with the functions p, ps and expressions to some specific charge distributions and field points, for which the inte- Q , grals ' . can be evaluated analytically. It is very important that we develop analytical solving these (and similar), true three-dimensional (3-D), vector problems. skills for There is not a unique recipe for an optimal solution algorithm. A general advice, however, is to use superposition whenever possible - to break up a complex problem into simpler ones, and then add up (integrate) their solutions to get the solution to the original problem. In doing this, sometimes we use directly the expressions in Eqs. (1.37)-(1.39), which essentially imply breaking up the structure into equivalent point charges. Often, on the other hand, we do not go directly all the way down to point charges. Instead, we decompose the structure and apply superposition “layer by layer,” modularly, going down toward simpler structures level by level. In this, we always try to incorporate already known solutions to relevant simpler problems into the solution to the problem under consideration. Example 1.6 Charged Ring A line charge of uniform charge density Q of radius a in its air. Find the electric is distributed around the circumference of a ring vector along the axis of the ring normal to field intensity plane. We subdivide the ring (contour C) into elemental segments of length Solution and apply Eq. (1.39). The contribution to the (z-axis) by each charged segment is in Fig. 1.11, axis dE = with the position of P along the 4neoR z R= R, axis being defined total electric field intensity vector is \Jz electric field at a point 2 + a2 d /, as shown P on the ring (1 , by the coordinate .40) z(-oo<z<oo). The obtained as (1.41) Due to symmetry, the horizontal (radial) element d / there is opposite horizontal electric and hence component of the vector E is zero (for every a corresponding diametrically opposite element that gives E has a vertical an equal but field component, so that the two contributions cancel each other), (axial) component only E = EZ Substituting the expression for Ez d E7 dE from Eq. (Fig. 1.11), = dE cos a = dE (1.40) into Eq. (1.42), d1 2eoR 3 47reoR 3 Jc axis of a along the its plane; for .6. (1.42) R we field charged ring normal to Example 1 = = 2na 1 Figure 1.11 Evaluation of the electric get (1 .43) (1 .44) or 47T£o (z 2 + a2 where Q = Q'2na is the total charge of the ring. 2 2 2 Note that for |z| a, z + a and Eq. z , E« (\z\ » a). electric field of charge (1.44) yields Q 4neoZ 2 3/2 ) (1.45) due to a ring 11 a 12 Chapter 1 Electrostatic Field in Free . Space This means that far away from the ring, at its center. In other words, when its charge is equivalent to a point charge the distance of the field point from the ring is Q located much larger than the ring dimensions, the ring can be considered as a point charge and the actual shape of the ring (or any other charged body) does not matter. This in fact is the definition of a point charge or a small charged body. Example A line Semicircular Line Charge 1.7 charge in the form of a semicircle of radius a is situated in free space, as shown in The line charge density is Q' Compute the electric field intensity vector at an Fig. 1.12(a). arbitrary point along the z-axis. Solution The electric field intensity vector along the z-axis due to an elemental charge on the semicircle [Fig. 1.12(a)] is given by the expression in Eq. (1.40). To be able to perform the integration in Eq. (1.39), we need to decompose this vector into convenient components. With reference to Fig. 1.12(b), we first represent dE by its horizontal and vertical components, <2'd/ dE = dEh + d£ z z, d Ez = = d£h d£sina, sin dEh the semicircle), = dE* x we need dEcosa, = cos a + dEy y, dE* (the range for the azimuthal angle the relationship d/ = ad0 0 decompose to is (segment d / —n/2 < 0 < d Ex - Figure 1.12 Evaluation of the electric field charge in due to a line the form of a semicircle; for Example 1.7. = E3 Q'az_ 1 47T£ 0 (i.e., as the point P' = - dEh sin 0 d£y 7r/2). From (1-47) the above expressions and cos 0 d0 =- a Q"-3 2 ItteqR 5 , (1.48) sin0d0 = 0. (1.49) J J-jt/2 —IT/ r n/2 ’ E J- r/2 3 its rx/2 2 / 47T£ 0 evaluated further [Fig. 1.12(c)], x/2 Q’a 2 O' is .46) we obtain AtTSqR} /_ x/2 dEy -Jf =- (1 R an arc of radius a defined by the angle d0, and length thus equals the radius times the angle), / it = - dEh cos0, is = R' Since the direction of the vector dEh varies as the integral moves along a y 4£ 0 E 3 ’ (1.50) s ) Section Electric Field Intensity .5 1 so that the final expression for the electric field vector E= Example 2 0 R3 V l is R = Vz 2 +a 2 . E an arbitrary point field at in space due to a straight line of medium uniformly charged with a total charge Q. The ambient Solution (1.51) . / Straight Line Charge of Finite Length 1.8 Find the expression for the length | 2 tr IB Vector due to Given Charge Distributions Let the line charge and the is air. point (P) be in the plane of drawing (Fig. 1.13). field The geometry of this problem can be defined using three parameters: angles 9\ and 62 and the perpendicular distance from the line to the point, d (for the particular position of the point P , shown in Fig. 1.13, 6\ < > 0 and 62 = In Eq. (1.39), Q' 0). R— Q/l, s/z 2 + d2 , dl = R = cos0 x — sin# z, where tively, z (zi < z < Zz) and 6 < ( 0\ 9 < dz, and (1-52) 62 ) are the length each determining the position of the source point, and angular coordinates, respecP', along the line. From Fig. 1.13, the following trigonometric relationship exists between these two coordinates: z = tand (1.53) d' and its differential form (by taking the dz cos 2 9 d d (. which, given that cos# = d/R, = const), d6 dz Finally, substituting the expressions O from Eqs. — (1.52) ( f 02 ( / (1.55) ~d’ cos 9 d9 x and t - ^ [(sin (1.55) into Eq. (1.39), sin 0 dO z J9l 6*2 - sin 4neold x + (cos 02 — cos0i) zl. An Infinite Line 1.9 infinite line electric field Charge charge of uniform density intensity vector at (1.56) an arbitrary point electric field due to a line charge of finite length Note that the expression in Eq. (1.56) can be combined for computing the due to an arbitrary structure assembled from straight line segments of charge. Example we have ®1 / Unsold \J9l E= (1.54) equivalent to is K2 = E = both sides of the equation) reads differential of d0 Q resides in Determine the air. electric field in space. Figure 1.13 Evaluation of the electric field charge of Example 1 due to a line finite length; for .8. 14 Chapter 1 Electrostatic Field in Free Solution Space Let r be the perpendicular (radial) distance of a (observation) point, field P, from the line charge, and r the unit vector along this distance directed from the line to the point We use and d= the expression for the field due to a finite line charge in Eq. (1.56) and n/2 (the line extends to and x = r. What we obtain 82 -» z -*• r, is with density electric field line due to an — oo and let 9\ z -> oo, respectively), as well as P. -> —tc/2 Q/l = Q', exactly the field expression for the infinite line charge Q Q' E= infinite (1.57) 2neor charge We see that the vector E is radial with respect to the line charge axis, and its magnitude varies in space as a function of the radial distance r only, with E being inversely proportional to r. z., Disk of Charge Example 1.10 .,dE Consider a very thin charged disk (i.e., a circular sheet of charge), of radius a and a uniform surface charge density p s in free space. Calculate the electric field intensity vector along the disk axis normal to its surface. , Solution Instead of directly applying Eq. (1.38), of width dr, as shown along the z-axis is in Fig. 1.14. The field due R = yr2 + z 2 4neoR i d<2 The surface area of the By = P = R2 = RdR = so that the substitution of rdr by PsZ 2 e0 disk where strip of length 2nrdr. 2eo Taking the differential of the relationship E= (1.59) equal (1.60) superposition, Js a charged a) at a point p s dS. r E =/dE=M S to < (1.58) can be computed as the area of a thin and width dr, that is, dS due r , ring, dS, to the ring circumference, 2nr, electric field < with d Q standing for the charge of the ring, given by the electric field due to a charged disk; for Example 1 .10. subdivide the disk into elemental rings [see Eq. (1.44)] dE=-^%z, Figure 1.14 Evaluation of we to a ring of radius r (0 7?| r Lf _o Example = 0 dR R2 yfz 1.11 Find the electric 2 i— RdR in 2 1 2e 0 \ r) r=0 + z 2 we , obtain (1.62) r dr, Eq. (1.61) yields \ E to allow a negative z as well Infinite (1.61) R7 a PsZ / |z|, r Jr = o * J z ( 2eo \|z| (-oo < z < yja 2 (1.63) + z2 / oo). Sheet of Charge field intensity vector due to an infinite sheet of charge of density p s in free space. Solution We note that the infinite sheet of charge can be obtained from the disk in Fig. 1.14 by extending its radius to infinity. Consequently, the field due to the infinite sheet can be obtained from the field due to the disk by letting a —> oo in Eq. (1.63). With this, the second term in parentheses in the final field expression in Eq. (1.63) vanishes. The first term, z/|z|, o Section is either 1 (for z > Electric Field Intensity .5 15 Vector due to Given Charge Distributions -1 (for z < 0), and we conclude that the field intensity around the uniform (constant) in both half-spaces cut by the sheet, and given by 0) or charge infinite sheet of 1 is (1.64) due electric field to an infinite sheet of charge with respect to the reference directions indicated in Fig. 1.15. Namely, for a positive charge (p s > 0) of the sheet, the actual orientations of E are those in Fig. 1.15 (field lines are directed outward from the positive charge), while the situation for a negative ps field vector points toward the negative charge). Example 1.12 A is air. E just opposite (the Hemispherical Surface Charge hemispherical surface of radius a medium is Compute uniformly charged with a charge density p s is . The the electric field intensity vector at the center of the hemisphere (center of the corresponding full sphere). Figure 1.15 Solution We perform a similar procedure as in Example The radius of a into thin rings, as depicted in Fig. 1.16. is defined by an angle 9 (0 < 9 < n/2) dQ = is ax ps dS = a sin ring 9. Its and subdivide the hemisphere whose position on the hemisphere 1.10 charge sheet of is = p 2nasm6 ad9, (1 s C Infinite charge; for Example 1.11. .65) d/ r r C r and d/r denote the ring circumference and width, respectively. Using Eq. (1.44), the electric field intensity dE at the point O due to the charge d Q is found as follows: where Zr = — a cos 9„ Rn = andj — a Jl7 = dE > — d<2(— flCOS#) r n cc\ (1.66) „ z, dE with z T being the local z-coordinate of the point O with respect to the ring center (Fig. 1.16). Figure 1.16 Evaluation of Therefore, the resultant field vector amounts to the electric /•tt/2 sin 9 cos 9 Example 1.13 is its Solution From that is, unit is (1.67) z. hemispherical surface charge; for Example s is distributed in free space over a surface in the a, as shown positioned along the semicylinder in Fig. 1.17(a). 1 .1 2. form of a half A line charge of uniform Find the electric force on the axis. line charge length. Eqs. (1.23) and (1.31), the per-unit-length electric force on the line charge, on one meter (unit length) of the structure (line charge) divided by 1 m (the given by the force N/m), is (1.68) per-unit-length on a with E axis. We standing for the electric field intensity vector due to the charged semicylinder at subdivide the semicylinder into elemental superposition principle. The field strips, of width d /, and find E its using the due to each long line charge with charge density dQ' = strip can be approximated by that of a very ps d /; namely, the charge of the elemental strip of length h, p s h d /, must be equal to the charge of the equivalent line charge of the same length, so d Q'h, which yields this expression for dQ'. Having in mind Eq. (1.57), the electric field the Force on a Line Charge due to a Charged Semicylinder A uniform charge of density p per unit of = (sin20)/2 of a very long circular cylinder of radius density Q' d9 z 4 £Q Je = field at center (point O) of a Ps vector due to the line charge semicylinder axis [Fig. 1.17(b)] is radial with respect to the line, comes out dE = to and its magnitude at the be dQ! 2neoa _ ps d <f> 2neo (1.69) ’ line Coulomb charge (unit: force N/m) E 16 Chapter! Electrostatic Field in Free = since r Space a and computed = dl Ex = f d Jl where denotes the / By symmetry, ad<p. the resultant field has an .^-component only, as x = f d£cos0 = Ji costf>d <p = 27r£o J (p = — 7r /2 — (1.70) , rr £q representing the semicylinder cross section. Finally, line (semicircle) Eqs. (1.68) and (1.70) result in w F„ = Q'Ps - (1.71) x. 7T£ 0 Companion Website): Problems'. 1.13-1.28; Conceptual Questions (on MATLAB DEFINITION OF THE ELECTRIC SCALAR POTENTIAL .6 1 The 1.3-1. 6; Companion Website). Exercises (on electric scalar potential tric field intensity the potential is a scalar quantity that can be used instead of the elec- Dealing with Three difgeneral for the evaluation of E, one integration vector for the description of the electrostatic mathematically simpler than dealing with the is field. field vector. needed in component, while a single integration is required for the potential, can easily be found from the potential by differentiation. In addition, ferent integrations are for each vector d/ and E, in turn, using the electric potential (b) The Figure 1.17 Evaluation of the electric force on a line charge positioned along the axis of a charged semicylinder: (a) three- we are able to connect the electric field with the voltage, fundamental bridge between the as a electric scalar potential field, that is, by the electric force, dW done by F e while moving 8 the dot product of F e and dl, Fe Qp e in , moving a test point charge, Qp . The work along an elementary vector distance dl equals dW = dimensional view showing theory and the circuit theory. field defined through the work done by the electric is Fe e dl = Fe d/ coso?, (1.72) the structure geometry and (b) cross-sectional field for view for computations; Example 1 .1 where a We 3. d | is the angle between the two vectors in the product, as dW note that W e is | on Eq. shown in Fig. 1.18. can be negative (for n/ 2 < a < n), meaning that the work e being performed against the electric (1.72), the electric scalar potential, defined as the work done by the field in V field at a , moving by an external agent. Based P in an electric field is point a test charge from P to a reference point 7Z (Fig. 1.18), W = e divided by Qp . Having mind Eq. V= definition of the electric potential (unit: in W— — c = Qp V) Thus, the potential to not 8 | depend on V Qp (1.23), this r - becomes — -dl= r / n E • dl. ip Qp P with respect and to be equal to the (1.73) dl, n Fe / ip at the point Fe to the reference point 7Z turns out line integral of vector The dot product (also known as the scalar product) of a being the angle between a and b. vectors a and b is E from P to 1Z? b = to a point B, is a scalar given by a • a b cos a, y I j | The line integral of a vector function (field) a defined as fpa dl = a dl. where dl is along a line (curve) /. from a point A the differential length vector tangential to the curve (as in Section Obviously, the potential at the reference point is Definition of the Electric Scalar Potential .6 1 zero (integral from P = 17 TZ to TV). and hence the unit V/m for the electric field intensity. Note that <f> is also used to denote the electric potential. By the principle of conservation of energy, the net work done by the electrostatic field in moving Q from a point A to some point B and then moving it back to p A along a different path is zero (because after the round trip, the system is the same The unit for the potential is volt (abbreviated V), means as at the beginning). This that the line integral of the electric field intensity vector along an arbitrary closed path (contour) is zero, Figure 1.18 Displacement of a test charge in an electrostatic field. £ which constitutes Maxwell’s first E • dl = (1.75) 0, electrostatic field equation for the electrostatic circulation (closed line integral) of E conservative nature of the in electrostatics is field. We see that the always zero, and hence the electrostatic field belongs to a class of conservative vector fields. 10 Eq. (1.75) can alternatively be derived from Coulomb’s law, i.e., from the expression for the electric field intensity vector due to a single point charge, Q, given by Eq. (1.24) and Fig. 1.7. To do this, we break the contour C up into elemental segments ( dl -* 0) parallel and normal to E, as indicated in Fig. 1.19. We realize that contributions to the overall line integral in Eq. (1.75) occur only for the seg90° ments parallel to E, while no contribution for the segments normal to E (a = and E • dl = 0). In specific, along the segments parallel to Q E respect to the charge ), • dl equals either Edl resulting in a zero net line integral (sum) along C. (a By E (segments radial with — 0) or —Edl (a = 180°), superposition, this result is 90° Figure 1.19 Derivation of Maxwell's first equation for the electrostatic field starting with Coulomb's Fig. 1.18) oriented from A contour (and usually mark of a along C. 10 By The reference toward B. it If the line is closed (for example, a circle or a square), and the corresponding line integral, §c a • dl, is we call it termed the circulation direction of dl coincides with the orientation of the contour. definition, a vector field arbitrary shape. C), is said to be conservative when its circulation is zero for a closed path of law. 18 Chapter Electrostatic Field in Free 1 Space also valid for any charge distribution (which, as we know, can be represented as a system of equivalent point charges, and for which the resultant field can be obtained as a vector sum of individual fields due to point charges). The line integral of E between two points in an electrostatic depend on the path of integration. To prove this, let us refer to Fig. the circulation of E <£ E C • dl electrostatic field vector between two points is the same for any path of path independence line integral of E • dl E -/ J AnB the is same does not and write • dl + E f dl • vBnA 0-76) dl, A to B along the path containing the point so that, from Eq. (1.75), the integral from m integration. dl= f E J AmB = f JA AmBnA -L 1.20 C as along a contour Jc Figure 1.20 For the proof that the line integral of an field as the integral along the path with the point n, for the MmB E w II E f • dl, (1.77) 7 AnB means that, for an adopted reference point, the any point in the field is a uniquely determined quantity, having the same value for any path of moving the test charge. Finally, let us see what happens with the potential at an arbitrary point P in the or along any other path. This also electric potential at field after a new reference point, 11', is pTZ plZ' V= E / new adopted. The • dl = is rlZ' E / • dl Jp Jp potential + E / dl, (1.78) Jiz ' ' v V and thus the change change new in in potential amounts AV = potential due to a to V-V = (1.79) reference point very important to note that It is in the field; if the the same constant AV does not depend on the position of the point P is changed, the potential by an additive constant. reference point value, i.e., at all points changes by Problems 1.29 and 1.30; Conceptual Questions (on Companion Website): and 1.8; MATLAB Exercises (on Companion Website). : 1 .7 1.7 ELECTRIC POTENTIAL DUE TO GIVEN CHARGE DISTRIBUTIONS Let us determine the electric scalar potential due to a point charge general, it is customary to adopt the reference point, when 1Z, at infinity As we in free space. In whenever possi- an example, this choice is impossible, or at least inadequate, for infinite charge distributions, such as an infinite straight line charge. From Fig. 1.21 and Eqs. (1.74) and (1.24), the potential at a distance R from the charge Q, and with respect to the reference point at ble, namely, the charge distribution is finite. shall see in infinity, is a point charge due to v= E JP poo rOO plZ electric potential • dl = / Jx=R Edx= I JR Q r 47r£ 0 * z dr V= (1.80) 4n£()R . Section R Q E P 1 Electric Potential .7 x as charge potential function (of (finite) Shown same in free due to a point space. obtained for a reference point at an arbitrary distance from the charge differs from this result by an additive constant, determined by Eq. vector R) 19 Distributions Figure 1.21 Evaluation of the electric potential The due to Given Charge E is (1.79). in Fig. 1.22 is tangential at potential, V= a sketch of electric-field lines (in general, lines to which all points) const, at all and equipotential surfaces (surfaces having the Of points) around the point charge in Fig. 1.21. “beams” starting at the charge and equipotentials are spherical surfaces (graphed as circles - equipotential course, based lines - on Eqs. in the figure) (1.24) and (1.80), field lines are radial centered at the charge, respectively. Field lines are perpen- and this conclusion holds always, for an arbitrary Namely, since a movement of a probe charge Q p (in Fig. 1.18) over an equipotential surface results in no work by electric forces (no change in V), E does not have a component along that movement, meaning that it is perpendicdicular to equipotential surfaces, electrostatic field. ular to the surface. The density the equipotential surface) it is of field lines (the number of lines per unit area of representative of the magnitude of the field vector, decreases as 1/R 2 (the area of equipotential surfaces increases as i.e., R 2 ) away from the charge in Fig. 1.22. Starting with Eq. (1.80) and applying the superposition principle, the expression for the resultant electric potential, similar to the field expression in Eq. (1.25), obtained for the system of is N point charges, which reads N (1.81) as well as for the three characteristic continuous charge distributions, corresponding to Eqs. (1.37)-(1.39) for the field vector, v= J_ 47T£() f Jv (1 R .82) electric potential volume charge V - const Figure 1.22 Field and equipotential lines for a point charge in Fig. 1 .21 due to 20 Chapter electric potential 1 due Electrostatic Field in Free Space f Ps 1 1/- to electric potential due v— to line rQ'di 1 (1.84) R h 47re 0 charge (1.83) ’ R 47T £ 0 Is surface charge Obviously, these integrals are substantially simpler than the respective field inte- and the same holds true for the resulting solutions for the potential due to charge distributions on various characteristic geometries, for which we have already grals, evaluated the electric Example 1.14 field vector in Section due to Electric Potential a 1.5. Charged Ring Find the electric scalar potential along the axis of a uniformly charged ring in free space normal to the ring plane. The line charge density of the ring is Q' and its radius is a. Solution (for — oo < From Eq. (1.84) and z < oo) is given by Fig. 1.11, the potential at J_I & 4nE 0 Jc Example 1.15 Compute Potential d/ R = Q' an arbitrary point P on the z-axis due to an Q' a 1 dl= 4ne0 RJc (1.85) 2e 0 s/z 2 Infinite Line + a2 Charge the electric potential at an arbitrary point in space due to an infinite uniform line charge of density Q' Solution in air. The expressions that the reference point is for potential computations given in Eqs. (1 .81 )— ( 1 .84) all evaluation of the potential due to to an infinite charge distribution with respect to the reference point taken at infinity nite). In such cases, we invoke Eq. straight line charge, the field of (1.74) instead is r, we apply Eq. (1.74) - to compute the potential, is given by Eq. (1.57), is this charge, at a point a typical, whose radial distance R Figure 1.23 Evaluation of the due to an charge - cross electric potential infinite line section of the structure; for Example 1.15. and an is infi- infinite and a very important, from the to a convenient integration path shown in Fig. 1.23, which which example. To find the potential due to charge line imply These expressions cannot therefore be used for the infinite charge distributions (because the potential due at infinity. Section 1.8 an arc between points P and consists of 21 Voltage M and a straight line segment between points M and and obtain 1Z, /> . E = dl JP The rH rM 72 V= / E dl+ / • = dl JM JP line integral rrn E between points P and / dx. A=r M is qi rr-jz Edx= zero because E is (1.86) 2 ;t£o* Jr perpendicular to dl along the from the line charge with respect to the reference comes out to be path. Thus, the potential at a distance r point that is a distance rn away from it rR Q! V= Note r V— oo. >• is in space is and (1.79)]. In an infi- given by Eq. (1.87) r) plus an infinite constant, as a consequence of the change of the reference point from Fig. 1.23 to infinity [see Eqs. (1.78) to also “correct” in a sense can be understood as the potential distribution (function of it due charge which implies that Note, furthermore, that this result, so an nite potential regardless of the location of the observation point, that electric potential infinite line that the adoption of the reference point at infinity in this case, oo in Eq. (1.87), would result in rji (1.87) In 2ti eq other words, the potential at all 7Z in points changed by the same constant value after the new reference point (at infinity) is infinite (AC -> oo), masks the correct potential distribution. However, adopted, which, being this result, although “correct,” is useless for the analysis, since the actual function C(r) can- not be extracted from an infinite additive constant, and that is why we say that in potential evaluations due to infinite charge distributions reference point cannot be taken at Problems: 1.31-1.36; MATLAB Exercises (on Companion Website). VOLTAGE 1.8 By infinity. between two points is the potential difference between The voltage between points A and B is denoted by Tab, so definition, the voltage them. The unit is V. we have ^AB = where V\ and Cb are the potentials Va~ Cb at point A (1.88) , same reference point. Combining Eqs. (1.74) and (1.88), we and point definition of voltage (unit: V) B, respectively, with respect to the Vab — / E • dl — • get dl = JA E • dl + rE • dl (1.89) or Eab= Edl. (1 .90) JA Therefore, the voltage between two points in an electrostatic field equals, in turn, the line integral of the electric field intensity vector along any path between these points. Obviously, new Cba = -Cab- adopted for the electric potential in a system, voltTo prove this statement, we recall that the potential at all points in the system changes by the same constant value (AC) after the new reference point is adopted [see Eq. (1.79)]. Since a voltage is the difference of the potential values at the respective points in the system, Eq. (1.88), the new voltage between points A and B is given by If a reference point is ages in the system remain unchanged. ^ab = i.e., it - ^b - (^A + AC) - (CB + AC) = CA - Cb = Cab, equals the old voltage between these points. (1.91) voltage via a line integral of E 1 22 Chapter Electrostatic Field in Free 1 Note Kirchhoff's voltage Space that, in terms of voltages, Eq. (1.75) can be written as E law along which, if (1.92) applied to a closed path in a dc circuit, represents Kirchhoff’s circuital law for voltages. This law closed path in a circuit is v'=°C is tells zero. 11 us that the algebraic sum of voltages around any Kirchhoff’s voltage law for dc circuits, Eq. (1.92), therefore just a special form of Maxwell’s first equation for the electrostatic field, Eq. (1.75). We shall see in a later chapter that with some restrictions and approximations, Kirchhoff’s voltage law in the form in Eq. (1.92) can also be used for the analysis of ac circuits. Problems: 1.37; Conceptual Questions (on Companion Website): 1.9. HISTORICAL ASIDE The Gustav Robert Kirchhoff SI unit for the vol- tage, electric and electromotive force (to be studied in a who is Berlin, Breslau, famous for his invention of the the electricity and voltage made — the first voltmeter. a series of electrochemical cells as a pile of alternating silver and zinc plates separated by cardboard disks soaked in salty water (electrolyte). (Portrait: Edgar Fahs Smith Collection, and Hei- Kirchhoff rec- eived his doctoral degree from the University of Konigsberg under experimental of ments, he used his tongue to sense and measure was chemist, taught at Universities of delberg. electric battery (voltaic pile) in 1800. In experi- Volta’s battery and was named in honor of Alessandro Volta (17451827), an Italian physicist and inventor, a prophysics at the University of Pavia, physicist later chapter), volt (V), fessor German (1824-1887), a potential, in Neumann 1847, (1798- Extending the Ohm (17891854), he formulated, in 1850, the fundamental relations between currents and voltages in an electric circuit with multiple loops, which we call Kirchhoff’s circuital laws for currents and voltages. 1895). work Kirchhoff also of made seminal contributions troscopy and thermal emission. (Portrait: to spec- Edgar Fahs Smith Collection, University of Pennsylvania Libraries) University of Pennsylvania Libraries) 1 .9 DIFFERENTIAL RELATIONSHIP BETWEEN THE FIELD AND POTENTIAL IN ELECTROSTATICS Eq. (1.74) represents an integral relationship between the electric field intensity vecand the potential in electrostatics, which enables us to determine V if we know tor E. In this section, we shall derive these two quantities, and then use 1 Algebraic the voltage sum means is an equivalent, differential, relationship between it for evaluating that the voltages in the sum can be E from V. of arbitrary sign, where the sign with which taken depends on the agreement or disagreement of the orientation (polarity) of the voltage with the orientation of the contour. Section 1.10 A Gradient 23 which the potential is EA Let — ck along the x-axis, to a point B, as shown in Fig. 1.24. The resulting change in the potential amounts to the potential at B (new potential) minus the potential at A (starting potential), that is, Consider a point us move from in an electrostatic field at . that point for an elementary distance dl dE=EB -EA (1.93) . d/ = dx Er 1 1 On the other hand, Eq. (1.90) tells us that the potential difference (voltage) between points A and B is related to the electric field intensity vector as La — Lb = E • dl = E dl cos a = E cos a dx — Ex dr, (1 .94) between E and V in electrostatics. ~~e7~ with Figure 1.24 Derivation of the differential relationship Ex standing for the x-component of E, which equals the projection of E on the E cos a. Combining Eqs. (1.93) and (1.94), we have x-axis, (1.95) 7 -D differential relationship between Similarly, the projections of the vector E on the other two axes of the Cartesian coordinate system are obtained as Ey = M E = 7 dy is „ V 9x dV (1.97) dy an important general means In an electrostatic system, the potential Cartesian x-axis, while it ones because the potential is a V= L(x, y, z). This diffor computing the field E from V2 Since , 1.38; constant in any individual plane normal to the = V0 + V^ + L 2 and d are constants. Find the dE/dy only, according to Exercises (on is varies along that axis as V(x) 1.10 — xH 1-D Problem Example 1.16 Problems: I E in electrostatics. the potential Solution given by three coordinates (multivariable function), ferential relationship E is partial derivatives instead of ordinary all where Eq, Vi, E / 3L E = Ex x + Ey y + Ez z = function of (1.96) ' dz Hence, the complete vector expression for where we use — dE = 0 and dE/d z = 0, (1.98) ^, electric field intensity vector in the system. we are left with an x-component of the vector Eq. (1.97), which comes out to be dE _ Ei 2 E2 x dx d cP (1.99) ' Conceptual Questions (on Companion Website): 1.10; MATLAB Companion Website). GRADIENT The expression function (E). in the parentheses in It is Eq. (1.97) is called the gradient of the scalar sometimes written as grad E, but much more frequently we write E and V / 24 Chapter Space Electrostatic Field in Free 1 it using the so-called del operator or nabla operator, defined as V = del operator So, E from V — 8x a a „ a „ yH J x-l dy „ z. (1 dz . 100 ) we have E = — grad V = — VF, in electrostatics where, in the Cartesian coordinate system (Fig. grad gradient in Cartesian V = VV = dv — A + x dx coordinates (1.101) 1.2), dv — „ y dy + (1.102) The other two best-known and most commonly used coordinate systems are the and the spherical. An arbitrary point (M) in the cylindrical coordinate system is represented as (r, 0, z), where r is the radial distance from the z-axis to the point M, 0 the azimuthal angle measured from the x-axis in the xy-plane, and z the same as in the Cartesian coordinate system, as shown in Fig. 1.25. The ranges cylindrical of the coordinates are 0 < r < 0 oo, < 0 < 2n (or — 7r < 0 < — oo < it), z < oo. (1 .103) r z The coordinate unit vectors, increasing and r, 0, r, 0, and z, respectively. are, by definition, directed in directions of They are all mutually perpendicular (the cylin- z, system belongs to the class of orthogonal coordinate systems), and cylindrical coordinates can be expressed as drical coordinate the vector E in E = Er r + X Figure 1.25 Point M(r. and coordinate in <p, z) unit vectors the cylindrical coordinate -f- £ z z. (1.104) Since 0 is not a length coordinate but an angular one, an incremental distance d corresponding to an elementary increment in the coordinate, d0, equals dl = rdrf) and this exactly is the computing the change in potential dV in Eqs. (1.93)— (1.95) now in the 0 direction. Therefore, the 0-component of the electric field vector at the point in Fig. 1.25 equals E ^ = — dV/dl = — dV/(rdrp) and not just — dF/d0. The other two cylindrical coordinates, r and z are length coordinates, so no modification is needed, Er — — dV /dr and E z — — dV /d z. Consequently, having in mind Eqs. (1.104) and (1.101), the gradient of V — F(r, 0,z) in the cylindrical coordinate system is, in place of Eq. (1.102), given by (the length of an arc of radius r defined by the angle d0), displacement dl system. E<p§ in Fig. 1.24 in M , gradient grad in cylindrical F = VF = dV — „ r + dr coordinates In the spherical l r dv 90 M coordinate system, a point dv 4> T H „ z. (1.105) dz is defined by (r, 0,0), with r being the radial distance from the coordinate origin (O) to M, 9 the zenith angle between the z-axis and the position vector of M, and 0 the same as coordinate system, as illustrated in Fig. 1.26. 0</-<oo, O<0<7r, and the component representation of the vector O<0< 2n, E r, 0, in directions and 4> (1 are .106) reads E = Er i + E0 Q + E^. where in the cylindrical The ranges of spherical coordinates (1.107) are mutually perpendicular coordinate unit vectors (directed of increasing r, 9, and 0, respectively). As depicted in Fig. 1.10, the Section 1.10 25 Gradient corresponding incremental distances dl along these vectors equal to dr, rdd, and rsin0 d0, and hence the following expression for the gradient of V = V(r, 6, <p) in the spherical coordinate system: grad 3F„ — V = VF = important to note that there It is 1 3F- r 36 is — 1 - 0 H r H dr : r sin no equivalent dV — tions So, V all (1.108) gradient in spherical coordinates in the cylindrical or spheri- cal coordinates to the expression for the del operator, V, in because 7 4> 9 30 Eq. (1.100), essentially the unit vectors in Figs. 1.25 and 1.26 except z (of course, their direc- and not magnitudes) depend on the location (on coordinates) of the point M. can be formally treated as a vector, given by Eq. (1.100), only in Cartesian coordinates. The expression in Eq. (1.102) and the corresponding expressions associated VF for any multivariable scalar with other coordinate systems enable us to calculate function F We (not necessarily the electric potential). shall now derive the rela- tionship between the gradient and so-called directional derivative of a scalar field, which will new provide us with a physical interpretation of the gradient operation in general. x Combining Eqs. dF (1.94), (1.95), Ex = E • x and = — VF (1.101), we have dF • x can now = VF-x. (1.109) dr due We Figure 1.26 Point M(r, 6, 0) and coordinate unit vectors in the spherical coordinate system. formally proclaim the Cartesian x-axis to be an arbitrarily positioned linear axis in space, and call it an /-axis, with respect to which (Fig. 1.27) — = VF-1= V | F| cos (1 /3, . 110 ) directional derivative dl where 1 is the associated unit vector (|1| = 1) and f5 the angle between the vector VF and the /-axis. The derivative dF/d / is referred to as the directional derivative of F in the / direction. It represents the rate of change of F in this direction, and equals the projection (component) of the vector VF on (along) the /-axis. Eq. (1.110) is an equivalent mathematical definition of the gradient of a scalar, and, although obtained specifically for the electrostatic potential, it is valid for any scalar field F (for which the necessary derivatives exist). In addition, Eq. (1.110) has a very important physical meaning and practical implications. is We a fixed vector. note that for a given scalar By field steering the /-axis around = 0, VF at a point dF ~dd = |VF| P in Fig. 1.27 by varying the angle /}, we For = ±90°, dF/d/ = 0 [from dF/d / at the location P. however, we reach the maximum obtain different derivatives Eq. (1.110)]; for which becomes F, P, i.e., 08 =0 in dF/d / (cos ft (1 ). = the directional derivative of 1), a scalar field. . 111 ) max VF VF maximum gradient of a scalar field which F space rate of change in F is F (fi = changes most rapidly and the magnitude of the all 0). We conclude that the a vector that provides us with both the direction in change. This property of the gradient operation applications, in physical gradient This means that (1) the magnitude of equals the maximum space rate of change in the function F per unit distance [|VF| = (dF/d/) max ] and (2) points in the direction of the Figure 1.27 Relationship between the gradient and is areas of science and engineering. maximum space rate of extensively used in numerous meaning of the 26 Chapter 1 Electrostatic Field in Free Space Field From from Potential, Charged Ring the electric potential due to a charged ring given by Eq. (1.85), find the electric field intensity vector along the ring axis By symmetry, Solution normal the vector E to plane. its along the z-axis that a combination of Eqs. (1.101), (1.105), and (Fig. 1.11) has a z-component only, so (1.85) gives (1 which, of course, the is same result as in Eq. . 112 ) (1.44). Conversely, according to Eq. (1.74), the expression for the potential due to the ring can be found from the expression for the the point P field, in Eq. (1.44), by integration along the z-axis from — (with a coordinate z) to the reference point at infinity (z oo), as (1.113) and it is a simple matter to verify that the solution of this integral is exactly the potential in Eq. (1.85). Problems: 1.39-1.45; Conceptual Questions (on Companion Website): MATLAB Exercises (on Companion Website). An D AND 2-D ELECTRIC DIPOLES 3 1.11 1.11; is a very important, fundamental electrostatic system consisting of two point charges Q of opposite polarities separated by a distance d. We alterna- electric dipole tively refer to this system as a three-dimensional (3-D) dipole, to distinguish it from line charges, which will be analyzed later in this an analogous 2-D system combining derive the expressions for the electrostatic scalar potential section. So, let us first and vector due to a 3-D dipole at large distances compared to field intensity Introducing a spherical coordinate system whose origin as shown in Fig. 1.28 and using Eq. is d. at the dipole center (1.80) for the electric potential due to a single point charge and superposition, the potential due to the dipole at a point P is (1.114) Since r d, /*2 — r\ ^ dcosO electric dipole potential and /q/q Q d cos 0 4;i£o r2 d d Figure 1.28 Electric dipole. r 2 , p and Eq. (1.114) becomes cos 6 p r ' Ansor 2 An e^r2 (1.115) Section 3-D and 2-D 1 .1 1 Electric Dipoles Figure 1.29 Cross section of a line dipole. (1.116) is the dipole moment = (jp Qd). The unit for p Applying the formula for the gradient the expression for V in Eq. 1 3 1/r , E and • m. in spherical coordinates, Eq. (1.108), to dV * 2 cos 6 e r note that C moment (1.115) yields E = — VE = We is electric dipole E due 4k sqT' dQ + sin 0 0 r (1.117) electric dipole field 1 to an electric dipole vary with distance as 1 /r 1 and respectively. We fundamental for the concept is used in electro- shall see in the next chapter that electric dipoles are discussion of dielectric polarization. In addition, a dipole magnetic interference (EMI) considerations, since the quasistatic (low-frequency) electric field produced by an electrical device can often be approximated by the field due to a single electric dipole, given in Eq. (1.117). The combination of two equal parallel infinite line charges Q' of opposite polarities separated by a distance d constitutes a line or two-dimensional (2-D) electric dipole. Let us find the potential due to this system. Fig. 1.29 shows the cross section of a line dipole with the following per-unitlength dipole moment: = P The system cylindrical coordinate (1.118) <2 d. adopted such that the z-axis is is normal to the plane of drawing and coincides with the central line of the dipole. Taking the point O (r = 0) for the reference point for potential and applying Eq. E= where /fti = d/2 and charges 1 and In 2k so ^2 Q! . In h ’ —= *R2 , In , r\ = r 2;r£o n+d cos 0 O' n 2k Sq <<C 1, Q'd. , In / 1 + dcosd) - (1 . 120 ) the final expression for the electric potential due to a out to be P' cos 0 (1 27r£or = (1.119) r\ d/2 are the distances of the reference point from large distances from the dipole (r^>d), v^ with p' in's 27T£q ri we have At 2k sq + x) ^ x for x line dipole turns -Q! rnx i 2, respectively. V Since ln(l Q! (1.87), . 121 ) line dipole potential 27 28 Chapter Space Electrostatic Field in Free 1 Note that for quasistatic electric field evaluations, transmission lines (telecom- munication power lines, lines, computer buses, can sometimes be approx- etc.) imately modeled by a single line electric dipole, with the potential computed by Eq. (1.121). Such evaluations are important between transmission Problems 1.46-1.50; ' : lines, as well as MATLAB Exercises (on in studying undesired couplings EMI associated field levels. Companion Website). FORMULATION AND PROOF OF GAUSS' LAW 1.12 Gauss’ law represents one of the fundamental laws of electromagnetism. It is an Coulomb’s law and a direct consequence of the mathe- alternative statement of matical form of the electric field intensity vector due to a point charge. Gauss’ law involves the flux (surface integral) of the vector 12 and can equivalently be formulated E through a closed mathemati- form, which is based on a differential operation called divergence on E. This important equation, in either form, provides an easy means of calculating the electric field due to highly symmetrical charge distributions, including problems with spherical, cylindrical, and planar symmetry, respectively. Gauss’ law states that the outward flux of the electric field intensity vector through any closed surface in free space is equal to the total charge enclosed by that surface, divided by sq. To prove it, let us consider first a single point charge Q in free space and evaluate the flux of vector E through an arbitrarily positioned cal surface, in a differential surface element (differentially small patch) of area dS, dS, as shown This elementary flux in Fig. 1.30. dV E = EdS, where n is is dS whose surface area vector is given by = dSn, (1.122) the unit vector normal to d S directed from the charge outward. Using Eq. (1.24), we have Figure 1.30 Evaluation of the flux of the vector to a point charge through surface element d.V, area vector, dS, is d^/.' E due from the charge outward. - — (1.123) ^r> 7 47r£ 0 R2 a whose oriented = EdScosa = EdS n = dS n standing with (Fig. 1.30). at the charge. The solid angle of the cone dft elementary solid angle [a solid dS on the plane perpendicular for the projection of to R This projection can be considered as the base of a cone with the apex angle is measured in — by definition. is, d (1.124) R2 steradians (sr or srad)], yielding d4> £ = -^- dQ. (1.125) 4jT£() Assume now depicted l2 The (to in Fig. that the charge 1.31(a). flux of a vector function a be discussed in this Q From Eq. section) is is enclosed by an arbitrary closed surface 5, as outward flux of the (1.125), by integration, the through an open or closed surface S is defined as fs a the vector element of the surface perpendicular to accordance to the orientation of the surface. it, dS, where dS and directed in , Section 1 Formulation and Proof of Gauss' Law .12 29 HISTORICAL ASIDE Johann Karl Friedrich Gauss (1777-1855) was a German mathematician and a director of the Gottingen Observatory. Gauss was born in Braunschweig (Brunswick) as a son of a gardener and a servant girl, and he showed his mathematical genius very early. At the age of seven, to the astonishment of his elementary school teacher, he summed the integers from 1 to 100 within seconds by realizing that the sum equaled 50 pairs of numbers from opposite ends of the list, each pair adding up to 101 (1 + 100 = 101, 2 + = 5,050. He attended 99 = 101, .), so 50 x 101 Universities of Braunschweig, Gottingen, and Helmstedt, and received his doctoral degree in . vector E . through the entire surface S (n '!'£ = <£ dvI/ £ JS is 1799. Gauss made numerous seminal contribu- number tions to the greatest mathematicians of <k d£2 all time, others being and Newton developed in 1813 the divergence theorem, providing the mathematical form for the famous law of electricity that relates the flux of the electric field intensity vector through a closed surface to the enclosed net charge and now carries his name - the law which first came out from experiments with charged concentric metallic spheres by Faraday (1791-1867) and was later translated into the Gaussian mathematical form by Maxwell (1831-1879). Upon Weber’s (1804-1891) arrival at Gottingen in 1831, they worked together on a variety of topics in electricity and magnetism (see Weber’s biography in Section 4.8). Archimedes (287 B.C.-212 B.C.) He (1642-1727). directed from the surface outward) = -0— theory, algebra, geometry, and calculus, proved several fundamental theorems in different branches of mathematics, and is considered by many to be one of the three = is (1.126) 4tt£q 47T£ 0 Js where £2 is the full solid angle, which, in turn, can be interpreted as the angular measure for a spherical surface of arbitrary radius r and area So = 4 nr2 resulting in (for spherical surfaces, dS n = dS) , (1.127) As the full solid angle turns out to equal 4;r -i EdS = — £o (sr), the total flux for a point charge full solid angle becomes Q enclosed by S. (1.128) Figure 1.31 Evaluation of the outward flux of E due to a point charge through a closed surface S, the charge being enclosed by S outside S (b). (a) or located 30 Chapter 1 Space Electrostatic Field in Free Q For a point charge fluxes through the outside the surface S, two surface elements shown we elementary which have opposite realize that the in Fig. 1.31(b), orientations, are d vl ( , = £;)through dSj Q ^ and ^£2 ( Q = d'Ff) through £ /\j[£ respectively, so that their contributions to the flux integral cancel each other, the total flux of E through S = Figure 1.32 Arbitrary closed surface containing a volume charge distribution By means E dS = • <j) and is for a point charge outside S. 0, (1.130) of the superposition principle, Eq. (1.128) can then readily be generalized to the case of N point charges enclosed by a (closed) surface S: in free ® E dS = ® space. • Js = <£ El • dS + Js where E <£ E 2 dS + • • • • + Js - ( is (Ei + E2 + (f E n dS = + Eyy) • dS Js —+—+ Js £() the field due to a charge Qi,i = Qn + £0 £() (1.131) ’ 1,2,..., N. With the notation N Qs = J^Qi, (1.132) i=l we have Gauss' law (1.133) There may also be charges located outside the surface S; recall from Eq. (1.130), however, that the contribution to the total flux resulting from such charges is zero. As we know, any (continuous or discontinuous) charge distribution can be represented as a system of equivalent point charges. This means that Eq. (1.133) holds true for any charge distribution (in free space), where Qs = Eq. (1.133) total is charge enclosed by S (arbitrary charge distribution). E-dS = volume charge distributions is the volume charge which Gauss’ law can be written as distribution, Fig. 1.32, in terms of for 34) the mathematical formulation of Gauss’ law. The most general case of continuous charge Gauss' law (1 .1 1 (1.135) £(J with v denoting the volume enclosed by the surface S and p the volume charge density. This particular form of Gauss’ law is usually referred to as Maxwell’s third equation for the electrostatic The field in free space. principle of conservation of energy in the electrostatic field, Eq. (1.75), and Gauss’ law, Eq. (1.135), are the two fundamental equations) describing the electrostatic 13 Gauss’ law Nevertheless, tion in the in later is 13 As shown here and the second equation out of a total of two Maxwell’s equations used we term complete chapters. integral equations (Maxwell’s field in free space. it Maxwell’s third equation because set of four general it is in electrostatics. customarily positioned as the third equa- Maxwell’s equations for the electromagnetic field, as we shall see , . Section in Fig. 1.19, both equations can be derived from Coulomb’s law, i.e., 1 .1 3 Applications of Gauss' Law 31 from the expression for the electrostatic field due to a point charge in free space. Problems'. 1.51 and 1.52; Conceptual Questions (on Companion Website): 1.12-1.16. LAW APPLICATIONS OF GAUSS' 1.13 now Let us discuss procedures for using Gauss’ law to determine the electric field is known. Gauss’ law is always true, and we can any charge distribution and any problem. However, it enables us to analytically solve only for the field due to highly symmetrical charge distributions. To understand this, note that Gauss’ law is mathematically formulated by an integral equation [Eq. (1.133)] in which the unknown quantity to be determined (E) appears inside the integral. We can use the law to obtain a solution, therefore, only in cases in which we are able to bring the field intensity, E, outside the integral sign, and solve for it. These cases involve highly symmetrical charge distributions, for which we are able to choose a closed surface S, known as a Gaussian surface, that satisfies two conditions: (1) E is everywhere either normal or tangential to S and (2) E = const on the portion of S on which E is normal. When E is tangential to the surface, E dS in Eq. (1.133) becomes zero. When E is normal to the surface, E dS becomes E dS, and since we have also that E is constant, it can be brought outside the integral sign in Eq. (1.133). We shall now apply these basic ideas to problems with spherical, cylindrical, and planar symmetry, respectively, in four characteristic intensity apply it if the charge distribution to • • examples. Example of Example 1.18 The class of first Problem with Spherical Symmetry a problems for which Gauss’ law can be used to solve for the field involve charge distributions that depend only on the radial coordinate in the spherical coordinate system. vector Due to spherical symmetry, only the radial present, is and this component these facts, the solution procedure radius a with a uniform distribution both inside is is component of the electric field intensity a function of the radial coordinate only. Based simple to perform. volume charge density p, in and outside the sphere and As an example, and determine free space, on consider a sphere of (a) the field (b) the electric scalar potential at the sphere center. with spherical symmetry - Solution (a) The Figure 1.33 Application of Gauss' law to a problem Example electric field vector at an arbitrary point in space E= is 1 E{r) r (1.136) E in a problem with spherical symmetry where r is vector, as the radial coordinate in the spherical coordinate system and r shown in Fig. 1.33. centered at the origin. dS= The outward • S is is the radial unit a spherical surface of radius r E dS = Er dSi = EdSr dSr dS surface On S, • flux of the vector ®E Js The Gaussian =® JS E(r) r = EdS. (1.137) E through S is hence dS = E{r) (p dS = E{r) S .18. of the form = 1 E{r) 4 nr (0 < r < oo) Js (1.138) Since p = const (uniform charge distribution), the charge enclosed by S is computed, according to Eq. (1.30), as p times the corresponding volume, which is that of either the i 32 Chapter 1 Electrostatic Field in Free entire Space domain < inside S, for r outside this sphere), for r > or the charged sphere of radius a (there a, Qs amounts so a, is no charge to pAnr3 /3 for r p\nc? j'b for r < > a (1.139) a I According to Gauss’ law, 4>£ = Qs/so yielding , E(r) Note that the field for r > a pr/Oso ) pa 3 /(3e0 r2 ) = for r for r < > a (1.140) a ' identical to that of a point charge is p47ra 3 /3 (the total Q= charge of the sphere) placed at the sphere center. This means that the point charge and the charged sphere are equivalent sources with respect to the region outside the sphere. Concept of equivalent sources (b) To is often used in electromagnetics. find the electric potential at the sphere center (with respect to the reference point at infinity 14 ), we first conclude that the potential outside the charged sphere is identical to the potential of the equivalent point charge, because the fields are identical, so that the potential at the sphere surface (r = a) is obtained directly from Eq. (1.80): V(a) Q = (1.141) Aneoa Invoking Eq. (1.90), the potential difference (voltage) between the sphere center and E from the center to the surface, which results in the surface equals the line integral of following for the potential at the center: E(0) = Example of Example 1.19 E(r) dr a + = V(a) 3 Q _ 8nsoa pa 2 (1.142) 2 eq Problem with Cylindrical Symmetry problems we deal with using Gauss’ law involve infinitely long charge depend only on the radial coordinate in the cylindrical coordinate system. As an example, consider an infinitely long charged cylinder of radius a and volume charge The second class of distributions that density r 2 pW = Po -y a (1-143) 1 in free space, where po is a constant and r the radial coordinate cylinder axis), and find the electric field everywhere. Solution Shown of the form given in associated unit vector. Due the distance from the symmetry and Eq. (1.136), r being here the radial cylindrical coordinate and r the The Gaussian surface is a cylinder of radius r and height (length) h, in Fig. 1.34 is a cross section of the cylinder. of the charge distribution in this case, the field is (i.e., is to cylindrical radial with respect to the cylinder axis positioned coaxially with the charged cylinder. Figure 1.34 Application of Gauss' law to a problem with cylindrical symmetry - Example 1 To determine E inside the charge distribution, we apply Gauss’ Gaussian cylinder of radius r < a. law, Eq. (1.135), to a 2 E(r) 2nrh .19. -if —E po £() Jr’={ a* 2jiE d Eh, (1.144) dv Sc p 14 Whenever infinity, [e.g., the reference point for the potential to be determined is not specified, we assume except for infinite charge distributions, where such assumption would give us an Eq. (1.87) for n r ->• oo]. that it is at infinite potential Section where Sc is the lateral surface area of the cylinder, and dv 1 shell of radius r (0 < explained in Section < 1 r 1.4). the is volume of 1 .1 3 Applications of Gauss' Law 33 a thin cylindrical thickness d F, and height h (we adopt as large as possible dv, as r), This volume is computed as that of a thin flat rectangular slab with 1 edges equal to 2 nr (circumference of the cylindrical shell), h and d/ (for the purpose of the volume computation, the shell is flattened into a rectangular slab). The flux of the vector E through the top and bottom surfaces of the Gaussian cylinder is zero since E is tangential to those surfaces (E dS = 0). By integration in Eq. (1.144), E(r) Qr = - ^z , , forr<n. (1.145) 4eo« For a field point outside the charged cylinder ( r Eq. (1.144) becomes a, which > a), the upper limit in the integral in a. (1.146) results in 2 E(r) = for r > 4e0 r Note that this field is identical to the field due to an infinite line charge positioned along the cylinder axis with the same charge per unit length O' as the cylinder [see Eqs. (1.31) and (1.57)]. Example of Example 1.20 a Problem with Planar Symmetry and the following example illustrate how problems involving charge distributions depend on one Cartesian coordinate only - problems with planar symmetry - can be solved using Gauss’ law. Consider first a layer of volume charges in free space with the volume charge density being the following even function of the Cartesian coordinate x\ Finally, this that P(x) where po and a (a > 0) are constants intensity vector everywhere, Solution = i.e., M Po < (1.147) a. and p(x) = 0 for |x| > a. Determine the and outside the charge layer. electric field inside shows the charge distribution. Due to planar symmetry, vector E at an x-component only, and Ex depends on the coordinate x only any plane x = const), so we can write Fig. 1.35 arbitrary point of space has an (it is constant in E = Ex {x) x. In addition, since the charge distribution tor E in the plane defined by x' = x (x > is (1.148) symmetrical with respect to the plane x 0) in Fig. 1.35 is = 0, vec- equal in magnitude and opposite in Figure 1.35 Application of Gauss' law to a problem with planar symmetry - Example 1 .20. E in a problem with planar symmetry : 34 Chapter 1 ' Electrostatic Field in Free Space direction to that in the plane x' (rectangular box) that is = —x. The Gaussian surface — x and x' = —x, field intensity Ex (x) areas So, positioned in the planes x' a right-angled parallelepiped is = cut symmetrically by the plane x 0 and has two of faces, with its respectively. Let us first determine the inside the charge layer (— a < x < a). Noting that E and dS are oriented in the same direction at both faces So, as well as that E and dS are mutually perpendicular on the remaining portions of the Gaussian surface, we have Ex (x) S 0 = 2 — p(x')S0 dx\ f (1.149) v— - £o Jx’=-x dv where dv volume of a the is thin slice of the parallelepiped with thickness dx'. The integration in x' yields E*(x) now Let us If we then nothing start will = ^ ^1 - ~ for < |x| (1.150) a. j, bring the parallelepiped faces So onto the boundaries of the charge layer. moving them toward change infinity, in the application of symmetrically with respect to the plane x Gauss’ law, since there is = 0, no charge beyond the layer boundaries. This reasoning gives us directly the field intensity outside the charge layer: Ex (x) = Ex (a) — - —— =— 2poa — - tor , x > (1.151) a, 3e 0 Ex (x) = Ex (—a) 2po« f for x , < —a. (1.152) 3e 0 Example Planar Symmetry, Antisymmetrical Charge Distribution 1.21 Repeat the previous example but for the following odd function of x as the charge density of the 2n-thick layer: p(x) = p0 ~, \x\ <a\ (1.153) a there is no charge outside the Solution As this is layer. = 0, zero, which implies that the electric field outside the layer is an antisymmetrical charge distribution with respect to the plane x the total charge of the layer is zero; namely, this field due is as if to an equivalent infinite sheet of charge (Fig. 1.15) with zero surface charge density (p s = 0). In other words, if we subdivide the charge layer described by Eq. (1.153) into differentially thin layers of thicknesses dx' and add (integrate) the fields, given by Eq. (1.64), due to all these thin layers (observed outside and negatives exactly cancel out, and the result place of Eqs. (1.151) and (1.152) we thus have p(x') dx' £ =0 (total field for Ex (x) = Positioning the Gaussian rectangular surface in Fig. 1 0 .35 all (|x| of them), the positives —a x < > > or x a) is zero. In (1.154) a). such that one of its parallel faces with area So is in the region x' < —a (on the left of the charge layer) and the other is in the plane x' — x where — a < x < a (inside the layer), Gauss’ law gives [instead of Eqs. (1.149) and (1.150)] £*(x)So = !/' Po e 0 J x '=-a since the flux of E through volume integration — So dx' (*) 2 = Je~a a the face outside the charge layer effectively starts at x' — (* is - fl2 0 155 ) - ) zero, from Eq. (1.154), Problems 1.53-1.64; Conceptual Questions (on Companion Website): 1.18; MATLAB Exercises (on and the -a, where the charge distribution begins. Companion Website). 1.17 and Section 1.14 1 .14 Differential Form 35 Law FORM OF GAUSS' LAW DIFFERENTIAL r an integral equation in the spatial domain (the integrations involved are carried out with respect to spatial coordinates), and, in the general case, it repGauss’ law of Gauss' is resents an integral relationship between the electric field intensity vector, E, E*(x) _____ Ex (x + dx) and the volume charge density, p [Eq. (1.135)]. In this section, we shall derive an equivalent, differential, relationship between E and p, that is, the differential form of x + dx x x dx Gauss’ law. Assume first that p D a function of the Cartesian coordinate x only, p charge distribution), so that the only present component of E is x = is E — p(x) (1Ex (x). Let S us apply Gauss’ law, Eq. (1.135), to a rectangular closed surface S, the dimension of dx and the sides normal to that direction are So in area, field being constant on both surfaces So, no integration is needed on the left-hand side of Eq. (1.135). In addition, since dx is differentially small, we can take p(x) as constant in the volume enclosed by S, so that no integration is needed on the right-hand side of Eq. (1.135) either. Noting that on the left-hand side of S, E and dS are directed in opposite directions, we have which in the x direction as indicated in Fig. 1.36. is The —Ex (x) So + Ex (x + dx) So = —1 p(x) So dx. (1 .1 Figure 1.36 For derivation of the one-dimensional Gauss' law in differential form. 56) £o The differential of Ex corresponding to the displacement dEx = Ex (x+ divided by dx, Eq. (1.156), is, by dx) - Ex (x), definition, the derivative of this derivative comes out dx, Ex (1.157) with respect to x. From to be d Ex _ dx p_ (1.158) 7 -D differential Gauss' law £q is a differential equation in the spatial domain, and it represents the onedimensional Gauss’ law in differential form. It states that, in cases when the charge distribution changes with a single linear spatial coordinate in free space, the rate of This change of the electric field intensity with that coordinate equals the local density of volume charge, divided by £oThe generalization of Eq. (1.158) to the three-dimensional case is straightforward. The charge density is now a function of all three coordinates, p = p(x, y, z), and the Gaussian surface S has to be differentially small in all three dimensions, as shown in Fig. 1.37. We break the flux integral over S in Eq. (1.135) up into three pairs of integrals, each pair being carried out over two parallel sides of S. All sides being differentially small, the flux over each of them can be approximated by taking a constant value of the field component normal to the side and multiplying it by the side area. This gives us the result for each individual pair of integrals that has same form as in the one-dimensional case above [Eqs. (1.156)— (1.158)]. For the first pair of integrals, we have that the outward flux of E through sides normal to the x direction (front side and back side in Fig. 1.37) is exactly the Figure 1.37 For derivation of Gauss' law in differential form / J front E dS + / E • dS = (change of Ex across dx) x ( dy d z) 7 back BEx dx (dy dz), dx for an arbitrary charge distribution. (1.159) - 36 Chapter 1 Electrostatic Field in Free , Space and adding the results for other two pairs of dE — E dS = dE v - cLt:(dydz) 3* 5 we integrals, obtain the total flux dE dy (dxdz) H dz (dvdy). H (1.160) dz 3y — dx dy dz, with which the above equation becomes dE y dE z / 3 Ex Av (Av — 0). = - + ^ + (1.161) \dx dy dz The volume enclosed by S is A v _ E dS • ( 5 Av Since is — — — very small, the total charge in it can be found with no integration in Eq. (1.135), simply as Qs — p Av = p (dvdy dz). Finally, interconnecting the flux Gauss' law in differential form (1.162) and charge by means of Eq. dEx dEy dx dy 3 Ez (1.135), we have p (1.163) ' This dz Gauss’ law in differential form, is for the electrostatic field in free space. i.e., It is £q Maxwell’s third differential equation a partial differential equation or PDE (partial derivatives with respect to individual coordinates enter into the equation) in unknowns three (three components of the vector E). This equation provides us with way E varies in space. It relates the rate of change valuable information about the components with of field spatial coordinates to the local charge density. We see components along the direction of that component only and not along y and z, etc.) contribute in this relationship. that changes of individual (change of Ex Problems 1.65 : along and x, 1.66; Conceptual Questions (on Companion Website): 1.19; MATLAB Exercises (on Companion Website). DIVERGENCE 1.15 The expression on the left-hand side of Eq. (1.163) vector function (E), and is is called the divergence of a written as div E. Applying formally the formula for the dot product of two vectors in the Cartesian (rectangular) coordinate system, a • b = (a x x + ay y + a z z) • (b x x to the del operator, Eq. (1.100), + by y + b z z) = ax bx + ay by + a z b z , (1 .1 64) and vector E, we get V E= • (1.165) and this is exactly div E, in Eq. (1.163). Note that the divergence is an operation that performed on a vector, but the result is a scalar. The differential Gauss’ law now can be written in a short form is divE Gauss' law using divergence = or V E= • where, in coordinates Cartesian in the — (1.166) eo £o notation divergence — Cartesian coordinate system, div E= V E= dEx dEy dx dy • (1.167) ) Section 1 .15 Divergence In nonrectangular coordinate systems, the corresponding formulas are more complex, due to curvature (or nonrectangularity) of differential volume cells in place of the one in Fig. 1.37. For instance, if, in analogy to the situation in Fig. 1.36, we consider a one-dimensional spherical volume charge distribution with p — p{r), r being the radial spherical coordinate in Fig. 1.26, the only component of E is as in Eq. (1.136), and applying Gauss’ law, Eq. (1.135), to the inner Er = Er (r), and outer (radius (radius r r + dr) surfaces of a thin spherical shell in Fig. 1.9, Eq. (1.156) becomes — Er (r ) A nr2 + Er (r + dr) An(r + dr) — p(r) Anr = 2 2 dr, (1 .1 68) £0 where the use is made of the expression for dv in Eq. (1.33). After eliminating An on both sides of the equation, the expression on the left-hand side is the differential Er corresponding to the coordinate increment dr, which divided by dr represents the derivative of r 2 E r with respect to r, and hence [see Eq. (1.157)] of the function r 2 d(r is Er _ ) p_ (1 ’ r2 analogously to Eq. (1.158). This 2 dr . 169 ) £q a one-dimensional differential Gauss’ law in the 3-D case of p = p(r,6,(p) mimics the derivation in Eqs. (1.159)— (1.163) accommodated to the elemental spherical cuboid in Fig. 1.10, and similarly for p — p(r, 0, z) (Fig. 1.25). The resulting and spherical coordinate system, its generalization to the formula for the cylindrical coordinate system divE =V E= • 1 - —9 ( is rEr ) r or and that 1 dE, r o<p - </> + dEz (1 dz . 170 ) divergence in cylindrical coordinates for spherical coordinates divE As + =V E= • \ rz (^Er) or \ + (sin rsin# / 6Ee + ) ot) the vector expression for the operator V Cartesian coordinate system, the dot product in dE, -4 <\> rsu sin0 in Eq. (1.100) V E • actually is 90 (1 . 171 ) divergence in spherical coordinates valid only in the makes sense only in we shall still use extensively the notation V E in cylindrical and spherical coordinate systems as well - to simply mean the divergence rectangular coordinates. However, • of E (div E) and a symbolic representation of the respective formulas in Eqs. (1.170) and (1.171). In general, we shall often draw conclusions about gradient, divergence, and other operations involving vector field quantities by treating V as a vector, Eq. (1.100), and performing formally vector operations with it, like in Eq. (1.165). Most importantly, these conclusions, properties, and relations, although derived in Cartesian coordinates, hold true (as identities) in all possible (cylindrical, spherical, and other) coordinate systems, because the properties of a physical quantity and relations between two or more quantities are the same regardless of the choice of a coordinate system. Combining Eqs. (1.161) and div (1.167), we E=V E= • obtain the equation lim Av—>0 which tells us that the divergence of E at a LE — dS (1 Av given point is . 172 ) physical meaning of the divergence the net outflow of the flux of E per unit volume as the volume shrinks about the point. Eq. (1.172) is an equivalent 37 : 38 Chapter 1 Electrostatic Field in Free Space mathematical definition of the divergence of a vector. From it, we may regard div E measure of how much the field diverges from the point. In general, a positive divergence of any vector field indicates a local source of the field at that point producing radial field components with respect to the point - the outward flow of flux is positive, which means that the vector diverges (spreads out) at the point. Possible field components that are tangential to S do not contribute to the flux, and physically as a are not related to the divergence. Similarly, a negative divergence indicates a sink at the point (local negative source) - the outward flux flow is negative, and the vector converges at the point. Finally, the divergence is zero where there is neither source nor sink of locally radial field components. Gauss’ law simply tells us that field these positive and negative sources in the case of the electric field, E, are positive and negative charges, p Av. Quantitatively, the divergence represents the volume density of sources, and in our case this density is p. Let us now replace p in the integral form of Gauss’ law, Eq. (1.135), by its equal, £qV E - from the differential form of the law, Eq. (1.166): • E iS = divergence theorem V Edv. - fs (1.173) • l is called the divergence theorem (or Gauss-Ostrogradsky theorem). Although we have obtained it specifically for the electrostatic field in free space, the theorem is true for any vector field (for which the appropriate partial derivatives exist) and is one of the basic theorems of vector calculus. It applies to an arbitrary closed surface S and, in words, states that the net outward flux of a vector field through 5 equals the volume integral of the divergence of that field throughout the volume v bounded by S. This equation Problem with Spherical Symmetry Using Example 1.22 Redo Example 1.18, part (a), but now employing Gauss’ law Differential Gauss' in differential Law form. Solution As we already know, from the spherical symmetry of the problem, that the electric vector both inside the charged sphere and outside it has the form in Eq. (1.136), the formula for the divergence in spherical coordinates, given by Eq. (1.171), retains only the first term. For a point inside the charge distribution in Fig. 1.33, Gauss' law in differential field form, Eq. (1.166), thus reduces to the following differential equation in a single coordinate, r V E= • — f^Efr)] = — rL dr L J which can be directly solved by integration. Since p p-E(f) = — e J[ 2 dr r ~+ + C\ = — < r const, — C\ 3e 0 0 (0 < a) (1.1 74) , £o E(r) we have = + 3£ 0 (1.175) "Tm and the integration constant, Ci, is found from the “initial” condition E(0) = 0 at the center = 0). Namely, given that there is not a point charge ( Qo ) at the point O in Fig. 1.33, the field at this point is zero, and hence C\ = 0 [otherwise, the constant would amount to C\ = 0o/(4tt£o), from Eq. (1.24)], which yields the result for E in Eq. (1.140). Outside the charge distribution in Fig. 1.33, p = 0, which substituted in Eq. (1.174) of the sphere (for r results in E(r)j = 0 = C2 2 r E(r) — » E(r) = (a < r < oo). (1.176) integration, C 2 is determined by means of the “boundary” condition boundary of the charged sphere, r = a; the field intensity on the inner side of the The new constant of at the — , Conductors Section 1.16 in boundary, computed from Eq. (1.175), must be the same (since there exist no surface charges on the boundary) as that on the outer side of it, from Eq. (1.176), = E(a-) E(a + ) = 3 — E{a) ~. C2 = (1.177) 3eo With this, the result for (1.176) agrees with the solution in Eq. (1.140). Problem with Planar Symmetry by the Example 1.23 Redo Example E in Eq. Differential Gauss' Law 1.20 but with the use of the differential Gauss’ law. Solution As this is a problem with planar symmetry, the electric field intensity vector everywhere is of the form in Eq.(1.148). Hence, Gauss’ law given by the differential equation in Eq. (1.158) applies, which we solve by integration with respect to x (also see the previous example). For the observation point inside the charge layer {—a <x<a) in Fig. 1.35, Ex (x) = p(x)dx + Ci = — eq Jf Because of symmetry — e0 [l - + Q, -^r-J 3a 1 for |x| < (1.178) f \ (Fig. 1.35), Ex (a) = -Ex (-a) — — _ + Cl = 3eo 3eo Cl — Q= (1.179) 0, which, substituted in Eq. (1.178), gives the same result as in Eq. (1.150). For the point outside the layer (|x| > a), p = 0, so that Eq. (1.158) yields Matching this field dition at the layer Ex {x) = C2 . value to the one from Eqs. (1.178) and (1.179) in the “boundary” con- boundary defined by x = a, Ex (a + = Ex (a ~ = Ex (a), we ) ) obtain C2 = Ex {x) = 2po«/(3eo) f° r x > a, the same as in Eq. (1.151). On the other side of the layer, the vector E has this same magnitude but opposite direction, resulting in 2poa/(3Eo), and thus the field expression in Eq. (1.152). Note that the application of Gauss’ law in differential form to a problem with planar symmetry and an antisymmetrical charge distribution is presented in Example 2.6. Problems'. 1.67-1.74; 1.16 So CONDUCTORS IN THE ELECTROSTATIC FIELD we have considered electrostatic fields in free space (a vacuum or air). We now extend our theory to electrostatic fields in the presence of materials. As far, shall we MATLAB Exercises (on Companion Website). shall see, most of the formulas derived and solution techniques developed and used for the electrostatic field in free space are directly applicable to the analysis of electrostatic fields in material space, although some require modification. Materials can broadly be classified in terms of their electrical properties as conductors (which conduct electric current) and dielectrics (insulators). In the rest of this chapter, we shall study the interaction of the electrostatic field with conductors, in case essentially no theoretical modification whereas the behavior of is needed to the which electrostatic equations, dielectrics in the electrostatic field will be discussed in the next chapter. Conductors have a large proportion of freely movable electrons and ions) that make electric charges (free the electric conductivity (ability to conduct electric current) of the material. Best conductors (with highest conductivity) are metals (such as silver, copper, gold, aluminum, etc.). In many applications, we consider the Electrostatic Field 39 p 40 Chapter 1 Electrostatic Field in Free Space metallic conductors as perfect electric conductors (with infinite conductivity). There many are also other, less conductive conductors such as water, earth (ground), so-called semiconductors (glass, and germanium), (e.g., silicon some paper, rubber, etc.) have etc. Practically all insulators (usually extremely low) conductivity, and thus theoretically are (very poor) conductors, although almost always (in electrostatics) may be considered they as perfect dielectrics (with zero conductivity), conductors. In our studies of electrostatic i.e., nonby conductor we normally mean a fields, metallic conductor. Consider an isolated conductor, shown in uncharged or electrically neutral (its net charge Eext Eext trostatic field E exl electric force [see + - E=0 + + “Ps zero). Assume When that it is an external elec- applied, the free charges in the conductor are influenced by the Eq. (1.23)] = GE ext Fe (1.180) . Ps + — is 1.38(a). Fig. is -f + This force pushes the positive charges along the direction of to right, while the negative charges move E ex t, that is from left opposite direction. Consequently, in the becomes progressively more positive, and the more negative. In fact, for a metallic conductor, what actually happens is movement of free electrons (negative charges) toward the left side leaving an equal number of positive charges, namely, a deficiency of electrons, on the other side of the conductor. Accumulated free charges form two layers of surface charge, of densities — s (ps > 0) and ps on the conductor sides. Creation the right-hand side of the conductor left-hand side progressively (b) Figure 1.38 (a) A conductor in an external electrostatic field, (b) After a transitional process, there is no , of surplus charges in the body caused by an external electrostatic field electrostatic field inside the electrostatic induction. conductor. tric field in The induced charges, the conductor, Ej nt which , is in turn, set is called the up an internal induced elec- directed from the positive to the negative oppositely to E ex t- As p s increases, Ej n becomes progressively stronger, opposes the migration of charges from left to right. In the equilibrium, Ej nt completely cancels out E ex in the conductor, so that the total field E in the conductor is zero, and the motion of charges stops, as illustrated in Fig. 1.38(b). Note that the conductor remains uncharged as a whole. The entire transitional process is extremely fast, and the electrostatic steady state is established practically instantaneously. In fact, based on the length of the time needed for this process of movement layer, and i.e., t it t of charges to the surface of a material body, that the total electric field inside the material is i.e., way we determine whether a their redistribution in such a body becomes zero, we shall see in a later chapter, commonly most used metallic conductor - a conductor or dielectric. For example, as the time to reach the equilibrium for the copper - is as brief as ~ 10~ 19 s, whereas it takes as long as ~ 50 days for the charge rearrangement across a piece of fused quartz (very good insulator). In the case of a conductor that had been charged (with a positive or negative excess charge) prior to being situated in the external field, a similar process takes place. All free charges (for a metallic conductor, free electrons of the conductor, which abundantly exist in the material also when it is electrically neutral as a whole, 15 plus excess charge ) are exposed to the force Fe and produce the internal field that cancels out the externally applied field 15 Nole that excess charge on a metallic ative excess charge) or by taking number some the electrostatic equilibrium. body may be produced by bringing electrons of of these extra or missing electrons of the body. in its is free electrons always much away to the (positive excess charge), body (negwhere the smaller than the total count of free electrons Section We .16 Conductors conclude that under electrostatic conditions, there cannot be electric in 41 the Electrostatic Field field conductor, in a E= This it, 1 the is we first derive all (1.181) 0. fundamental property of conductors in electrostatics. Starting from other fundamental conclusions about the behavior of conductors in the electrostatic no electrostatic a conductor field inside field. According to Eqs. and (1.181), (1.90), (1.88), the voltage points in the conductor, including points on conductor is an equipotential body, conductor and on its surface, i.e., its surface, the potential V= is is the between any two means that a same everywhere in the zero. This const. (1.182) From Eq. (1.181), V E = 0 in a conductor, implying that [Eq. (1.166)] there cannot be surplus volume charges inside it, (1.183) interior and surface of a conductor are equipotential no volume charge inside a conductor So, any locally surplus charge of a conductor (whether it neutral as a whole or is not) must be located at the surface of the conductor. isfy in a Let us now derive so-called boundary conditions that the electric field must sata conductor surface. The electric field intensity vector E near the conductor on vacuum can be decomposed into the respect to the boundary surface, as En respectively, where a is shown = £cosa the angle that normal and tangential components with The two components are in Fig. 1.39(a). = and E makes Esina, (1.184) with the normal to the surface. We apply Eq. (1.75) to the narrow rectangular elementary contour C in Fig. 1.39(a). The field is zero along the lower side of C (E = 0 in conductors), and we let the contour side Ah shrink to zero pressing the sides A/ tightly onto the boundary surface, so is E AI along the upper that the only contribution to the line integral in Eq. (1.75) C side of (no integration E Hence, there is is • needed, because dl =E AI A / is • small). = EAl sin a = £ t A/ = (1.185) 0. no tangential component of E over the surface of a conducting body in electrostatics, zero tangential electric on a conductor surface (a) (b) Figure 1.39 Deriving boundary conditions for the electrostatic field (E) near a conductor surface: (a) narrow rectangular elementary contour (used for the boundary condition for the tangential component of E) and (b) pillbox elementary closed surface (for the boundary condition for the normal component of E). field 42 Chapter 1 Electrostatic Field in Free Space In other words, the electric field intensity vector on the surface of a conductor is always normal to the surface, E = £ n n, (1.187) where n is the normal unit vector on the surface, directed from the surface outward. To obtain the boundary condition for the normal (the only existing) component of E, we apply Eq. AS (1.133) to the pillbox Gaussian surface, with bases and height Ah (shrinking to zero), shown For similar reasons as in obtaining Eq. (1.185), the flux in Eq. (1.133) reduces to E AS over the upper side of S. Because the charge enclosed by S is p s AS, in Fig. 1.39(b). • (t E dS = E AS = (E n n) • • • = En AS = (ASn) —p JS s AS, (1 .1 88) £0 providing the relationship between the normal component of the electric field and the surface charge density on the intensity vector near a conductor surface surface: normal electric field (1.189) component on a conductor surface The lines of the electric field are normal to the surface of a conductor. should always remember that the normal component En in Eq. (1.189) is We defined n. When p s > 0, the field lines start from the whereas they end on it (£„ < 0) when ps < 0. In analyzing complex conducting structures, we usually do not know in advance with respect to the outward normal conductor (E n > 0), the orientation of the electric field intensity vector at specific portions of conducting surfaces. In such cases, the following expression for obtained noting that En = n • E from Eq. (1.187), ps Example 1.24 An = £0 is ps in terms of the field vector, useful: n E. (1.190) • Metallic Sphere in a Uniform Electrostatic Field uncharged metallic sphere is brought into a uniform electrostatic field, around the sphere after electrostatic equilibrium is reached. in air. Sketch the field lines Figure 1.40 Uncharged Solution The field lines in the new electrostatic state are sketched in Fig. 1.40. Because the due to induced charges on the sphere surface (this field exists both inside and outside the sphere) is superimposed to the external field, the field inside the sphere becomes zero, and that outside it is not uniform any more. Negative induced charges are sinks of the field lines on the left-hand side of the sphere, whereas the positive induced charges are sources of the field lines on the right-hand side. The field lines on both sides are normal to the sphere surface, and they therefore bend near the sphere. At points in air close to the left- and metallic sphere in a uniform right-hand side of the sphere, the electric field external electrostatic the remaining space. This field for Example 1 .24. field; sphere, in tive air, is is stronger (the field lines are denser) than in obvious as well from noting that near the left-hand side of the the field due to negative induced charges dominates over the field due to posi- charges on the opposite side of the sphere, adds to the external field intensity. The field it is due directed toward the negative charges, and to positive induced charges the right-hand side of the sphere, which results in the at these points in air. is The field at dominates near same strengthening of the external field distances from the sphere a few times the sphere diameter practically equal to the external field (the field due to induced charges is negligible). . Evaluation of the Electric Field and Potential due to Charged Conductors Section 1.17 43 EVALUATION OF THE ELECTRIC FIELD AND POTENTIAL DUE TO CHARGED CONDUCTORS 1.17 Assume that we know the charge distribution ps over the surface of a conductor situated in free space. The electric field intensity at points close to the conductor surface can be evaluated from Eq. (1.189). How do we obtain the electric field and potential at an arbitrary point in space? The answer is straightforward. Because E = 0 inside the conductor, nothing will change, as far as the field outside the conductor cerned, if we remove the conductor and fill the space previously occupied by is con- it with a vacuum, keeping the charge distribution ps on the surface unchanged. With this useful equivalence, we are left with the problem of evaluating the field and potential due known surface charge distribution to a (1.83), (1.101), (1.133), Example 1.25 A and (1.165) in free space, and we can use Eqs. (1.38), to solve the problem. Charged Metallic Sphere metallic sphere of radius a is situated in air and charged with a charge Q. Find (a) the charge distribution of the sphere, (b) the electric field intensity vector in air, and (c) the potential of the sphere. Solution Due (a) to symmetry, the charge distribution over the sphere surface is uniform, and hence the associated surface charge density turns out to be /9s _Q So _ Q2 (1.191) ' Ana where So stands for the surface area of the sphere. (b) The around the sphere is radial, and has the form given by Eq. (1.136). Eq. (1.133), to a spherical surface of radius r (a < r < oo), positioned concentrically with the metallic sphere [see Eqs. (1.137) and (1.138)], we obtain electric field Applying Gauss’ law, E(r) Note Q = which We that is in agreement with Eq. due to a point charge sphere, the same Example 1.26 (1.189). at Q r (1.192) oo). = a is Q Ps Ansoa 2 (1.193) £o’ 16 thus given by Eq. (1.141). This any point of its interior and surface Charged Cylindrical + 0) [E(a + 8), 8 is is is identical to the potential. The the potential of the [see Eq. (1.182)]. Conductor infinitely -+ 0] designates the electric field due metallic sphere the charge per unit length of the conductor £(a + ) or E(a = a. < placed at the sphere center, and so Repeat the previous example but for an 16 r realize that the field outside the charged metallic sphere, Eq. (1.192), potential at the surface r air, if < that E{a + ) (c) (a AnsQr2 is long cylindrical conductor of radius a in Q field in air very close to the conductor surface at to a charged 44 Chapter 1 Space Electrostatic Field in Free Solution (a) Using Eq. (1.31), the charge per length h of the conductor Qh = is Q'h, (1 . (1 . 194 ) so the surface charge density amounts to Q' = Qh = Qh ^ 2 nah So where So (b) The is 2na' the surface area of that part of the conductor. electric field is radial (with respect to the conductor axis). [Eq. (1.133)] applied to the cylindrical surface of radius r(a h, coaxial with the electric field cylindrical due to 195 ) < r < From Gauss’ law oo) and height (length) conductor [see the left-hand side of Eq. (1.144)], a charged E(r) 1 Qh Q' 2nrh £q 2tre 0 r = (a conductor < r < oo). (1 . 196 ) is an infinite charge distribution, and the reference point for the potential cannot be adopted at infinity. The field in Eq. (1.196) being identical to the field due to an infinite line charge of density Q\ Eq. (1.57), the potential due to the charged cylindrical conductor is given by the expression in Eq. (1.87). In particular, this expression for r = a This (c) represents the potential of the conductor: cylindrical Q' V= . T/ potential due to a charged In 2tt £q conductor where r-n ( rji > m — , (r ^ > , a) ^conductor — r a) is In—— ~ 2ne o A metallic sphere, of radius a and charge Q (Q Solution is air. As Find the potential > 0), is enclosed by an uncharged concentric at the As we know, and outer surface of the Qh and shell, respectively. Qc tive The medium is induced surface charge on the denote the charge on the inner surface of the total induced charges on the inner On the surface of the sphere, Qa = line of the electric field originating at a positive surfaces c). there cannot be volume charges in the metal under electro- Eq. (1.183). Let static conditions, < b < center of the sphere. a result of the electrostatic induction, there surfaces of the shell. (1.197) a), Shell metallic spherical shell, of inner radius b and outer radius c (a everywhere — the distance of the reference point from the conductor axis. Charged Sphere Enclosed by an Uncharged Example 1.27 {r a Q. Since every charge on the sphere terminates at a nega- shell, the relationship between total charges on two is Qb = -Qa- (1-198) This can be obtained also from Gauss’ law, Eq. (1.133), applied to a closed surface that is = 0 [because of Eq. (1.181)], which implies entirely inside the metal of the shell, so that that Qs = 0, i.e., Q a + Qb — 0- O n the other hand, since the shell Qh + <2c = 0, is uncharged, (1.199) Q c = —Qh = Qa Because of symmetry, charges on individual surfaces are disis in the form given by Eq. (1.136). Distributions of the charge and field in the system are sketched in Fig. 1.41. By means of Gauss’ law, the electric field for a < r < b and c < r < oo is and hence tributed uniformly, and the electric field everywhere in air E(r) = (in air), - (1.200) 47T£o'' whereas for 0 < r < a and b < r < c, E(r) = 0 (in metal). (1.201) ) Section Figure 1.41 Charge and 1 .1 7 Evaluation of the Electric Field and Potential due to Charged Conductors field distributions in the system of Example The .27. 1 potential at the point V= O in Fig. E(r) d r 1.41 is thus [Eq. (1.74)] -O-fS-l + l) = \a 47reo Q(bc — ac b c + ab) (1.202) 4neoabc Example 1.28 Consider Five Parallel Large Flat Electrodes five parallel large metallic electrodes situated in air, as shown in Fig. 1.42(a). The thickness of each electrode, as well as the distance between each two adjacent electrodes, The surface area of sides of electrodes facing each other is S = 1 m 2 The first, and fifth electrodes are grounded, the potential of the second electrode with respect to the ground is V = 2 kV, and the charge of the third electrode is Q = 2 iiC. Find the electric field intensity between the electrodes. is d = 2 cm. . fourth, Solution Since the dimensions of the electrodes are much larger than the separation between them, we can neglect the fringing effects, i.e., we can ignore the nonuniformity of the electric field near the electrode edges, as well as the existence (“leakage”) of the outside the spaces between electrodes. in spaces field in field lines We assume, thus, that the electric field is localized only between the electrode sides S, that vector E is normal to those sides, and that the is uniform. Charge distributions over electrode sides are also uniform. In each space ddddddddd So Figure 1.42 Electrostatic analysis of a system of five large geometry and (b) charges on electrodes and fields between them; for Example 1 .28. electrodes in air: (a) of the system (b) 45 46 Chapter 1 Electrostatic Field in Free Space each space between electrodes, neighboring sides of electrodes must be charged with charges of equal amounts, but of opposite polarities, as indicated in Fig. 1.42(b). The potential can be expressed electrode, which is V in is (which we know) of the second electrode with respect to the ground terms of the line integral of grounded potential (its E from The zero). is that electrode, to the field is to the left, first uniform, and the line integral simply first electrode v= Edl = -Eid. (1.203) ./second electrode Hence, the field — 100 kV/m. On E to the right, to between the intensity the other side, the same and second electrodes first the fourth electrode, which a similar token, the field same — -V/d = Ei,d. between the fourth and (1 fifth electrodes is the associated charges on the sides of electrodes facing each other electrodes are at the E\ also grounded, yielding is V = E2d + By is potential can be expressed as the line integral of zero, (Q 4 £4 — 0), = 0, .204) and so are because these (zero) potential. £ 3 we need one more equation with them as unknowns. Applying Gauss’ law, Eq. (1.133), to the surface So that encloses the third electrode (the charge of which is known), we obtain To solve for field intensities £2 and , -£2 + £3 = —z- (1.205) e0 S The solution of the system of equations -62.94 kV/m and £3 = 162.94 composed from Eqs. Problems'. 1.75-1.82; Conceptual Questions (on MATLAB 1.18 Exercises (on it, Because there bounded by shell cavity is is no field field Faraday cage. in Companion Website): £2 = 1.20-1.22; in the external electrostatic field throughout the sphere outside the sphere. We a metallic shell (Fig. 1.43). This perfectly protected (isolated) thickness of the shell can be arbitrary, and Figure 1.43 Metallic shell an electrostatic field - is ELECTROSTATIC SHIELDING without affecting the field, and (1.205) Companion Website). Let us consider again the metallic sphere Fig. 1.40. (1.204) kV/m. interior, thus obtain a means in domain with no that the space inside the from the external its shown we can remove electrostatic field. The shape does not need to be spherical. Hence, an arbitrary closed conducting shell represents a perfect electrostatic shield or screen for its interior domain. We call such a shield a Faraday cage. If the field outside the cage is changed, the charge on the cage walls will redistribute itself so that the field inside will remain zero. We see that a Faraday cage provides an absolute protection to its interior from an external electrostatic field. Let us now reverse the problem. Can an electrostatic field be encapsulated by a metallic shell so that the domain outside the shell is protected from the sources inside it? The answer to this important question is twofold. Let us get to it by analyzing two simple examples. Consider first a single positive point charge Q positioned arbitrarily inside an uncharged spherical conducting shell. The distribution of induced charges on the shell surfaces and the field lines are sketched in Fig. 1.44(a). The total induced charges on the inner and outer shell surfaces are —Q and Q, respectively (see Section 1.18 Electrostatic Shielding Figure 1.44 Single point charge (a) and two point charges amounting to a zero total (b) (a) Example 1.27). The concentration of induced negative charges of the inner surface closer to the point charge. Because there on the outer wall, the positive charge shell surface is is is cage. higher on the side no field in the shell distributed irrespective of the means uniformly in this case smooth and symmetrical). Let us now add another point charge in the cavity, and let it be exactly — Q as shown in Fig. 1.44(b). The total negative and total positive induced charges on the inner surface of the shell are both less than Q in magnitude, because some field lines originating at the positive point charge end at the negative one inside the cavity, but mutually they are equal in magnitude and opposite in polarity. The net induced charge on the inner surface is therefore zero, implying that there is no charges whatsoever on the outer shell surface. This means, in turn, that there is no field outside position of the point charge inside the cavity, which (the surface is , the shell, which is also in agreement with Gauss’ law, applied to a spherical surface enclosing the shell. We conclude that a Faraday cage can completely encapsulate an interior elecif the total charge inside the cage is trostatic field, with a zero field outside, only zero. This is true for any interior charge distribution, provided that the object or the system of objects (devices) inside the cage as a whole. In cases when is electrically neutral the total interior charge is (uncharged) not zero, the exterior domain induced charges on the outer surface of independent of the distribution of interior sources. Its relative distribution in space depends only on the shape of the outer cage surface, whereas its absolute values at individual points in space are also (and neighboring objects) the cage. The is in the field of exterior field, however, is totally proportional to the total amount of interior charge. It is interesting to note that trostatic shields, either in the one illustrated in Fig. 1.44(b). even very thin metallic mode shells represent ideal elec- of operation illustrated in Fig. 1.43 or in the However, as we shall see in a later chapter, this is not where the effectiveness of a shield of a given thickness depends on the metal conductivity and on the rate with which the necessarily the case with time-varying fields, field varies in time (that is, frequency in the case of time-harmonic Conceptual Questions (on Companion Website): 1.23 and 1.24. fields). charge (b) in a Faraday 7 48 Chapter 1 Space Electrostatic Field in Free CHARGE DISTRIBUTION ON METALLIC BODIES OF 1.19 ARBITRARY SHAPES body of an arbitrary shape, the charge body surface is not uniform. The determination of this distribution for a given body with nonsymmetrical and/or nonsmooth surface is a rather complex problem. In this section, we shall get some qualitative insight about how In the general case of a charged metallic distribution over the the charge is distributed over the surface of an arbitrarily shaped isolated conduct- we shall present a general numerical method for determining approximately the charge density function over conducting objects. ing body, and in the next section Consider a system composed of two charged metallic spheres of different radii, whose centers are a distance d apart, in free space. Let the spheres be connected by a very thin conductor, as shown in Fig. 1.45. Assume, for simplicity, that d a,b, so that the electric potential of each sphere can be evaluated as if the other one were not present. In addition, we assume that the charge along the connecting a and b, » conductor is zero, because the conductor is very thin, and ignore its 1.25 and Eq. (1.141)] Qa y‘ = and Anega respectively, on Example influence the field between the spheres. Therefore, the potentials of spheres are [see Qa where and Qb (1.206) 4 The are the associated total charges of spheres. spheres being galvanically connected together, and thus representing a single conducting body, which must be equipotential, Eq. (1.182), these potentials are the same, Va = Vb So, by equating the expressions in Eq. (1.206), Qa _ a ~ b Qb Using Eq. (1.191), the sphere charges 0 20 . - we obtain (1.208) ’ can be expressed ing surface charge densities, with which Eq. (1.208) charge density oc surface Psa b Psb a ) in terms of the correspond- becomes (1.209) curvature and, from Eq. (1.193), the corresponding relationship between the electric field intensities near the surfaces of spheres turns out to be local field intensity a Ea surface Eb curvature We = b (1 see from Eqs. (1.209) and (1.210) that the charge two spheres in Fig. Figure 1.45 Two 1 .45 in same potential. 210 ) is distributed between the such a way that the surface charge density on and electric metallic spheres of different . a' radii at the 2 < , Jfhb Qb , , Section field intensity sphere radius. 1 .20 Method of Moments for Numerical Analysis of Charged Metallic Bodies 49 near the surface of individual spheres is inversely proportional to the The surface charge is denser and the field stronger on the smaller sphere. The importance of Eqs. (1.209) and (1.210) is much beyond the particular system in Fig. 1.45. They imply a general conclusion that the surface charge density and the nearby field intensity at different parts of the surface of an arbitrarily shaped conducting body are approximately proportional to the local curvature of the sur17 This means, generally, that the largest concentration of face, as long as it is convex. charge and the strongest electric Note we that this phenomenon is field are shall see in the next chapter. Problems'. 1.83; Conceptual Questions (on 1 around sharp parts of conducting bodies. essential for the operation of lightning arresters, as .20 Companion Website): 1.25-1.27. METHOD OF MOMENTS FOR NUMERICAL ANALYSIS OF CHARGED METALLIC BODIES Consider a charged metallic body of an arbitrary shape situated in free space. Let the electric potential of the body with respect to the reference point at infinity be Vo- Our goal is to determine the charge distribution of the body. The potential at an arbitrary point on the body surface, S, can be expressed in terms of the surface charge density, ps over the entire S [Eq. (1.83)]. On the other hand, this potential equals Vo (the body is equipotential), and hence , Ps 4nso This is L dS = R Vq (at an arbitrary point on S ). surface integral equation for (1 .21 1 ) charge distribution an integral equation with the function p s over S as unknown quantity, to be determined. Eq. (1.211) cannot be solved analytically - in a closed form, but only numerwith the aid of a computer. The method of moments (MoM) is a common numerical technique used in solving integral equations such as Eq. (1.211) in electromagnetics and in other disciplines of science and engineering. can be implemented in numerous ways, but the simplest solution in this case implies subdivision of the surface S into small patches A 5), i = 1, 2, N, with a constant approximation of the unknown function ps on each patch. That is, we assume that each patch is uniformly charged, ically, MoM MoM . Ps ~ psi (on A Si), t = . . l,2, (1 . 212 ) piece-wise constant approximation for charge With this approximation, we reduce Eq. N E= (1.211) to dS = its approximate form: Vo, f in which the unknown quantities are N charge-distribution coefficients, p i PsN- 17 If Shown in Fig. 1.46 is we JA A Si 4ne 0 R an example of the application of the surface of a conducting body incurvature, 1 is this concave (curved inward), the effect method is s i, pS 2 , • • • to a metallic deep and a decrease of the just opposite; for a actually have a partial effect of a Faraday cage (cavity), Fig.1.43, local field intensity. (1.213) 8 50 Chapter 1 ] Space Electrostatic Field in Free yyyy Psl Ps2 AS, as2 • • • Figure 1.46 Method of moments (MoM) for analysis charged metallic bodies: of approximation of the surface charge distribution on a cube AS,- by means of N small square patches with constant charge densities. cube, where square patches are used [for the particular subdivision figure, N=6 By x (5 x 5) = 150, which is shown in the a rather coarse model]. stipulating that Eq. (1.213) be satisfied at centers of every small patch, individually, we obtain 18 Anpsi +-4i2Ps2 H A 21 Psi + ^22Ps2 H + A iN psN = h MnPsN = Vo (at the center of AS,), Vo (at the center of AS 2 ), (1 +A N2 p 2 H AmPst This is \-A NN psN S a system of N = V0 (at the center of linear algebraic equations in N . 214 ) A S N ). unknowns. Psl 5 Ps2> • • • > PsN- In matrix form. MoM matrix equation [A][ps ] where [p s ] a is = [B], Elements of matrix r A id = JASi patches . 215 ) column matrix whose elements are the unknown charge-distribution elements of the column matrix B are known and all equal Vo- coefficients, while point-matching at centers of (1 [ [A], which is a square matrix, are given by ds 4n£ 0 R (at the center of AS*), k,i = 1,2, (1 . 216 ) and they can be computed irrespective of the particular charge distribution. Physically, Aid is the potential at the center of patch AS* due to patch AS, that is uni- C/m 2 ) surface charge density. In the case of commonly used square or triangular flat patches, this potential can be evaluated analytically (exactly), while it is evaluated numerically (approximately) if some other surface elements are used (for example, curvilinear quadrilateral or triangular patches). Once matrix [A] is filled, i.e., all its elements computed, we can use matrix inversion, formly charged with unit (1 [Ps] = [A]- 1 [B], (1 . 217 ) method for solving systems of linear algebraic equations Gaussian elimination method), to obtain the numerical results for the chargedistribution coefficients, which constitute an approximate surface charge distribution of the body, and a numerical solution to the integral equation, Eq. (1.211). or any other standard (e.g., method of moments in which the left-hand side and the right-hand side of an integral “matched" to be equal at specific points of the definition domain of the equation (surface S in our case) is called the point-matching method. The idea of point-matching is similar to the concept of taking moments in mechanics, and hence the generic name method of moments. 1 The variant of the equation [in our case, Eq. (1.21 l)j are l Section 1.21 The number of subdivisions, N, the more accurate (but more computademanding in terms of computer resources) solution. The crudest approximation in computing the elements of matrix [A] is given by larger the tionally _ ASi/(4ne 0 R ki ~ ( ^ASi/a^eo) \ ) Aki for for k^i k= U^ ' i all nondiagonal elements of [A] ( k ^ i ) are evaluated by approximating the 2 charged patch AS, by an equivalent point charge, A <2, = 1 (C/m ) x AS/, placed at the patch center, and using the expression for the electric potential due to a point charge in free space, Eq. (1.80), with Rki being the distance between centers of patches A S^ and AS/ (Fig. 1.46). In filling diagonal elements (self terms) of [/l] (k = i), the potential due to a (square or triangular) patch AS/ at the center of that same patch is evaluated by approximating the patch by the equivalent circular patch of the same surface area and radius y/ A Sj/jr, and employing the potential expression given in Problem 1.34 with z = 0 and ps = Here, 1 C/m 2 . from the result for [ps ], we can now obtain any other quantity of interand field at any point in space, etc.). For instance, the total charge of the body can be found using the approximate version of the surface integral Starting est (potential expression in Eqs. (1.29): N Q= J^Psi AS/. (1.219) ;= Problems 1.84-1.86; : 1.21 MATLAB Exercises (on Companion Website). IMAGE THEORY Often, electrostatic systems include charge configurations in the presence of grounded conducting planes. Examples are charged conductors near grounded metallic plates or large flat bodies, transmission lines in which one of the conductors is a ground plane (such as microstrip transmission lines), various charged objects (power lines, charged clouds, charged airplanes, lightning rods, etc.) above the earth’s surface, and so on. There is a very useful theory (theorem) by means of which we can remove the conducting plane from the system, and replace it by the equivalent charge distribution in free space. In this section, this theorem, and apply it we shall derive to problems which otherwise could not be analytically solved. Consider two point charges of equal magnitudes and opposite polarities, Q and in free space. By symmetry, the total electric field intensity vector due to the charges is normal to the plane of symmetry of the charges (the tangential components due to individual charges are of equal magnitudes and opposite directions, so they cancel each other out), as shown in Fig. 1.47(a). The plane of symmetry is —Q, and at the potential V — 0 same magnitudes and opposite due to individual charges and they cancel out). Hence, nothing will change in the entire space if we insert an infinite grounded metallic foil (conducting plane) in the plane of symmetry, as is done in Fig. 1.47(b), because the boundary condition in Eq. (1.186) is automatically satisfied and V = 0 over the plane of symmetry (the foil is grounded). In the new system, surface charges will be induced on both sides of the foil, according to Eq. (1.190). We note that the foil actually separates the entire space onto two completely independent half-spaces, i.e., it equipotential, (the potentials are of the signs, Image Theory 51 r 52 Chapter 1 Electrostatic Field in Free , Space Figure 1.47 Deriving image theory: systems (a), (b), and (c) are equivalent with respect to the electric field in the upper half-space. between the two half-spaces. change in the upper half-space if we, furthermore, remove the point charge — Q and the induced charge on the lower side of the foil from the system. (Note that we can put whatever we want below the foil, and the field above it will remain the same, in the electrostatic state.) We are thus left with the point charge Q above the foil and the induced charge on its upper side (and nothing in the lower half-space), as depicted in Fig. 1.47(c). We conclude that, as far as the electrostatic field in the upper half-space is concerned, systems in Figs. 1.47(a) and (c) are equivalent. This is so-called image theory, which, generalized to more than one point charge, i.e., to a (discrete or continuous) charge distribution, states that an arbitrary charge configuration above an infinite grounded conducting plane can be replaced by a new charge configuration in free space consisting of the original charge configuration itself and its negative image in the (former) conducting plane. The equivalence is with respect to the electric field above the conducting plane, whose component due to the induced charge on the acts as a perfect electrostatic screen (see Section 1.18) Because of (a) plane is that, equal to the Example 1.29 A nothing point charge Q will field of the image. Induced Charge Distribution on is a Conducting Plane placed in air at a height h above a grounded conducting plane, (a) Determine the density of induced surface charges the total induced charge on the plane. at an arbitrary point on the plane, (b) Find Solution (a) (b) Figure 1.48 Computation of the induced surface charge density, pS( on a conducting plane underneath a point charge Q : (a) original The surface charge density ps at a point M on the conducting plane, in Fig. 1.48(a). is given by Eq. (1.190), with the unit vector n being vertical and directed from the plane upward, and E representing the electric field intensity vector in air, at a point that is M from its upper side. This field is produced by the point charge Q and the induced surface charges on the conducting plane. According to the image theory, however, the field due to the induced charges equals that due to the negative image of Q in free space, as shown in Fig. 1.48(b). Let the position of the point be defined by a radial distance r from the projection of Q on the plane (point O). Vector E is given by “glued” to the point M system and (b) equivalent system using image theory; for Example 1 .29. E— Eq = Eoriginai Q 4jr eoR 2 + Ej ma ge R= — 2Eq cosa yj 2 + h2 , ( n) cos a = — A (1 . 220 ) 53 Problems and the charge density comes out to be ps (r) = e 0 n-E = 2tt (r2 (b) The induced charge on the conducting plane total 2ind = J ps (r)2nrdr = as Qh rdr f ~&~~ Qh Lo (1.221) 2 + h2 ) is f 00 dR _ ~ Qk 1=0 00 1 R = -G, (1.222) r=0 where dS is the surface area of an elemental ring of width dr and radius r (0 < r < oc) around the point O (see Fig. 1.14), and the use is made of Eq. (1.62) to change variables in integration. The result in Eq. (1.222) is expected, because all field lines terminating on the image, — Q, in the equivalent system [(Fig. 1.48(b)] terminate on the surface charges of the conducting plane in the original system [Fig. 1.48(a)]. Example 1.30 An Infinite Line Charge above charge of uniform density Q' infinite line conducting plane at a distance h from unit of its it. is a Conducting Plane situated in air Compute and is parallel to a grounded the electric force on the line charge per length. Solution Under the influence of the electric field of the line charge, surface charges are induced on the conducting plane. The electric force on each meter of the line charge is therefore [Eq. (1.68)] K = Q'E where E2 (1.223) 2, represents the electric field vector at points along the line charge due to the induced charges on the plane. By image theory, this field is equal to the field due to a line charge in free space obtained as a negative image of Q', Fig. 1.49. and hence the following for and the per-unit-length [see Eq. (1.57)] E2 = Q 2nso(2h) its intensity force: 2 = Q ’ K = Q’E 2 (1 .224) 4jT£()h Figure 1.49 Force on a charge above (the distance between the original and the image is 2 h). The force is plane; for attractive. a line conducting Example 1 .30. Problems: 1.87-1.89; Conceptual Questions (on Companion Website): 1.28-1.30; MATLAB Exercises (on Companion Website). Problems 1.1. Three unequal charges in a triangle. Repeat Example 1.1 but assuming that one of the three charges in Fig. 1.3(a) amounts to (a) 3 Q and (b) —3 Q, 1.2. respectively. Three charges in equilibrium. The distance between point charges Q\ = 36 pC and Qi = 9 pC system are in the electrostatic equilibrium, that the resultant Coulomb force on each charge is zero. this i.e., is placed D = 3 cm. at the line If O 3 is Q 2 at a the third charge, connecting Q\ and Qi D , , shown in Fig. 1.50, find 03 and d which ensure that all the charges in distance d from Qi, as Q3 Qi Figure 1.50 Three point charges along a line; for Problem 1 .2. / 54 1.3. Chapter Electrostatic Field in Free 1 Space Four charges at rectangle vertices. Four small charged bodies of equal charges Q = — 1 nC are placed at four vertices of a rectangle with sides <7 = 4 cm and b = 2 cm. Determine the direction and magnitude of the electric force on 1.11. face charge Five charges in equilibrium. Four small charged balls of equal charges Q\ — 5 pC are positioned at four vertices of a square, whereas the fifth ball of unknown charge Qi is at the square center. Find Q2 such that all the balls 1.6. sum of the three forces, all 1.12. Q —Q Compute 1.13. Field 1.14. Point circle. A fifth charge the electric force is E z (z), —00 < q and x is the length coor- at the axis of a ring, (a) Q > 0, For and electric field intensity along maximum, z < 00 (b) Plot the function . charge equivalent to a charged semi- Show that far away along the z-axis, the semicircular line charge in Fig. 1.12(a) is equivalent to a point charge with the same amount of charge located on the top at the coordinate origin. charged bodies of equal charges Q exist at the vertices of a cube with sides of length a, in free space. Find the magnitude and direction of the Charged contour of complex shape. Fig. 1.51 shows a contour consisting of two semicircular parts, of radii a and b (a < b ), and two linear parts, each of length b situated in air on one of the charges. due = a constant which the the z-axis 1.15. electric force is charge along a rod. A rod of charged with a line charge of Q' [ 1 — sin(7rx//)] (0 < x < /), maximum find z for Eight charges at cube vertices. Eight small 1.8. Electric field line in air is the charged ring in Fig. 1.11, assume i charge. 1.7. The charge density dinate along the rod. Calculate the total charge mid. All sides of the pyramid have the same length, a. / density Q'(x) positioned at the top vertex of the pyra- is Nonuniform where Q'0 are positioned in air at the corners of the square base of a pyramid. sur- — ps {r) length Fe + F e 2 + Fe 3 ? Five charges at pyramid vertices. Four point charges A of the rod. Three point charges in space, (a) For the three charges from Example 1.3, find the resultant electric force on the charge Q2 (F e 2 ). (b) Determine the force F e 3 (on Q3). (c) What is the disk. distributed in free space over a 2 Pso^/a (0 < r < a), where r is the radial distance from the disk center, and p s 0 is a constant. Obtain the total charge of the disk. is are in the electrostatic equilibrium. 1.5. is circular disk of radius a. each of the bodies. 1.4. Nonuniform surface charge on a and distributed along tric field intensity to three point charges in — a. The carries a charge its length. contour is Q uniformly Compute the elec- vector at the contour center (point O). from Example 1.3, determine the magnitude and direction of the space. For the three charges electric field intensity vector at (a) the coor- dinate origin and (b) the point at the z-axis defined by z 1.9. = 100 m. Nonuniform volume charge infinitely is in a cylinder. An long cylinder of radius a in free space charged with a volume charge density p{r) — po(a and r r) a (0 < the radial < — where po is a constant distance from the cylinder axis. r a), Figure 1.51 Uniformly charged contour with two semicircular and two linear parts; for Problem 1.15. Find the charge per unit length of the cylinder. 1.10. Nonuniform volume charge in a cube. of edge space length a in free is A cube charged volume with a charge density p(x) = sin(7TJc/a), 0 < x < a, where po is a constant Po and x is the normal distance from one of over its the cube sides. the cube. Compute the total charge of 1.16. Nonuniform line charge along a semicircle. Consider the geometry in Fig. 1.12(a), and assume that the charge along the semicircle is nonuniform, given by Q\4>) = Q' s\n4> < tt/2), where Q' is a constant, (— n/2 < () <f) (a) 0 Find the total charge of the semicircle. 55 Problems (b) Prove that the charged over electric field intensity vector along the z-axis equals E= 2 —Q'qCi y/[8eo(z 2 + a 2 ) 3 / 2 ]. 1.17. is 3/4 of a circle of radius a The total charge of the arc is situated in total charge of the sphere field intensity charged sheet with a circular hole. An charge with a constant density p s has a hole of radius a in it. The sheet is in the xy-plane of the Cartesian coordinate system and the center of the hole is at the coordinate origin. The ambient medium is air. Under these circumstances, determine the electric field intensity vector at an arbitrary point along the z-axis - in the following A of a circle of radius Determine the a. elec- vector at an arbitrary point along the axis that contains the arc center and is normal to the arc plane. 1.19. Semi-infinite A charge. line uniform charge density Q' line is two ways, respectively: (a) integrating the fields due to elementary rings as in Fig. 1.14 and (b) combining the results of Examples 1.11 (infinite sheet of charge, with no hole) and 1.10 charge of distributed in space along the negative part of the free x-axis in 0). system coordinate Cartesian the (— oo < x < (charged disk). Find the expression for the electric field intensity vector at vector at the sphere center. infinite sheet of Line charge along a quarter of a circle. charge of density Q' in free space is distributed uniformly along an arc representing a quarter tric field intensity (a) the (b) the electric 1.25. Infinite electric field intensity vector at the arc center. 1.18. Compute and air. Q. Calculate the is surface such that the charge is defined as in Fig. 1.10 or 1.16. A Line charge along three-quarters of a circle. uniform line charge in the form of an arc that is its given by p s (0) = ps o sin 20, where p s o a constant and the angle 0 (0 < 0 < n) is density an arbitrary 1.26. point in the xy-plane. Force on a charged semicylinder due to a line charge. For the structure composed from a line charge and a charged semicylinder shown in 1.20. Half-positive, half-negative infinite line charge. A line charge in free space is Fig. 1.17(a) distributed along and described in Example 1.13, find the force per unit length on the semicylinder. the x-axis in the Cartesian coordinate system. The line — oo < charge density > Q' (Q' is x < 0 and —Q' for 0 < x < oo. 0) for 1.27. Derive the expression for the electric field intensity vector at a point where d > 1.21. M with coordinates problem as in Fig. 4.11 in 0. Find the electric is Consider an infinitely long unistrip of width a and surface charge density p s in air. Using the geometristrip. cal representation of the cross section of the (0, d, 0), field intensity at a distance a given by Eqs. (4.43) and (4.44), obtain the expression for the E field at an arbitrary point in space vector at a point from each of the square Chapter 4 (also see and change of integration variables Fig. 1.13) Charged square contour. A line charge of uniform charge density Q' is distributed along a square contour a on a side. The medium is air. that Charged formly charged 1.28. Two due to this charge. parallel oppositely charged strips. parallel, vertices. with charge densities ps and 1.22. Point charge equivalent to a charged disk. Consider the charged disk in show is that for |z| a, 2 ps 7ta placed 1.14, and Eq. (1.63) of a point charge Q = the equivalent to the field E Fig. field in (ps 0). shown the same The —ps , respectively cross section of the structure in Fig. 1.52. The width of the as the distance and the medium is air. between them ( a Find the is strips is — d), electric field at the disk center. a due to a nonuniformly charged disk. Consider the disk with a nonuniform charge 1.23. Field distribution > Two very long strips are uniformly charged from Problem 1.11, and find the Ps electric field intensity vector along the disk axis normal to its • plane. A d Figure 1.52 Cross section two parallel, very long of 1.24. Nonuniformly charged spherical surface. A sphere of radius a in free space is nonuniformly _ f I charged strips; for Problem 1.28. a 56 ^ Chapter Space Electrostatic Field in Free 1 intensity vector at the center of the cross sec- 1.29. in an electrostatic field. What is the work done by electric forces in moving a charge Q = 1 nC from the coordinate origin to the point (1 m, 1 m, 1 m) in the electrostatic field given m) = 2 — in the (x x the straight line joining the 1.30. V/m z) surface center of the nonuniformly charged spherical 1.37. M marked 2 in 1.38. Sketch field potential M, V from in a in Fig. 1.54. The potential. region point electrostatic a function of a sin- is gle rectangular coordinate x, P~ Two Q\—l /rC and Q 2 = —3 q,C, are located at the two nonadjacent vertices of a square contour a = 15 cm on a side. Find the voltage between any of the remaining two vertices of the square and the square center. Find the work done by charge Qi = — 1 nC the figure. Voltage due to two point charges. charges, electric forces in carrying a from the point Mi to the point due to a nonuniform spherical surface surface from Problem 1.24. A in Fig. 1.53. 0). charge. Determine the electric potential at the two points? in the field of a point charge. point charge Q\ = 10 nC is positioned at the center of a square contour a = 10 cm on a side, shown = 1.36. Potential (x, y, z in Work as (z +y y Cartesian coordinate system along y, z) hemispherical a to Consider the hemispherical surface charge from Example 1.12, and find the electric scalar potential at the hemisphere center charge. Work by E(x, due 1.35. Potential tion (point A). V (x) and Sketch the electric is shown field intensity 1 Ex {x) & i Figure 1.53 \ a Q2 of a charge q field of a in this region. Movement in the charge Q\ positioned at the center of a square contour; for M2 a 1.31. Electric potential due Problem 1.30. to three point charges in space. For the three charges from Example 1.3, calculate the electric potential at points defined by (0,0,2 m) (a) and (b) (1 m, 1 m, 1 m), respectively. Figure 1.54 1-D potential distribution; for 1.32. Point charge and an arbitrary reference point. Derive the expression for the potential at a distance r from a point charge Q in free space with respect to the reference point which is an arbitrary (finite) distance rji away from the Problem 1.39. Field charge. 1.33. 1.34. Q' / (4eoV z 2 + a2 1.40. Field following expression for potential along the e-axis + z2 - |z|)/(2e 0 )- the electric (—00 < z scalar < 00 ): V = from V in potential, charged semicircle. For (a) obtain the expression for Ez Example in 1.7, Eq. (1.50) from the expression for V given in Problem 1.33 and (b) explain why it is impossible to obtain the expression for Ex in Eq. (1.48) from this same expression for V. ). due to a charged disk. For the charged disk from Example 1.10, derive the 2 obtain the expression for the semicircular line charge from Potential Ps(%/« in free space, Eq. (1.24) from the expression for Eq. (1.80). Potential due to a semicircular line charge. Prove that the electric scalar potential at an arbitrary point along the e-axis in the field of the semicircular line charge shown in Fig. 1.12(a) and described in Example 1.7 is V= in .38. from potential, point charge. For a point charge E 1 1.41. from potential, charged charged disk from Example Field the expression for expression for E V given disk. For the 1.10, obtain in Eq. (1.63) from the in Problem 1.34. 57 Problems from potential, charged hemisphere. For the hemispherical surface charge from Example 1.12, explain why we cannot obtain the expression for E at the hemisphere center (z = 0), given in Eq. (1.67), from the expres- 1.42. Field sion for 1.43. V computed in Problem 1.49. in 1.50. 1.35. due to a line the expression for V Eq. (1.121). Near and far potential Two dipole. Q and field due to a infinite line charges, line with densi- 100 pC/m and Q'2 = —100 pC/m, are positioned along lines defined by (1 m, 0, z) and are perpendicular to equipotential surfaces (as coordinate in Fig. 1.22). Calculate ties h(x,y ) lOOxlny [m] (x, y in km), where x and y are coordinates in the horizontal plane and 1 km < x, y < 10 km. (a) What is the direction km, of the steepest ascent at (3 steep, in degrees, Maximum The 3 km)? (b) (—1 m, a maximum Q\ 1 1.52. 0), respectively. Flux for a different placement of the point charge. If the point charge ous problem is Q from the previ- placed at the center of a side of the electric field vector due to the charge m, 1 electric dipole. Two nC and Q 2 — — 1 nC, at points through the surface composed of the remaining point are five sides of the cube. sit- along the z-axis of a point charge from Gauss’ law. Using Gauss’ law, derive the expression for the elec- 1.53. Field tric field intensity vector at the point defined by Cartesian (0, 0, 0), (100 m, 100 m, 100 m), and by (b) The cube edges are a outward flux of the electric field intensity vector due to this charge through each of the cube sides. m 1.47. Potential and long. Find the defined by z = 1 and z = — 1 m, respectively. Compute the electric potential and field inten- coordinates (a) 0, 0) of the cube, determine the total outward flux uated in free space sity at the point defined m, —1 m). Large and small = E increase in the electric scalar potential at a point (1 charges, and , A by E(x, y, z) = (4 x — z 2 y + 2yzz)V/m (x,y,z in m). Find the direction of the V < 00 in the Cartesian The medium is air. of a cube in free space. How intensity vector in z system. (100 m, 100 m, given is —00 < 1.51. Flux of the electric field vector through a cube side. point charge Q is located at the center increase in electrostatic potential. electrostatic field region 0, z), Cartesian coordinates (a) (2 m, the ascent in (a)? is = x given by a function is = field (b) (0, 1 m, 0), and (c) vector of a point charge in free space [Eq. (1.24)]. 1.54. respectively. Uniformly charged thin spherical shell. An infinitely thin spherical shell of radius a in free due to a small space electric An electric dipole with a moment = 1 pCm z is located at the origin of a spheri- uniformly charged over is its surface with dipole. a total charge Q. Determine: (a) the electric p field cal is coordinate system. The length of the dipole d = 1 cm. Find V and E m, n/2, n/2), (c) (1 m, (d) m, jt/ (e) (1 (10 m, zr/4, 0), 4,0), (f) (100 m, jr/4, 0). n, 0), nonuniform line (1 m, 0, 0), (b) (1 Dipole equivalent to a 1.55. far and charge density given by Eq. (1.32). an infinite line charge from Gauss’ law. Using Gauss’ law, derive the expression for the electric field intensity vector of an infinite line 1.56. Field of from Problem away along the z- (|z|»a) this charge distribution can be replaced by an equivalent electric dipole moment, at the coordinate origin. p, of the equivalent dipole. and outside the sphere with the volume inside charge in free space [Eq. axis located Sphere with a nonuniform volume charge. Find the distribution of the electric scalar potential distribution along the semicircle and show that and outside the the potential of the shell, and (c) the potential at the shell center. charge. Consider the nonuniform line charge 1.16, intensity vector inside shell, (b) at the following points defined by spherical coordinates: (a) 1.48. field Angle between field lines and equipotential surfaces. Using the concept of gradient, prove that in an arbitrary electrostatic field, field lines elevation in a region 1.46. E from the expression for 1.44. Direction of the steepest ascent. The terrain 1.45. Expression for the electric dipole. For the line dipole in Fig. 1.29, obtain Find the 1.57. (1.57)]. Uniformly charged thin cylindrical shell. An infinitely long and infinitely thin cylindrical shell of radius a shell is is situated in free space. charged over its The surface with a uniform 58 Chapter Electrostatic Field in Free 1 Space charge density ps Find the electric . sity 1.58. field inten- 1.66. vector inside and outside the shell. with Cylinder Compute volume uniform charge. the voltage between the surface and is the axis of a uniformly charged infinite cylinder of radius a in free space, density in the cylinder is if the volume charge 1.67. of an infinite sheet of charge from Gauss’ Using Gauss’ law, derive the expression for the electric field intensity vector of an infinite the 1.68. Charge from field, cylindrical symmetry. From component Two only given by Eqs. (1.145) and (1.146), obtain the corresponding charge distribution in free charged sheets. parallel oppositely —p s Two space [Eq. (1.143)]. are situated in free space, (a) Find between the sheets, the voltage between the sheets? side the space (b) What 1.69. Charge from field, spherical symmetry. Using Gauss’ law in differential form, show that the field with a radial spherical component only is layer of charge in free space has a uniform given by Eq. (1.140) is produced by a uniformly charged sphere of radius a and charge density volume charge density p and thickness p in free space. Equivalent sheet of charge. An infinitely large d. (a) Compute the electric field vector inside the layer, (b) Show side the layer is 1.70. that, as far as the field out- concerned, the layer can be charge, and find the surface charge density, p s of this sheet. , Layer with a cosine volume charge distribution. density of a volume charge in free space depends on the Cartesian coordinate x only and is given by p{x) = pocosinx/a) (|x| < a) and p(x) — 0 (|jc| > a), where po and a (a > 0) are constants, (a) Determine the electric field intensity vector in the entire space, (b) Find the voltage between planes x — —a and x = a. The the differential Gauss’ law. 1.71. layer). space A is volume charge nate system as p(x) 0, distribution in free described in the rectangular coordi- and p(x) — poQ x ^a = — po e - *? 0 for x for > 0, x < 0, p(0) = with po and a being positive constants. Calculate the electric field intensity Uniform is vector for — oo < x < oo. volume charge density in that What region? symmetry by differen- 1.72. Problem with planar symmetry using differenGauss’ law. Redo Problem 1.61 employing tial differential Gauss’ law. Antisymmetrical charge, differential Gauss’ law. Redo Example 1.21 applying differential Gauss’ law. 1.74. Gauss’ law in differential and integral form. In is given by a certain region, the electric field (4xyx -(- 2x 2 y + z) V/m (x, y in m). The medium is air. (a) Calculate the charge density, (b) From the result in (a), find the total charge enclosed in a cube situated dinate octant (x,y,z > 0), in the first coor- with one vertex at the coordinate origin, and the edges, of length 1 m, parallel to coordinate axes, (c) Confirm the validity of Gauss’ law in integral form and electric field. In a certain region, there a uniform electric field, Eo- cylindrical Gauss’ law. Redo Example 1.19 but with the use of Gauss’ law in differential form. E= Exponential charge distribution in the entire space. Problem with tial 1.73. Layer with a sine charge distribution. Repeat the previous problem but for the following charge density function: p(x) = po sin(7rx/o) for |jc| < a (there is no charge outside the Nonuniformly charged sphere using differenGauss’ law. For the nonuniform volume charge distribution in a sphere defined by Eq. (1.32) and analyzed in Problem 1.55 (based on Gauss’ law in integral form), compute the electric field intensity vector everywhere using tial replaced by an equivalent infinite sheet of 1.65. From the field with a radial cylindrical the electric field intensity vector inside and out- 1.64. planar symmetry. field, expressions in Eqs. (1.150)— (1.152), obtain sheet of charge in free space [Eq. (1.64)]. p s and 1.63. Charge from the corresponding charge distribution in free parallel infinite sheets of charge with densities 1.62. . space [Eq. (1.147)]. law. 1.61. e0 field p. 1.59. Field 1.60. Charge distribution from 1-D field distribuFind the volume charge density p(x) in the electrostatic system from Example 1.16, assuming that the permittivity of the medium tion. is the the divergence theorem by evaluating the net outward flux of E through the surface of the cube defined in (b). 59 Problems 1.75. Excentric charged sphere inside an unchar- ged Consider shell. Example the and assume that the sphere 1.27, moved toward unit length of the first and the third conductor, Q\ and from structure is the shell wall so that the centers of the sphere and the shell are separated by a distance d. Find the potential of the shell in the new (b) electrostatic state = d b — is = d (a) if a (the sphere (b — a)/2 and pressed against the shell wall). charge inside a charged 1.76. Point 2Q charge shell. A point placed at the center of an is air- spherical metallic shell, charged with filled and situated in air. The inner and outer < of the shell are a and b (a b). (a) Q radii What is the total charge on the inner and on the outer surface of the shell, respectively? (b) Find the Figure 1.55 Detail of the cross section of a system potential of the shell. of four cylindrical conductors; for 1.77. Three concentric shells, concentric spherical metallic shells are situ- ated in is a = air. 30 The outer radius of the inner mm, and its Q— charge 1.80. 1 .79. 10 nC. Three concentric conductors, one grounded. Shown shell in Fig. 1.56 is a system consisting of three concentric spherical conductors (the The inner and outer radii of the middle shell are inner conductor = 50 mm and c = 60 mm, and its potential V = 1 kV with respect to the reference point at remaining two are spherical shells). The radius of the inner conductor is a — 2 mm. The inner radius of the middle conductor is b — 5 mm, b infinity. The inner and outer shell are d = 90 radii of the outer mm and e = 100 mm, and and (b) the voltage between the inner and the outer shell. dle shell tial. etry shells, two at the that the charges of the inner are Q\ = 2 nC and and outer Q 3 = —2 nC, = V3 ). Under c — 6 mm. ground are V\ — 15 V and V2 = 10 V, respectively. Determine total charges of the inner and middle conductors, Q\ and 02to the shells respectively, as well as that their potentials are the (V\ a solid sphere, while the the inner and middle conductors with respect same potenConsider a structure with the same geomas in the previous problem, and assume Three concentric is The inner radius of the outer conductor is d = 8 mm. The space between the conductors is air-filled. The outer conductor is grounded, and the potentials of and outer it is uncharged. Calculate (a) the charge of the mid- 1.78. Problem one uncharged. Three same these circumstances, com( Q 2 ) and and the middle pute (a) the charge of the middle shell (b) the potentials of the inner shells (V\ point at 1.79. and V2 ) with respect to the reference infinity. Figure 1.56 System Four coaxial cylindrical conductors. Four very long conductors, each in the form of a cylindrical shell with thickness d= 1 cm, are posi- of three concentric spherical conductors; for tioned in air coaxially with respect to each other, as indicated in Fig. 1.55, detail of the cross section of the system. first The and the fourth conductor are grounded, and the potential of the third conductor with respect to the ground is V 3 = 1 kV. The second conductor is Problem 1 .80. which shows a uncharged. Find the charges per 1.81. Charged metallic metallic foil is foil. An infinitely large flat situated in air and charged uni- formly with the surface charge density ps = 2 1 nC/m Find the electric field intensity vector everywhere. . N 60 Chapter 1 . 82 . Two Electrostatic Field in Free 1 metallic slabs. An slab of thickness d = 1 Space infinitely large metallic cm is situated in air 1.86. and charged such that the surface charge density 2 at each of the slab surfaces is p s = 1 p C/m field intensity . Fig. 1.46), 3 cm. In the new all four surfaces of the slabs, (b) electric field intensity vector (c) the 1 . 83 . dicular to everywhere, and and metallic spheres at the same potential. Consider the system in Fig. 1.45, and assume that a — 5 cm, b = 1 cm, and d = 1 m, as well as that the total charge of the two spheres 600 pC. Find (a) the potential of the spheres and (b) the electric field intensities 1.87. Ea 1.88. plate into a = 1 m N Subdivide the square patches, and assume that and Vq = 1 V. (a) Tabulate and plot 85 . 100). (c) image. Find its Q in Imaging a For the structure defined determine the distribution of induced surface charges on the conducting line charge. Example 1.30, plane. 1.89. Charged wire shows Fig. 1.57 parallel to corner screen. a a cross section of the structure consisting of a metallic wire of radius a 90° corner metallic screen in air. and a The distance of the wire from both the horizontal and ver- each dimension), (b) Compute the total charge where h (i) N = 9, N — 25, N = 49, and (iv) N = 100, respectively. MoM computation for a charged cube. Write a of the plate, taking . N— the results for the surface charge density (p s ) of the patches, taking 100 (ten partitions in N= 1 plane at points that are a/2, 2a, Force on a point charge due to in determine the charge distribution on a very thin charged square plate of edge length a at a potential Vq, in free space. its the electric force on the point charge sented in Section 1.20, write a computer proto approx- Fig. 1.48(a). MoM-based computer program for a charged plate. Using the method of moments as pregram is center. and Eb near the surfaces of the spheres. 1 . 84 . the cube in 100fl, respectively, distant surface (for Two is (e.g., distribution from the plate Also compute the electric field inside the cube from the previous problem (Problem 1.85), at a quarter of its space diagonal (body diagonal) and at its voltage between the slabs. Q— whose charge program, compute the electric field along the axis of the plate from Problem 1.84 perpen- elec- trostatic state, calculate (a) the surface charge densities at vector at an arbitrary point in imately described by Eq. (1.212). (b) Using the expression in (a) and the associated computer distance between the surfaces of the two slabs D— Write the approximate inte- space due to a charged body placed parallel to the charged slab such that the is integral expression for the elec- gral expression for the evaluation of the electric Another metallic slab of the same thickness, which is uncharged, is then introduced and facing each other Approximate tric field vector, (a) computer program (ii) for the (iii) tical half-planes constituting the screen » a. If the wire is is h, charged with Q' per unit length, calculate the voltage between the wire and the screen. method-of-moments analysis of a charged metallic cube, Fig. 1.46, = m, and compute the = 1 V and ten, or as many as possible (given available computational resources), subdivisions per cube = 600 if ten subdivisions per edge are edge ( section of a charged adopted). for with edge length a total 1 charge of the cube for Vq Figure 1 .57 Cross metallic wire parallel to a metallic corner screen; Problem 1 .89. Dielectrics, Capacitance, and Electric Energy r Introduction: n chapter, this I fields we shall analyze electrostatic vector and the electric potential due to polarized and study dielectrics using free-space formulas and techniques from the previous chapter. Gauss’ law will be generalized for an electrostatic system that includes arbitrary media (conductors and dielectrics), and in the presence of dielectrics several important related topics. Dielectrics or insulators are little nonconducting materials, having very free charges inside them (theoretically, per- have no free charges). In addition, redistribution of any free charges (e.g., electrons) fect dielectrics deposited inside the material lasts much longer than in metallic conductors, a typical example, as already indicated in Section 1.16, being the charge rearrangement time of ~ 50 days for fused quartz compared to ~ 10~ 19 s for copper. We shall see, however, that another type of charges, called bound or polarization charges, exist in a polarized dielectric. We shall first investigate the mechanisms of the polarization of dielectrics, caused by an external electric field. By introducing the macroscopic quantities such as the polarization vector, and the volume and surface density of bound charges, is it possible to evaluate the electric field intensity characterization of dielectric materials in terms of their linearity, homogeneity, and isotropy will be presented. Dielectric-dielectric boundary conditions will be derived and used. The chapter will also introduce Poisson’s and Laplace’s second-order differential equations for the potential and their solution. Having both conductors and dielectrics, we then put them together to form capacitors and related electrostatic systems. The capacitor is a fundamental element in electrical engineering. Its basic shall property itors is its capacitance. We shall analyze capac- with electrodes of different shapes and with different types of dielectrics, and evaluate capac- itance per unit length of various two-conductor 61 62 Chapter 2 Dielectrics, Capacitance, and Electric Energy we transmission lines as well. In addition, shall of tric conducting bodies, operation of the structure prior to an eventual breakdown. ties for the safe and other systems capacitors, transmission lines, and introduce the maximal permissible extents of quanti- line) define evaluate the electric energy contained in charged The elec- energy density to help us find and quantify electrostatic transmission lines, to analysis of capacitors and determine their capacitance, the localization and distribution of the energy in energy, and such systems. Dielectric breakdown, occurring for a culmination of the theory of the electrostatic exceedingly strong electric fields in a dielectric field. It material causing it to become conducting, many will also be discussed for various electrostatic structures. We shall analyze structures with electric fields close to breakdown breakdown represents, characteristics, represents on the other hand, a gateway to practical applications of this theory. Finally, a clear understanding of concepts that will be pre- sented in this chapter levels in order to predict critical val- is essential for many similar, ues of voltages and other circuit quantities for the and analogous concepts in other areas of electromagnetics, which are to be introduced later in structure at breakdown. Such parameters the text. breakdown voltage of (e.g., dual, the a capacitor or a transmission POLARIZATION OF DIELECTRICS 2.1 Each atom or molecule in a dielectric is electrically neutral. For most centers of “gravity” of the positive and negative charges in an coincide - in the absence of the external electric When field. dielectrics, atom or molecule a dielectric is placed in E ex t. however, the positive and negative charges shift in opposite directions against their mutual attraction, and produce a small electric dipole (Fig. 1.28), which is aligned with the electric field lines. The moment of this equivalent dipole is given by p = Qd [Eq. (1.116)], where Q is the positive charge of the atom or molecule (-Q is the negative charge), and d is the vector displacement an external of field, of intensity Q with respect to — Q. The charges are displaced from their equilibrium positions by forces Eel respectively, and thus = £?E ex and t Fe2 = — £?E ex t, (2.1) - Q shifts in the direction of E ext , while —Q moves in the oppo- so that p and E ex t are collinear and have the same direction. The displacement d is very small, smaller than the dimensions of atoms and molecules. site direction, Q and — Q bound by atomic and molecular forces and can field. So, the two charges in an equivalent small dipole cannot separate one from the other and migrate across the material in opposite directions run by the electric field. Hence, these charges are The charges only are shift positions slightly in called bound charges Some in place response to the external (in contrast to free charges). such as water, have molecules with a permanent displacement between the centers of the positive and negative charge, so that they act as small electric dipoles even with no applied electric field. Such molecules are known as polar molecules, and the dielectrics are called polar dielectrics (those with no dielectrics, built-in dipoles are nonpolar dipoles are oriented in a electric dipole, is dielectrics). In the random way. If absence of the electric a polar molecule, which field, all the we model by an brought into an electric field, however, the forces on the two dipole The torques (moments) of charges, given in Eq. (2.1), act as indicated in Fig. 2.1. forces with respect to the center of the dipole (point Tj = ri x F e] and T2 = O) r2 are x F e2 , (2.2) Section 2.2 ri and r2 center. We notice that with 63 Polarization Vector Q and — Q with respect to the dipole and hence the resultant torque on the dipole denoting the position vectors of r\ — t2 = d, turns out to be Ton dipole — Tj + T2 — Q{*\ — *2) X E ex — Q d t X Eext — p X E e xt» (2.3) torque on an electric dipole an external electric field where we assume that E ex is practically uniform along the dipole. Vector T on dipole is normal to the plane of p and E ex t (the plane of drawing in Fig. 2.1), and its magnitude amounts to t Ton We dipole — |P x E ex — P Eext sina. (2.4) t | see that the torque given by Eq. (2.3) tends to rotate the dipole about the through the dipole center and being normal to the dipole and the plane axis passing about the vector Ton dipole- The action of such torques in the dielecrandom intermolecular thermic forces, and is to align the dipoles, to some extent, in the same direction - toward the field lines. We also see that the stronger the field, the larger Ton dipole and the larger the component of the resultant of drawing, tric is i.e., against > moment of all the molecules, J2 P> along the direction of E ex A sufficiently strong field may even produce an additional displacement between the positive and Figure 2.1 Polarization of negative charges in a polar molecule, resulting in a larger p. a polar molecule in an dipole t- We conclude that both an unpolar and polar dielectric in an electric field can be viewed as an arrangement of (more or less) oriented microscopic electric The process of making atoms and molecules in a dielectric behave as dipoles and orienting the dipoles toward the direction of the external field is termed the polarization of the dielectric, and bound charges are sometimes referred to as polarization charges. This process is extremely fast, practically instantaneous, and the dielectric in the new electrostatic state is said to be polarized or in the polarized state. For almost all materials, the removal of the external electric field results in the return to their normal, unpolarized, state. A very few dielectrics, called electrets, remain permanently polarized in the absence of an applied electric field (an example is a strained piezoelectric crystal). dipoles. Conceptual Questions (on Companion Website): 2.2 POLARIZATION VECTOR When polarized (by an external electric and the field), 2.1 and 2.2. a dielectric is a source of its own an arbitrary point in space (inside or outside the dielectric) is a sum of the external (primary) field and the field due to the polarized dielectric (secondary field). To determine the secondary field, we replace the dielectric by a collection of equivalent small dipoles, which can be considered to be in a vacuum, as the rest of the material does not produce any field. Theoretically, we could use the expression for the electric field due to an electric dipole, Eq. (1.117), and obtain the field due to a polarized dielectric by superposition. However, as many atoms or molecules in a dielectric body that many equivalent small dipoles in it, and, with the “microscopic” approach to the evaluation of the field due to the polarized dielectric, we would need to consider every single dipole, which is practically impossible [there is on the order of as many as 10 30 atoms per unit volume (1 m 3 ) in solid and liquid dielectrics]. We rather adopt a “macroscopic” approach, and introduce a macroscopic quantity called the polarization vector to describe the polarized state of a dielectric and electric field, total field at polar dielectrics: model external electric field. of in 64 Chapter 2 Dielectrics, Capacitance, and Electric Energy the resulting average dipole moment We first average dipole moments in field. P)in dv in (2.5) ^in dv an elementary volume of a polarized dielectric and then multiply this average by the concentration of dipoles of atoms or molecules in the dielectric), which equals /v, What we get is, by = (i.e., concentration (2.6) dv definition, the polarization vector: P— polarization vector (unit: C/m an elementary volume dv, P)in dv Nv Pav (2.7) dv 2 ) Note that P would represent the resultant dipole moment in a unit volume (1 it were polarized uniformly (equally) throughout the volume. Note also that Pdv= (]C p ) indl is the dipole moment ized dielectric, i.e., to m3 ) if (2- 8 > , of an electric dipole equivalent to an element dv of the polarall the dipoles within The 1 it. unit for P C/m 2 is In any dielectric material, the polarization vector at a point is . a function of the (total) electric field intensity vector at that point, P = For linear Xe - electric susceptibility of P(E). (2.9) (in the electrical sense) materials, this relationship P= a is linear, i.e., Xe^oE, ( 2 10 ) . linear dielectric where Xe is the electric susceptibility of the dielectric. It is a pure number, i.e., dimensionless quantity, obtained by measurements on individual materials, and always nonnegative (x e 2.3 We > 0). For a vacuum, Xe = 0, whereas Xe ^ 0 for air. BOUND VOLUME AND SURFACE CHARGE shall of excess now DENSITIES derive the expressions for calculating the macroscopic distribution bound charges body from a given distribution of obtained by averaging the microscopic the dielectric material. These expressions will be used in the next section in a polarized dielectric the polarization vector, P, which, in turn, dipoles in a is is for free-space evaluations of the electric field due to polarized dielectrics. Let us first find the total bound (polarization) charge Q p s enclosed by an arbitrary imaginary closed surface S situated (totally or partly) inside a polarized dielectric body, as shown in Fig. 2.2. Knowing of a vast collection of small electric dipoles, 1 An elementary volume dv, as we use sense, and cannot be vector, for instance, that “on average,” but yet it infinitely small in a means that dv in macroscopic electromagnetic theory, is small in a physical mathematical sense. Within the definition of the polarization enough to contain many small dipoles to be treated P can be considered constant in dv from the macroscopic number (millions) of atoms or molecules. is large sufficiently small so that point of view. Such dv still contains a vast bound charge actually consists each dipole being composed from a that Bound Volume and Surface Charge Section 2.3 Figure 2.2 Closed surface S a polarized dielectric and a negative —Q, we realize that all the dipoles that appear inside S Q and — Q, as well as dipoles that are totally outside S, contribute with zero net charge to Q s Only dipoles whose one end is inside S (and the v other end outside S ) contribute actually to the total bound charge in S. (We notice right away that Q $ = 0 when S encloses the entire dielectric body.) To evaluate p positive Q with both their ends, QpS we therefore count the dipoles that cross the we count the contribution of such dipoles as either Q or — Q (in the general case), In doing that, Q, generally, differ from dipole to dipole), surface S. (note that by inspecting which end of the dipole is inside S. Consider an element dS of S and the case when the angle a between the vector p av ) and vector dS, which is oriented from S outward, is less than 90°, as depicted in Fig. 2.3(a). Note that negative ends of dipoles that extend across dS with one (negative) end inside S are in a cylinder with bases dS and height P (or vector h so that the number = dcosa, (2.11) of these dipoles equals the concentration of dipoles, Nw , times dv = dSh. The dipole ends on the inner side of d S being all negative, and with an assumption that all dipoles in dv are with the same moments and charge, the corresponding bound charge is given by the volume of the cylinder, dQ p = In the case when a > Nw dSdcosa(-Q) (0 <a < 90°). 90°, portrayed in Fig. 2.3(b), h =d cos (7T — a) = d (— cosa), (2.1 3) and, because the ends of dipoles on the inner side of dS are dQ p = (2.12) jVv dSd(— cos a) Q which turns out to be the same result as in Eq. (90° <a < (2.12). all positive, 180°), (2.14) body. Densities in 65 66 Chapter 2 Dielectrics, Capacitance, and Electric Energy Figure 2.3 Element of surface 5 in Fig. 2.2, in two cases with regards to the angle a between P and (b) 90° dS: (a) 0 < a < <a < 90° and 180°. For unpolar where not all dielectrics, dipole p av moments = and P p, = Nv p = Nv Qd. are mutually parallel, we can For polar dielectrics, consider the dipoles in moments p av = Hence, for an arbitrary a (0 < a < 180°), we have small cylinders in Figs. 2.3(a) and (b) to be equivalent dipoles with Qd, so that P = 7Vv pav = Nv Qd. dQp = — NvQddS cos a = — PdScosa = — P dS (2.15) • (note that the boundary case, a we total bound = 90° and d Q p = 0, is Qp formula). Finally, by integrating the result for d this also properly included in over the entire surface S, obtain = QpS (polarization) P • dS. (2.16) charge enclosed by a closed surface S This is an integral equation similar form to Gauss’ in law, Eq. (1.133). It tells us that the outward flux of the polarization vector through an arbitrary closed surface in an electrostatic system that includes dielectric materials is equal to the total bound (polarization) charge enclosed by that surface, multiplied by —1. Eq. (2.16) is true for any closed surface S. Let us now apply it to the surface S enclosing an elementary volume Av inside a polarized dielectric: - ptin Av (Qp) LP dS (Av Av Av (2.17) 0), The expression on the volume bound charge, represents the density of excess with both sides of the equation being also divided by Av. left-hand side of Eq. (2.17) Pp while the expression on — (Gp)in Av (2.18) Av by definition [Eq. of the divergence of the polarization vector. Hence, bound volume charge its right-hand side PP density If P = in a uniformly polarized dielectric — — div P = —V • (1.172)], the negative P. (2.19) const inside the dielectric (uniformly polarized dielectric), derivatives of no volume bound charge is, P all spatial are zero, and using Eq. (2.19), P = const PP = o. ( 2 20 ) . . Bound Volume and Surface Charge Section 2.3 67 Densities Figure 2.4 Elementary closed surface used for deriving the boundary condition for the P on a surface vector dielectric-free space, Eg. (2.23). If P^ volume bound charge exists only if the polarizavolume of the dielectric (nonuniformly polarized const, however, then excess tion vector varies throughout the way that its divergence is nonzero, otherwise p p = 0. On the surface of a polarized dielectric, there always exists excess surface bound dielectric) in a charge (there are ends of dipoles pressed onto the surface that cannot be compensated by oppositely charged ends of neighboring dipoles), except on parts of the where P and the dipoles are tangential to the surface. To determine the bound (or polarization) surface charge density, pps we apply Eq. (2.16) to a small pillbox surface, with bases AS and height Ah (Ah -> 0), shown in Fig. 2.4. There is no polarization in free space (a vacuum or air), surface associated , P= so that the flux of vector S, and we have P in Eq. (2.16) is (2.21) 0, reduced to P AS over the lower side of Eq. (1.188)] no polarization in a vacuum or air [also see the similar derivation in pps AS = -P AS. ( With Ad denoting the normal unit vector oriented from the AS = — ASAd, which yields Pps — Ad • dielectric 2 22 ) . body outward, bound P. (2.23) fid surface charge density; outward normal on a dielectric surface This is the boundary condition for the vector P on a surface dielectric-free space, connecting the polarization vector in the dielectric near the boundary surface and the bound surface charge density on the component of P contributes to p ps surface. Note that only the normal . Example The 2.1 Nonuniformly Polarized Dielectric Cube polarization vector in a dielectric cube shown P(x,y) in Fig. 2.5 is given by = P0 ^x, 1 (2.24) a where Pq is a constant. The surrounding medium is air. Find the distribution of bound charges of the cube. Solution Using Eqs. (2.19) and (1.167), the bound volume charge density inside the cube Pp - dPx T0 y is (2.25) dx whereas Eq. (2.23) tells us that bound surface charge and its density amounts to exists on the front side of the cube only, Figure 2.5 Dielectric cube with polarization P(x, Pps = x-P(fl-,y) = (2.26) Example 2.1 y); for 68 Chapter 2 Dielectrics, Capacitance, with P(n _ and Electric Energy ,y) denoting the polarization vector in the dielectric very close to the the back side, p ps = 0 since P(0 + y) 0 because fid and P are mutually perpendicular. surface at x = On a. = 0, , pps = Note that, with v designating the volume of the cube and S sides Q P = f Ppdv+ (fpps dS = [ f — —&y-') a 2 dy + f Jv Jy = 0 \ JS Jy = 0 / as expected (the total bound charge of a dielectric body is boundary while on the remaining four its boundary — ady surface, = 0, (2.27) & always zero), where, in accordance with our general integration strategy explained in Section the cube of thickness dy, and dS a strip of width dy 1.4, dv is adopted to be a on the front cube side. slice of Problems'. 2.1. EVALUATION OF THE ELECTRIC FIELD AND POTENTIAL DUE TO POLARIZED DIELECTRICS 2.4 In this section, we scalar potential due assume that the state of polarization of a dielectric shall evaluate the electric field intensity vector and electric to polarized dielectric bodies in several characteristic cases. body is We described by a given dis- From P, using Eqs. (2.19) and volume and surface bound charge densities, p p and p ps throughout the body volume and over its surface, respectively. Then, we calculate the field E and potential V (and any other related quantity of interest) using the appropriate free-space formulas and equations [Eqs. (1.37), (1.38), (1.82), (1.83), (1.133), etc.] and solution techniques suitable to specific geometries and source distributions. tribution of the polarization vector, P, inside the body. (2.23), we first find the distribution of , Example 2.2 A Uniformly Polarized Dielectric Disk dielectric disk of radius a polarized throughout its magnitude being its and thickness d is situated in free space. The disk is uniformly volume, the polarization vector being normal to the disk bases and P. Find (a) the distribution of bound charges of the disk and (b) the electric field intensity vector at the disk center. Solution (a) Eq. (2.20) tells us that there is no bound volume charge inside the disk. According to Eq. (2.23) and Fig. 2.6, the bound surface charge densities on the upper and lower disk bases are Ppsi = fidi p = p and Pp S 2 -P, (2.28) on the side disk surface, p ps 3 = fid 3 P = 0. due to the polarized disk equals the field due to two circular sheets of charge with densities p ps i and p pS 2 in free space. We use the expression for the field due to a circular sheet of charge (thin charged disk) in Eq. (1.63) and the superposition principle to add up the fields due to two sheets, Ej and E 2 The charge densities are p s = ±P and the distance from each sheet at the disk center (point O) is d/2, so the two fields are the same, and the total field comes out to be respectively, while (b) = fid2 p = The • electric field . E= E] + E2 = 2Ei ' eo (2.29) ' . L 2 s/a 2 + (d/2) 2 Evaluation of the Electric Field Section 2.4 and Potential due P P ^dl' p e0 Ppsl fid3 < »o Pps3 £0 fi A charge on d2 dielectric sphere of radius a, in free space, Compute P. is a uniformly polarized dielectric disk; for Example 2.2. Uniformly Polarized Dielectric Sphere Example 2.3 vector Figure 2.6 Bound surface a Pps2 scalar potential at the sphere center, and is uniformly polarized, and the polarization bound charge of the sphere, (a) the distribution of (c) the electric field intensity (b) the electric vector at the sphere center. Solution (a) Let us adopt a spherical coordinate system with the origin z-axis parallel to the vector P, as pp = 0. The bound the angle 6, We now replace E surface charge density at a (b) = nd P = • -Pcos Z(n d P) , From Eq. (1.83) and is because the Due i total a bound charge of the to symmetry, vector is 0 < 6 < n. the function of 0 given in Eq. (2.30), and Fig. 2.7, the potential at the point 4nSp Js which = Pcos<9, (2.30) compute V and sphere center (point O) using free-space concepts and equations. F= (c) sphere center and the the polarized sphere by a nonuniform spherical sheet of charge in free whose charge density at the at the The bound volume charge density is point M on the sphere surface, defined by in Fig. 2.7. is pps space, shown computed E / i (2.31) 4nsoa J s O 4ire 0 a ^ sphere, <2 p at the point essentially in the O turns out to be , is zero. in Fig. 2.7 has a (negative) same way surface into thin rings and integrating the fields z-component only, as in Fig. 1.16, subdividing the sphere dE due to individual rings, the only two is now a function of 9, Eq. (2.30), and differences being that the surface charge density 1 Figure 2.7 Dielectric sphere with a uniform polarization; for Example 2.3. to Polarized Dielectrics 69 t 70 Chapter 2 Dielectrics, Capacitance, that the and upper Electric Energy limit in the integration E=(f>dE — —^— is now = 9 n. With p ps (9) sin# cos# d9 z [ this, = Eq. (1.67) becomes f 2eo Je = o Js electric field inside ' 2eo Jo — cos 2 —— 0sin0d0z -'' ' -t a uniformly (2.32) polarized dielectric sphere where the substitution given by u = cos 9 is used to solve the integral in 9. It can be shown that E has this same (constant) value at any point inside the polarized sphere. Nonuniformly Polarized Example 2.4 Dielectric Sphere A nonuniformly polarized dielectric sphere, of radius a, is situated in free space. In a spherical is coordinate system whose origin coincides with the sphere center, the polarization vector given by the expression =P V{r) r {r)i = Po- (2.33) a Pq is a constant). Determine (a) the bound volume and surface charge densities of the sphere and (b) the electric scalar potential inside and outside the sphere. ( Solution (a) Using the expression for the divergence in spherical coordinates, Eq. volume charge density of the sphere amounts to (it is same the at all points inside the sphere). = r P(0 = P0 pps (b) The field > outside the sphere (for r equivalent point charge a) is is = 0, V(r) = V(r) —J is a Exercises (on dielectrics. in free Gauss' law for conductors and a system with dielectrics < r < , 0 (2.36) oo. is < given by r < (2.37) a. 1 E(/) dr = ® Companion - r — a - (2-38) Website): 2.3 and 2.4; Companion Website). We now consider the most and placed at the sphere center. thus GENERALIZED GAUSS' 2.5 identical to the field of the it is eoa Problems'. 2.2-2. 6; Conceptual Questions (on MATLAB is also zero, ^ = -— 3eo potential inside the sphere bound (2.35) . zero, because (1.140), the electric field inside the sphere E(r) The surface charge density Q p = 0 (total charge of the sphere) Hence, the potential outside the sphere From Eq. The bound (1.171), the LAW general electrostatic system containing both conductors The equivalent field sources are space, and Gauss’ law, Eq. (1.133), Qs + Qps E dS = • l now both free and bound charges, becomes £0 (2.39) 71 Characterization of Dielectric Materials Section 2.6 where Qs and Q v s are the total free charge and the total bound charge, respectively, enclosed by an arbitrary closed surface S. Multiplying this equation by £q, moving Q ps to the left-hand side of the equation, then substituting it by the negative of the flux of the polarization vector, P, from Eq. (2.16), and finally joining the two integrals over S into a single integral, we obtain the equivalent integral equation: E j> (£ 0 To shorten the we writing, + P) new define a = Qs dS • (2.40) . vector quantity, D = £qE T P, which is (2.41) called the electric flux density vector (also ment vector or electric flux known (unit: as the electric displace- electric induction vector). Accordingly, the flux of (symbolized by electric flux density vector D C/m 2 ) termed the is fl>), 41 = f D • dS, C) (2.42) electric flux (unit: (2.43) generalized Gauss' law JS' where any designated surface (open or closed). In place of Eq. S' is D • dS = Qs (2.40), . is an equivalent form of Gauss’ law for electrostatic fields in arbitrary media, which is more convenient than the form in Eq. (2.39) because it has only free charges on the right-hand side of the integral equation, and not the bound charges, and thus This is simpler to use. states that the referred to as the generalized Gauss’ law, and, in words, It is outward electric flux system including conductors and dielectrics equals the total the surface. density, D, From Eq. is C/m2 (2.43), the unit for the electric flux is it any electrostatic free charge enclosed by through any closed surface in C, so that the unit for its . In the general case, free charge is represented by means of the volume charge density, p, yielding generalized Gauss' law (2.44) in terms of the volume charge density with v denoting the volume bounded by less of the choice of v, S. Since this integral relation is true regard- the divergence theorem, Eq. (1.173), gives the differential form of the generalized Gauss’ law: V-D = p. (2.45) generalized differential Gauss' law Problems'. 2.7-2.11; Conceptual Questions (on 2.6 Companion Website): 2.5. CHARACTERIZATION OF DIELECTRIC MATERIALS The polarization properties of materials can be described by the relationship between the polarization vector, P, and the electric field intensity vector, E, Eq. (2.9). We now employ the electric flux density vector, D, and substituting Eq. (2.9) into Eq. (2.41), obtain the equivalent relationship D = e 0 E + P(E) = D(E), (2.46) constitutive equation of an arbitrary (nonlinear) dielectric 72 Chapter 2 Dielectrics, Capacitance, and Electric Energy which is more often used for characterization of dielectric materials and is termed a constitutive equation of the material. For linear dielectrics, Eq. (2.10) applies, and Eq. (2.46) becomes D= constitutive equation of a linear dielectric where £ is the permittivity (Xe + and =£ 1 )£qE r £oE D = eE, or (2.47) e T the relative permittivity of the medium (£ r is some- times referred to as the dielectric constant of the material). The unit for e per meter (F/m), while £ r is dimensionless, £r is farad obtained as = Xe + £r > 1. (2.48) 1, and hence The value of eT £ permittivity of a linear dielectric (unit: F/m) is (2.49) shows how much the permittivity of a — (2.50) £r£()> higher than the permittivity of free space (vacuum), given in Eq. space and nondielectric materials (such as metals), e r = 1 (1.2). For free and D = £qE. constitutive equation for free space dielectric material, (2.51) Table 2.1 shows values of the relative permittivity of a number of selected materials, for electrostatic or low-frequency time-varying (time-harmonic) applied electric 2 at room temperature (20°C). For nonlinear dielectrics, the constitutive relation between fields, is nonlinear. This also on the means E electric field intensity, independent of E D and E, Eq. (2.46), that the polarization properties of the material (for linear dielectrics, Xe and depend e are constants, ). In so-called ferroelectric materials, Eq. (2.46) is not only nonlinear, but also shows hysteresis effects. The function D(E) has multiple branches, so that D is not uniquely determined by a value of E but it depends also on the history of polarization of the material, i.e., on its previous states. A notable example is barium titanate (BaTiC^), used in ceramic capacitors and various microwave devices (e.g., ceramic filters and multiplexers). Another concept in characterization of materials is homogeneity. A material is said to be homogeneous when its properties do not change from point to point , region being considered. In a linear homogeneous dielectric, £ is a constant independent of spatial coordinates. Otherwise, the material is inhomogeneous [e.g., in the e = e(x, y, z) Finally, in the region]. we introduce the concept of isotropy in classifying dielectric materials. Generally, properties of isotropic media are independent of direction. In a linear isotropic dielectric, £ in the same is a scalar quantity, and hence D and E are always collinear and however, individual components of D depend differently on of E, so that Eq. (2.47) becomes a matrix equation, ~D X [e] ~ Dy - permittivity tensor of an anisotropic dielectric LDzJ 2 medium, components direction, regardless of the orientation of E. In an anisotropic ~ = different - £ xx £ xy £xz ~EX £yx £ yy £ yz Ey _ £ zx £ zy £ zz _ lEzi (2.52) At higher frequencies, when viewed over very wide frequency ranges, the permittivity generally most materials) is not a constant, but depends on the operating frequency of electromagnetic waves propagating through the material. (for Section 2.6 Table 2.1 Characterization of Dielectric Materials Relative permittivity of selected materials* . Material St Material St Quartz 5 5-6 Vacuum 1 Freon 1 Air 1.0005 Diamond Wet soil Styrofoam 1.03 Mica (ruby) 5.4 Polyurethane foam 1.1 Steatite 5.8 Paper 1.3-3 Sodium chloride (NaCl) 5.9 Wood 2-5 Porcelain 6 Dry 2-6 Neoprene Paraffin 2.1 Silicon nitride (Si 3 Teflon 2.1 Marble Vaseline 2.16 Polyethylene 2.25 Alumina (AI 2 O 3 ) Animal and human muscle Oil 2.3 Silicon (Si) Rubber 2.4-3 Gallium arsenide 13 Polystyrene 2.56 Germanium 16 PVC 2.7 Ammonia 22 Amber 2.7 Alcohol (ethyl) 25 Plexiglass 3.4 Tantalum pentoxide 25 Nylon 3. 6-4.5 Glycerin 50 soil Fused 02 ) 5-15 6.6 N4 7.2 ) 8 8.8 10 11.9 (liquid) 3.8 Ice 75 Sulfur 4 Water 81 Glass 4-10 Rutile (Ti 02 ) Bakelite 4.74 Barium * For silica (Si static or low-frequency applied electric Thus, instead of a single scalar e, fields, at titanate 89-173 (BaTiOa) 1,200 room temperature. we have a tensor [e] (permittivity tensor), i.e., nine (generally different) scalars corresponding to different pairs of spatial components of D and E. Crystalline dielectric materials, in general, are anisotropic; the moments to be formed and oriented by much more easily along the crystal axes than in periodic nature of crystals causes dipole means of the applied other directions. electric field An example is rutile (Ti02), whose relative permittivity is e = 173 and e = 89 at right angles. For many r in the direction parallel to a crystal axis r change in permittivity with direction is small. For example, quartz has and it is customary to adopt a rounded value e T = 5 for its average relative permittivity and treat the material as isotropic. The theory of dielectrics we have discussed so far assumes normal designed regimes of operation of electrical systems - when the electric field intensity, E, in crystals the £r = 4.7 — 5.1, individual dielectric parts of a system the intensity E in a dielectric cannot is below a certain “breakdown” be increased exceeded, the dielectric becomes conducting. It indefinitely: if level. Namely, a certain value is temporarily or permanently loses down. The breaking field value, i.e., the an individual dielectric material can withstand without breakdown, is termed the dielectric strength of the material. We denote it by Ecr (critical field intensity). The values of Ecx for different materials are obtained by measurement. For air, its insulating property, maximum electric field and is said to break intensity that £C rO = 3 MV/m. (2.53) dielectric strength of air 73 r 74 Chapter 2 Dielectrics, Capacitance, and Energy Electric In gaseous dielectrics, like air, because of a very strong applied electric field, the and ions are accelerated, by Coulomb forces free electrons ities [see Eq. (1.23)], to veloc- high enough that in collisions with neutral molecules, they are able to knock electrons out of the molecule (so-called impact ionization). The newly created electrons and positively charged ions are also accelerated by the field, free they collide more electrons, and the result is an avalanche process of impact ionization and very rapid generation of a vast number of free electrons that with molecules, liberate constitute a substantial electric current in the gas (usually sparking occurs as well). In other words, the gas, normally a very into an excellent conductor. time in thunderstorms all Note that good many air insulator, is suddenly transformed breakdowns occur spheric electric fields (fields due to charged clouds), reaching the in at any instant of over the earth. Basically, they are caused by large atmo- breakdown value Eq. (2.53), and their most obvious manifestation is, of course, lightning. Similar avalanche processes occur at high enough electric field intensities liquid and solid dielectrics. For of the dielectric strength (Ec ) solids, these in processes are enhanced and the value of the particular piece of a dielectric is lowered by impurities and structural defects in the material, by certain ways the material is manufactured, and even by microscopic air-filled cracks and voids in the material. In addition, when, under the influence of a strong electric field, the local heat due to leakage currents flowing in lossy (low-loss) dielectrics is generated faster than it can temperature may cause a change and lead to a so-called thermal breakdown of the dielectric. Such breakdown processes depend on the duration of the applied strong field and the ambient temperature. Breakdowns in solid dielectrics most often cause a permanent damage to the material (e.g., formation of highly conductive channels of molten material, sometimes including carbonized matter, that irreversibly damage be dissipated in the material, the resulting rise of in the material (melting) the texture of the dielectric). The values of Table in 2.2. Dielectric ECT for some selected dielectric materials are presented in strengths of dielectrics other than air are larger than the value Eq. (2.53). Note that, by definition, the dielectric strength of a vacuum Conceptual Questions (on Companion Website): Table 2.2. is infinite. 2.6. Dielectric strength of selected materials* £C (MV/m) Material r Ect (MV/m) Material Air (atmospheric pressure) 3 Bakelite 25 Barium 7.5 Glass (plate) 30 Freon ~8 Paraffin Germanium -10 Silicon (Si) titanate (BaTiOi) Porcelain 11 Gallium arsenide ~30 —30 ~35 —40 Oil (mineral) 15 Polyethylene 47 Paper (impregnated) 15 Mica 200 Polystyrene 20 Fused quartz (Si02) Teflon 20 Silicon nitride Rubber (hard) 25 Vacuum Wood (douglas fir) At room temperature. ~ 10 Alumina (S^N,^ -1000 -1000 oo Section 2.8 2.7 Electrostatic Field in Linear, Isotropic, 75 and Homogeneous Media MAXWELL'S EQUATIONS FOR THE ELECTROSTATIC FIELD We note that Maxwell’s first equation for the electrostatic field, Eq. depend on the material properties, and is same the in all (1.75), does not kinds of dielectrics as it Eq. (2.44) is Maxwell’s third equation, and we now write down the full set of Maxwell’s equations for the electrostatic field in an arbitrary medium, together with the constitutive equation, Eq. (2.46) or (2.47): is in free space. §c E • dl = 0 Maxwell's first equation in electrostatics &D.dS = /v pdv D = D(E) We [D = (2.54) . eE] Maxwell's third equation constitutive equation for shall see later in this text that these equations represent a subset of the full set of Maxwell’s equations for the electromagnetic static case. In the field, specialized for the electro- general case, the set contains four Maxwell’s equations and three constitutive equations. same form As we shall see, the third equation (generalized Gauss’ law) under nonstatic conditions. Constitutive equations are not Maxwell’s equations, but are associated with them and are needed to supply the information about the materials involved. retains this 2.8 also ELECTROSTATIC FIELD IN LINEAR, ISOTROPIC, AND HOMOGENEOUS MEDIA Most often we deal with and homogeneous dielectrics, in which Eq. (2.47) applies, and the permittivity e is independent of the intensity of the applied field, is the same for all directions, and does not change from point to point. For such media, we can bring e outside the integral sign in the integral form of the linear, isotropic, generalized Gauss’ law, Eq. (2.43), E dS = Qs Gauss' law for a (2.55) • or outside the operator (div) sign in the differential generalized Gauss’ law, Eq. (2.45), V E= — (2.56) . £ We notice that Eqs. (2.55) and (2.56) are identical to the corresponding free-space laws, Eqs. (1.133) and (1.165), except for £o being substituted by e. Recall that the expression for the electric field intensity vector due to a point charge in free space, and with it also Coulomb’s law, can be derived from Gauss’ law (see Problem 1.53). Based on this, we can now reconsider all charge distributions in free space we have considered so far, and all structures with conductors in free space we have analyzed, and by merely replacing £o with £ in all the equations, obtain the solutions for the same (free) charge distributions and the same conducting structures situated in a homogeneous dielectric of permittivity e? This is the power of the concept 3 In what follows we shall always assume linear and isotropic media, medium under consideration is nonlinear and/or anisotropic. (in this entire text), explicitly specify that the except when we homogeneous dielectric D 76 Chapter 2 Dielectrics, Capacitance, and Electric Energy of dielectric permittivity. sity We emphasize again with using the electric flux den- that, we are left to deal with free charges in of bound charges to the field is properly vector and the dielectric permittivity, the system only, while the contribution added through e. Thus, for example, Eq. (1.82) implies that the potential due to a free volume charge distribution in a homogeneous dielectric with permittivity e is given by v= _L fe±. (2.57) R Ane Jv Also, the free surface charge density on the surface of a conductor surrounded by a dielectric with permittivity e is [from Eq. (1.190)] Ps = e fi • E (2.58) (boundary condition for the normal component of E), and so on. Note, however, boundary condition for the tangential component of E near a conductor surface, Eq. (1.186), is always the same, irrespective of the properties (e) of the surrounding dielectric. that the tric, Once we we can and (2.47)] find the electric field in a structure filled with a homogeneous dielec- calculate the polarization vector in the dielectric as [Eqs. (2.41) D — £qE = P= polarization vector in a linear (e — £q)E, (2.59) dielectric and then the distribution of volume and surface bound charges of the dielectric using Eqs. (2.19) and (2.23). Note that, from Eqs. (2.19), (2.59), (2.56), and (2.50), the bound volume charge density, p p at a point in the dielectric can be obtained directly from the free volume , charge density, p, Pp at that point as = —V P„ = -(£ • In an analogous manner, - £o)V _ = E — e -—— p eT £() — — 1 - (2.60) p. we derive the relationship between the bound and free surface charge densities on the surface of a conductor surrounded by a dielectric with relative permittivity e x . Shown in Fig. 2.8 is a detail of the surface. Eqs. (2.23), (2.59), (2.58), and (2.50), and noting that directed from the dielectric outward; in Eq. (2.58), outward], we fi is =— Combining Eq. (2.23), Ad is directed from the conductor fid fi [in obtain ii dielectric pps = fi d • P = -(£ - £o)fi E = • — - (2.61) ps. £r conductor fid Figure 2.8 Detail of a conductor-linear dielectric Although the free surface charge density, p s is actually localized on the conductor boundary surface and the bound surface charge density, p ps is localized on the dielectric side of the surface, they can be treated as a single sheet of charge , side of the , with the total density surface. Pstot Example 2.5 A Dielectric — Ps + Pps — (2.62) Sphere with Free Volume Charge homogeneous dielectric sphere, of radius a and relative permittivity e r is situated in There is a free volume charge density p(r) = por/a (0 < r < a) throughout the sphere volume, where r is the distance from the sphere center (spherical radial coordinate) and po is a constant. Determine (a) the electric scalar potential for 0 < r < oo and (b) the bound air. charge distribution of the sphere. , Section 2.8 and Homogeneous Media Electrostatic Field in Linear, Isotropic, Solution (a) Because of spherical symmetry of the problem, the electric flux density vector, D, is purely radial and depends only on r. From the generalized Gauss’ law [Eq. (2.44)], Example applied in a similar fashion to that in symmetry (see also Example D(r) 2 Por /(4a) = Poa /(4r electric field intensity vector = E(r) The potential at a distance r V(r) ) [also see V(r) = a for r > a D(r)/(e T eo ) for r D(r)/eo r>a for r°° 1 is — 1 Dir f (2.64) hence: for r ’ > (2.65) a. 4e 0 r ) dr* + Ppa = V{a) 1 (2.60), the r - — £r , 1 for r . . P(r) and < (2.66) a. inside the sphere amounts - 1V --P0(£r = to (2.67) sra St (2.23), (2.59), (2.47), 3 3e r bound volume charge density PpW = , 1 + 4e 0 Jr, =r According to Eq. Using Eqs. given by is a Poa = d/ £>(/-') / < magnitude its Eq. (1.142)] ^•r^O (b) < same form, and £0 Jr'=r and for r from the sphere center — = of the is for spherical D is found to be (2.63) 2 3 The accommodated 1.19 magnitude of 1.18), the (2.63), the bound surface charge density on the sphere surface comes out to be Pps D = P(a , ) = sr - 1 D(a = ) po(e r Sketched in Model of a Fig. 2.9(a) is a 1 )a . ( 4e r £r Example 2.6 - 2 68 ) . pn junction pn junction between two semiconducting half-spaces, doped p-type and «-type, respectively. The volume charge distribution in the semiconductor can be approximated by the following function: p{x) - —po for x < 0 for x =0 -*/ 0 for x po e > 0 (2.69) , 0 where po and a are positive constants. The permittivity of the semiconductor is s. Find (a) the electric field intensity vector in the semiconductor, (b) the electric scalar potential in the semiconductor, and (c) the voltage between the ends of the semiconductor, from the end on the n-type side to the end on the p-type side of the junction. Solution (a) This is a problem with planar symmetry (see Examples electric field intensity vector in the The semiconductor is differential generalized Gauss’ law, Eq. (2.56), d Ex (x) p(x) dx e 1.20, 1.21, given by and 1.23), E = Ex (x)x and the [Eq. (1.148)]. becomes (2.70) 77 > 78 Chapter 2 and Dielectrics, Capacitance, 1 Energy Electric Figure 2.9 Model of a pn volume charge junction: (a) density, (b) electric field intensity, and potential; for This to x is (c) electric scalar Example 2.6. a first-order differential equation in [as in x and we solve , it by integrating with respect Eq. (1.178)]: X EAx) where C x -> — oo. is = p(x')dx’ + C, -f £ Jx'=-oo the constant of integration, which represents the field We meaning Ex in the plane note that p(x') field (2.71) that the total charge of the cLc' = 0, semiconductor can exist far from the junction [see also Eq. Ex (x — =poo) ( 1 = . is (2.72) zero. This means, in turn, that no 54)], 0, (2.73) e Section 2.9 and hence C= 0. Dielectric-Dielectric Substituting Eq. (2.69), the integration for the field points in the p-type region of the semiconductor yields Ex {x) = — — In the «-type region, Ex (x) = Fig. 2.9(b) field is — we have x '/ a dx' x/a — <x < (— oo (2.74) 0). & up to break the integration e^dx' + J shows the Q f J — oo £ e~x J = dx'^ Ex (x) electric field intensity into e~ in the two parts: x la (0 <x < semiconductor. oo). We (2.75) see that the oriented from the n-type doped region to the p-type doped region. This field is pn junction, as it exists in the junction even when an external voltage is not applied (e.g., when the terminals of a pn diode are not connected to an external voltage source). In the equilibrium state established after the pn junction is called the built-in field of a formed, the built-in right (b) field prevents further diffusion of positive charges (holes) to the and negative charges (electrons) to the across the junction. 4 Let us arbitrarily adopt our reference point for potential (7Z) at the center of the junction, x = 0, so that the potential at points in the p- type region is [Eq. (1.74)] pH V(x) = /»0 E / dl = JP qX /ci _ ( In the n-type region, we n V{x) ^ e ) _oo < x < , - r0 y/fl e / dx' Jx (2.76) 0. nX /»P E • dl =— / E • dl Jn Jp The i n n reverse the direction of integration for convenience, = _ Ex (x') dx' = — / Jx'=x Po a _ (c) left Po« /" £ Jo e -*7a fat =— Ex (x') dx' / Jo _ _ e -x/a\ o <x < oo. distribution of the potential, V(x), along the semiconductor is shown in Fig. 2.9(c). The voltage between the n-type and p-type ends of the semiconductor turns out V{x -> oo) (2.77) £ — V{x -» — oo) = . to be (2.78) e This voltage is called the built-in voltage of a pn junction (diode). Problems 2.12-2.15; : 2.9 So far, MATLAB Exercises (on Companion Website). DIELECTRIC-DIELECTRIC we have considered boundary BOUNDARY CONDITIONS surfaces conductor-free space (Fig. 1.39), and conductor-dielectric (Fig. 2.8), and analyzed the fields close to surfaces and surface charge densities on the surfaces [Eqs. (1.186), (1.189), (2.23), and (2.58)]. Let us now consider a dielectric-dielectric boundary dielectric-free space (Fig. 2.4), 4 Note that initially there are excess holes to the left of the plane x = 0 (p-type semiconductor is positively its entire volume) and excess electrons to the right (n-type semiconductor is negatively charged by doping). In a brief transient, a diffusion of holes occurs from the p-region into the n-region and electrons diffuse in the opposite direction, until an electric field is built up in such a direction that charged over the diffusion current drops to zero. Boundary Conditions 79 2 I 80 Chapter 2 and Dielectrics, Capacitance, Electric Energy Figure 2.10 Dielectric-dielectric boundary surface: deriving boundary conditions for (b) (a) tangential components of E and normal components of D. surface, shown in Fig. near the surface. 2.10, We and derive the boundary conditions for field components apply the same technique as in deriving the corresponding boundary conditions for the conductor-free space case, Eqs. (1.186) and (1.189). The main difference is that now the field exists at both sides of the boundary. Let Ei and Dj be, respectively, the electric field intensity vector and electric flux density vector close to the boundary in medium 1, whereas E 2 and D 2 stand for the same medium quantities in 2. Applying Eq. (1.75) to a narrow rectangular elementary contour C, we Fig. 2.10(a), obtain continuity of the tangential i component of E E • dl = Eu A — E A — This boundary condition on the two On / tells 0 / 2x us that the tangential sides of the boundary, i.e., that E t is = E2t E\t (2.79) . components of E are the same continuous across the boundary. the other hand, an application of Eq. (2.43) to a pillbox Gaussian elemen- tary surface, Fig. 2.10(b), gives l D • dS = D\ n AS — D 2n A S — (we employ vector ponents of to region 1, D ps AS D 2n — D\n (2.80) ps D to avoid dealing with bound charges), where the normal com- are defined with respect to the unit normal n directed from region 2 and ps is the free surface charge density that may exist on the surface. In the absence of charge, continuity of the normal component of D, charge-free surface ^ln — D 2n (Ps — (2.81) 0). D be the same boundary with no free charge on it. In other words, D n is continuous across the boundary free of charge. Relationships in Eqs. (2.79) and (2.80) represent two primary boundary conditions for the electrostatic field at the interface between two arbitrary media. When the dielectrics 1 and 2 are linear, D) = £]Ei and D 2 = e 2 E 2 so we obtain an additional pair of (secondary) boundary conditions - for the tangential components of D and normal components of E. From Eq. (2.79), the boundary condition for D reads This boundary condition enforces that the normal components of on the two sides of a , t D it D 2l (2.82) £\ £ Section 2.9 Dielectric-Dielectric 81 Boundary Conditions Figure 2.11 Refraction of electric field lines at a dielectric-dielectric interface. and we see that D is discontinuous across the boundary. Similarly, Eq. the boundary condition for E n (if p s = 0 at the interface): (2.81) yields t eiE ln = e 2 E 2n (2.83) , which shows that E n is also discontinuous across the boundary. Note that the boundary conditions for E and D, that is, for Eqs. (2.79) and (2.80) can be written in vector form: n x Ei — n x n Di E2 = - n D2 = E and D n t 0, (ii ps directed from region 2 to region , in (2.84) boundary condition for E (2.85) boundary condition for Dn 1), , t which is often more convenient for use in analyzing complex structures. Let us consider again the interface between two dielectric media and the angles a\ and a 2 that field lines in region 1 and region 2 make with the normal to the interface, as depicted in Fig. 2.11. The tangents of these angles can be expressed as tanai Eu = —— tan a 2 E\n We divide the E2 = —— E2n 1 and (2.86) . tangents and use Eqs. (2.79) and (2.83) to get tanai e\ tan a 2 s2 (2.87) This is law of refraction of electric field lines the law of refraction of the electric field lines at a dielectric-dielectric bound- is free of charge (p s = 0). Bending of field lines unequal bound charges on the two sides of the boundary. ary that Finally, let us find the distribution of From Eq. dielectric interface (Fig. 2.10). bound is, essentially, a result of on a dielectricbound charge densities that surface charges (2.23), the accumulate on the two sides of the interface are Ppsi = nd i • Pi and p ps2 = n d2 P 2 (2.88) , = — n and n d2 = where n d i n [n d in Eq. (2.23) is directed from the dielectric outward]. Hence, by adding p ps i and p ps2 together, the total bound surface charge density at the interface is given by p ps This is = n P2 — n • • (2.89) Pi. the boundary condition for the normal components of vector P at a dielectric- dielectric boundary. Problems 2.16-2.18; Conceptual Questions (on Companion Website): : MATLAB Exercises (on Companion Website). 2.7 and 2.8; boundary condition for Pn 82 Chapter 2 Dielectrics, Capacitance, and Electric Energy AND LAPLACE POISSON'S 2.10 EQUATIONS S Consider the electric field intensity vector, E, and scalar potential, V, in a homogeneous dielectric region of permittivity e. The spatial derivatives of E at a point in the region are related to the free point by Eq. (2.56), which volume charge may density, p, that exist at that a first-order differential equation. E, in turn, is V to the spatial derivatives of through Eq. (1.101). It is related is obvious, then, that the second-order spatial derivatives of V are related to p by differential equation. This equation is Poisson’s equation, which means of is a second-order easily derived by substituting Eq. (1.101) into Eq. (2.56): V.(VV0 = --. (2.90) £ The double-V operation equation in Poisson’s performed by evaluating is the first gradient of V, and then the divergence of the result. In the Cartesian coordi- nate system, gradient and divergence are calculated using Eqs. (1.102) and (1.167), and we have respectively, , = V (VL) • 9 - fdv\ div „ = V- |^— grad E) fdv\ a dv dv A x+— y+ — zj fsv JTr ( fdv\ 9 d + „ 2 v 9? + d 2 v W + 2 v a?' d (Z9,) We note that the same result would have been obtained by applying formally the formula for the dot product in the Cartesian coordinate system, Eq. (1.164), to the dot product of two identical vectors (V) in (V V)V, where V is expressed as in Eq. (1.100). Hence, we can write • V (VV) = The operator V V becomes • is abbreviated V2 (V V)V. • (2.92) (“del squared”), and Poisson’s equation V2 V = Poisson's equation (2.93) £ where vv = , Lapiacian in Cartesian coordinates In a charge-free region ( p — d 2 v ^ + d known as Laplace’s equation. (2 ' 94) (2.95) o, The V 2 operator see that the Lapiacian operates on a scalar -p/e or 0). The expressions 2 v a?' d + a7 v2 f = is v 0), Laplace's equation which 2 (e.g., is called the Lapiacian. V), and the result is We another scalar (e.g., for V2 V in the cylindrical coordinate system spherical coordinate system (Fig. 1.26) can be obtained in the first taking the gradient of [Eqs. (1.170) V= Lapiacian in cylindrical coordinates V(r, (p, and V [Eqs. (1.105) and (Fig. 1.25) and the i.e., by same manner, and then the divergence (1.108)], (1.171)] of the result. In cylindrical coordinates, the Lapiacian of z ) thus comes out v, V = to be 19/ 9VA 7TrV17) 1 + ? d 2 V w 2 W d + V (2.96) 0 Section 2.10 Poisson's 83 and Laplace's Equations HISTORICAL ASIDE and magnetism. In 1813, he published a general- Simeon Denis Poisson (1781-1840), French mathwas a student of Laplace (17491827) and Lagrange (1736-1813), and successor ization of Laplace’s differential equation for the of Fourier (1768-1830) as a professor, at Ecole so inside a solid, in mechanics or for a nonzero He is best known for his work charge density, so inside a charge distribution, in ematician, Polytechnique, Paris. on probability (Poisson’s distribution), potential theory, valid for a nonzero and his con- mass density, electromagnetics. tributions to mathematics as applied to electricity Pierre Simon de Laplace (1749-1827), French mathematician and astronomer, was President of Academie des Sciences (French Academy). His work in mathematical astronomy was summed up in a monumental five-volume book on partial appeared. on the the- under Napoleon, and was made a marquis by Louis XVIII. V(r,6,(f)) in spherical coordinates —— V2 V = him, Laplace’s equation, also wrote a treatise of substances. Laplace was a minister and senator celes- mechanics ( Mecanique Celeste ), published between 1799 and 1825, in which the second-order V= He after ory of probability, and worked on specific heats tial and that of equation for potential that differential we now name r2 dr Example 2.7 / 3 1 + ; I r2 sin0 3 0 . sin 9 — 30 ) \ d 1 + r2 2 sin 6 3 2 V (2.97) 2 La placia n in spherical coordinates Application of a 1-D Poisson's Equation A free volume charge of a uniform density p exists in a homogeneous dielectric, of permittivbetween two flat metallic electrodes, as shown in Fig. 2.12. The electrodes are connected and the distance between them is d. Neglecting the fringing effects, find (a) the electric potential and (b) the electric field intensity vector in the dielectric. ity e, to a voltage Vo, Solution (a) is equivalent to assuming that the electrodes are infinitely which case the potential in the dielectric varies with the distance from the electrodes only. Let x be the normal distance from the left electrode (Fig. 2.12). Poisson’s equation, given by Eqs. (2.93) and (2.94), becomes Neglecting the fringing effects V(x) large, in d 2 V (x) dx 2 By integrating it twice, we —— (0 <x < (2.98) d). e get 2 V(x) =— 2s + C\x + C2 (2.99) , where C\ and Figure 2.12 Uniform results in volume charge between pd/(2s) — C2 are the constants of integration. The boundary condition F(0) = Vo C2 = Vo, and the condition on the other boundary, V(d) = 0, gives C\ = metallic plates: application Vo/d. Hence, of a 1-D Poisson's equation vW _£^ + v0 1 ( -£). ( 2 100 ) (fringing neglected); for . Example 2.7. 84 Chapter 2 Capacitance, and Electric Energy Dielectrics, Using Eqs. (1.101) and (b) E(x) (1.102), we obtain the electric dV(x) = -VF = = \p( x „ x b( dr Problems'. 2.19-2.22; Conceptual Questions (on 2.1 FINITE-DIFFERENCE 1 In many vector from the potential: d\ -2) + Vo] -d\ (2.101) X. Companion Website): 2.9. METHOD FOR NUMERICAL SOLUTION OF LAPLACE ically, field EQUATION S practical cases, Poisson’s or Laplace’s equation cannot be solved analyt- but only numerically. The most popular and perhaps the simplest numerical method for solving these equations (and other types of differential equations generally) is the finite-difference (FD) method. tives in the differential It consists of replacing the deriva- equation by their finite-difference approximations and solving the resulting algebraic equations. To illustrate this, coaxial cable with conductors of square cross section, and b cross-sectional dimensions of the cable be a (a consider an air-filled shown in Fig. 2.13(a). Let the < b). The cable is charged by Va and Vb, are known. V in the space between time-invariant charges, and the potentials of the conductors, Our goal is to determine the distribution of the potential the cable conductors. This is a two-dimensional electrostatic problem for which, according to Fig. 2.13(b), Laplace’s equation, Eqs. (2.95) V2 F = d 2 v dx We discretize the region of sides d [Fig. 2.13(b)], compute the and d by v ( 3y2 2 2 102 ) . between the conductors by introducing a grid with cells and employ the FD method in order to approximately potentials at the grid points (nodes). Obviously, the accuracy of the computation depends on the grid resolution, d, the 2 (2.94), is given more accurate i.e., the smaller the spacing of the grid, (but computationally slower) the solution. y d va 2 43 i a y Yb E0 X b (a) O O (b) (c) Figure 2.13 Finite-difference analysis of a coaxial cable of square cross section: (a) structure geometry, (b) nodes with unknowns, and (c) detail of the grid for approximating Laplace's equation in terms of finite discrete potential values as differences. , Section 2.1 . Finite-Difference 1 Method for 85 Numerical Solution of Laplace's Equation shows a detail of the grid in Fig. 2.13(b). At the node 1, the backward-difference approximation for the first partial derivative of V with respect to the x-coordinate turns out to be Fig. 2.13(c) dV dx V!-V2 „ (2.103) d ' 1 finite-difference (FD) ’ approximation of a derivative which, combined with the forward-difference approximation for the second partial derivative with respect to x, yields d 2 v dx /3V\ _9_ 2 dx \ dx ) 1 dV 1 d\dx 1 \ 1 ( dx 3 1 J V3 d \ -V V!-V2 1 d V2 + V3 - 2V! (2.104) d2 Analogously, d 2 v dy v4 + V5 - 2V! (2.105) d2 2 1 FD so that the approximation for Laplace’s differential equation at point 1 in Fig. 2.13(c) is V2 + V3 + V4 + F5 - 4Vi =0 2 VZ V Vi d = i (V2 + V3 + V4 + V5 ) FD approximation of Laplace's equation (2.106) aid The simplest technique to solve the above finite-difference equation with the of a computer is an iterative technique expressed as v[ When some k+1) = ( k) ^ [v2 +v {k) k= 0,1,... (2.107) iterative solution to the FD Laplace's equation of the nodes and 5 belong to one of the surfaces of conduc- 2, 3, 4, the potential at such nodes tors, {k) +v ] + is Va in all iteration steps equal to the respective For the initial solution, at the zeroth at all nodes between the conductors. By traversing the grid in a systematic manner, node by node, the average of the four neighboring potentials is computed at the (k 4 l)th step for each node and is used to replace the potential at that node, Eq. (2.107), and thus improve the solution from the kth step. This procedure is repeated until the changes of the solution (residuals) with respect to the previous iteration at all nodes are small enough, i.e., until a final given potential of the conductor (k — 0) iteration step, ( we can adopt or V' (0) Vt> ). =0 - set of values for the unknown potentials consistent with the criterion T/(&+l) V l is obtained, where Once <5^ T/(fc) V 1 < 8y (2.108) stands for the specified tolerance of the potential. is known, numerinodes can be obtained Eqs. (1.101) and (1.102) in the approximate solution for the potential distribution cal results for the electric field intensity vector, E, at the grid by approximating the gradient operator involved in terms of finite differences. For example, E at the node approximately as Ei =- VVIj dV =dx 1 in Fig. 2.13(c) is V2-V3 x 2d + computed T4-F5 2d y (2.109) FD approximation of the electric field vector ) 86 Chapter 2 Dielectrics, Capacitance, and Electric Energy (central-difference approximation). Additionally, the surface charge density, ps on the conducting surfaces can be found by means of the boundary condition in Eq. (1.190). For example, assuming that the node 4 in Fig. 2.13(c) belongs to the surface of the inner conductor in Fig. 2.13(a), ps at that point can be approximately , evaluated as Ps4 = £0 y ' E4 = £q Ey4 = dV -£Q so dy V 1 -V4 Va-Vi £() ( 2 110 ) . charge per unit length of each of the conductors, Eq. (1.31), can be found by numerically integrating ps that is, summing pS( along the individual conductor contours. Thus, the per-unit-length charge of the inner conductor is given by Finally, the total , Q' a r = , Na psi d Jc° (2.1 , where Na denotes the total number of nodes along the contour and similarly for the outer conductor. Problems: 2.23 and 2.24; 2.12 MATLAB 1 1 i=\ Exercises (on Ca of the conductor, Companion Website). DEFINITION OF THE CAPACITANCE OF A CAPACITOR shows a system consisting of two metallic bodies (electrodes) embedded in and charged with equal charges of opposite polarities, Q and —Q. This system is referred to as a capacitor. The principal property of a capacitor is its capability to store the charge. The potential difference between the electrode carrying Q and the one with — Q is called the capacitor voltage. The capacitor is linear if its dielectric is linear. The dielectric can be homogeneous or inhomogeneous. Fig. 2.14 a dielectric, In linear capacitors, the capacitor charge, capacitor voltage, V, Q oc Q, is linearly proportional to the To prove this, assume for the moment that the 2 Q. The surface charge density, ps will remain V. changed to the same way over the surfaces of electrodes, and capacitor charge distributed in i.e., is dielectric Figure 2.14 Capacitor. , its magnitude will Section 2.1 2 double everywhere. The electric dielectric also becomes twice its field intensity vector at an arbitrary point in the previous value; for homogeneous dielectrics this is obvious from Eq. (1.38) with eo replaced by e, while for inhomogeneous dielectrics we first note that the vector D doubles [e.g., generalized Gauss’ law, Eq. (2.43)], Ea D any point of the dielectric [Eq. (2.47)], we find the same for the vector E. Finally, because the capacitor voltage equals the line integral of E (along any path) between the electrodes [Eq. (1.90)], we conclude that the voltage doubles as well. Summarily, V doubles because Q is doubled, meaning that Q and V are linearly proportional to each other. This proportionality is customarily and then, since 87 Definition of the Capacitance of a Capacitor Q c —r— V ~Q at Figure 2.15 Circuit-theory representation of a capacitor. written as Q= CV, ( from which the constant C, termed the capacitance of the capacitor, is 2 112 ) . defined as ( 2 113 ) . capacitance of a capacitor (unit: The capacitance is F) a measure of the ability of a capacitor to hold the charge - per between the electrodes. It depends on the shape, size, and mutual position of the electrodes, and on the properties of the dielectric of the capacitor. For nonlinear capacitors, however, the capacitance depends also on the a volt of the applied voltage applied voltage, C= C(V), ( 2 114 ) . nonlinear capacitor and a notable example is a varactor diode. The capacitance of a capacitor is always (C > 0), and the unit is the farad (F). For two-conductor transmission lines (two-body systems with very long conductors of uniform cross section), we define the capacitance per unit length of the line, that is, the capacitance for one meter (unit length) of the structure divided by 1 m, positive C—C v- where C, /, and Q' are the Shown (i.e., it is in Fig. 2.15 is U i c — — Q_ The ( v’ total capacitance, length, the structure [see Eq. (1.31)]. theory, '-.p unit for C is 2 115 ) . capacitance p.u.l. of a transmission line (unit: F/m) and charge per unit length of F/m. the circuit-theory representation of a capacitor. In circuit assumed that the charge is stored only in the capacitors in a circuit on the capacitor electrodes), while the connecting conductors are considered as ideal short-circuiting elements with zero capacitance (and also with zero resistance and inductance, as we shall see in later chapters). Finally, let us introduce metallic body another related concept, the capacitance of an isolated situated in a linear dielectric. It is defined as ( 2 116 ) . ^isolated body where Q is the charge of the body and Violated body is its potential with respect to the reference point at infinity. Note that this definition can also be regarded as a special case of the definition in Eq. (2.113), for the capacitance of a two-body system, with the assumption that the second body, carrying —Q, is at infinity. capacitance of an isolated metallic body 88 Chapter 2 Dielectrics, Capacitance, and Electric Energy ANALYSIS OF CAPACITORS WITH HOMOGENEOUS 2.1 3 DIELECTRICS We now consider various examples of the dielectrics. The examples cover analysis of capacitors with homogeneous several characteristic types of capacitors and trans- mission lines of both theoretical and practical importance. The analysis of capacitors and transmission presented in the Example 2.8 lines with different types of inhomogeneous dielectrics will be next section. Spherical Capacitor Consider a spherical capacitor, which consists of two concentric spherical conductors, as in Fig. 2.16. Let the radius of the inner conductor be a and the inner radius of the shown outer conductor be b (b homogeneous Solution +Q). Due > a). Find the capacitance of the capacitor dielectric of permittivity Assume that the capacitor is if it is filled with a e. charged with a charge Q (the inner electrode carries to spherical symmetry, the electric field in the dielectric is radial and has the form given by Eq. (1.136). Applying the generalized Gauss’ law in Eq. (2.55) to the spherical surface S of radius r (a < r < b) positioned concentrically with the capacitor electrodes [see the flux computation in Eq. (1.138)], we E(r) The capacitor voltage is obtain = Q Ansr2 (a < < r b). (2.117) [Eq. (1.90)] (2.118) so that the capacitance comes out to be ^= — Q = Aneab C V b-a capacitance of a spherical capacitor Example 2.9 . (2.119) Capacitance of an Isolated Spherical Conductor Determine the expression for the capacitance of a metallic sphere of radius a situated Figure 2.16 Spherical capacitor with a for homogeneous Example 2.8. dielectric; in air. Analysis of Capacitors with Section 2.1 3 We Solution use the definition of the capacitance of an isolated body, Eq. (2.116). potential of the sphere if it carries a charge capacitance Dielectrics 89 The Q is (see Example 1.25) ^sphere K Its Homogeneous = (2.120) ~T~~ 4neoa hence is C= Q 4neoa. (2.121) ^sphere capacitance of an isolated metallic sphere in air Note that an isolated sphere can be regarded as the inner electrode of a spherical capacwhose outer electrode has an infinite radius, and that the same result for the sphere capacitance is obtained from Eq. (2.119) with b oo. itor Coaxial Cable Example 2.10 A coaxial cable a transmission line consisting of an inner cylindrical conductor (a wire) is and an outer hollow cylindrical (tubular) conductor, with the conductors being coaxial with Note that this structure is sometimes referred The inner conductor has a radius a, and the outer conductor has an inner radius b (b > a). The cable is filled with a homogeneous dielectric of permittivity e. Find the capacitance per unit length, C, of the cable. respect to each other, as depicted in Fig. 2.17. to as a cylindrical capacitor. Solution This is a problem with to the cable axis (Fig. 2.17). surface we S positioned cylindrical symmetry. By means where Q is = electric field in the dielectric coaxially with the cable obtain the field intensity, and from E(r) The O' (a < r < is radial with respect of the generalized Gauss’ law applied to a cylindrical it conductors (see Example 1.26), the voltage between the conductors, V= b) 2ner r b E(r) dr Ja = — b O' 2ne In - , (2.122) a the charge per unit length of the cable. Hence, the capacitance per unit length of the cable, Eq. (2.115), equals Q_ V _ 2ne (2.1 In (b/a)' 23) capacitance p.u.l. cable Figure 2.17 Coaxial cable (cylindrical capacitor) homogeneous Example 2.10. with a dielectric; for of a coaxial 1 . 90 Chapter 2 Dielectrics, Capacitance, and Electric Energy We electric field in — ’ note that, combining together Eqs. (2.122), V = E(r) a coaxial cable (2.124) r\n(b/a ) “^Gauss which Q is a very useful expression for the field intensity (in terms of the voltage) in a coaxial cable. + + + + + + + +i+ + + Example Parallel-Plate Capacitor 2.1 Consider a parallel-plate capacitor, which consists of two parallel metallic I* area S, charged with neous d, is Q and —Q. Let the space between the plates be dielectric of permittivity e, as shown in Fig. 2.18(a). Assume filled plates, each of with a homoge- that the plate separation, very small compared to the dimensions of the plates, so that the fringing effects can be (a) \ \ / Q S +++++++++++ neglected. Calculate the capacitance of the capacitor. Solution Under given assumptions, the electric field in the dielectric is uniform, and there no field outside the dielectric. Applying the generalized Gauss’ law to a rectangular closed surface that encloses the upper plate [Fig. 2.18(a)], we get [see the left-hand side of Eq. (1.155)] is -Q ES = — (2.125) , e (b) Figure 2.18 and hence E= Q/(eS). The voltage between the plates V= Parallel-plate capacitor with a homogeneous and er = for (2.126) dielectric (b) fringing field for 1; is (a) Example 2.1 and the capacitance of the capacitor 1 capacitance of a parallel-plate (2.127) capacitor, fringing neglected For arbitrary capacitor dimensions, there is a considerable fringing field extending far outside the capacitor and the field between the capacitor plates close to plate edges form (edge is Electrodes of an from Eq. Electric Forces Example 2.12 air-filled between the plates a = effects), as illustrated in Fig. 2.18(b) for e r larger than the value obtained is d, With 1. this, is not uni- the actual capacitance (2.127). on Capacitor Plates capacitor are parallel square plates of edge lengths where d <&a. The capacitor voltage is a. The distance V. Find the electric forces on electrodes. on the lower plate (Fig. 2.19), we subdivide it into and add up (integrate) the forces on individual patches that are due to the electric field of the other charged plate (the upper plate). Each charged patch can be considered as a point charge and the force dFe 2 on it can be calculated To Solution find the net electric force differentially small patches of surface areas dS, using Eq. (1.23), so that dFe 2 = / (2.128) Ps2dSEi. Js 2 — - d <?2 the surface charge density of the lower plate and Ej Figure 2.19 Evaluation of Here, p s 2 the electric force on the vector due to the charge of the upper plate. Since lower electrode of an and these quantities are is d is the electric field intensity <& a, the fringing effects can be neglected, air-filled parallel-plate capacitor; for Example 2.12. Ps2 = -^ S and E] = x 2s[)S ( S = a 2 ). (2.129) ii Analysis of Capacitors with Section 2.1 3 We infinite Ei equals a half of the total electric field intensity between the plates (the to the charge of the lower plate) and can also be obtained as the field of an sheet of charge, Eq. (1.64). Q is the charge of the capacitor, is due Q = CV = C is its capacitance. The force on s 0 SV the lower plate definition, this is opposite. The The 5 2d2 (2.131) (2.132) ‘ Fe 2 on Pa is „ 2 eoV' unit for pressure hence forces are attractive. Fe 2 the pressure of the force is the electric pressure. is 2 £o a V Fe2 = - The force on the upper electrode Note that, from Eq. (2.131), (2.130) ' d 2 By Dielectrics realize that other half and Homogeneous ing the total electric field intensity of the capacitor, We term it = N/m 2 By introduc- the surface of the plate. (pascal), E= where Pa . V/d, the electric pressure can be expressed as (2.133) This expression valid for is any conducting surface, with intensity in the dielectric (air) near the surface. electric pressure (unit: Pa) E standing for the local electric field The pressure acts from the conductor toward the dielectric. Microstrip Transmission Line Example 2.13 Consider the transmission resting on line shown in Fig. 2.20. It consists of a a dielectric substrate of permittivity s the substrate, and is and thickness h, conducting strip of width w, and a ground plane beneath called a microstrip line. Neglecting the fringing effects, determine the capacitance per unit length of this line. Without taking into account the fringing effects, the electric field of the line is in the dielectric below the strip only. The capacitance of the part of the with length / is, from Eq. (2.127), Solution uniform and localized line C = e~, (2.134) h and the capacitance per unit length of the line comes out to be (2.135) capacitance microstrip Note, however, that this expression Example is is accurate only for h <&w. Namely, as pointed out in 2.11, the actual electric field distribution in the structure with quite different from that in Fig. 2.20. In a later chapter (on of a fringing neglected an arbitrary ratio w/h field analysis of h p.u.i. line, transmission Figure 2.20 Microstrip transmission 2.13. line; for Example 91 92 Chapter 2 Dielectrics, Capacitance, and Electric Energy Figure 2.21 Strip transmission line; for Example 2.14. lines), we shall present accurate empirical formulas for the electrostatic analysis of microstrip lines for all values of Example 2.14 w/h. Strip Transmission Line The transmission line consisting of a strip conductor between two conducting planes (large plates) at the same potential, as depicted in Fig. 2.21, is called a strip line. Let the width of the strip be w and its distance from each of the planes be h. The permittivity of the dielectric is £. Assuming that h <5C w, find the capacitance per unit length of the line. Solution Neglecting the fringing fields (as h to a rectangular closed surface of length we have l <£w) and applying the generalized Gauss’ law (along the line) that encloses the strip (Fig. 2.21), [see the left-hand side of Eq. (1.149)] 2 O'! Ewl=—, (2.136) £ E is the electric field intensity between the line conductors and Q' the charge per unit The voltage between the conductors is V = Eh, from which the per-unitlength capacitance of the line amounts to where length of the capacitance line, p.u.l. line. (2.137) of a strip fringing neglected Accurate analysis for an arbitrary ratio w/h Example 2.15 will be presented in a later chapter. Thin Symmetrical Two-Wire Transmission Line shows a thin symmetrical two-wire transmission line in air. The charge per unit is Q and the radii of the wire conductors, a, are much smaller than the distance between the conductor axes, d. Under these circumstances, compute the capacitance Fig. 2.22 length of the line ' , per unit length of the Solution By line. superposition, the total electric field intensity vector in air E= Ej + E2, is given by (2.138) I Figure 2.22 Two-wire transmission line with d air; for Example 2.1 5. » a in Homogeneous Analysis of Capacitors with Section 2.1 3 Dielectrics where Ei and E 2 are the fields due to the individual charged conductors. Since d^> a, these fields can be evaluated independently from each other, so we use the expression for the field of an isolated charged wire conductor in air, Eq. (1.196). At a point the axes of conductors, Fig. 2.22, vectors Ei and intensity turns out to E2 M in the plane containing are collinear, and thus the resultant field be E= + £2 E\ (7.1 3° 5 where x stands for the coordinate defining the position of the point M. The voltage between the conductors is d -° V= f Eix Jx=a d ±-t Jk-\r ya % 27re 0 —x a d (d ) d—x Ja | 2^ V aA = — Q' of the line is made d - lnW - d-a 0' a 71 £q ln ( ~ ln zh) (2.140) In , Cl d-a of the approximate relation ^ d. The capacitance per unit length is C-IO V Let us potential -ir d In tceq where the use -~“ now due TZEO (2.141) ln (d/a) two-wire obtain the same result in a different way, using the expression for the to an isolated charged wire conductor in air, Eq. (1.197) for r > a, and the We adopt a point M 2 on the surface of the right conductor (Fig. 2.22) to be the reference point for potential (Vm = 0). The potential at a point Mi on the surface superposition principle. 2 of the other conductor (with respect to the reference point) VMl = Q! , d-a ln 2n£o + -Q! amounts to a , ln (2.142) , d 2 tzbq a —a with the two terms representing the potentials [Eq. (1.197)] due to the conductors with perunit-length charges Q' and — 0', respectively. The capacitance per unit length of the line is = C' which yields the same expression Example 2.16 Three as in O' ^ = — VV O' (2.143) , Eq. (2.141). Parallel Equidistant Wires Three parallel thin wire conductors are situated in air, as depicted in Fig. 2.23. The wire radii are a = 1 and the distance between the axes of conductors is d = 50 mm. Two wires are galvanically connected to each other. Determine the capacitance per unit length of this mm system. Solution nite length. is This The is a two-conductor transmission line, which first the isolated (free) wire (wire 1), we analyze as a capacitor of infi- conductor of the transmission line) while the other electrode (the other line conductor) con- electrode of the capacitor (the first of the two short-circuited wires (wires 2 and 3), which are at the same potential. Let the electrodes be charged by Q' and —Q' per unit of their length, respectively. Because of symsists metry, the charge —0' of the second electrode as indicated in Fig. 2.23. is distributed equally (Fig. 2.23), first between wires 2 and 3, We assume that there is no charge on the short-circuiting conductor. Let us proclaim the second electrode of the capacitor to be potential of the electrode, i.e., the potential at the point with respect to the reference point M 2 capacitance at a Mi on zero potential. The the surface of wire taken on the surface of wire 3 (Fm 2 = 1 0) p.u.l. line of a thin 93 7 94 Chapter 2 Dielectrics, Capacitance, and Electric Energy Figure 2.23 Transmission line consisting of an isolated (free) wire and two galvanically interconnected wires for is the in air; Example 2.16. sum of the corresponding potentials due to wires Eq. (1.197) for r > VMl = In 2neo The capacitance per - + a (a) 2.1 shown In + —(272 — = Q! and 3, respectively. By means of d = 3 Q' d In 47reo -. a (2.144) is 47T£q _ VMl a In 2neo unit length of the line (capacitor) - 9.48 pF/m. (2.145) 3 In (d/a) Thin Wire Conductor above a Ground Plane Consider a transmission ing plane, as d d Q'/2 2tteq C' Example 1, 2, a. line that consists of a thin in Fig. 2.24(a). The medium height of the wire axis with respect to the plane wire conductor and a grounded conduct- is air, is h, the wire is parallel to the plane, the and the wire radius is a (a <5C h). Find the capacitance per unit length of this line. Solution Assume that the charge per unit length of the line is Q', namely, that the wire is charged by Q' and the plane by — Q' per unit length. By image theory (Figs. 1.47 and 1.49), we can replace the conducting plane by a negative image of the charged wire and obtain the equivalent thin two-wire line in Fig. 2.24(b), with the distance between the conductor axes d = 2 h. Of course, the two systems are equivalent only in the upper half-space. The voltage between the conductors in the original system (wire-plane) equals a half of the voltage in the equivalent system (wire-wire). Hence, the capacitance per unit length of the wire-plane transmission line comes out to be (see Fig. 2.24) & (b) Figure 2.24 V (a) Thin wire conductor above a grounded conducting plane in air and (b) equivalent two-wire for Example 2.1 7. where C'c from Eq. is _ Q' Vc /2 = 2 — = V c 2 C e = 2jr£ ° (2.146) ln(2A /a)' the capacitance per unit length of the equivalent two-wire line, which is computed (2.141). line; Problems 2.25-2.41; Conceptual Questions (on Companion Website): 2.10-2.18; : MATLAB Exercises (on Companion Website). Section 2.14 Analysis of Capacitors with Inhomogeneous Dielectrics ANALYSIS OF CAPACITORS WITH INHOMOGENEOUS 2.14 DIELECTRICS In this section, we deal with systems (capacitors) containing dielectrics, and, specifically, mogeneity. The first class includes (abruptly or continuously system if air-filled. 5 ) inhomogeneous with two basic classes of systems with dielectric inho- systems in which the dielectric permittivity varies in the direction of the electric field lines of the same In the second class of systems, the dielectric permittivity varies normal to the electric field lines of the air-filled system. As we shall which the electric flux density vector, D, varies in the first class of dielectrics is the same as in the corresponding air-filled system, while in the second class of dielectrics this is true for the electric field intensity vector, E. Hence, the two classes of systems are referred to as D- and E-systems, respectively. To illustrate these concepts, consider the two parallel-plate capacitors with in the direction see, the way piece-wise tric in dielectrics shown in Fig. 2.25. The permittivities of dielecAssume that the fringing effects are negligible, and that the homogeneous pieces are s\ and £ 2 field in each dielectric piece is uniform. The capacitor in Fig. 2.25(a) belongs to the first class of systems with dielectric inhomogeneity - the dielectric permittivity (abruptly) changes in the direction of the field lines, perpendicularly to the capacitor plates, so it is a D- system. From the generalized Gauss’ law, Eq. (2.43), applied to a rectangular closed surface enclosing the plate charged with Q, with the right-hand side positioned in either one of the dielectrics, we have D\ = £>2 = D = Q (2.147) s' D = const, Fig. capacitor in 2.25(a) where S is the surface area of the plates. That D\ = £>2 is also obvious from the boundary condition for the normal components of D, Eq. (2.81), applied to the interface between the two dielectrics. The electric field intensities in the dielectrics are not the same, Ei — D — Q= — £1 £10 D — E2 = and Q = — The voltage of the capacitor comes out to be V = Eidi+E2 d2 = fS (2.149) \s 1 e2 J with d\ and d2 standing for thicknesses of the layers. The capacitance c_ Q = V (2.148) e2 S £2 is £1^2 S ' e 2 d\ + s\d2 In the capacitor in Fig. 2.25(b), the change of dielectric characteristics normal to the - (2.150) is in the an E-system (the second class of systems). By means of Eq. (1.90) applied to a straight path between the capacitor plates, positioned in either one of the dielectrics, direction field lines E\ 5 Inhomogeneous - this is E2 — E V 1 ' composed of a number of homogeneous pieces of different permittivities homogeneous dielectrics; continuously inhomogeneous dielectrics, on the other dielectrics are called piece-wise side, are characterized (2.151) by continuous spatial variations of permittivity. E= Fig. const, capacitor in 2.25(b) 95 96 Chapter 2 Dielectrics, Capacitance, and Electric Energy Figure 2.25 Capacitors with two-layer dielectrics representing a D-system (a) and £-system (a) (b). where is V\ + -Q Q V2 + is the separation between plates [E\ from the boundary condition D\ c\ d the capacitor voltage and also obtained for tangential = E2 components of E, Eq. (2.79), applied to the dielectric-dielectric interface]. The electric flux density is discontinuous across the interface, -Q Q V (b) c2 = e,V — — = e\E e2 D 2 = e 2 E= and d V (2.152) d Applying the generalized Gauss’ law to a rectangular surface positioned about the 0— Mb plate with the charge Q yields V + Q=D (a) 1 S1 + D 2 S2 = (eiSi + £ 2 ^2 ) — d (2.1 , 53) with S\ and S2 being surface areas of the parts of the plate interfacing the individual dielectric layers, and the capacitance is Q _ V~ The systems view. sufficient for Note Figure 2.26 Equivalent two capacitors and parallel (b). series (a) some is usually + e 2 S2 (2.154) d can be explained also from the circuit-theory point of much simpler than the full field-theory view, and is evaluations. Let us compute the capacitances of capacitors in Fig. 2.25 using the associated equivalent circuits. (b) circuits for in Fig. 2.25 Such a view £iSj that the interface between the dielectric layers of the capacitor in equipotential, and nothing will change in the system Fig. 2.25(a) is this surface, i.e., in insert a metallic foil between the layers. We if we metalize thus obtain two capac- itors in series, as indicated in the equivalent circuit in Fig. 2.26(a). The charges of both capacitors are the same, and therefore v= from which the equivalent capacitance of two capacitors in series Vi Q Q + v2 = ^- + ^- = Q L| L2 total capacitance of the 1 + (2.155) C2 system amounts to c=e = V 1 C[ C\C2 C\ + c2 (2.156) Section 2.14 With the use of Eq. S On allel, Inhomogeneous Dielectrics 97 (2.127), the capacitances of individual capacitors are Ci=ei— a\ same expression resulting in the Analysis of Capacitors with £2 S — a (2.157) , 2 C as for C2 = and Eq. (2.150). in the other hand, the system in Fig. 2.25(b) represents two capacitors in par- the equivalent circuit of which is shown The voltages in Fig. 2.26(b). of both capacitors are the same, so that Q= The QV + c + 02 = Qi total capacitance of the 2 system v= + c2 (Ci v. ) (2.158) is C=^ = Ci + C 2 (2.159) , equivalent capacitance of two capacitors in parallel where, Ci=ei-j C2 = and a which gives the same result as in A spherical capacitor is filled Two Concentric Dielectric Layers with two concentric dielectric layers. = 2. The radius of the inner electrode a is = The relative permittivity = 4, and that of the other layer 10 mm, the radius of the boundary surface of the layer near the inner electrode of the capacitor e T2 (2.160) -f, a Eq. (2.154). Spherical Capacitor with Example 2.18 S2 e2 is e r\ between the layers is b = 25 mm, and the inner radius of the outer electrode is c = 35 mm. If the voltage between the inner and outer electrode is V = 10 V, what are (a) the charge of the capacitor and (b) the total bound charge on the interface between the layers? Solution (a) This is a D-system, and the electric flux density vector D = D(r) where r and r are is of the form (2.161) r, the radial coordinate and the radial unit vector in the spherical coordi- nate system with the origin at the capacitor center, as shown in Fig. 2.27. Gauss’ law applied to a spherical surface placed in either the first The generalized or the second layer tells us that D(r) where = ® a 4nrz < r < (2.162) c, Q is the capacitor charge - to be determined. The electric field intensity vector the second one. The capacitor voltage = Ei is D/ei in the first layer and E2 = D /e 2 in given by is = [ Ei(r)dr+ f E2 (r)dr = Q_ 4n Ja Jb 1 . £l l 1 , U ( U bj *YI c) J (2.163) from which Q = 47T60 /b — a r \e r i ab + c — b\ 1 —r oc ) V=CVr= 53.7 pC, (2.164) e x2 where C = 5.37 pF is the capacitance of the capacitor. Note that the capacitor charge can also be found by computing C as the equivalent total capacitance of a series connection of two spherical capacitors with homogeneous dielectrics, Eq. (2.156), where, using Eq. (2.119), Ci = 4ns T i£oab b —a 7.4 pF and C2 = —= 4jr£ r2 £ohc 19.5 pF. (2.165) 98 Chapter 2 Dielectrics, Capacitance, and Electric Energy Figure 2.27 Spherical capacitor with two concentric homogeneous for (b) Example From dielectric layers; 2.1 8. Eqs. (2.59) and (2.47), the intensities of the polarization vector in the dielectric layers with respect to the positive (outward) radial direction (Fig. 2.27) are Pi ( r ) e = J±Z± D(r) P2 (r) = and e rl By means of Eq. (2.89), the between the layers turns out Pps = n P2 - and hence the Qp = total Pi D(r). (2.1 66) bound surface charge density on the boundary surface Sb to be = -P2 (b + + Pi(b~) = ) ( V £ £ -^~ - -^l) D(b), e rl e r2 (2.1 67) (2.1 68) / bound charge on the surface PpsSb = P P s4nb 2 = - ( E 2 J -Zl) Q = Spherical Capacitor with a Continuously Example 2.19 A n ^ £ r2 13.43 pC. Inhomogeneous Dielectric is filled with a continuously inhomogeneous dielectric, whose perdepends on the distance r from the capacitor center and is given by the function e(r) = 3eo b/r, a < r < b, where a and b are radii of the inner and outer capacitor electrode, respectively. The outer electrode is grounded and the potential of the inner electrode is V. Find (a) the capacitance of the capacitor and (b) the bound charge distribution of the spherical capacitor mittivity dielectric, (c) By the dielectric is integrating the charge densities in (b), prove that the total bound charge of zero. Solution (a) Let us subdivide the dielectric into thin spherical concentric layers of radii nesses d r, a < r < b, as shown in Fig. 2.28. Each r and thick- thin layer can be considered as being now obvious that this capacitor represents a genwhich has only two such layers. Therefore, the electric flux density vector in the two capacitors is the same, given by Eqs. (2.161) and (2.162). Here, E = D/e(r), and the potential of the inner electrode with respect to the ground (outer electrode) and the capacitance of the capacitor can be obtained as homogeneous, of permittivity eralization of that in Fig. 2.27 h V= f E(r) dr = Q_ 4tt f Ja e(r). It is , dr r 2 e(r) Specifically, for the given function e(r), C Q = 47T v C = Unepb/ In (b/a). r 1 b dr Mr) (2.169) b Section 2.14 Inhomogeneous Analysis of Capacitors with Dielectrics 99 Figure 2.28 Spherical capacitor with a continuously inhomogeneous dielectric of permittivity s(r), subdivided into differentially thin homogeneous for (b) The polarization vector in the dielectric P{r) - e(r) = is Example radial, with intensity [Eqs. (2.59) £p D(r) e 0 (3 b = and (2.170) bound volume charge density of Eqs. (2.19) and (1.171), the (2.47)] -r)V r 2 In (b/a) e(r) By means layers; 2.1 9. inside the amounts to dielectric (2.171) In (b/a) while, from Eq. (2.23), the bound surface charge densities ’ on the surfaces of the dielectric near the inner and outer capacitor electrodes are Ppsfl p a +,) P( — , £0(3 b _ — - a)V and 2 Ppsb '~ p!,D = P(b" * ) v 2s 0 V = ' a \n(b/a ) (2.172) b\n(b/a)' respectively. (c) To verify that the total bound charge f Qp = of the dielectric I dv 4Treo(3b volume of the — a)V In (b/a) In (b/a) the we write 2 Sb 5a — a)V 4neo(b is zero, 4na 2 +pps b 4n dr Tppsa Pp(r) a where dv is b + 8 nepbV = 0 (2.173) , In (b/a) thin layer of radius r in Fig. 2.28, while Sa and Sb are the areas of surfaces of inner and outer capacitor electrodes, respectively. Spherical Capacitor Half Filled with a Liquid Dielectric Example 2.20 A spherical capacitor has a liquid dielectric occupying a half of the space between the elecThe radii of electrodes are a and b (a < b ), and the permittivity of the Determine the capacitance of the capacitor. trodes. Solution Assume that the capacitor charged by Q, as indicated is E-system, and the electric field intensity vector, E, is entirely radial. dielectric in Fig. 2.29. This From is e. is an the boundary con- components of the electric field intensity vector, Eq. (2.79), its magnitude depends on the radial spherical coordinate, r, only, i.e., it is the same at all points of a sphere dition for tangential of radius r (a The < r < b), shown D, in Fig. 2.29. electric flux density vector is Di = e r eoE in the dielectric Applying the generalized Gauss’ law to the sphere of radius and r (Fig. 2.29), D 2 = eoE we in air. obtain Figure 2.29 Analysis of a spherical capacitor half filled with a liquid dielectric; for 1 1 Dilnr + D22nr = Q — -> (e r + 2 l)£oE(r)27rr = Q, (2.174) Example 2.20. . 100 Chapter 2 Dielectrics, Capacitance, and Electric Energy and hence ^ ' + 27t(e T l)£ 0 r 2 The capacitor voltage and capacitance then come out b v= We a E(rHr Jfa 2n(e T _ + note that this expression for b) \a l)e 0 C — 2n£ r £oab/(b — r _Q_ V ~ ~ and C2 = in Fig. 2.30(a) is ( 2 176 ) . 2jr(s r + 1 )e 0 ab b-a . represents the equivalent total capacitance of a parallel 2n£oab/(b — dielectrics, Eq. (2.159), where a). Coaxial Cable with a Continuously Example 2.21 Shown a) 2 175 ) to be connection of two hemispherical capacitors with homogeneous C\ ( Inhomogeneous Dielectric a cross section of a coaxial cable with a continuously inhomogeneous which is given by a function £(<p) = (3 + sin0)fo, e [0, 2n]. The voltage between the inner and outer conductor of the cable is V. Compute (a) the capacitance per unit length of the cable and (b) the distribution of free charges on the cable dielectric, the permittivity of conductors. Solution (a) Let us subdivide the dielectric into thin sectors (slices) defined by elemental azimuthal Each such sector can be considered as being homogeneous, of permittivity e(</>). This way it becomes obvious that dielectric inhomogeneities in the systems in Figs. 2.29 and 2.30 are of the same type. angles d0, as depicted in Fig. 2.30(b). Because of symmetry, the electric field intensity vector in the dielectric = and, based on the boundary condition in Eq. (2.79), Ei [Fig. 2.30(b)]. b ), i.e., it is vector is This means that E is <p) = = E3, E3 radial, is = E4, constant on a cylindrical surface of radius r (a a function of the radial cylindrical coordinate, D(r, E2, E2 r, only. The < ... r < electric flux density e(0)E(r). Generalized Gauss’ law applied to a cylinder of radius r and height (length) h gives r2n I — — = Q'h £(<p)E(r) rd<ph 70=0' v - D dS Qs — > E{r) = — q' ^ r J0 f , n £((/)) ( 2 177 ) . d<p is the area of an elemental strip of width rdcp and height h, and Q is the charge per unit length of the cable. The voltage between the cable conductors and the where dS Figure 2.30 Analysis of a coaxial cable with a continuously inhomogeneous dielectric of permittivity e(0): (a) cross-sectional structure and view of the (b) subdivision of the dielectric into differentially thin homogeneous Example 2.21 sectors; for 1 ' Section 2.14 Inhomogeneous Analysis of Capacitors with capacitance per unit length are then obtained as V= E{r)dr f = Q'\n(b/a) fo* (b) e(<f>) fo* s(cp) d(p d0 From Eqs. (2.177) and (2.178), the electric field intensity, E(r), filled coaxial cable, Eq. (2.124), which Eq. (2.58), the free surface (2.178) In (b/a)' is expected, because this is 6jt£q _ In (b/a) the is same as in the air- an E-system. Using charge density on the inner conductor of the cable = Psa((p) ,, NEV s(4>)E(a +v + ) = is + sin <j>)V eo(3 (2.179) , a In (b/a) while on the outer conductor, Psb(</>) ^ srvI.- s -e(4>)E(b ) = Two-Wire Line with Example 2.22 e 0 (3 = , Coated Conductors Each conductor of a thin symmetrical two-wire transmission the conductor axes d and conductor permittivity e and thickness a. (d^> ni9m (2.1 80) . b In (b/a) Dielectrically radii a — + sm<p)V a), is line, with the distance between coated by a coaxial dielectric layer of Find the expression for the capacitance per unit length of this line if situated in air. Solution Let the line Q per unit of be charged by its length, as shown in Fig. 2.31. Similarly to the analysis of a thin-wire line with bare conductors in Fig. 2.22, the electric fields due to the individual charged conductors can be evaluated independently from each other, because the line point is thin. Thus, the electric flux densities due to the conductors 1 and 2 evaluated at the M in Fig. 2.31 are D2 = and Q' 2n(d —x (2.181) ) The total electric flux density is D = D\ + D 2 To simplify the computation, however, we take that D & D\ in the first dielectric coating (the second conductor is far away) and D D 2 in the second coating. Having in mind that the electric field intensity in respectively. . dielectric coatings is E= D/e, while E = D/eo in air, the voltage between the conductors is given by nd—a V= / r2a j Didx-| = |:ri ln2+ \_s i(ln sq \ rd—2a / (£>i + D2 ) dx + e 0 J2a & Ja Ja 2n j Edx=- ln d rd—a D 2 dx / £ Jd—2 -la ^_ ^) 2a ^ - — 2a J + Q' ( i.„2l^( — 1 ( s J tc T 7 \£ \ , 1 „ In 2 , In ft —d 2a Eq (2.182) [see the integration in Eq. (2.140)]. C' = The capacitance per unit length of the line d Q! — = 7r[-ln2+ — — 1 . In eg is (2.183) 2a Figure 2.31 Cross section of a thin two-wire line with dielectric coatings conductors, over in air; for Example 2.22. Dielectrics 101 102 Chapter 2 Dielectrics, Capacitance, and Electric Energy Example 2.23 Spherical Capacitor with a Nonlinear Dielectric Consider a spherical capacitor with short-circuited, and there steady state (Fig. 2.32). where Po a constant is field intensity a and b (a < b), filled with a fer- is a The and remanent (residual) polarization polarization vector r is in the dielectric in the radial, with is magnitude P(r) = new Pob/r, the distance from the capacitor center. Calculate the electric vector in the dielectric. Since this Solution Figure 2.32 Nonlinear radii of electrodes After being connected to a voltage generator, the capacitor electrodes are roelectric. is a nonlinear capacitor (ferroelectrics are nonlinear materials), rela- tionship in Eq. (2.112) does not hold, and there them a charge, is Q new and —Q, on the capacitor state. From spherical sym- spherical capacitor with electrodes although the voltage between short-circuited electrodes metry and the given form of the vector P in the dielectric, the electric flux density vector in the dielectric, D, has that same form (radial vector that depends on r only), as in Eq. (2.161). This means that the capacitor in Fig. 2.32 represents a D-system. Then, using the generalized Gauss’ law, Eq. (2.43), which is true for arbitrary (including nonlinear) media, the magnitude of the vector D is found to be that in Eq. (2.162). To determine the electric field intensity vector, E, from D, however, we cannot use the relationship in Eq. (2.47), as the dielectric is not linear, but we can use the definition of the vector D in Eq. (2.41), which yields and remanent polarization; for Example 2.23. is zero in the D-P (2.184) £o From the condition V= 0, f E(r) dr = that — £0 J r=a f [D{r) — P(r)] dr = 0, (2.185) Ja is, P0 b In - = a Hence, Q= AtzPq In (b/a) ab 2 /(b — a) E= 0. (2.186) and ab Pob e0 r |_ ( In (b/a) b-a)r (2.187) Problems 2.42-2.56; Conceptual Questions (on Companion Website): 2.19-2.26; : MATLAB 2.15 Exercises (on Companion Website). ENERGY OF AN ELECTROSTATIC SYSTEM Every charged capacitor and every system of charged conducting bodies contain a amount of energy, which, by the principle of conservation of energy, equals the work done in the process of charging the system. This energy is called the electric energy and is related to the charges and potentials of the conducting bodies in the system. To find this relationship, we perform a numerical experiment in which a system that was initially uncharged is being gradually charged, by bringing elementary charges to the conducting bodies by an external agent. We evaluate the net work done against the electric forces while the charges of the bodies are being changed from zero to their final values. Consider first a linear capacitor of capacitance C. Let the charges of the electrodes of the capacitor be Q and —Q, and their potentials with respect to a reference point be V\ and V2 respectively. The capacitor voltage is certain , Q V=V\-V2 = ^. (2.188) ) Section 2.1 5 103 Energy of an Electrostatic System we add an elementary positive charge d Q to the first electrode and — dQ to the second one, the change of the energy of the capacitor is equal to the work done against the electric forces in moving d Q from the reference point (i.e., from the zero potential level) to the first electrode (which is at the potential Vi) and — d Q to the second electrode (at the potential V2 ). By the definition of the electric scalar potential, the potential of a point on either one of the electrodes equals the work done by the electric field in moving a charge from the electrode to the reference point, or, conversely, the work done against the electric field in moving a charge from the reference point to the electrode, divided by the charge [Eq. (1.74)]. This is exactly what we have in our experiment, so the elementary work done while moving If d Q and — dQ is given by dWe = dQ El + (- d Q) V2 (2.189) , or, equivalently, dWe = dQ(V -V2 = dQV = dQ^. When (2.190) ) 1 uncharged, the capacitor has no energy. The total energy stored when in the charged with Q is therefore equal to the net work done in changing the capacitor charge from q = 0 to q = Q, and is obtained by adding up all elementary works dWe in Eq. (2.190): capacitor it is W e ‘ C (2.191) 2 Hence, the equivalent expressions for the energy of a capacitor are We = The unit for the electric energy of an isolated charged metallic is ^ 2C l 2 the joule body is its C here is the capacitance (J). - CV2 (2.192) . energy of a capacitor (unit: J 2 As a special case, the electric energy in a linear dielectric is C ^isolated body’ 9 where QV ^ = (2.193) energy of an isolated metallic body of the single body, given by Eq. 2.116, Violated body and two additional potential with respect to the reference point at infinity, equivalent expressions, like in Eq. 2.192, can be written as well. Note that the energy of a capacitor can also be expressed in terms of the charges and potentials of individual electrodes as We = - QV = - Q(V 1 -V2 ) Generally, for a linear multibody system with — ~ (Q\Vi + Q 2 V2 ). (2.1 94) N charged conducting bodies, (2.195) energy of a multibody electrostatic Finally, tric if a system also includes a between the conductors, volume charge distributed throughout the dielec- in addition to surface charges over the conductors, the system d . 104 Chapter 2 Dielectrics, Capacitance, and Electric Energy total electric energy of the system is N 1 1 2 2 We energy of a multibody system with volume charge in the l P V dv, (2.196) dielectric where p is the charge density and V the electric scalar potential in the dielectric, and energy (p dv) V /2 of an elemental charge d Q = p d V is integrated throughout the volume v of the dielectric. ELECTRIC ENERGY DENSITY 2.16 Expression in Eq. (2.196) enables us to find the total energy associated with the As the charges are sources of the charge distribution of an electrostatic system. electric field, it turns out that the energy of the system can be expressed also in terms of the electric field intensity tion that the electric energy is (which can be in the dielectric throughout the system. This leads to an assumpand therefore actually localized in the electric field, air and a vacuum) between the conductors of an we electrostatic system. Quantitatively, as energy shall see, the concentration (density) of at specific locations in the dielectric is proportional to the local electric field intensity squared. Let us consider first a simple case of a homogeneous electric field in the dielectric of a parallel-plate capacitor, with plate areas 5, plate separation is compared small [Fig. 2.18(a)]. From d = e 1 - S 1 9 - e-(Ed) 2 2 d CV 92 = 2 = -1 eE29 Sd = 2 ( dielectric permittivity e Eqs. (2.192), (2.127), and (2.126), the energy of the capacitor W where and to the dimensions of the plates), 1 is 9 - eE2 v„ (2.197) 2 E is the electric field intensity in the dielectric and v — Sd is the volume of the i.e., the volume of the domain where the electric field exists. We define dielectric, the electric energy density as we = W = — e or, electric energy density by employing Eq. we = 1 — eE29 = 2 J/m 3 ) unit in the is (2.198) , (2.47), (unit: The 1 - sE29 2 v 1 - D — 2 ED = (2.199) 2e 2 J/m 3 (volume density). The energy of the capacitor can now be written form W =w e We shall now energy contained (2.200) e v. generalize this result to obtain a field-based expression for the in an arbitrary capacitor (Fig. 2.14) and an arbitrary electrostatic system. Let us subdivide the domain between the conducting bodies of a system into elementary flux tubes containing the lines of vector D, that start nate at surfaces of the bodies. cells of volume then cut each tube along dv, such that the interfaces pendicular to the field, We field lines. We its and termi- length into small between the neighboring cells are per- can metalize these interfaces, without changing the and get an array of small parallel-plate capacitors along the tube. The entire i.e., the domain with the electric space between the conducting bodies of the system, Section 2.16 field, is thus partitioned into volume parallel-plate capacitor with a between the plates. cells, and each homogeneous The energy contained in cell dielectric in all capacitors, that is, is Energy Density 105 represents an elementary and uniform electric field each capacitor is dWe = w e dv, where the energy density Electric (2.201) given by Eq. (2.199). Summing the energies contained dWe throughout the entire dielectric integrating the energy between the conducting bodies of the system (volume v), we obtain the electric energy of the entire system: l The expressions ED dv. (2.202) and (2.202) are equivaenergy of an electrostatic system. namely, they give the same result for the total The latter expression, however, implies that the actual localization of electric energy is throughout the electric field, that is, in the dielectric between the conductors of the system (even if the dielectric is a vacuum), and not in the conductors, nor on their surfaces. It also provides us with a means, the energy density in Eq. (2.199), for evaluating and analyzing the exact distribution of energy throughout the dielectric. On might the other hand, reconsidering Eq. (2.196) and reinspecting now come up its terms, we with an alternative viewpoint on the actual energy localization which implies that the stored electric energy of a system resides system charge, and not the field. Accordingly, pV / 2 might be considered to be the volume energy density at points where p # 0 throughout the volume of the dielectric [from the second term of the energy expression in Eq. (2.196)]. The in electrostatics, in the corresponding surface energy density, equal to ps V /2, would then quantify energy localization in the surface charge distribution over the surfaces of the conductors in Eq. (2.196)]. Both viewpoints have merit and are equally “correct.” Nevertheless, the assumption that the system [constituting the the energy is first term of the energy expression actually “contained” in the field, turns out to be much better and not in the in charge producing it, suited for energy considerations in the analysis of elec- tromagnetic waves (to be done in a later chapter). Namely, an electromagnetic wave consists of time-varying electric and magnetic fields which travel through space (even in a vacuum) and carry energy independent of the sources (charges and currents) that it is much produced them (the sources might not even simpler and more than to associate of time. This is it any more). Therefore, fields, which change in time and travel in space, with the stationary distribution of sources at previous instants why we have adopted in the first place exist natural to describe the energy distribution of a system with traveling waves in terms of the the field-based energy localization approach and the associated energy density expressions, those in Eq. (2.199), to quantify the energy distribution in an arbitrary (linear) system (with static or time-varying charges and Example 2.24 fields). Energy of Parallel-Plate Capacitors with Two Dielectric Layers Find the energy of each of the capacitors in Fig. 2.25, neglecting the fringing effects. Express the energy in terms of the capacitor charge in case (a) and the capacitor voltage in case (b). Solution Based on Eqs. energy of an electromagnetic system, via for the electric energy in Eqs. (2.196) lent, electric (2.202), (2.147), (2.192), (2.157), and (2.150), the energy of the capacitor in Fig. 2.25(a) can be found using either (1) the electric flux density in the dielectric, or (2) the capacitances of two capacitors corresponding to individual dielectric layers, or the energy density V 106 Chapter 2 Dielectrics, Capacitance, and Electric Energy (3) the equivalent total capacitance of a series connection of capacitors with homogeneous dielectrics: D — D —Sd 2s 2 W e = Q ~+~ 2 Sell + 2e\ = Q 2 2C2 2C\ 2 (1 ) ( 2 {e 2 d\ C +e\d2 )Q 2 (2.203) 2 e\e 2 S (3) 2) energy of the capacitor Similarly, the & 2 2 2.25(b) can be in Fig. computed by employing two capac- either (1) the electric field intensity in the dielectric, or (2) the capacitances of the itors with homogeneous dielectrics, or (3) the total capacitance of a parallel connection of capacitors: W £\ r2 o . = —z— S\d e s2 -| E—2 Sn 2 dj = QV — 2 0) ( [see Eqs. (2.151), (2.160), Determine + £ 2 S 2 )V~ (£[5] (2.204) 2) (3) (2.154)]. Energy Per Unit Length of a Coaxial Cable Example 2.25 conductor and CV 2 C2 V—2 1 energy density and (b) the energy per unit length of a coaxial cable, with and b (a < b) and dielectric permittivity £, if the voltage between the cable (a) the radii a conductors V. is Solution (a) The electric field intensity in the dielectric of the cable is given by Eq. (2.124), and hence the electric energy density at a distance r from the cable axis 1 we = - sE(r) = 2 electric The electric a transmission J/m) K= We (We)p.„.l. W e is \L / line (unit: where < (a r we (2.206) dv the energy contained in a part of the cable of length J/m. Because the energy density is W' = C energy via is, r and width b J r=a = /, S ix e w e (r) 2nrdr dr, 2 we adopt dS is in the Eq. (1.60), and obtain r h dr txeV 2 (2.207) 2 r In (b/a) Ja dS the cross-sectional is d SI. The unit for W' a function of the coordinate r only, form of an elementary ring of radius p.u.l. electric (2.205) b). — d SI area of the dielectric between the cable conductors, and dv This result < energy per unit length of the cable can be evaluated as energy per unit length of 2 r 2 2r 2 ln (b/a) 2 (b) ——eV is In (b/a)' of course, in agreement with the expression WL = Ic'v 2 p.u.l. (2.208) , capacitance where C' is the capacitance per unit length of the cable, given in Eq. (2.123). Example 2.26 Shown in Fig. 2.33 Energy is in a Coaxial Cable with a Dielectric Spacer a cross section of a coaxial cable that of relative permittivity e x . The dielectric is in is partly filled with a dielectric the form of a spacer between the cable conductors defined by an angle a. The remaining space between the conductors is air-filled. The conductor radii are a and b (a < b), and the voltage between the conductors is V. Find f 1 . Section 2.1 6 Electric dS2 Figure 2.33 Coaxial cable with a dielectric spacer; for Example 2.26. a for which the energy contained in the dielectric equals a half of the total energy in electric the cable. = Wei ^e r soE2 where the electric field intensity between the dielectric and air. in the air-filled coaxial cable, This is air part of the ^£o E w e2 = and respectively, By and Electric energy densities in the dielectric Solution cable interior are 2 (2.209) , E in the cable is continuous across the interface 2.21), and E is the same as an E-system (see Example Eq. (2.124). the requirement in the statement of the example, the energy per unit length of the cable contained in the dielectric spacer and that in the air are stipulated to be the same, W'tl = W' 2 which means that , f (Fig. 2.33) l-e T e 0 E(r) 2 ardr = J r=a 2 dSj ^s 0 E(r) 2 (2n - a)rdr f (2.210) J r =a 2 dS2 We 2 Wel Here, dSi and dS2 are surface areas of the parts of a thin ring of radius r and width dr determined by angles a and 2n — a, respectively. Even without any integration, Eq. (2.210) gives eT a 2 .7 =2n — a a er Example 2.27 dielectric Example (0 < r < and a) where the electric flux density, = / w e (r)47rr2 dr+ Jo where dv Eq. (1.33). is w eo = D, is D(r) — — (a < r < (2.212) oo), 2 £o given in Eq. (2.63). Hence, the electric is /• t energy densities in the electric 2 D(r) 2 energy of the system W ) air are 2e r £o respectively, . 2.5. The This system does not contain conductors. sphere and we = 2 211 Energy of a System with Volume Charges Calculate the electric energy of the system from Solution ( + adopted / OO Weo(r) Anr2 dr = Ja in the form of a thin spherical p 2 — 7V 56£ 0 1 \ £r / (7-1 \ ), shell of radius r (2.213) and thickness dr, Energy Density 107 108 Chapter 2 Dielectrics, Capacitance, and Electric Energy The above result can also Wc = lj P Vdv= be obtained employing the expression l -l" p ( r)VM W dr ^ = (7 in Eq. (2.196), + i) (2.21 4) , with v being the volume of the sphere, p the volume charge density given in Example V the electric scalar potential in Eq. (2.66). Note that the integration in Eq. (2.213) 2.5, and over both the dielectric sphere and the air, is because the electric field exists in the entire space, whereas the volume charge occupies the volume of the sphere only. while that Eq. (2.214) in is is over the sphere only. This Problems'. 2.57-2.68; Conceptual Questions (on MATLAB DIELECTRIC 2.17 Companion Website): 2.27-2.31; Companion Website). Exercises (on BREAKDOWN IN ELECTROSTATIC SYSTEMS We shall now analyze electrostatic systems in high-voltage applications, uations where the electric field in the dielectric of dielectric down breakdown occurs when in the system. As so strong that there is i.e., is in sit- a danger discussed in Section 2.6, dielectric break- the largest field intensity in the dielectric reaches the critical ECT Under value for that particular material - dielectric strength of the material, the influence of such strong electric fields, a dielectric material is . suddenly trans- formed from an insulator into a very good conductor, causing an intense current to flow. While systems with gaseous and liquid dielectrics can completely recover after a breakdown, those with solid dielectrics are most often permanently damaged by breakdown fields, as insulating properties of the dielectric are irreversibly degraded. In this section, our goal is not to analyze local breakdown processes in materials nor the overall behavior of systems resulting from these processes (these phenomena are largely nonlinear and are not electrostatic), but to evaluate the systems and their performance in linear electrostatic states close to the breakdown occurrences. In fact, our goal here is basically to determine the maximum extent of values of various quantities in a system (or device) that are “permitted” for its safe operation prior to an eventual breakdown. In systems (devices) with nonuniform electric field distributions, the principal task is to identify the most vulnerable spot for dielectric breakdown, and to relate the corresponding largest electric field intensity to the charges or potentials of conducting bodies transmission is called the line, in the system. In the case of a capacitor or a two-conductor the voltage that corresponds to the critical field in the dielectric breakdown voltage of the capacitor or voltage that can be applied to the system (before it line. This is the highest possible breaks down), and also referred to as the voltage rating of the system. 6 In some is sometimes applications, we the radius of the inner conductor optimize individual parameters of a system (e.g., 6 vary considerably depending on the actual con- Since the dielectric strength of a given material ditions under which the material is is may used, as well as the manufactured, the voltage rating and critical way the particular piece of a solid dielectric values of other quantities for a system in practice are always defined with a certain safety factor included. For example, with a safety factor of rating of a given capacitor equals a tenth of the voltage that conditions and assumptions. would lead to a 10, the voltage breakdown under ideal Section 2.1 7 of a coaxial cable) such that the breakdown voltage is Dielectric maximum. Breakdown It is in Electrostatic also of (breakdown) values for the capacitor charge and energy, and the corresponding values for forces on conductors (charge, energy, and forces per unit length for transmission lines), as these are the largest possible charge, energy, and forces for that particular system. For systems filled with a homogeneous dielectric, the strongest electric field in the dielectric is most frequently right next to one of the conducting bodies of the system. In addition, this is most likely in the vicinity of sharp parts of a conducting surface [see Eq. (1.210)]. In systems with heterogeneous dielectrics, on the other side, the most vulnerable spots can be close to a boundary surface between dielectric parts with different permittivities, where the normal component of the electric field interest to evaluate the critical intensity is discontinuous. In such systems, furthermore, the largest field intensity in a given dielectric part is not necessarily the breakdown field, because the dielectric strength vary from material to material, and only an electrostatic analysis (or an experiment) can tell which of the voltage of critical value is dielectric parts would break down first after a applied to the system. For a system of conducting bodies situated in air (or any other gas), the region, where the air is ionized by an avalanche breakdown mechanism (see Section 2.6) and behaves like a conducting material, is usually localized only in the immediate vicinity of “hot” spots on conductor surfaces, as the field farther away from the conductor is not sufficiently strong to sustain the breakdown. Due to an avalanche of impact ionization of air molecules, vast numbers of free electrons and positive ions are created, so that the air near the conductor becomes more and more conducting. The ionized air might even glow (at night), appearing as a luminous “crown” around the conductor, and hence such a local discharge close to a conductor surface is referred to as a corona discharge. A corona discharge on a conductor is equivalent to an enlargement of the conductor, since a layer of conducting material (ionized air) is added over the conductor surface. It substantially increases losses on power transmission lines and also emits electromagnetic waves that can interfere with nearby communication devices and systems. On some occasions, the field intensities in the system are so high that breakdowns occur even away from the conductors. A continuously ionized path is formed from a part of a conductor surface with an exceedingly high charge density to the nearest conductor with opposite charge polarity. This path is apparent as a bright luminous arc carrying a current of a very large intensity (typically hundreds to thousands of amperes, sometimes as large as 100 kA). As a result, a very rapid and violent discharge of the conductors occurs, known as arc discharge. The most apparent and spectacular arc discharges certainly are intense cloud-to-ground lightning discharges in the form of giant sparks in thunderstorms. Lightning strikes are, as we know, a frequent cause of loss of human lives and property. Analysis of electrostatic systems with fields close to breakdown levels and predicting critical values of voltages, charges, and other quantities at breakdown require no new theory, and is just a matter of applying what we already know in an appropriate way and order, having in mind the above general comments. The rest breakdown of this section consists therefore of a Example 2.28 Breakdown Consider the capacitor number of characteristic examples. in a Parallel-Plate in Fig. 2.19 for a = 1 m Capacitor and d = 2 cm, and determine its breakdown voltage and the corresponding charge and energy, as well as the electric pressure and forces on capacitor electrodes at breakdown. Systems 109 > 110 Chapter 2 Dielectrics, ' Capacitance, and Electric Energy The Solution breakdown between the capacitor plates electric field intensity is approximately uniform, and its is E = Ecro = 3 MV/m (2.215) Eq. (2.53)], from which the breakdown voltage of the capacitor, [dielectric strength of air, Eq. (2.126), charge of the capacitor, Eq. (2.125), and capacitor energy, Eq. (2.192), come out to be V = Ecrod = 2 Q = £ 0 « ^cr0 = 26.56 pC, 60 kV, respectively. According to Eqs. (2.133) and on electrodes at breakdown amount to pe Note = above values that the = ^eo^cro W and e X = - QV = 0.8 J, and (2.132), the electric pressure 39 84 Pa Fe =p e a 2 = 39.84 and (2.216) electric forces N. (2.217) energy and force are quite small compared to energies and for forces of nonelectrical origin, and these are the largest possible values for this (quite large) capacitor. Maximum Breakdown Example 2.29 Conductor Voltage of a Coaxial Cable = 7 mm = of a coaxial cable are a and 6 radii (a with a homogeneous dielectric of relative permittivity e r ECI = MV/m 20 maximum, (b) (polystyrene), (a) Find a for which the What is the maximum breakdown < b). The cable filled is 2.56 and dielectric strength breakdown voltage of the cable is voltage of the cable? Solution (a) Assume between the cable conductors that the voltage in the cable is given by Eq. (2.124). Obviously, the V. is The electric field intensity field is the strongest right next to the inner conductor of the cable. maximum electric field of V E(a + ) a (2.218) a \n(b/a) coaxial cable meaning that dielectric breakdown occurs when this field intensity reaches the critical value - dielectric strength, E(a + ) For a given radius withstand, is a, breakdown the = ECI voltage, (2.219) . the largest voltage that the cable can i.e., therefore VCT fa) = (2.220) Eci-aln -. a To find the optimal radius a, for VCT which the voltage is the largest, we impose the condition dVcr = — 0 (2 da The solution - = maximum breakdown a e = 2.718 — a opt » Because the second derivative of Vo-fa) for a Ecr (a) has indeed The maximum breakdown function (b) 221 ) is coaxial cable radii ratio for a voltage . a maximum = — a 0 pt is 2.57 ^cr(^opt) mm. negative (equals (and not a minimum) voltage of the cable (^cr)max = — = (2.222) -ECT /a opl ), the at that point. is — Ecr fl Q pt — 51.5 kV. (2.223) Section 2.1 7 Dielectric Breakdown in Electrostatic Systems 111 HISTORICAL ASIDE The rod lightning was Stream invented around the mid- electricity, and proposed that upright of the eighteenth sharp-pointed iron rods be mounted on roofs century by dle Benjamin of buildings and connected by conducting wires Franklin (1706-1790), an to the iron bars buried in the earth (grounding American statesman and electrodes). tor in the area of electricity principle of (e.g., conservation labeling of charge, and positive negative charges, investigation of lightning, etc.), as well as in other areas of scientific inquiry (e.g., One of the first lightning rods with grounding conductors based on his design was that installed in 1752 on the Pennsylvania State House (now Independence Hall), in Philadelphia. With several improvements by Franklin and others in decades to follow, grounded lightning rods soon proved to be an efficient way of protecting buildings from lightning damage. (Portrait: Library of inven- scientist, a prolific sses, chart, etc.). Franklin speculated that light- ning was numerous bifocal gla- Congress) Franklin stove, improved printing press, Gulf Breakdown Voltage Example 2.30 of a Thin Two-Wire Line Consider a symmetrical two-wire line with conductor radii a = 1 mm and the distance between conductor axes d = 0.5 m. The line is situated in air. Compute (a) the breakdown voltage of the line and (b) the largest possible forces on line conductors per unit length. Solution (a) Let Q' > 0 and The —Q be the charges per unit length of the line conductors, as in electric field intensity of the line of the conductors. At Fig. 2.22. the largest very close to the surface of each is these points, the total field (due to both charged conductors), Eq. (2.138), can approximately be evaluated as the field due to a single isolated charged wire conductor, because the other conductor is far away (d^> a). Thus, by means of Eq. (1.196), Q! (2.224) 2tteq a — The breakdown voltage of the vcr = line —C — — 0cr In where C is - = 2aEcr0 In - = 37.29 kV, (2.226) a computed and as in Eqs. (1.223) Oa = = 2tt £o d the largest possible force for this Example 2.31 A (2.225) the corresponding intensity of electric forces on line conductors per unit FL is nC/m. the capacitance per unit length of the line [Eq. (2.141)]. «d, length can be which 167 to a jreo (b) Since a cr amounts Q'a = 2neoaECXQ = Q' = = 5 mm, is 1 and the result is mN/m, (2.227) line. Grounded Wire Conductor wire conductor, of radius a earth, at a height h (1.224), electric field of thin two-wire line and, at breakdown, the critical charge per unit length of the line turns out to be Emax = Ecr0 = 3 MV/m maximum as a Lightning Arrester positioned horizontally above the surface of the 6 m, as shown in Fig. 2.34. The conductor is grounded. A uniform a . 112 Chapter 2 ' Dielectrics, Capacitance, and Electric Energy atmospheric electric field of intensity Eq, due to a large charged cloud, exists above the and directed upward. 7 Assuming that Eq is known conductor per unit of its length and (b) the electric field intensity on the surface of the conductor, (c) Consider a cloud discharge to the ground and discuss the protection that the wire conductor provides to the space below it. earth’s surface. (£o > The vector Eo vertical is induced 0), find (a) the charge in the h Solution •2 fi 0 :3 I 1 V=0 (a) The earth’s surface represents a conducting plane. Let the potential of the plane be zero. Since the wire conductor is connected to the earth, a charge of opposite polarity to that of is, of positive polarity, is induced in the conductor (pulled the lower part of the cloud, that out from the earth) under the influence of the atmospheric Figure 2.34 Grounded wire conductor in a uniform atmospheric electric field; for Example 2.31 charge per unit length of the conductor ( Q > ' of the conductor, V\, can be expressed as the 0), as sum first plane, as in Fig. 1.49, and thus using Eq. (1.119) with ri = 2 h (since The this potential r\ — its image in the as conducting a (distance of the point at computed form the axis of the original wire equals the wire radius) 2 h 3> a). The other component of the potential can be found from is Eo from the conductor to the earth's surface along the uniform, the voltage is simply — Eq times the length potential of the conductor is given by Eq. (1.90), by integrating the straight, vertical path. As of the path, and the total field this field is V = In X ——E the condition that the conductor Vi = is (2.228) 0 h. a lit Eq From Q designate component can be obtained the potential at the surface of the wire due to both Q' and and Let of the potential due to charge Q' and the potential produced by the external field Eo- This which the potential field. indicated in Fig. 2.34. grounded, we Q' 0 = now have 27t£oEoh (2.229) ln(2 h/a) (b) The electric field intensity on the surface of the wire conductor (point 1 in Fig. 2.34) is approximately [Eq. (2.224)] Q' E0 h 2neoa a ln(2 h/a) Ei (c) On the other hand, having in conductor Fig. 2.34) at a height amounts mind Eq. = 154£0 (2.230) . (2.139), the electric field intensity below the H — 2 m, for instance, with respect to the ground level (point 2 in to N £2 = + -i--^fi )+£o = 0.71£ 2h — r 2tieq \ r () (2.231) ) (E2 is directed upward), where r = h — H = 4 m. Similarly, the field intensity below the conductor close to the earth’s surface (point 3) turns out to be £3 = —Q'/(neoh) + £0 = 0.743£0 (r = h). As the atmospheric field becomes stronger and stronger, in a thunderstorm, the field intensity E\ reaches the breakdown value [Eq. E\ The air ionizes 7 = £cro- and becomes conductive, so cloud flows through this conducting air (2.225)] (2.232) that a portion of the negative charge of the channel to the wire conductor and down to the and cause eventual lightning discharges, there is a positive in its base. The bottom buildup induces another positively charged area at the earth's surface, and these two charge layers, like a giant charged capacitor, generate an electric field above the earth that is directed vertically upward, as in Fig. 2.34. In cold clouds, that charge buildup at the contain water and ice, top of a cloud and a negative one — 2 Section 2.1 7 ground. The E2 field intensities £2 = 0. 71 E„ =—£, = Dielectric and £3, however, are much lower than — — £3 = 0.743E, = and £, Breakdown in Electrostatic Systems 113 critical, = fe, (2.233) meaning that a structure or a person that may be underneath the conductor is safe from breakdown and cloud discharge (or lightning strike). The grounded conductor protects 8 therefore the space below it and serves principally as a lightning arrester. Breakdown Example 2.32 The in a Spherical Capacitor with dielectric of a spherical capacitor consists of two concentric Two Dielectrics layers. The relative permit- = 2.5 and its dielectric strength £cr = 50 MV/m. For the outer layer, e 2 = 5 and £ cr2 = 30 MV /m. Electrode radii are a = 3 cm and c — 8 cm, and the radius of the boundary surface between the layers is b = 5 cm. Calculate (a) the breakdown tivity of the inner layer e ri is i X voltage and (b) the corresponding energy of the capacitor. Solution (a) Let be the charge of the capacitor. The electric flux density in the dielectric Q by Eq. (2.162). With reference is given to Fig. 2.35, the electric field intensities in individual layers are —-— E\(r) = E2 (r) = —4ne - =£l( fl+ )-2’ 47re r ieo ^1 < a r < (2.234) b, Figure 2.35 Evaluation of = E2 - T 2£or 2 + (b )^j, b < r < (2.235) c, r spherical capacitor with a where r is the radial spherical coordinate. Note that E\{a + ) is the largest field inten+ sity in the inner layer, while E2 (b ) represents the strongest field in the outer layer. Combining Eqs. (2.234) and (2.235), we arrive to the following relationship between these field intensities: £ Tl a We do not know in voltage of critical value possibility of a 2 E 1 (a + advance which is breakdown ) = the same time, in the inner layer, from Eq. As E2 {b + ) < £cr2, we conclude Note that this is would break down first first after a check the which implies that = £cr i. (2.237) gri « 1 £r2 b 2 2 = 9 MV/m. (2.238) that the electric field intensity at lower than the critical assumption of a breakdown occurring 8 (2.236) ). (2.236), ) is E2 (b + dielectric layer E2 (b + = £cr dielectric layer 2 applied across the capacitor electrodes. Let us £i(a+) At e r2 b all points in the outer value for that dielectric. This means that the in the inner dielectric layer the principle of operation of lightning arresters, in general. is correct. Grounded lightning rods, placed as high as possible, protect exposed objects (buildings or other structures, as well as humans and animals) and their immediate surroundings from violent atmospheric electrical discharges by enforcing breakdown leading to a lightning strike is “induced” near the surface of the some other part of the object. Then, the rod routes the lightning discharge through an that an eventual dielectric rod, and not at insulated wire conductor to a grounding electrode, rather than through the object. In addition to simple vertical metallic rods with pointed ends, designs of lightning arresters include rods with spherical metallic tops or umbrella-like wire extensions, which have capacity for carrying charge. the breakdown voltage of a much larger amounts of induced two-layer dielectric; for Example 2.32. 114 Chapter 2 Dielectrics, Capacitance, and Electric Energy The other which assumes that an eventual breakdown occurs possibility, in the outer layer, gives E2 (b + = £cr2 ) and E\(a + ) larger than (2.239) , = Ecr2 £ 2 b /(e ia = 166.7 MV /m. This is impossible, as E\ cannot be Ecx under the assumption that the inner layer is in a normal regime, while 2 2 ) T t \ the outer layer breaks down. With the the capacitor field-intensity values in Eqs. (2.237) + E2 (b + + 2 Ei(a )a f Tcr = and (2.238), the breakdown voltage of is b ) = 2 r- Ja f rL Jh e T \a 2 Ecrl = ( \£ r ab 769 kV. e x2 bc J i (2.240) This voltage can also be obtained by considering the critical value of the capac- charge for breakdown. Let us denote by itor an eventual breakdown in Q and ??’ the charge in the case of the inner and outer dielectric layer, respectively. Based on Eqs. (2.234), (2.235), (2.237), and (2.239), = 4ne Qct As 0 becomes and larger T ie 0 a 2 ECTl larger, the the two charges in Eqs. (2.241). The 0® = 4ne and x2 eob breakdown occurs when critical 2 Ecr2 it (2.241) . reaches the smaller of charge of the capacitor is thus Q cr = min (0^,0®). = 1 For the given numerical data, Q^x < 0c? [0^ occurs in the inner dielectric layer. Hence, 0 cr = The corresponding voltage C= where The pF 16.28 largest possible is = 1 ], meaning that the 2.242) breakdown 12.52 fiC. (2.243) 769 kV, (2.244) is Tcr = (b) 0<J> ( 0.30c? | ^= j the capacitance of the capacitor [Eq. (2.164)]. energy of the capacitor W t = l is CVl = - 4.81 (2.245) J. j Problems 2.69-2.81; Conceptual Questions (on Companion Website): 2.32-2.34; : MATLAB Exercises (on Companion Website). Problems 2.1. Nonuniformly A lelepiped. lelepiped is polarized dielectric rectangular dielectric situated in air in the first the Cartesian coordinate system (z/c) paral- densities of octant of (x, y, z > edges, of lengths a, b to coordinate axes x, y, The is , and and z, c, parallel respectively. polarization vector in the parallelepiped given by POc, y, z) — Po[{x/a) x + (y/b) y + z], of the parallelepiped, (b) Show that the total bound charge of the parallelepiped 0), with one vertex at the coordinate origin, and the where Pq is a constant, (a) Find the volume and surface bound charge paral- 2.2. is zero. Uniformly polarized disk on a conducting plane. A uniformly polarized dielectric disk surrounded by as shown air is lying at a conducting plane, in Fig. 2.36. The polarization vector 115 Problems in the disk is P = Pi, the disk radius and is a, thickness d. Calculate the electric field intensity vector along the disk axis normal to the con- ducting plane (z-axis). Figure 2.38 Very thin dielectric disk with a nonuniform polarization; for Problem and free space The is 2.4. lying at a conducting plane. polarization vector is P, and it is normal to the plane, as depicted in Fig. 2.39. Find (a) bound surface charge density the Figure 2.36 Dielectric disk with a uniform polarization lying at a . and and (b) the electric field intensity vector at the center of conducting plane; for Problem 2.2. the 2.3 at the flat spherical surfaces of the hemisphere Uniformly polarized hollow dielectric cylinder. A hollow dielectric cylinder of radii a and b, and height 2h, is uniformly polarized and situ- flat that surface (point it is on the O in the figure), assuming dielectric side of the boundary surface (dielectric-conductor), very close to the surface. ated in free space. The polarization vector, of magnitude P, as shown is parallel to the cylinder axis, in Fig. 2.37. Find the electric field intensity vector at the center of the cylinder (point O). Figure 2.39 Dielectric hemisphere with a uniform polarization lying at a conducting plane; for Problem 2.6. 2.5. Nonuniformly polarized large dielectric slab. An infinitely large dielectric slab of thickness d = 2a, shown in Fig. 2.40, is polarized such that the polarization vector is P = Pqx2 x/a 2 where Figure 2.37 Hollow dielectric cylinder , with a uniform The medium outside the slab is air. Compute (a) the distribution of volume and surface bound charge of the slab, (b) the Po polarization; for Problem 2.3. a constant. . A thickness d (d <£ a), is polarized throughout shown in Fig. 2.38, the polarization vector (c) the voltage of the slab. its volume. In the cylindrical coordinate system everywhere, and between the boundary surfaces electric field intensity vector 2 4 Nonuniformly polarized thin dielectric disk. very thin dielectric disk, of radius a and . is 2.7. Electric flux density vector. Find the electric flux density vector, is P = Pori /a, where The medium around the disk D, (a) at the center of the defined by the expression polarized dielectric sphere in Fig. 2.7 and (b) P0 along the axis of the polarized dielectric disk is a constant. Determine (a) the distribution of bound charge and (b) the electric field intensity vector (z-axis) in Fig. 2.36. is air. along the z-axis. . hemisphere of radius a Total (free plus bound) volume charge density. 2 5 . Uniformly polarized dielectric hemisphere. dielectric 2.8. is A situated in The electric field dielectric, E, is a intensity vector in a known function of spatial coordinates, (a) Prove that the total (free plus 116 Chapter 2 Dielectrics, Capacitance, and Electric Energy a closed surface S situated entirely inside the body. 2.12. Total enclosed bound and free charge. Con- S sider an imaginary closed surface homogeneous Infinitely large dielectric slab charge enclosed by S is is homogeneous medium. Prove that homogeneous linear medium with no volume charge, there is no bound volume 2.13. Charge-free with a nonuniform in a polarization; for Problem The Q$. What the total bound charge Q p s enclosed by S? total free Figure 2.40 inside a dielectric of permittivity s. free 2.6. charge either. bound) volume charge density tric, £oV ptot = P + Pp, in the dielec- can be obtained as p t0 E. (b) Specifically, find p to t for • E 1 = given as the following function of Cartesian coor- dinates: (2 x 2.9, 2 -f z y Uniform E(jc, y, z) 3 ) z] = [4xyz V/m (x, y, z x in field in a dielectric. + (2jc 2 z — y3 ) y + m). There is a uniform The volume charge density is p. Determine the bound volume charge density, p p free . Closed surface in a uniform field. Consider- ing an arbitrary closed surface in a uniform = 0) region prove the following vector identity: (Fig. 2.41), electric field, in a charge-free (p tot fs dS volume charge. A very long homogeneous dielectric cylinder, of radius a and relative permittivity e T , is charged uniformly with free charge density p throughout its volume. The cylinder is surrounded by air. electric field in a certain dielectric region. 2.10. 2.14. Dielectric cylinder with free (a) Calculate the voltage and the surface of the bound charge between the axis cylinder, (b) Find the distribution of the cylinder. volume charge distribution. Repeat Example 2.6 but for a model of a pn junction given by the volume charge density p{x ) = pq{x/o) e~ |j:|/a where po and a are posi- 2.15. Linear-exponential , tive constants, as shown in Fig. 2.42. = 0. Figure 2.42 Model of a pn junction with a linear-exponential charge distribution; for Problem 2.15. boundary 2.16, Dielectric-dielectric conditions. Assume that the plane z = 0 separates medium 1 (z > 0) and medium 2 (z < 0), with relative permittivities e x \ Figure 2.41 Closed surface electric field in a region with a uniform and no volume charge; for Problem 2.10. The volume charge 2.11. Flux of the electric field intensity vector. polarization vector, P, and free density, p, are tric known at every point of a dielec- body. Find the expression for the flux of the electric field intensity vector, through The = 4 and e T2 = 2, respectively. electric field intensity vector in medium near the boundary (for z = 0 + ) is Ei = (4x — 2y + 5 z) V/m. Find the electric field 1 medium 2 near the boundno free charge exists on the boundary (ps = 0) and (b) there is a sur2 face charge of density p s = 53.12 pC/m on the intensity vector in ary (for z boundary. = 0_ E2 ), , if (a) 117 Problems boundary conditions. and conductor- 2.17. Conductor-dielectric Obtain boundary conditions, Eqs. (1.186), and (1.190), from Eqs. (2.84) and (2.85). free space (2.58), boundary. Sketch the field emerging from water (s r = 80) into air, 2.18. Water-air “incident” angle (in water) 2.19. 2.20. a is = 2 23 . . ductors, 2.43 total 2 . 24 . . = 5 cm, and assume that a nonuniform free volume charge whose denthe function p(r) given in Eq. (1.32), with 3 Po = 3 C/m exists between the electrodes, for a < r < b, where the permittivity is £q. In addi, tion, let the potential of the inner electrode (sphere) with respect to the outer one (shell) be Vo = 10 V. Under these circumstances, com- pute (a) the electric potential and (b) the electric field 2.22. computer program - direct solution. vector at an arbitrary point between number of same quantities as in the previous problem, and compare the results obtained by the two programs. 2 25 . . Capacitance of the earth. Find the capacitance of the earth assuming that it is a conducting sphere of radius R= 6378 km (grounds and waters are conducting media). 2 . 26 . Capacitance of a person. C that the capacitance ducting body is in It can be shown of an arbitrary con- between the capacitance of body and that of the sphere inscribed in the the sphere overscribed about the body, that is [Eq. where i? spheres. (2.121)], 47r£ 0 m n and R m ax j Based on R m in < C < 4^£ 0 R m ax, are the radii of the two that, estimate the capaci- the electrodes. tance of an average Application of Laplace’s equation in spherical sues are conducting media). coordinates. Repeat the previous problem but As plot the . enclosed by a shell) as the one in Fig. 1.41, N= which equals the number of unknown potentials], The system of equations, in which known potentials at nodes on the surface of conductors appear on the right-hand side of equations, should be solved by the Gaussian elimination method (or by matrix inversion). Compute and Vacuum concentric metallic electrodes (a solid sphere and nodes in Fig. 2.13(b) as unknowns [applying Eq. (2.106) to each interior grid node, we get coordinates. Consider a system of two spherical sity is = a/N 12, respectively. equations with the potentials at interior grid anode and b FD and an alternative to the iterative technique based on Eq. (2.107), write a computer program for the FD analysis of a square coaxial cable by directly solving the system of linear algebraic Ps2^ = 1cm and the tolerance of V. (b) Compute the a set of simultaneous equations the 2 20 electric charge per unit length of the inner and 2, 3, 5, 7, 9, 10, Application of Poisson’s equation in spherical with a = a/10 = 0.01 the outer conductor, taking d diode; for Problem 2.21. and the space between the con- and the surface charge density on spacing to be d V=V0 d V, 1 V. (a) Plot the results for the the potential 8y , Figure 2.43 (2.107). the surfaces of conductors, taking the grid 0 < x < d, be described as V(x) = Vo (x/d) where d is the distance between the electrodes. Find (a) the volume charge density in the diode, (b) the surface charge density on the cathode, (c) the surface charge density on the anode, and (d) the total charge of the diode. 0 — —1 on Eq. = 3 cm, Va = cm, b 1 distribution of the potential 4/ 3 cathode Vt, = that a field intensity in the shows a vacuum two flat electrodes, the cathode and the anode, and a charge distribution in a vacuum between them. Let the potential of the cathode be zero and the potential of the anode Vo (Vo > 0). The dis- Psl computer program - iterative solution. Write a computer program for the finite- and diode, which consists of V=0 b). FD Assume 45°. tribution of the potential in the diode can < r difference analysis of a coaxial cable of square inhomogeneous media. (a) Derive Poisson’s equation for an inhomogeneous medium, (b) Write Laplace’s equation for an inhomogeneous medium. diode. Fig. < cross section (Fig. 2.13) based the Poisson’s equation for Vacuum 0 (no volume charge) between the electrodes (for a lines if — for p(r) conductor-dielectric human body (human tis- , 118 2.27. Chapter 2 Capacitance, and Electric Energy Dielectrics, Capacitance of a metallic cube, computed by the MoM. Find the capacitance of the metallic cube numerically analyzed by the method of moments in Problem plates in Fig. 2.19 are at potentials V\ and metallic spheres, respectively: (a) the sphere inscribed in the cube, (b) the sphere over- radius mean the arithmetic is spheres in (a) and , whose . of the radii of (b), (d) the sphere having same surface as the cube, and (e) the sphere with the same volume as the cube. the 2.28. RG-55/U ial 6 coaxial A cable. RG-55/U cable has conductor radii a = 3 2.25). mm. The dielectric is = 0.5 mm Determine the capacitance per . . , - - coax- and polyethylene (s r V on the two plates are equal in magnitude and opposite in polarity. With this, the unknowns in the procedure are charge densities p s i, p s 2 ps /v on the upper plate only, and matching points, at which the potentials are computed, are centers of the same patches; however, these potentials are due to pairs of patches on both plates, with charge densities p s; and — p S( (/ = 1,2,..., N). Using the MoM program, find C fora = 1 m and the following d/a ratios: (i) 0.1 (ii) 0.5, (iii) 1, (iv) 2, and (v) 10, and compare the result with capacitances of the following scribed about the cube, (c) the sphere 1 V, respectively, as well as that the charge densities of pairs of corresponding patches that are right above/below each other and compare 1.85, = V2 = —1 = the results with the corresponding unit C , values obtained from Eq. (2.127), which neglects the length of the cable. fringing effects. 2.29. Capacitance FD analysis. p.u.l. of a square coaxial cable, Compute the capacitance per unit 2.32. line. Derive the length of the coaxial cable of square cross sec- expression for the capacitance per unit length by a finite-difference technique in Problems 2.23 and 2.24, and compare the result (using the grid spacing of d = of a nonsymmetrical thin two-wire transmis- tion numerically analyzed sion line in a (standard) coaxial cable (of circular cross section) having the radii ( b/a = 3) and same 2.33. conductor as the square cloud. A model of its a thunder- electrical properties are concerned, as a parallel-plate capacitor vertical separation d— = km 2 and km. Assume that the charge Q = 300 C and with horizontal plates of area S 15 conductors are Thick symmetrical two-wire line. Consider a symmetrical two-wire line with conductors of is — 1}. Take d/a to be 3, 5, 10, 20, and using this expression and and calculate s/[d/(2a)] 2 1 100, C upper plate has a total the lower plate an equal amount of negative the expression in Eq. (2.141), obtained for thin charge. Neglecting the fringing effects, find (a) two-wire lines. Compare the two sets of results and evaluate the error due to thin-wire approx- the capacitance of this capacitor, (b) the volt- age between the top and bottom of the cloud, imation of the line for individual distance to and radius ratios. | (c) the electric field intensity in the cloud. MoM numerical analysis of a parallel-plate 2.34. small metallic spheres in radii, a, and write a computer program based on the method of moments to evaluate its capacitance (C). In specific, subdivide each of the plates into N = 10 x 10 = 100 square patches, and assume constant charge densities on individual patches (as in Fig. 1.46). Additionally, assume that the upper and lower Two air. A capacitor consists of two small metallic spheres of equal capacitor. Consider the parallel-plate capacitor in Fig. 2.19, I not necessarily small compared to the distance between wire axes, d) in air. Using a version of image theory for line charges in the vicinity of conducting cylinders, it can be shown that the capacitance per unit length of this line is given by C' — ttsq/ \n{d/(2a) + thundercloud can be approximately represented, as far as radii of the b), arbitrary thickness (radius of wires, a, cable. 2.30. Parallel-plate capacitor The and the distance between the ^ axes of conductors is d (d ~S> a, b). ratio of dielectric (air) air. a and b (a a/ 10) with the per-unit-length capacitance of 2.31. Nonsymmetrical thin two-wire placed tance d (d » in air at a a). center-to-center dis- Find the capacitance of this capacitor. 2.35. Four parallel wires in air. shown The transmission line two pairs of gal- in Fig. 2.44 consists of vanically interconnected thin wire conductors, situated in air. The distance between the axes l £ 119 Problems of adjacent wires radii are a = 1 is d = 200 mm mm. Compute per unit length of the and the wire O the capacitance oo Qd Qb Qc Qa line. Cab Figure 2.46 C-cd Equivalent circuit for + Vcd + vab + the system Fig. 1/ 1 .41 in for ; Problem 2.38. and solving an equivalent circuit with three Repeat Problem 1.78 using the same circuit and the corresponding set of con- Figure 2.44 Cross section of a transmission line consisting of two capacitors, (b) pairs of short-circuited wires in air; for ditions. Problem 2.35. 2.40. 2.36. Two wires at the same A short-circuited two-wire large flat metallic foil, and a potential line is foil. parallel to a Equivalent circuit with parallel-plate capacitors. Repeat Example 1.28 by solving the equivshown in Fig. 2.47. alent circuit as portrayed in Fig. 2.45. = 1 mm, h = mm, and d = 30 mm, and the medium is air. (a) What is the capacitance per unit length of Q The geometry parameters are a 20 -Q\ Qi\ Q\ 2 l O on the II 4 II 3 c3 c2 c'1 + at the central -Q\ | [ whose one conductor is the two-wire line and the other conductor is the foil? (b) If the voltage between the line and the foil is V = 20 V, find the point —02 03 # a two-conductor transmission line induced surface charge density ' V Figure 2.47 Equivalent circuit for the system foil. Fig. 1 .42; for in Problem 2.40. 2.41. Equivalent circuit with cylindrical capacitors. Figure 2.45 Repeat Problem Short-circuited an equivalent by generating and solving 1.79 circuit with capacitors. horizontal two-wire line metallic foil; for layer Consider the system composed of a wire parallel to a corner screen in Fig. 1.57, and assume that a — 2 mm and h — 10 cm. Calculate the capacitance per unit length of this system (transmission 2.38. line). Equivalent circuit with two spherical capacitors. (a) Consider the system described in Example 1.27, and show that it can be replaced by the equivalent circuit given in Fig. 2.46. (b) What are the capacitances of the capaci- is the outer layer is oil ( r2 = 2.3). The geometri- — 8 cm, and c = connected to a source is — of voltage V 100 V. The source is then disconnected, and the oil is drained from the capacitor. Find the voltage between the elec- cal parameters are a 16 cm. = 2 cm, b The capacitor new trodes of the capacitor in the electrostatic state. without disconnecting the source. 2.43. Oil drain If the oil in problem is the capacitor from the previous drained without disconnecting the Obtain the voltage source, determine the flow of electricity expression for the potential at the center of the through the source circuit (that is, the difference in the charge of the capacitor) between tors in this schematic diagram? (c) structure in Fig. 1.41, Eq. (1.202), that Example 2.39. and assume that the inner dielectric made from mica (e r i = 5.4), whereas Fig. 2.27 Capacitance per unit length of a wire-corner line. Consider the spherical capacitor in dielectric. Problem 2.36. 2.37. capacitor with a solid and liquid 2.42. Spherical over a large 1.27, by solving is, repeat this circuit. Equivalent circuit with three spherical capacitors. (a) Repeat Problem 1.77 by generating the two electrostatic states. 2.44. Metallic sphere with dielectric metallic sphere of radius a = 1 coating. cm is A covered 120 Chapter 2 Dielectrics, Capacitance, and Electric Energy with a concentric dielectric layer of relative permittivity £ r = situated in air, tial — 4 and thickness b as shown a = cm and 2 The poten- in Fig. 2.48. of the sphere with respect to the reference point at infinity is V= 1 kV. Compute (a) the cable with two coaxial dielectric layers. The geometry parameters are a — 1 mm, b = 2 mm, and c — 4 mm. The dielectric parameters are £ r = 5 and e T 2 = 2. Calculate the capacitance i per unit length of the cable. capacitance of the sphere, (b) the free surface charge density on the sphere surface, (c) the bound volume charge density in the dielectric, and (d) the bound surface charge densities on dielectric interfaces. Figure 2.50 Cross section of a coaxial cable with Figure 2.48 Metallic sphere with a dielectric coating for two dielectric layers; for Problem 2.48. in air; Problem 2.44. 2.49. Coaxial cable with a radial variation of permit- 2.45. Charge densities in a half-filled spherical capac- Consider the half-filled spherical capacitor from Fig. 2.29, and assume that a — 2 cm, b — 10 cm, £ r = 3, and Q — 10 nC. Find the distributions of (a) free surface charges on metallic surfaces and (b) bound surface charges on itor. Empty and air-filled radii a = half-filled spherical capacitor. 3 cm and trostatic state An = 15 cm is connected to V = 15 kV. After an elec- is established, the source is dis- is then half filled with a liquid dielectric of relative permittivity 2.5. What is the new voltage between the A sphere half embedded metallic sphere of radius a given by the following function = <r < r/a (a = b ), where a and b the cable conductor radii. (a) the capacitance If and (b) the 5a are the potential difis V, find per unit length of the cable bound charge distribution of the dielectric. 2.50. Coaxial cable with four dielectric sectors. is in a dielectric. pressed into a up volume, as shown in medium in the upper half-space Fig. 2.49. The (a) Find its is air. the capacitance of the sphere, (b) If to a the sphere A with a piece-wise homogeneous dielectric composed of four 90° sectoral coaxial cable is filled parts with different permittivities, as Fig. 2.51. shown in Let the relative permittivities of the = 6, e r 2 = 10, the radii of the cable dielectric half-space of permittivity £ half of £ r (r) sectors be e r \ electrodes of the capacitor? 2.47. Metallic is of the radial distance r from the cable axis: b connected. The capacitor = permittivity spherical capacitor with conductor a source of voltage er Consider a coaxial cable with a continu- ously inhomogeneous dielectric, whose relative ference between the cable conductors dielectric surfaces. 2.46. tivity. 2, £ r 3 = 1, and conductors a = £ r4 2 = mm and b = 1 mm, and the potential of the outer conductor with respect to the inner conductor V = 25 V. Compute (a) the capacitance per unit length of the cable and (b) the free charge density at an arbitrary point on the surface of the inner conductor. charged with Q, determine what portion of this charge is located on the upper half of the sphere surface. is £q Q e Figure 2.49 Charged Figure 2.51 metallic sphere half Cross section embedded of a coaxial in a dielectric half-space; for Problem 2.47. cable with four 90° dielectric 2.48. Coaxial cable with ers. Fig. 2.50 shows two coaxial dielectric lay- a cross section of a coaxial sectors; for Problem 2.50. 121 Problems 2.51. Charge distribution for two coated Consider the two-wire ings in Fig. 2.31, of thickness d. and assume that a = 1 mm, 4, and V = 10 V. Under these permittivity of the dielec- = 2(1 + 3x/d)eo (0 < x < d). Neglecting fringing, calculate the capacitance charge and (b) bound charge in the system. of this capacitor. = 25 mm, £ r = ^-coordinate: e(x) Two metallic spheres with dielectric coating. If 2.55. Permittivity two identical metallic spheres with dielectric plates. gradient Assume dielectric are placed in air so that the distance between y-coordinate, e(y) is d = 1 y < m, determine the capaci- tance of such a capacitor. Two 5 mm itor. spheres half metallic dielectric. Two are half embedded in a metallic spheres of radii a = embedded 2.56. = 4, as air. Fig. capacitor 2.53 is a function of the = 2[1 + 3sin(7ry/b)]eo < Neglect fringing. shows a is layer. parallel-plate capacitor that half filled with a ferroelectric. tor (0 find the capacitance of the capac- part of the capacitor The distance between the centers is d = 30 cm. The upper medium to Capacitor with a nonlinear dielectric is shown of the spheres is and Fig. 2.54 in a dielectric half- space of relative permittivity e r in Fig. 2.52. b), in parallel that the permittivity of the coating as the one described in Problem 2.44 their centers 2.53. The given by the following function of the tric is circumstances, find the distribution of (a) free d 2.52. b and continuously inhomogeneous dielectric wires. line with dielectric coat- is air-filled. The other The capaci- charged by being connected to a voltage The source is then disconnected, and source. The spheres are charged with charges the capacitor electrodes are short-circuited. In between the spheres new electrostatic state, there is a remanent uniform polarization throughout the volume 200 V. For such a capacitor, find (a) the of the dielectric, with the polarization vector capacitance, (b) the distribution of free charges being normal to the capacitor plates and its magnitude being P. Determine the electric field intensity vector between the capacitor plates in (a) air and (b) dielectric. Fringing can be of equal magnitudes and opposite polarities. The is potential difference V= over conductors, and (c) distribution of the bound charges in the dielectric. d neglected. £o e Figure 2.54 Short-circuited Figure 2.52 Two charged metallic spheres parallel-plate pressed into a dielectric half-space; for capacitor containing Problem 2.53. 2.54. Permittivity a nonlinear dielectric gradient plates. Fig. 2.53 shows normal layer with a uniform to capacitor remanent a parallel-plate capaci- tor with rectangular plates of dimensions a polarization; for and Problem 2.56. 2.57. Energy of a spherical capacitor with two layers. For the spherical capacitor with two concentric dielectric layers from Example 2.18, compute the electric energy stored in each of the layers by (a) integrating the electric energy density Figure 2.53 over the volume of each layer and (b) rep- Parallel-plate resenting the capacitor as a series connection capacitor with a of continuously inhomogeneous dielectric; for Problem 2.54. two spherical capacitors with homogeneous dielectrics. 2.58. Change in energy of a spherical capacitor. For the spherical capacitor with solid and liquid 122 Chapter 2 Dielectrics, Capacitance, and Electric Energy from Problem dielectric layers electrodes from Example 1.28 two ways: (a) using the electric field intensities between electrodes (found in Example 1.28) and the corresponding energy densities and (b) using the equivalent circuit in Fig. 2.47 and the involved capacitances, respec- 2.42, find the parallel large energy of the capacitor between the electrostatic state with the capacitor connected change to the voltage source and the final electrostatic state. 2.59. Energy of a coated metallic sphere. For the charged metallic sphere with a dielectric coating from Problem 2.44, determine the radius b such that 1/2 of the total energy of the system is tively. 2.67. Energy of a coaxial cable with two coaxial layers. For the coaxial cable with two coaxial dielectric layers from Problem 2.48, assume that the voltage between the cable conductors is V == 100 V and find the per-unit-length electric energy contained in each of the layers by energy density over the cross section of each layer and (b) representing the cable as a series connection of (a) integrating the electric from Problem 2.68. Example elec- pn junction from in the assuming that the area of the 2.6, junction cross section perpendicular to the xaxis is S. For liq- surface. 2.69. Example energy stored is 2.20, calculate the in the liquid, if the 2.70. charge Q. Problem total electric stored y = what percentage of the energy of the capacitor, if charged, is the lower half of the dielectric, from in 0 toy = = 5 10 cm (a < b ). is It filled and 2.71. dielectric strength voltage of the capacitor is the ECT = Find a for which the breakis maximum, maximum breakdown the capacitor? (c) What is (b) voltage of the energy of the breakdown? Breakdown in a wire-plane transmission line. (a) Compute the breakdown voltage of the wire-plane transmission line assuming that a is the maximum — 1 cm and h 2.24(a), Fig. in = 1 m. (b) What energy per unit length of this Determine the largest possible elecforce on the wire conductor per unit of its line? (c) For the parallel-plate capacitor with from = (a) capacitor at Energy of a capacitor with an inhomogeneous a permittivity variation parallel to plates mittivity e r 25MV/m. down What Energy of a capacitor with a variable permittivConsider the parallel-plate capacitor with a permittivity variation normal to plates from Problem 2.54. Determine what percentage of the total electric energy of the capacitor, when charged, is contained in the first half of the dielectric, from x = 0 to x — d/2. voltage of a spherical spherical capacitor has electrodes with a homogeneous dielectric of relative per- sectors. ity- 2.55, find A of radii a and b Energy of a coaxial cable with four dielectric. Maximum breakdown capacitor. For the coaxial cable with four dielectric sectors from Problem 2.50, find the per-unit-length electric energy contained in each of the sectors. 2.65. energy contained half-filled spherical capacitor. Energy of a of the capacitor 2.64. 2.14. the spherical capacitor half filled with a electric 2.63. volume charge. Breakdown charge and energy of the earth. Determine the maximum possible charge and electric energy that could be stored on the earth, as described in Problem 2.25, and in the electric field around it, limited by an eventual dielectric breakdown of air near the earth’s uid dielectric from 2.62. free Energy of a pn junction. Calculate the tric two coaxial cylindrical capacitors with homogeneous dielectrics. 2.61. Energy of a system with Find the electric energy per unit length of the system with a volume free charge distribution stored in the coating. 2.60. flat in the following in tric length. 2.72. Grounded arrester. metallic sphere Repeat Example as 2.31 a lightning but for a grounded small metallic sphere (instead of the b/ 2. Energy of a system of spherical conductors. wire) in a uniform atmospheric electric field Calculate the energy of the system of three above the surface of the earth. Adopt the same radius and the same height of the conductor. spherical conductors from 2.66. Energy of a system of flat Problem 1.80. electrodes. Compute the electric energy stored in the system of five two dielectric layConsider the parallel-plate capacitor with 2.73. Parallel-plate capacitor with ers. 123 Problems two dielectric layers in Fig. 2.25(a), that = s r2 d\ — Ecr2 = 11 mm, d2 = 4 mm, and assume — 3, and 2.78. Ecr = 20 MV /m two concentric are MV/m, respectively. Find the break- \ , down voltage of the capacitor. Neglect fringing. 2 . 74 . dielectric layers in Fig. 2.27. Let the dielectric strengths of the inner and outer layer be Ecr \ and Ecr2 respectively. Find the relationship between the parameters of and layers Simultaneous breakdown in two spherical Consider the spherical capacitor with layers. er\ well as that the dielectric strengths 5, as the for 2 this capacitor two dielectric secRepeat the previous problem but for the parallel-plate capacitor with two dielectric sectors from Fig. 2.25(b) and d = d\ + d 2 Parallel-plate capacitor with ( a b, , e r \, e r 2 c, , Ecr \, and FC 2 ) r such that, for sufficiently large capacitor volt- tors. age, dielectric breakdown will occur in both dielectric layers simultaneously. - 2 . 75 . Breakdown in a spherical capacitor with two 2.79. (Fig. 2.50). tric layers state (outer layer The is air). 2.80. Metallic dielectric strengths for mica and oil are ECT \ = 200 15 /m, respectively. dielectric. MV /m and E 2 = . Breakdown potential of a sphere, (a) Determine the ct . coated metallic breakdown potential maximum potential of this sphere such that the from Problem 2.44 breakdown will not occur after it is removed from the liquid and raised high above maximum . A half of the metallic sphere with a dielectric coating the coating 2 77 sphere half immersed in a liquid metallic sphere of radius a — 2 cm immersed in a liquid dielectric, as in Fig. 2.49. The relative permittivity of the dielectric is £ r = 3 and its dielectric strength is E cx = 20 MV/m. The upper medium is air. Calculate is MV 2 . 76 lay- Repeat the previous problem but for a coaxial cable with two coaxial dielectric layers Find the breakdown voltage of the spherical capacitor with two concentric dielec- layers. from Problem 2.42 in (a) the first (outer layer is oil) and (b) the second state Simultaneous breakdown in two coaxial ers. if Ecr — the dielectric strength for 30 energy of Breakdown MV /m. (b) What is the the interface. this structure? in a coaxial cable with two layers, Find the breakdown voltage of the coax- (a) ial is cable with two coaxial dielectric layers from Problem 2.48. The dielectric strengths of the inner and outer layer are ECI 1 = 40 MV /m and ECT2 = the 20 MV /m, maximum per unit of its respectively, energy that length. dielectric (b) this cable Calculate can store 2.81. Breakdown in a coaxial cable with a dielectric spacer. Consider the coaxial cable with a dielectric 2 spacer in Fig. 2.33, and assume that a mm, b = 6 mm, a — 60°, and er = as that the dielectric strength for the spacer Ect = 200 = 5, as well is MV/m. Find (a) the breakdown volt- age of the cable and (b) the electric energy per unit length of the cable at breakdown. Steady Electric Currents Introduction: S o far, we have dealt with electrostatic fields, associated with time-invariant charges at rest. We now consider the charges in an organized macroscopic motion, which constitute an electric current. Our focus in this chapter is on the steady flow of free charges in conducting materials, i.e., on steady (time-invariant) electric currents, whose macroscopic characteristics (like the amount of current through a wire conductor) do not vary with time. Steady currents are also called direct currents, abbreviated dc. The subject of steady electric currents links field theory to several important concepts of circuit theory, such as Ohm’s law, Kirchhoff’s current and voltage laws, Joule’s law, resistance, conductance, voltage and current generators, and power balance in a circuit. Discussions regime, and help us further develop and understand the concept of a transmisin a time-invariant (dc) sion line as a circuit with distributed parameters. We shall derive and discuss integral and dif- ferential field equations for steady electric currents and their electric field, along with the correspond- boundary conditions. We shall also study the mechanism of conduction for various materials, introduce models of energy sources, and derive expressions for power and energy calculations. These equations and concepts will enable us to develop and demonstrate techniques for analysis ing of several general configurations with steady currents, such as resistors of various composition and shapes, capacitors with imperfect inhomogeneous dielectrics, transmission lines with imperfect con- of steady electric currents will bring us to the field ductors and imperfect inhomogeneous dielectrics, and and grounding electrodes buried 124 circuit analysis of transmission lines with losses in the earth. Section 3.1 Current Density Vector and Current Intensity 125 CURRENT DENSITY VECTOR AND CURRENT INTENSITY 3.1 In the absence of an externally applied electric random field, free charges in a conduc- motion due to their thermal energy. This is so-called thermal motion of charges. The corresponding velocity is the thermal velocity, denoted as v t Generally, v is rather large. In metallic conductors, free elec5 trons move at thermal velocities on the order of v ~ 10 m/s at room temperature, between collisions with the atoms and with one another. Because of the entirely random nature of thermal motion of charges (without any external electric field applied), there is no net macroscopic motion in any given direction, i.e., the macroscopic average vectorial resultant of thermal velocities of individual charges at any point in the conductor is zero. For an electric current, defined as a macroscopic net flow of free charges, to exist, there must be a nonzero macroscopic average velocity of microscopic velocities of charges in some direction in a conductor. This can be achieved, as we shall see, by establishing and maintaining an external (i.e., externally tor are in a state of (chaotic) . t t 1 + I V applied) electric field within a conductor. Consider a conducting body whose two ends are connected to a generator of voltage V, as shown in Fig. 3.1. Because a potential difference is maintained between the conductor ends, there is an electric field of intensity E inside the conductor [the line integral of E, Eq. (1.90), through the conductor is nonzero]. Note from the situation in Fig. 1.38, where the transient redistribution of charge occurred and electrostatic equilibrium with zero field inside the conductor was established. Here, the conductor is not isolated but wired to a source of electromotive force (generator), providing a mechanism that forces the free charges to move and prevents them from piling up, which would tend to reduce the field in the conductor. Assume, for simplicity, that the free charge carthat this situation riers in the is conductor are electrons only (as in metallic conductors). The electric on each charge force essentially different is thus [Eq. (1.23)] Fe = -eE, (3.1) e (e > 0) is given in Eq. (1.3). This force compels the move through the conductor, between its ends. However, since the elec- where the charge amount electrons to trons are not in free space, they cannot accelerate indefinitely under the influence of the electric field. Before they can acquire any appreciable speed, the electrons collide with the atomic lattice the field E A tc ~ -14 ture), in a systematic is velocities . manner. This relatively slight systematic drift of the free electrons the basis of electric conduction (current). After a brief initial transient, the elec- trons acquire a steady-state average velocity, determined by the balance the accelerating force of the applied field, lisions by v d 1 vt Since s (typical average time interval between collisions at room temperacan change the random thermal velocities of the electrons only slightly, but 10 it and acquire new random essentially has to start accelerating the electrons all over again every The with the lattice. This velocity is Fe and , between the scattering effect of the col- termed the drift velocity and symbolized . free electrons in a metallic conductor are so-called conduction-band electrons (or valence elec- trons), which are very loosely bound to their atoms and are essentially free to move about the crystalline atom of copper has 29 electrons, 28 of which are bound structure of the metal. For example, each electrons (tightly bound in their shells), while the outermost one is a free electron. Figure 3.1 Conductor whose two ends maintained at difference. are a potential 126 Chapter 3 Steady Electric Currents In general, v d <£ v t , makes only a slight change in the was applied, and v d is a macro- since the electric field velocity distribution that existed before the field scopic resultant of microscopic velocities of free charges in a direction along the -4 electric field lines. In most cases, its magnitude is not larger than v d ~ 10 m/s in amounts of current carried metals, for reasonable The (as we proportional to the electric drift velocity is linearly Vd drift velocity = shall see in an example). field intensity vector, -/Z e E, (3.2) where the constant pt e is the so-called mobility of electrons in the given material. The mobility is measured in the units of m 2 /(Vs) and is positive by definition. For electrons, the direction of v d as well as the direction of Fe is opposite to the , , Good conductors have high now say that charge carriers in direction of E. We can mobility. the conductor with the macroscopic average velocity v d This . is move through volume its an organized, directive motion of charges, which constitute an electric current throughout the conductor volume. introduce then a density vector, J. new field quantity, By definition, J current density vector (unit: A/m We to describe the current at a point: the current = Ny (-e )\ d (3.3) , 2 ) where /Vv or per 1 is m 3 the concentration of charge carriers, (the unit is m~ 3 The current density can sity, /, which, in turn, surface current intensity current (unit: or, (e.g., is i.e., their number per unit volume 2 ). alternatively be defined by defined as a rate of movement means of the current inten- of charge passing through a cross section of a cylindrical conductor). That is, simply, (3.4) A) amount of charge that flows through the surface dr. The unit for current intensity, which is current, is ampere or amp (A), equal to C/s. The cur- In other words, I equals the total during an elementary time usually referred to as, rent density vector is the current current current density vs. lines. If dr, simply, divided by directed along the macroscopic motion of charges, i.e., along perpendicular elementary surface of area dS to the set an 3.2(a), the magnitude of the current density vector is given by we lines, as in Fig. intensity (3.5) where d I is the current flowing through dS. We see that the unit for J is A/m 2 which means that it actually represents a surface density of a volume current. , Figure 3.2 (a) Definition of the means of the current density by current intensity, (b) Evaluation of the total current through a surface. 2 For example, the concentration of conduction electrons in copper is Nv = 8.45 x number of copper atoms per unit volume, since copper has one conduction-band The number of atoms per unit volume is approximately the same for all solids. the It) 28 m -3 . It equals electron per atom. 1 To show equivalent, that the definitions of current density, J, in Eqs. (3.3) we amount realize that the total time interval d t 127 Current Density Vector and Current Intensity Section 3.1 of charge that crosses and (3.5) are dS during the is dQ = dSVd dr (— e) (3.6) Av time interval (in the volume Av dr, a charge moves a distance v^dr, and all charges in the Ny times Av, pass = dSVd dr, the number of which is the concentration through dS). Using the definition of J in Eq. (3.3), this becomes dQ=JdSdt. Dividing by dr, we is dS obtain the current intensity d I through the same terms of /: in = JdS, dI which indeed (3.7) (3.8) as in Eq. (3.5). In the case where the current density vector element, the current through the element dl is not perpendicular to the surface is — J dS n = JdScosa = J • dS, (3.9) where dSn is the projection of dS on the plane normal to J and a is the angle between J and dS (see Fig. 1.30). Hence, the total current through an arbitrary surface S, Fig. 3.2(b), equals the flux of the current density vector through the surface, 1 If = J J dS. (3.10) total current through a surface there are several types of free charge carriers in a conductor drifting with different average velocities, the resultant current density vector is a vector sum of current densities in Eq. (3.3), M J — 'y ' (3.11) i= that correspond to individual types of carriers (e.g., electrons conductor). Equivalently, the net charge flow current intensity in Eq. (3.4). Positive charges is and holes move (3.10) are therefore valid for dl in the direction of E, negative charges in the opposite direction, but both add to the total current. Eqs. and in a semi- taken in evaluating the resultant (3.4), (3.5), any conductor and any combination of charge carriers. In many situations, current flow infinitely thin) film current, described N localized in a very thin (theoretically 3.3. This is by the surface current density vector, J s which , Js where is over a surface, as shown in Fig. = N q\d s , so-called surface is Figure 3.3 Surface current density vector (J s ). defined as (3.12) surface current density vector (unit: A/m) the surface concentration of charge carriers (number of carriers per -2 unit surface area, in ). Note that the surface current density vector is somes is m times denoted as K. In terms of the current intensity, the surface current density is given by (3.13) surface current density current intensity vs. 128 Chapter 3 Steady Electric Currents where d I the current flowing across a line element d / set normal to the current is The unit for J s which represents A/m. For example, the surface current density flow (Fig. 3.3). is with a current I/w and J s , Example = / 10 A. directed parallel to the strip axis. 3.1 Electron Drift along a 1-km of length The current which the electrons Solution As = / km 1 and radius a tv, strip (foil) equals 7S drift = 3 mm carries a steady current of intensity the current is distributed uniformly across the cross section (S) of the wire, the is [Eq. (3.10)] = = -4z = (3.3), the drift velocity of electrons v<j 3.54 x 10 A/m 2 5 na S = — = (3.14) . is 2.62 x 10 -5 m/s, (3.1 5) /Vv e /V v = 28 8.45 x 10 m -3 is the concentration of conduction electrons in copper and e the absolute value of the charge of an electron, Eq. (1.3). along the wire drift is in along the wire. J where = Copper Wire uniformly distributed across the wire cross section. Find the time is current density in the wire From Eq. aluminum of a very thin uniformly distributed across the strip width, I that is is A copper wire a line density of a surface current, , The time it is takes for an electron to hence t = — =3.82 x 10 7 (3.16) s, Vd which is 3.2 approximately 442 days. 3 OHM S LAW IN CONDUCTIVITY AND LOCAL FORM Consider again a metallic conductor (the charge is carried by electrons), with the J. Substituting Eq. (3.2) into Eq. (3.3), we obtain current density J current density in a metallic Ny efx e E. (3.17) = ctE, (3.18) N^ene, (3.19) conductor This equation can be rewritten as Ohm 's law in local J form where the proportionality constant, o conductivity of metallic conductors (unit: S/m) is a = macroscopic parameter of the medium called conductivity. It is always positive, conduct electric reciprocal of a meter (S/m). The current. The unit for conductivity is siemens per x meter (£2m). resistivity. The unit is ohm is denoted by the symbol p and termed and represents, 3 a We know in general, a we do that, of course, 1-km long transmission line. measure of the ability of materials to not need to wait 442 days to receive a communication signal sent via We shall see in a later chapter that time-varying signals traveling along transmission lines propagate as electromagnetic waves outside the conductors that constitute the line, motion of electrons within the conductors. The conductors actually serve as guides for the waves along the line. That is why signals travel at the velocity of electromagnetic waves in the medium surrounding the line conductors. If the medium is air, the velocity is 3 x 10* m/s (speed of and not light in a via the drifting vacuum), and the travel time is only 3.33 ps for a 1-km line. J J J Conductivity and Section 3.2 Using resistivity, Ohm's Law in Local 129 Form Eq. (3.18) becomes E= 1 pJ, p (3.20) o p - resistivity (unit: fimj Both Eqs. (3.18) and (3.20) are known as Ohm’s law in local or point form. Note that Eq. (3.18) is one of the three general electromagnetic constitutive equations for characterization of materials [another one being Eq. (2.46)]. It can be written in the following form: J = J(E), (3.21) constitutive equation for J, for an arbitrary material to encompass all possible conducting properties of materials. However, in terms of most conducting materials are linear and isotropic, i.e., J(E) = independent of electric field intensity and current density (the property of linearity), and is the same for all directions (isotropy). In homogeneous conductors, a does not change from point to point in the region being considered. For inhomogeneous conductors, on the other hand, a is a function of spatial their conductivity, crE, where a is [e.g., a = <j(x, y, z) in the region]. Almost always, the conductivity is a function of temperature, few exceptions is an alloy called constantan (55% copper, 45% coordinates conductivity is practically constant in a temperature range One of the nickel), whose T. 0— 100° C. For metallic conductors, the mobility of electrons decreases with an increase in temperature (because the average time interval between collisions with vibrating atoms, decreases). Hence, the conductivity decreases perature rise. and resistivity increases Around a room temperature of Tq = 293 and we can write K A tc , with a tem- (20°C), the resistivity varies almost linearly with T, P(T) where po = = po[l + a(T- (3.22) To)], p( 7o). For most metals (copper, aluminum, silver, is approximately 0.4% per kelvin. etc.), the temperature coefficient of resistivity, a, For some materials the resistivity drops abruptly to zero below a certain temperature: p( J) =0 for J < Jcr , (3.23) where Jcr is called the critical temperature of the material. This property, discovered by Kamerlingh Onnes in 1911, is called superconductivity, and the materials are said to behave like superconductors. Most superconductors are metallic elements that exhibit transition into superconducting states at a temperature of a few kelvin. Examples are aluminum ( cr = 1.2 K), lead ( cr = 7.2 K), and niobium ( cr = 9.2 K), as well as their alloys and compounds. More recently, new ceramic materials were discovered that become superconducting at considerably higher (and thus less expensive to produce and maintain) temperatures. For example, yttriumbarium-copper oxide (YBa 2 Cu 3 C> 7 ), discovered in 1986, has Jcr = 80 K, so its superconductivity can be utilized by cooling with liquid nitrogen. Interestingly, some of the best of the normal conductors, such as silver and copper, cannot become superconducting at any temperature, while the ceramic superconductors are normally good insulators - when they are not at low enough temperatures to be in a superconducting state. Ohm’s law in local form holds also for conductors with more than one type of charge carriers [see Eq. (3.11)]. In plasmas and gases, the charge carriers are electrons and positive ions (electron-deficient atoms or molecules). In liquid conductors, called electrolytes, the charge is carried by positive and negative ions. In all cases, superconductors 1 30 Chapter 3 Steady Electric Currents both positively and negatively charged particles (ions and electrons) contribute to the conductivity. and germanium), vacancies in the atomic crysby electrons, called holes, can move from atom to atom and behave like positive charge carriers, each hole carrying charge e. The conductivity of a semiconductor is therefore In semiconductors (e.g., silicon tal lattice left conductivity of a = Nwe ep, e +Nyh en h (3.24) , semiconductors 4 where the first term represents the contribution to the conductivity from electrons, moving opposite to the field E, while the second term represents the contribution from holes, which move with E. The concentrations /Vve and Nv h rapidly increase with an increase in temperature (temperature rise accelerates generation of free electrons and holes). Consequently, the conductivity of semiconductors increases with increasing the temperature, which is opposite to the temperature behavior of metallic conductors. By adding very small amounts of impurities to pure (intrinsic) semiconduc- may be increased dramatically. Impurities called donors (e.g., phosphorus) provide additional electrons and form n-type semiconductors, while acceptors (e.g., boron) introduce extra holes, forming p-type semiconductors. This procedure is known as doping of semiconductors. Note that the boundary between p-type and n-type parts of a single semiconductor crystal forms a junction region, called pn junction (see Fig. 2.9), which is utilized in semiconductor devices (diodes tors, and the conductivity transistors). Unlike the relative permittivity (e r ), shown in Table 2.1, the conductivity of materials varies over an extremely wide range of values, as we go from the best insulators to semiconductors, to the finest conductors. In S/m, o (at room temperature) ranges from around 10 water, and 2.2 for from quartz ductivity it germanium goes to infinity for to silver -17 for fused quartz, 10 to 6.17 x 10 is 7 for silver. -9 We as large as 25 orders of superconductors. Table 3.1 lists for bakelite, 10 -2 for fresh see that the range in con- magnitude (10 25 ), and then values of the conductivity of selected materials. Copper (Cu), the most commonly used metallic conductor, conductivity of copper, at cr Cu = 58 has a conductivity of MS/m. (3.25) 20° C many In electric applications, we consider copper and other metallic conductors as perfect conductors (PEC), with (3.26) oo. perfect electric conductor (PEC) Of course, superconductors also (3.26), no electric body field inside we conclude fall under this category. From Eqs. (3.18) and that a PEC (3.27) a 4 Generally, the current density in semiconductor devices = is composed of two components: a drift current which depends on the gradient of the concentration of charge carriers in a material and, therefore, does not satisfy Ohm’s law in local form. Consequently, the relationship between the total current density vector and the electric field intensity vector, Eq. (3.21), density. J in crE, and a diffusion current density, Jjjf, semiconductor devices is, in general, nonlinear. Conductivity and Section 3.2 a (S/m) Material - 17 ~ io ~ io - 17 ~ 1(T 16 ~ 10~ 15 ~ io - 15 ~ 1(T 15 ~ 1CT 15 ~ KT 14 2 x KT 13 ~ IO" 12 Quartz (fused) Wax Polystyrene Sulfur Mica Paraffin Rubber (hard) Porcelain Carbon (diamond) Glass Polyethylene 1.5 io Bakelite ~ KT Marble 10-8 - 12 IO "8 9 6 1(T 10“ 4 Granite soil KT 4 2 x Distilled water Silicon (intrinsic) 4.4 Clay 5 x 1(T 1(T Fresh water ~ soil x 1(T 4 3 2 io Material o (S/m) Carbon 7.14 x 10 4 (graphite) Bismuth 8.70 Cast iron ~ -2 10- 2 1.04 x 10 x 10 Stainless steel 1.1 Silicon steel 2 x 106 Titanium 2.09 x 10 6 Constantan (45% Ni) 2.26 x 10 6 German 6 3 x 10 silver Lead 4.56 x 10 6 Solder 7 x 10 6 Niobium 6 8.06 x 10 Tin 8.7 Platinum 9.52 x 10 x 10 6 Bronze 10 Iron 1.03 x 10 7 Nickel 1.45 x 107 Brass (30% Zn) 1.5 Zinc 7 1.67 x 10 Tungsten 7 1.83 x 10 7 2.17 x 10 x 10 7 0.4 Aluminum 3.5 x 10 Animal blood** 0.7 Gold 4.1 Germanium 7 5.8 x 10 (|| to fiber)** 2.24 x 10 3 x 10 2.2 Copper Seawater 3-5 Silver Ferrite 10 2 Mercury (at <4.1 K) DO Tellurium ~ Niobium (at <9.2 K) OO (intrinsic) 5 x 102 1.18 x 10 Silicon (doped) For dc or low-frequency currents, at 7 7 7 x 10 7 7 6.17 x 10 YBa 2 Cu 3 0 7 3 6 7 0.22 8 x IO" (_L to fiber)** 2 6 6 Animal muscle ** 131 x 10 5 Animal, body (average)** 4 x fat** Animal muscle * Form 10 6 Sodium Magnesium Duralumin Animal Local 106 Nichrome Mercury (liquid) x 1(T - 11 Wood Wet in Conductivity of selected materials* Table 3.1. Dry Ohm's Law (at <80 K) OO room temperature. Also for humans. i.e., the electric field is always zero in perfect conductors. This, in turn, implies that the voltage between any two points of a perfect conductor [Eq. (1.90)] is zero. Finally, so-called convection currents, which are the result of the motion of posin a vacuum or rarefied gas (where a = 0), Examples are electron beams in a cathode ray tube and a violent motion of charged particles in a thunderstorm. The convection current density is given by itively or negatively charged particles are not governed by Ohm’s law. J = p\, (3.28) where v is the velocity of particles and p is the volume density of charges (charge per unit volume) in the vacuum or rarefied gas. Noting that p = Nw q, Ny being the concentration of particles and q an elementary charge, we observe the equivalency of the definitions of convection and conduction current densities given by Eqs. (3.28) convection current density 1 32 Chapter 3 Steady Electric and Currents (3.3), respectively. (vj) of charges and Eq. However, the velocity v (3.2) in Eq. (3.28) is not a drift velocity not satisfied. is Conceptual Questions (on Companion Website): 3.1 and 3.2. HISTORICAL ASIDE Kamerlingh Onnes (1853-1926), Dutch and professor at the University of Leyden, was awarded a Nobel Prize in Physics which was a fascinating result at that time. the first to produce liquid helium (in 1908). Onnes demonstrated in 1911 that the resistivity of mercury absolutely disappears at temperatures below about 4 K, and thus discovered supercon- Heike 0.9 K, physicist He was in 1913 for his investigations of the properties of matter at extremely low temperatures. His experiments led him as close to absolute zero as ductivity. The SI unit of power, the watt, was named in honor of James Watt (1736-1819), a Scottish mechanical engineer and inventor, who is famous for his rev- the 1760s, which led to great advancements in the olutionary improvements of the steam engine in his Industrial Revolution, ing the “horsepower” steam engines. AND JOULE'S LAW LOSSES IN CONDUCTORS 3.3 and is also known for devis- to measure the power of IN LOCAL FORM now consider the current flow in a conductor from As we know, free charge carriers (e.g., electrons) are Let us the energy point of view. accelerated on their paths between collisions with vibrating atoms, and at every collision they lose their acquired kinetic energy. Energy is thus transmitted from the electric field, E, and ultimately resulting in a higher temperature of the conductor. This means that in a conductor with electric current, electric energy is constantly converted into heat. We wish to derive the expression for the rate (power) of this energy transformation. We start with the electric force on a charge d Q given in Eq. (3.7), which equals dQ E. The work done by this force in moving d Q a distance d / along the field lines via charge carriers to the atoms, enhancing their thermal vibration is [Eq. (1.72)] dWe = dQEdl = JdSdtEdl = JEdvdt where dv to heat, is an elementary volume known power of Joule’s volume dv is called the in the The volume Joule's law ohmic power W/m 3 ) form; p\ density (unit: in local density of this rate of this conversion, losses or dPi power Pi = conductor. This work in the The as Joule’s heat. (dv=dSdl), ohmic = d W e losses. is (3.29) converted dWe /dr (J/s), is (lost) power, Thus, the power of Joule’s losses =JEdv. (3.30) is dPj /2 = JE= — =oE (3.31) dv and this is known as Joule’s law in local (point) form. The unit for power (W), and hence the unit for power density is W/m 3 . is watt Section 3.4 Continuity Equation 133 The total power of Joule’s losses (the electric power that is lost to heat) in a domain of volume v (e.g., in the entire conducting body) is obtained by integration of power dPj throughout v: Pj = l JEdv. (3.32) power of joule's (ohmic) losses (unit: W) dPj Conceptual Questions (on Companion Website): 3.3. CONTINUITY EQUATION 3.4 We now consider one of the fundamental principles of electromagnetics - the continuity equation, which is the mathematical expression of the principle of con- and cannot be lost or created. It can nowhere nor disappear. Let an arbitrary closed surface, S, enclosing volume v, be situated in a region with timevarying currents, as shown in Fig. 3.4(a), and let Qs and Qs + d Qs denote the net charges in v at instants of time t and t + dr, respectively. The change in charge, d Qs, cannot be created in v, but only brought in from the domain outside the surface S, and it is brought by the current flowing through S. So, the charge dQs passes the surface S during the time dr, and this is exactly what we have in the definition of servation of charge. Charge move from is indestructible, place to place, but can never appear from current intensity in Eq. (3.4). Therefore, dQs (3.33) dr is the intensity of the current flowing across the surface current crossing the surface in the opposite direction, region, is S i.e., into the region v. The the current leaving the hence Jout — dQs ~hi (3.34) ‘ dr On (b) the other hand, by virtue of Eq. (3.10), the current leaving v across S equals the total outward flux of the current density vector through S (which is a closed Figure 3.4 Arbitrary closed surface), that surface is, in a region with currents: (a) general case /out = ^J-dS. (3.35) Combining the two preceding equations, we obtain (f Js This is J-dS = node. --^. (3.36) dr the continuity equation (in integral form). It tells is us that the outward flux of equal to the negative of the derivative in time of the total charge enclosed by that surface. By expressing the charge in terms of the volume charge density, p, the becomes continuity equation, for currents of any time dependence the current density vector through any closed surface continuity equation and (b) current flow through wires meeting at a 1 1 34 Chapter 3 Steady Electric If Currents S does not change the surface the volume time derivative can be in time, the moved inside integral, yielding continuity equation in terms (3.38) of the volume charge density The ordinary derivative is replaced by a partial derivative because p, generally, is a multivariable function of time and space coordinates. By applying the divergence theorem, Eq. (1.173), to Eq. (3.38) or simply by analogy to the differential form of the generalized Gauss’ law, Eq. (2.45), we get the differential form of the continuity equation: V continuity equation in differential J • = - dp (3.39) dt' form It tells us that the divergence of J at a given point equals the negative of the time rate of variation in charge density at that point, and is also called the continuity equation at a point. For steady (time-invariant) currents, the charge density is constant = 0, and the integral form of the continuity equation reduces to in time, dp/dt = dS continuity equation for steady (3.40) 0. currents, integral form The differential equation of continuity of steady currents is given by differential continuity V equation for steady currents J = (3.41) 0. Thus, time-invariant current density vector has zero divergence everywhere, and under all We circumstances. solenoidal. say that steady electric currents are divergenceless or The zero divergence of a vector field indicates that there are no sources means upon themselves (steady currents must or sinks in the field for the lines of flux to originate from or terminate on. This that the streamlines of steady currents close flow in closed loops), unlike the streamlines of the electrostatic field intensity, which originate and end If on charges. the steady current is carried into and out of the volume v by a number (TV) of wire conductors meeting at a point, as indicated in Fig. 3.4(b), then Eq. (3.40) implies that the algebraic sum of all the currents leaving the junction is zero. N Kirchhoff's current y>=°. law (3.42) k= For the situation and notation in Fig. 3.4(b), I\ —h—h+h — 0- Eq. (3.42), if applied to a node in a dc circuit, represents Kirchhoff’s circuital law for currents. Like Kirchhoff’s voltage law, Eq. (1.92), Kirchhoff’s current law also applies to time-varying situations assumptions, which will and ac be discussed later In studying steady current fields, circuits, meaning conservative nature of field E in in mind field, that time-invariant which is a conservative that the line integral (circulation) of the electric field intensity vector, E, along an arbitrary contour (closed path) steady current above form in this text. we always have currents in a conductor are produced by a static electric field, in the with certain restrictions and l E dl is = zero, 0. (3.43) ' Section 3.4 We also have in mind that J and E 135 Continuity Equation are related at any point in the conductor by the constitutive equation for the current density, Eq. (3.21) [or Eq. (3.18) for linear media]. Example 3.2 Element Law for a Capacitor Prove that the current through a capacitor equals the product of the capacitor capacitance and the time rate of change of the voltage drop across the capacitor. Solution Fig. 3.5 shows a capacitor of capacitance C connected to a time-varying voltage 5 v. We assume that the capacitor is ideal, i.e., its dielectric is perfect (nonconducting), as well as that there is no excess charge along the connecting conductors in the circuit (charge is localized only on the capacitor electrodes). To relate the current through the capacitor, i, to the capacitor charge, Q, we apply the continuity equation for time-varying currents in integral form, Eq. (3.36), to a surface S enclosing completely only the upper electrode of the capacitor (Fig. 3.5). The total current leaving the 0 c — Q c enclosed domain (left-hand side of the , ) -Q equation) equals —i, whereas the total enclosed charge (appearing on the right-hand side of the equation) equals the charge of the upper electrode, i.e., the capacitor charge: Qs = Q. Thus, Eq. (3.36) becomes 0 (3.44) time-varying current; for Example Substituting Q= Cv =C dv (3.45) dt see that i is linearly proportional to the rate of proportionality constant. This itor, which is 3.2. [Eq. (2.112)] yields i We 1 Figure 3.5 Capacitor with a is element law (current-voltage characteristic) for change of v in time, with C as the a capacitor Wlul a time-varying current the element law (current-voltage characteristic) for a capac- widely used in circuit theory, in conjunction with Kirchhoff’s laws and other element laws. Spherical Capacitor with an Imperfect Dielectric A spherical capacitor is filled with an imperfect homogeneous dielectric of conductivity a. The radius of the inner electrode is a and the inner radius of the outer electrode is b (b > a). The electrodes have a conductivity that is much larger than a, so that they can be considered as perfect conductors. The capacitor is connected to a generator of time-invariant voltage V. Find (a) the current density vector in the dielectric and (b) the power of Joule’s losses in the capacitor. Solution (a) Since a 0, cated in Fig. there exists a steady current of intensity / in the capacitor circuit, as indi3.6. To determine the current density vector in the dielectric, we apply the continuity equation for steady currents in integral form, Eq. (3.40). In general, appli- cation of the continuity equation law, Eq. 2.43. This current density vector is is analogous to application of the generalized Gauss’ one with spherical symmetry (see Example of the form problem is J where r is We The surface S for applying the continuity equation and the (3.46) the radial coordinate in the spherical coordinate system and vector (Fig. 3.6). 5 = J{r) r, 1.18), is f is the radial unit a spherical surface use lowercase (small letter) symbols for the voltage and current here to emphasize that those are time-varying quantities. volume current distribution with spherical symmetry 1 36 Chapter 3 Steady Electric Currents Figure 3.6 Spherical capacitor with an imperfect homogeneous dielectric and time-invariant current; for Example 3.3. of radius r (a < r < b) centered at the origin (the surface in electrostatics, outward direction is e.g., that in Fig. 2.16). zero, that The same as the corresponding total current flowing Gaussian through S in the is. JOAnr2 -1 = 0. (3.47) This means that the outward flux of J through S equals the current / through the capacitor terminals. Hence, J{r) From Eq. = ( a <r < b )• 3.18, the electric field intensity in the dielectric (3-48) is I (3.49) Anar2 The voltage between the electrodes given by is v = v,-n = where Va because li Using Eq. power of Joule’s r " [ dv= , Pj in power equipotential = aabV (b — a)r 2 losses in the dielectric — Wdr=- "-' J(r) 2 . ° 9 r dv (3.51) , Anaa l2 b 2l V 2z (b - ry2 a) r b / Ja is — = AnaabV — dr r 2 b 2 (3.52) a the form of a thin spherical shell of radius r and thickness dr [Eq. (1.33)]. There are overall loss 6 J r=a Jv with dv adopted is » (3.32), the Pj= no losses in the electrodes (perfect conductors), so this is the in the capacitor. Conceptual Questions (on Companion Website): 3.4-3. 6; Companion (3.50) ^(l-l} and Vb are the potentials of the electrodes (each electrode a). Combining Eqs. (3.48) and (3.50), we get cr e ie C trodes 7(r) (b) E *r = Website). MATLAB Exercises (on Section 3.5 3.5 Boundary Conditions 137 for Steady Currents BOUNDARY CONDITIONS FOR STEADY CURRENTS Applications of steady current fields involve considerations of interfaces between conducting media of different conductivity. In this section, we shall formulate the boundary conditions that govern the manner in which the current density vector, J, and the electric field intensity vector, E, behave across such interfaces. Comparing the integral form of the continuity equation for time-varying currents, Eq. (3.38), to the integral form of the generalized Gauss’ law, Eq. (2.44), we conclude that the boundary condition for normal components of the vector J is of the same form as the boundary condition for normal components of the vector D, Eq. (2.85). The only difference is on the right-hand side of the equation, where p s (the surface charge density that may exist on the surface) is replaced by —dp s /dt. With this. n • —n Ji dps = J2 • (3.53) dt boundary condition for J n/ time-varying currents is the normal unit vector on the surface, directed from region 2 to region 1. For steady currents, —dps /dt = 0 in Eq. (3.53), and the boundary condition that corresponds to Eq. (3.43) is that in Eq. (2.84) [Eq. (3.43) is the same as for the electrostatic field], so the complete set of boundary conditions for steady currents is given by where n n x Ei n • —n Ji E2 = x 0 — n-J 2 = E\ t or 0 or J\ n — E2 1 , =hn- E (3.54) boundary condition for (3.55) boundary condition for J n/ t regime We see that in the steady current field, the tangential component of the electric field intensity vector and the normal component of the current density vector, are both continuous across the boundary. boundary condition for E t in As we Eq. (3.54) has this E and J n t , shall see in a later chapter, the same form for fields of any time variation. By analogy to the procedure of deriving Eq. (2.87) for the electrostatic field, we obtain the law of refraction of the current density lines at a boundary between two linear conducting media of conductivities oq and tan oq oq tan oq a2 oq: (3.56) law of refraction of current streamlines where aq and 012 are the angles that current lines in region 1 and region 2 make with the normal to the interface, as shown in Fig. 3.7. Note that if medium 2 is a good conductor and medium 1 is a low-loss dielectric, then <72 oq, i.e., oq/oq ~ 0, and Eq. (3.56) gives tanoq « 0 for any 012 Therefore, » . oq meaning «0 (oq » oq), (3.57) that the current lines always leave (or enter) a good conductor at a right angle to the boundary (zero angle to the normal on the boundary). Note tan 012 also that if medium 1 is a perfect dielectric (e.g., air), then oq = 0, and -> 00 from which , oq = 90° (oq = (3.58) 0). means that the lines of current flow are always parallel to the surface of a conductor surrounded by a nonconducting medium. This Figure 3.7 Refraction of steady current lines at a conductor-conductor interface. Conceptual Questions (on Companion Website): 3.7 and 3.8. dc a 138 Chapter 3 Steady Electric Currents DISTRIBUTION OF CHARGE IN A STEADY CURRENT 3.6 FIELD The electric field intensity vector, E, in a steady current field is produced by station- ary excess charges in the system. In the general case, these charges exist not only on the surfaces of conductors, but also inside their volume. Distribution of charges in the system is conditioned by the distribution of currents, and can be determined only after the current distribution is determined first. The distribution of the current density vector, J, inside conductors can be obtained by solving the basic equations that govern steady current fields, which we summarize here: Maxwell's first equation for §c E static fields J dc continuity equation J constitutive equation for J Once the solution for J is =0 dS = 0 • = dl • J(E) (3.59) = [J known, the charge crE] distribution can be obtained from the generalized Gauss’ law in conjunction with the constitutive equation for the electric flux density vector, D: [&D-dS = Q S D = D(E) [D = eE] Maxwell's third equation constitutive equation for D (3.60) | Note that, assuming that the medium is linear, both D and J are linearly As a result, we have a linear relationship between D and J: proportional to E. duality of D D = £E = - J. and J (3.61) a This duality relationship will be used on many occasions. Starting from the differential form of the generalized Gauss’ law, Eq. (2.45), and using Eqs. volume charge in (3.61) and (3.41), the volume charge density in the conductor p-V.D-V.(£j)-[v(i)].J+i(V.J) = J.v(i). a steady is 6 (3.62) current field We see that V = J 0 does not imply that p = Volume charge 0. inhomogeneous conducting media where e/a in portional to the gradient of e/o. On cannot be volume excess charges (p ^ const, we the other hand, = 0) inside and is nonzero magnitude is pro- density its also conclude that there homogeneous media ( and e do not vary with position) with steady currents. The corresponding boundary condition D, Eq. for (2.85), in combination with Eqs. (3.61) and (3.55), gives the surface charge density on the interface between medium (with parameters o\ and £j) and medium 2 (with cr2 and £ 2 ): 1 surface charge in a steady Ps = r. Dl - n D? n • = —n *1 o\ current field ^Having in mind the • (/a) = (V/) a J, 1 —n 02 • J2 --W = V^i rule for calculating the derivative of a product of gence operator, which V « is + /(V (3.63) two functions, we apply the diverand a vector function, and get a differential operator, to a product of a scalar a). a2 J 0 Section 3.7 Relaxation Time n 3\ — n J2 is the normal component of the current density vector boundary (n is directed as in Figs. 2.10 and 3.7). Note that ps — 0 only for the special case of £\/o\ — 82 / 02 If both media are metallic conductors, we have approximately s\ = e 2 = £o> so that Eq. (3.63) becomes where Jn = • • across the - (3.64) Ps Finally, the distribution of bound (polarization) charge in the dielectric can be found by computing first the polarization vector, P, from Eq. (2.59), and then bound volume and surface charge densities, p p and pps using Eqs. (2.19) and (2.89), , respectively. Problems'. 3.1; Conceptual Questions (on Companion Website): 3.9 and 3.10; MATLAB Exercises (on Companion Website). 3.7 RELAXATION TIME We know from electrostatics that charge placed in the interior of a move to the conductor surface and redistribute itself in such way a conductor that E= will 0 in the conductor under electrostatic equilibrium conditions. With the continuity equation (for time-varying currents) now in hand, we can quantitatively analyze this nonelectrostatic transitional process and calculate the time it takes to reach an equilibrium. Consider a homogeneous conducting medium of conductivity 0 and permittiv- The current density vector, J, and electric flux density vector, D, in the medium interconnected by the duality relationship in Eq. (3.61). Combining the differ- ity e. are ential form of the generalized Gauss’ equation, Eq. (3.39), law, Eq. £ £ = V.D = V- (-j)\ = — V-J = \o / a / /0 where the ratio £ / — £ dp o (3.65) dt can be brought outside the divergence sign because it is a conRewriting, we have that the charge density, p, -\ dt is — 0 dp_ which and that of the continuity medium is homogeneous). medium satisfies the equation stant (the in the (2.45), we can write £ P — 0, (3.66) a first-order partial differential equation in time, p where po is = po e' ~(a/e)t Poe' e.g., Its solution is -t/T p — given by (3.67) the initial value of the charge density at function of the space coordinates as well, t. t — 0. In general, p p(x,y,z,t). Eq. (3.67) is redistribution of charge a tells us that the charge density at a given location in the conductor decreases with time exponentially, completely independent of any applied electric field. The time constant (3.68) is s) and it equals the time required any point to decay to 1/e (36.8%) of its initial value. It can easily be shown that p decreases to 1% of po after approximately 4.6r, while at _5 t = lOr, p % 4.5 x 10 p 0 he., p « 0. is referred to as the relaxation time (the unit for the charge density at , relaxation time 139 140 Chapter 3 Steady Electric Currents For metallic conductors, r is so short that it can hardly be measured or observed. For example, copper has rc u % 10 -19 s. Even much poorer conductors than metals -5 have very short relaxation times (e.g., rn 2 o % 10 s for distilled water). On the other hand, the relaxation time for good dielectrics (insulators) (e.g., r minute, rm 1 g ass i i ca ~ 15 hours, and rquar tz % is relatively long 50 days). devoted to time-invariant currents and fields, we note is also used to determine the electrical nature (in terms of conducting and dielectric properties) of materials for timevarying currents and fields. Namely, whether a material of parameters o and s is considered a good conductor or a good dielectric is decided on the basis of While chapter this is here that the concept of relaxation time the relaxation time, as compared to times of interest in a given application. Thus, for a time-harmonic (sinusoidal) field of frequency /, the relaxation time, given by Eq. is (3.68), classified as said to be a perfect electric conductor (PEC). considered a good dielectric (insulator) (lossless) dielectrics (a rial is classified as = 0). For all a quasi-conductor. that are considered as good = \/f. If x T, the medium = 0 (a -> oo), the material is compared to the time period, 7 T a good conductor. In particular, if r is if On the other hand, the material other (intermediate) values of We can good conductors now understand that at certain frequencies dielectrics at sufficiently higher frequencies (i.e., is oo for perfect T. In a limit, r r r, some the mate- materials may behave like shorter time periods). For = 10 MHz, and fi, = 30 GHz, average rural kHz, S/m, assuming no change in the parameters as a function of frequency) behaves like (1) a good conductor, (2) a quasi-conductor, and (3) a good dielectric, respectively. Further discussions of conducting and example, at frequencies ground (e r = 14 and a f\ = = 10 1 -2 dielectric properties of materials at different frequencies, in a context of general Maxwell’s equations and electromagnetic wave propagation, are provided in later chapters. 3.8 RESISTANCE, OHM'S LAW, AND JOULE'S LAW Consider an arbitrarily shaped conductor made from a linear (generally inhomogeneous) material of conductivity a, as shown in Fig. 3.8. Let the end surfaces 5 a and 5 b of the conductor be coated with perfectly conducting material (or with material of conductivity much greater than a). If the voltage V is applied between the conductor ends, the current, of density J, in the conductor flows normal to the end surfaces [Eq. (3.57)] and parallel to the sides of the resistor [Eq. (3.58)]. of the continuity equation, the same section of the conductor. This current total current is By virtue must pass through every cross given by (3.69) where E is any path On the other hand, same vector, E, along the electric field intensity vector in the conductor. the voltage between 5 a and 5 b equals the line integral of the conductor connecting these surfaces [Eq. in the (1.90)], that is, B (3.70) 7 The time period ( T) is defined as an interval after which a time-harmonic (or other time-periodic) over time, whereas the frequency (/) per unit time (one second), so that / times T equals unity. function repeats itself is the number of repetitions of the function Section 3.8 Resistance, Ohm's Law, and Joule's Law 141 Figure 3.8 Arbitrarily shaped conductor - Now, we note that not change shape, but and, by means if resistor. V is increased (for some reason), the electric field lines do E proportionally increases everywhere within the conductor, of Eq. (3.69), so does the current I. The ratio of V and I is thus a constant, (3.71) called the resistance of the conductor. stantial) resistance voltage, current, R is O) A conductor with two terminals and a (subThe relation between Ohm’s law: usually referred to as a resistor. and resistance of a resistance (unit: resistor is known as V — RI. the (3.72) Ohm's law The resistance is always nonnegative (R > 0), and the unit is the ohm (£2), equal to V/A. The value of R depends on the shape and size of the conductor (resistor), and on the conductivity a (or resistivity p) of the material. HISTORICAL ASIDE Georg Ohm analogies between the flow of electricity and flow German of heat. In a series of papers in 1825 and 1826, he gave a mathematical description of conduction in Simon (1789-1854), a and mathematiwas a professor physicist cian, at University the Munich. As a received a and matical tific child, fine of Ohm mathescien- education from his father, who was a lock- smith and entirely an man. In 1811, he received a doctorate mathematics from the University of Erlangen. He was also interested in physics, where he studied self-taught in modeled after Fourier’s study of He assumed that, just as the rate at which heat flowed between two points depended on the temperature difference between the points and on the ease with which heat was conducted by the material between the points, the electric current between two points should depend on the difference in electric potential between the points and on the electric conductivity of the material between the points. By experimenting with wires of different lengths and thicknesses, he found that electric circuits heat conduction. 142 Chapter 3 Steady Currents Electric the current intensity through a wire, for a given Analyzed Mathematically), published in first, Ohm’s work was received with little enthusiasm, and a full acknowledgement and recognition of his results, including Ohm’s law, did not come until 1841, when he was awarded the Copley Medal of the Royal Society and soon after became member of several European academies. Only in 1852, two years before his death, Ohm Circuit, between the wire ends, was inversely proportional to the length and directly 1827 in Berlin. At potential difference proportional to the cross-sectional area of the wire. As very knowledgeable of both mathematics and Ohm physics. was able to deduce mathematical on the experimental evidence relationships based that he a wire existed had tabulated. He defined the resistance of and showed, in 1827, that a simple relation among (voltage), and current now known achieved his lifelong ambition of being appointed the resistance, potential difference Ohm’s as intensity of a wire. This law. The fully to the chair of physics at the University of developed ohm presentation of his theory of electric conduction lished the appeared in his famous book “Die Galvanische Kette, Mathematisch Bearbeitet” (The Galvanic Edgar Fahs Smith Libraries) The siemens the mechanic Johann adopted as the SI unit for honor of brothers Werner and conductance in Wilhelm von Siemens, German engineers and inventors. Ernst Werner von Siemens (1816-1892) is contributed to then new the “Siemens Company” (ohm ter. Georg Halske (1814—1890), Halske Telegraph Construction in Berlin, later He later as “Siemens.” furnace and electric pyrome- became British subject, Sir Charles is called the conductance the siemens (S), where S is = 1 I R V = A/V. Note and symbolized by G. that sometimes the shows the circuit-theory representation of a resistor. assumed that the resistances are located only in the resistors In circuit theory, in a circuit, the interconnecting conductors are considered as perfectly conducting. necting conductor has therefore zero resistance and The only voltage drops in the Ohm’s law, Eq. (3.72), is valid also circuit). that of a resistor (v Figure 3.9 Circuit-theory representation of a resistor. elements = mho used instead of the siemens. Fig. 3.9 it is known William Siemens. of resistance spelled backwards) & the regenerative (unit: S) Its unit is (Portrait: Collection, University of Pennsylvania mechanical engineer by training and successful businessman. His most important inventions are discipline of electrical phy (needle telegraph), cable transmission (large undersea cables), and energy generation, and was one of the first great entrepreneurs in electrical industry. In 1847, he founded, together with conductance as the unit of resistance. Karl Wilhelm von Siemens (1823-1883) was a engineering with several inventions in telegra- The reciprocal From Ohm’s law, Munich. Ohm’s name was further immortalized in 1881, when the International Electrical Congress estab- is is while Each con- equipotential (acts as a short occur across circuit elements. Note for time-varying voltage (v) and current ( i ) circuit It represents the element law for a resistor, as one of the basic theory [another element law being Eq. (3.45) for a capacitor]. Ri). in circuit For linear resistors, R remains constant for different voltages and currents, V and / For nonlinear resistors, however, the conductivity of the material depends on the applied electric field, E, and R therefore depends on the applied voltage (or . current), nonlinear resistor R= R(V). (3.74) ' Section 3.8 Resistance, Ohm's Law, and Joule's Law 143 Figure 3.10 Evaluation of the power of Joule's (ohmic) losses a resistor of general shape. in Examples are semiconductor (pn ) diodes, whose current-voltage characteristics, I = I(V), are nonlinear functions. The total power of Eq. (3.32), where v is Joule’s or ohmic losses in the resistor in Fig. 3.8 the volume of the resistor. we first cut v into thin slices with bases chosen to be dicular to current lines). We then subdivide each small tubular cells of volume dv = is given by To carry out the volume integration, equipotential surfaces (perpen- slice d S d /, as depicted along the current lines into in Fig. 3.10, and write JEdv — f fjEdSdl. J A Js dv — Note same that E dl equals the voltage dV between (3.75) the bases of the cell, which is the between for all cells within a slice (because of the equipotentiality of interfaces adjacent Therefore, slices). dV is a constant for cross-sectional integration over S, and can be taken out of that integral. We have thus separated the overall integration throughout v into two independent integrals: -Us JdSEdl dV Pi (3.76) Figure 3.1 1 which equal, respectively, the voltage V [Eq. (3.70)] and the current I [Eq. (3.69)] of the resistor. Employing Eqs. (3.72) and (3.73), the equivalent expressions for the in a power of Joule’s section; for losses in a resistor are conductor of a uniform cross Example 3.4. 2 Pj This is known I = VI = RI2 G' as Joule’s law. Current Uniformity Example 3.4 A steady current is in Conductors with Uniform Cross Section flowing through a long conductor made of a homogeneous material and having a uniform cross section of an arbitrary shape. Prove that the current density same (3.77) in the entire is the conductor. With reference to Fig. 3.11, the current density vector in the conductor has a = Jz (x, y, z) z. From the continuity equation for steady currents in differential form, Eq. (3.41), and the expression for divergence in the Cartesian coordinate system, Eq. (1.167), we have Solution 2 -component only, J V which means that Jz is not a function of J = z, i.e., (3.78) it does not vary along the conductor. Steady current homogeneous Joule's law 144 Chapter 3 Steady Electric Currents HISTORICAL ASIDE Janies Prescott Joule was scientist, a produced by a current that the heat (1818-1889), an English tric circuit mem- in an elec- over a certain interval of time equals the square of the current intensity multiplied by ber of the Royal Society. the resistance of the circuit and time. This As to be called Joule’s law. a son of a wealthy He measured came the heat and was especially produced by every process he could think of, and studied the relationship between the amount of work entering the system and the amount of heat exiting the system. In his famous “paddle wheel” experiment in 1847, he used a falling weight to spin a paddle wheel in an insulated barrel of water and measured very precisely the increase in the water temperature produced by the friction of the wheel - to determine the mechanical equivalent interested in the efficiency of electric motors. After of the heat dissipated in the water. His general several attempts to design a superefficient electric conclusion was that heat was only one of would produce infinite power (this possibility had been suggested in previous papers) and thus offer an ideal alternative to steam engines, he realized that this goal was not achievable and became interested in studying the heat gener- forms of energy, which can be converted from one form to another but the total energy of a closed system remains constant. With this, he contributed fundamentally to the discovery and recognition of ated by electricity. Joule lacked in mathematical Joule also by using joule as the unit for brewer. Salford Joule was educated by private tutors, including the famous English chemist and physicist John Dalton (1766-1844). Later, while working in the family brewery, he studied in his spare time the subject of electricity motor (new at that time) that rigor, but was a fanatic experimentalist. the principle of conservation of energy. Based on energy. (Portrait: many We honor work and National Bureau of Standards Archives, extensive measurements of heat in electric motors courtesy AIP Emilio Segre Visual Archives, E. Scott Barr and other Collection) electric circuitry, he discovered, To prove in 1840, that J does not change Eq. (3.43) to a rectangular contour parallel to current lines (Fig. 3.11). <t C Using also Eq. E dl= Jc where a the is the conductivity of the medium is homogeneous ( a — in a cross section of the - (f dl dl 1 8), (f we can J-dl = write 0, (3.79) ° Jc 1/cr can be taken out of the integral because Hence, const). • (3. =- Jc ° medium, and J conductor either, we apply placed inside the conductor, with sides of length a set = J\a — J2 a = 0, (3.80) with J\ and Jj standing for the current densities along the two sides of the contour parallel to J. We conclude that these current densities are the same: J\=J2 Note that C can be translated to any position in . (3.81) the conductor, so that J\ and Ji can be associ- depend on x and y, namely, it is the same in the entire cross section of the conductor. Combined with our previous conclusion about the current uniformity in the z-direction, we have that ated to any pair of points dc current uniformity in in the conductor cross section. This implies that J does not a J homogeneous conductor of a uniform cross section everywhere inside the conductor. = const (3.82) . Section 3.8 Example Resistance, Ohm's Law, and Joule's 145 Law Resistance of a Resistor of a Uniform Cross Section 3.5 Determine the resistance of a homogeneous resistor with a uniform cross section of an arbitrary shape and surface area S. The length of the resistor is / and the conductivity of the material is a Eq. (3.82) Solution tells us that the current current density in the resistor is distributed uniformly in the resistor. is The thus J = (3.83) S' where I is the current intensity through the resistor, Eq. (3.69). Using Eq. (3.70), the voltage across the resistor is J/ // 11 a cro V S El=-l=—. (3.84) Hence, the resistance of the resistor comes out to be (3.85) oS resistance of a resistor with homogeneous a uniform cross section Example 3.6 Fig. 3.12(a) Two shows a Resistors in Series two parts of lengths cylindrical resistor of radius a consisting of Conductivities of the parts are ay and 02 (ay ^ 02 ). The voltage across the resistor l\ is and l2 . V. Find the current through the resistor. Solution The current through the two parts of the resistor represented as two homogeneous resistors connected in total resistance of the connection is the same, and they can be series, as shown in Fig. 3.12(b). The is R= R\ + R2 (3.86) , equivalent resistance of two resistors in series where, using Eq. (3.85), the resistances of individual resistors amount to Ri h = R„ 2 and G\na^ respectively. The current through the 1 Two g2 ttcij2a resistor in Fig. 3.12(a) = R\ Example 3.7 h - is (3.87) ’ hence 2 V + Ri nax a2 a V vih (3.88) + <f\h Resistors in Parallel A cylindrical resistor of length / consists of two coaxial layers of different conductivities, oy and G2 as shown in Fig. 3.13(a). The cross-sectional surface areas of the layers are S\ and S2 The current through the resistor is I. Compute (a) the current density in the resistor and (b) + Vi ,+ V2 . , ^WW\ the voltage across the resistor. R2 /?! it Solution (a) Eq. (3.54) tells us that the electric field intensities in the two layers are the same: E X =E 2 = E. V (3.89) Hence, the current densities in the layers are not the same, as total current through the resistor is given by J\ = g\E and J2 = g2 E. The I from which = Jx Si + J2 S2 = (giS\ + g2 S2 )E, (b) Figure 3.12 Cylindrical resistor (3.90) with two parts of different conductivities: (a) (b) for geometry and network representation; Example 3.6. p 146 Chapter 3 Steady Electric Currents and o\I = J\ CTlSl (b) The voltage across the resistor and + (7252 J2 < El 5 i (3.92) + O2 S 2 II = ctj5i V can 7i is V= Note ail = (3.93) + 02S2 found using the equivalent circuit shown in Fig. 3.13(b). Namely, two layers of the resistor is the same, they can be represented as two homogeneous resistors connected in parallel. The total conductance of the connection turns out to be that also be since the voltage across the G— equivalent conductance of two resistors in parallel that [Eq. (3.85)], is + G 2 ,\ (3.94) — — (3.95) G\ G= 01S1 The voltage of the resistors is V= Conductance of Example 3.8 + a2 Si I/G. a Spherical Capacitor Find the conductance of the spherical capacitor with an imperfect (conducting) dielectric (of conductivity cr) from Fig. 3.6. Solution Using Eqs. (3.73) and by definition, (3.50), the conductance of the nonideal spherical capacitor is, I leakage conductance of a V nonideal spherical capacitor This, actually, is = 4 naab (3.96) b-a' the so-called leakage conductance of the capacitor. (An ideal capacitor has a perfect dielectric and zero leakage conductance.) Note, however, that the system in Fig. 3.6 may also represent a spherical resistor, with resistance J_ _ P(b-a) G (3.97) 4ji ab where p = \/a is the resistivity of a (resistive) material between the electrodes. Note also that, employing the conductance (or resistance) and Joule’s law, Eq. the power of Joule’s losses in the capacitor (resistor) can now easily be found as PjJ which, of course, Problems : MATLAB 3.9 is the 3. 2-3. 9; same = GV 2 = result as in Eq. —R AnaabV 2 b — a (3.98) (3.52). Conceptual Questions (on Companion Website): Exercises (on (3.77), 3.1 1-3.17; Companion Website). DUALITY BETWEEN CONDUCTANCE AND CAPACITANCE (b) Figure 3.13 Resistor with two coaxial layers (a), which can be represented as two homogeneous resistors connected in parallel for Example 3.7. (b); Consider a pair of metallic bodies (electrodes) placed in a homogeneous conducting medium of conductivity o and permittivity e, as shown in Fig. 3.14. The conductivity of electrodes is much larger than cr. Let the voltage between the electrodes be V. From Eq. (3.62), there is no volume charges ( = 0) in the medium (a and e are constants). Having in mind Eq. (3.57), the current (and field) lines are normal to the surfaces of electrodes. We now use the duality relationship between the current Section 3.9 Duality between Conductance and Capacitance / / \ \ Figure 3.14 Two metallic homogeneous conducting medium. electrodes in a D, in the medium, Eq. (3.61), to and capacitance, C, between the electrodes. Applying the continuity equation for steady currents, Eq. (3.40), and generalized Gauss’ law, Eqs. (3.60), to an arbitrary surface S completely enclosing the density vector, J, and electric flux density vector, relate the conductance, G, positive electrode (Fig. 3.14) gives (3.99) where I is the total current leaving the positive electrode through the conducting medium (and entering the negative electrode), Q is the total charge of the positive electrode (the total charge of the negative electrode outside the integral sign because By dividing this equation by V, it is we a constant, is i.e., — Q), and o/e can be brought medium is homogeneous. 8 the obtain L= V °_Q £ (3.100) v' which, by means of Eqs. (3.73) and (2.113), yields the following duality relationship G and between C: (3.101) £ Note charges that, since the definition of C, Q Eq. (2.113), depends on the existence of is independent of whether or not a cur- and — Q on the electrodes and medium (of permittivity e), G and C in Eq. (3.101) do not necessarily represent the conductance and capacitance of a system with an imperfect dielectric of parameters a and e. They can also be associated with two independent dual systems, representing the capacitance (C) between a pair of rent also exists in the dielectric electrodes when placed in a perfect dielectric (of permittivity e, including the case = £o, and zero conductivity) and the conductance (G) between the same electrodes when placed in a conducting medium (of conductivity o and totally arbitrary £ permittivity), respectively. 8 Note, however, that this a/e = const. is true also for inhomogeneous media for which functions a and e are such that duality of G and C 147 148 Chapter 3 Steady Currents Electric R between In terms of the resistance, the electrodes, Eq. (3.101) can be , rewritten as R-C analogy (3.102) where C Q given by Eq. (3.68), is the relaxation time of the material between the elecshows the circuit-theory representation of the system in Fig. 3.14. a first-order capacitive circuit, whose time constant (r c = RC) is equal, by r, trodes. Fig. 3.15 -Q This is virtue of Eq. (3.102), to the relaxation time of the material (r The and (3.102) are very useful relations in Eqs. (3.101) in = e/o). deriving expressions conductance and resistance of electrode configurations for which we already have the capacitance, or vice versa. For example, from the expression for the capacitance of a spherical capacitor with dielectric permittivity e (and zero conductivity) and electrode radii a and b (a < b), given in Eq. (2.119), we can immediately determine the expression for the conductance of the same capacitor if filled with a conducting medium of conductivity a\ for Figure 3.15 Network representation of the system in Fig. 3.14. 4ncrab G= —C= b e [the same as in Eq. (3.96)]. Self-Discharging of a Nonideal Capacitor Example 3.9 A (3.103) —a parallel-plate capacitor is filled with an imperfect dielectric of relative permittivity e r -14 = 6 and conductivity a — 10 S/m. The capacitor was connected to an ideal battery and fully charged to a voltage of 20 V between its plates, and then disconnected from the battery. Find the time after which the voltage of the capacitor decays to 1 V. Solution The capacitor can be represented by the equivalent network in Fig. 3.15, where C and R are the capacitance and resistance between the capacitor plates, respectively. With no battery in the circuit, the current discharging C is also the current across R, which, with help of Eqs. (3.45) and (3.72), gives —dv— v =0, I dt where the time constant rc equation = RC is found from Eq. (3.104) rc (3.102). The solution of this differential is self-discharging of a capacitor v(/) = v(0) e _r/(/?C) = V (0)e~ (<7/S) '. (3.105) with an imperfect dielectric Finally, we we take the natural logarithms of both sides of the equation, and obtain the time seek: t =— where v(0) We = 20 V and v(f) = 1 = In er 265 minutes, (3.1 06) v(0) V. charged capacitor with homogeneous imperfect dielecis completely independent of its shape (geometry) and size, and completely analogous to redistribution of charge in a conductor described by Eq. (3.67). The capacitor voltage, v(t), and charge, Q(t) = Cv(r), decrease exponentially with time, at a rate that is set by the relaxation time of the lossy dielectric filling the space between tric that realize that discharging of a is left to itself (open-circuited) the capacitor electrodes. Conceptual Questions (on CD): 3.18-3.21; Website). MATLAB Exercises (on Companion External Electric Energy Section 3.10 3.10 149 Volume Sources and Generators EXTERNAL ELECTRIC ENERGY VOLUME SOURCES I AND GENERATORS We observed in connection with chemical battery) is Fig. 3.1 that an external voltage source (e.g., a required to maintain a steady current in a conductor. This source creates a rise in potential in the circuit and a potential difference between the ends of the conductor. From the energy standpoint, an external source of energy is necessary to maintain steady electric current flow through the circuit by continuously supplying the energy that is then dissipated in the conductor as Joule’s heat. In concrete situations, various forms of the external energy (chemical energy, mechanical energy, thermal energy, converted light energy, etc.) are and ultimately lost to heat. analogy to circuit-theory generators, we use two field-theory mod- to the energy of the electric field in conductors Generally, in els of volume-distributed energy sources capable of transmitting energy to electric and sources analogous to current charges: sources analogous to voltage generators generators. Consider a source region of volume v shown in Fig. 3.16(a) in which a nontermed the impressed force, acts on charge carriers (e.g., electric external force, Fj, and separates positive and negative excess charges. An example is the on electrons in a metallic wire moving in a magnetic field (as we shall see in a later chapter). We can formally divide F, by the charge of a carrier, q (q = — e for electrons), and what we get is a quantity expressed in V/m: electrons) force Ej = — (3.107) , q which we call the impressed electric field intensity vector. This although of the same dimension and unit as E, due to charges). Since there force on q is is is in the domain v, A however, not a true electric field (a due to charges also a field field, (b) field the total Figure 3.16 (a) External electric Ftot = F; + qE = (/(Ej + E), (3. 08) energy sources modeled by an impressed electric field, of intensity Ej. and compels the charges to move through the region. The the region is, therefore, given by this is the force that current density vector in J = <r(E + Ej), (3.109) (b) Voltage generator in circuit theory. Ej - impressed electric field intensity vector (unit: where a is the conductivity of the material in v. It is very important to note that Ej does not depend on J. The model with a volume distribution (throughout v) of energy sources modeled by an impressed electric field, of intensity Ej, can be used in many different electromagnetic situations, as we shall see later in this text. However, we note here that the particular situation in Fig. 3.16(a) actually represents a voltage generator in circuit theory. The generator is connected to a resistor of resistance R. Combining Eqs. (1.90) and (3.109), the voltage between the ends B and A of the source region (generator), can be written as dl. Reversing the order of integration FB a = limits, we have B f / JA r Ej - (3.110) dl - / B JA -i <r dl. (3.111) ’V/m) 150 Chapter 3 Steady Electric Currents The first term on the right-hand side of motive force (emf) of the generator: equation this is, by definition, the electro- B electromotive force (unit: £ - emf V) = f JA Ej-dl. (3.112) that £ equals the work done by the external force, Fj, in moving the charge q through the generator, from its negative terminal (A) to its positive terminal (B), divided by q. The unit for emf is V. The second integral in Eq. (3.111) is linearly Note proportional to the current / in the circuit: J = Rg I dl a where the constant of proportionality, R g generator. Hence, Eq. (3.111) becomes , is FBA = £ - voltage generator (3.113) , called the internal resistance of the RgI. (3.114) shows the equivalent circuit-theory representation of the generator. the generator terminals A and B are open-circuited (R —» oo), no current flows through the generator (/ = 0), and we can write [Fig. 3.16(b)] Fig. 3.16(b) Note that if (^BA) open-circuited Thus, the An no its emf of = ^CA =£ the generator also equals ideal voltage generator is its voltage (1 = (3.1 0). when internal losses, implying that the voltage of such a generator it. That is ( Rg = for all values of I impressed electric and all i.e., is, (3.116) values of R. This, essentially, field intensity, Ej, 0), always equal to Vba=S ideal voltage generator 5) open-circuited. one with a zero internal resistance emf, irrespective of the current flowing through 1 is a consequence of the being completely independent of the current density, J, in the generator. By Ohm’s law, Eq. (3.72), on the other hand, the voltage Tba in Fig- 3.16 can be written as F BA = RI, (3.117) which, combined with Eq. (3.114), gives £ = Rg I + RI. (3.118) many emf sources), we have In general, for a closed path in a circuit with (including the internal resistances of the Kirchhoff's voltage terms of emf's law sources and resistors M N — $ Rkhyi in and voltage y=l (3.119) k=l drops Rf This equation is an expression of Kirchhoff’s voltage law, Eq. (1.92). It states that sum of the emf’s (voltage rises) around a closed path in a circuit equals algebraic sum of the voltage drops (RI) across the resistances around the path. Multiplying both sides of Eq. (3.1 18) by /, we obtain the algebraic the £1 = Rg I 2 + RI 2 . (3.120) Section 3.10 By Joule’s law, Eq. (3.77), the expressions appearing External Electric Energy on the right-hand Volume Sources and Generators 151 side of this equation represent the power of Joule’s losses in the generator and that in the resistor of resistance R, respectively. Both these powers are lost to heat. Therefore, by the principle of conservation of energy, the expression on the left-hand side of the equation, = £I, Pi (3.121) must be the power generated by the emf of the generator. In local form, the volume 3 density (in W/m ) of the power of the impressed electric field of intensity Ej is [see Eq. (3.31)] (3 dv The power of sources total in the domain v - 122 ) thus [see Eq. (3.32)] is J dv. (3.123) power generated by impressed Let now field the charges in a source region be carried by an impressed nonconduc- tion current of density as depicted in Fig. 3.17(a), Ji, independent of the electric model of energy sources we use in 9 electromagnetics. The current density Jj does not enter Ohm’s law in local form, and the total current density vector in the region is given by 10 This field intensity, E. is the second general J — E + Jj, (3.124) cr Jj - impressed current density vector (unit: A/m 2 ) which, in scalar notation [for the situation in Fig. 3.17(a)], becomes J This is = —oE + J[. analogous to a current generator (3.125) in circuit theory, shown in Fig. 3.17(b) and described by I where Ig and respectively. circuited 0), is (V = -G g V + Ig = 0). that /g equals the current of the generator with its terminals short- An ideal current generator has a zero internal conductance (G g = or an infinite internal resistance. This means that the current of such a generator always constant, i.e., it is A = h. (3.127) independent of the generator terminal voltage (and of the conductance G). example classical is so-called Van de Graaff generator, where the impressed current consists of charges transported mechanically on a moving dielectric belt, so that the velocity at which the belt moves actually represents an equivalent drift velocity (vd) of charges. Here, however, the velocity of charges (i.e., the velocity of the belt) obviously does not E, and Eq. (3.2) does not 10 current generator G g are the current intensity and internal conductance of the generator, Note I 9 (3.126) , Note make depend on the electric field intensity vector, sense. that both Eq. (3.109), for a region with energy sources modeled by an impressed electric field, and Eq. (3.124), for a region with an impressed current, can be regarded as versions of the general constitutive equation for the current density, Eq. (3.21). In other words, Eq. (3.21) generally includes models of vol- ume energy of materials. sources based on impressed fields and impressed currents, along with conduction properties ideal current generator 5 152 Chapter 3 Steady Electric Currents The power density of sources represented by an impressed current of density amounts to Pi = —E • Jj (3.128) Jj, where the minus sign is necessary because vectors E (the electric field intensity vector, due to positive and negative charges) and J; (the impressed current density vector, which, like any other current density vector, carries positive charges in direction and negative charges For the situation in the = EJ in Fig. 3.17(a), p\ its opposite direction) are directed oppositely. . X The total E power generated by Jj power of sources in v is dv. (3.129) impressed current Note that the power generated by 7g in Fig. 3.17(b) Pi This power is = VIg is given by (3.130) . delivered to the rest of the circuit and dissipated to heat in the two resistors, the total power of Joule’s losses being G g V 2 + GV 2 . ANALYSIS OF CAPACITORS WITH IMPERFECT 3.1 1 INHOMOGENEOUS DIELECTRICS We now (a) / deal with steady current fields in capacitors containing piece-wise homo- geneous and continuously inhomogeneous imperfect (lossy) dielectrics. Generally, charges in these systems exist not only on the surfaces of capacitor electrodes, but also on the boundary surfaces between homogeneous dielectric layers and over the volume of continuously inhomogeneous dielectrics. The analysis starts with the evaluation of the current distribution, i.e., the computation of the current density vector, J, in the dielectric. The electric field intensity vector, E, is then determined by Ohm's law in local form. By integrating E through the dielectric, we find the voltage between the electrodes and the conductance (and resistance) of the capacitor. Finally, the distribution of charge in the system can be found from the electric flux density vector, D, using the generalized Gauss’ law in differential form and the corresponding boundary condition. As an illustration, consider a parallel-plate capacitor with two imperfect dielectric layers shown in Fig. 3.18(a). Let the permittivities of the layers be e\ and £ 2 their conductivities o\ and 02 and thicknesses d\ and d 2 respectively. The surface area of each of the plates is S. The plates can be considered as perfect conductors. Let us analyze this capacitor, assuming that it is connected to an ideal generator of , , , (b) Figure 3.17 (a) External electric energy sources modeled by an impressed electric current, of density Jj. (b) Current generator in circuit theory. time-invariant voltage V. The current density vector in the dielectric is normal to the plates [Eq. (3.57)] and uniform in each dielectric layer [Eq. (3.82)]. From the boundary condition for the normal components of J, Eq. (3.55), applied to the interface between the two dielectrics and the continuity equation for steady currents in integral form, Eq. (3.40), applied to a rectangular closed surface enclosing the positive plate [Fig. 3.18(a)], with the right-hand side positioned in either one of the dielectrics, we have . JX where / is -J2 =J = the current through the capacitor. ~, The (3.131) electric field intensities in the dielectrics are FEl = L = ±_ / a, / a, and E2 = -L_. — = —. (72 (72 S (3.132) Section 3.1 1 Analysis of Capacitors with Imperfect Inhomogeneous 153 Dielectrics Figure 3.18 Analysis of a capacitor with a two-layer imperfect dielectric: geometry (a) of the structure with vectors J, and D and (b) E, individual layers equivalent circuit of the (b) (a) The voltage across the capacitor V= is given by E\d\ S\o so that the conductance of the capacitor G_ The 11 V (3.1 33) 02 / to be crio2 S _ J_ + i comes out (3.134) o2 di+oid2 electric flux densities in the layers are D = e\E\ = system. s\GV and i 12 £2 D 2 = s 2 E2 = GV (3.135) o2 S CTlS Eq. (3.62) tells us that, since the dielectric layers are homogeneous, there is no volume charge in them. Using Eq. (3.63), the surface charge on the boundary surface between the layers amounts Psi2 to — D 2 - Di ( £2 V° 2 r _ £l \ 04 / j _ (£204 ~ &\d2 ei<*2)V -f- o2 d\ (3.136) where we take Jn = —J in Eq. (3.63) because the unit vector n in the corresponding boundary condition is directed from medium 2 to medium 1. Note that if £\/o\ — £ 2 /o2 Psl2 = 0. From Eq. (2.58), the surface charge densities on the plates are , Psi =D\ = The system £\Q2V o\d2 + o2 d\ and ps2 = -D 2 = - £204 0\d2 F + CT2^1 (3.137) can be explained and analyzed also invoking the and concepts. The interface between the dielectric layers is equipotential, and can, therefore, be metalized. We thus get two nonideal capacitors in series, each of them being represented by a parallel connection of an ideal in Fig. 3.18(a) circuit-theory point of view 11 Note that the current field in the resistor of Fig. 3.12(a) can be analyzed in the same way. Note that D\ ^ D 2 contrary to Eq. (2.147) for the same capacitor with two perfect dielectric layers, shown in Fig. 2.25(a). The capacitor with a perfect dielectric represents an electrostatic system (with no current), and the analysis starts with the generalized Gauss’ law. The capacitor with an imperfect 12 , dielectric represents a system with steady current, and the analysis starts with the continuity equation. in , 154 Chapter 3 Steady Electric Currents capacitor and an ideal resistor, as shown in the equivalent circuit in Fig. 3.18(b). The characteristics of the elements in the circuit are: Ci eiS = r C2 = £2 S = d2 r2 = aiS’ A, CT2 (3.138) S where the resistances R and R 2 can be obtained either from the corresponding capacitances, C\ and C2 [also see Eqs. (2.157)], using the relationship in Eq. (3.102), or from Eq. (3.85). Using the expression for the equivalent total resistance of a series connection i of resistors, Eq. (3.86), the conductance of the system 1 G= R\ The charges + R2 is = cr 2 d\ (3.139) + o\d2 of the capacitors in Fig. 3.18(b) are computed as C\R\I = C\R\GV = — GV, Qi = C2 V2 = C2 R 2 I = C2 R 2 GV = — GV a 01 = Ci V\ = (3.140) (3.141) , 2 with V\ and V2 standing for the voltages across individual dielectric layers. 13 Finally, the surface charge densities in Fig. 3.18(a) are given by the expressions 0i Psi = y. Psi2 which give the same results as Example 3.10 —— = 02-0i in Eqs. (3.136) A and i and -Qi =— — ps2 (3.137). Spherical Capacitor with a Continuously Inhomogeneous Imperfect Dielectric A is filled with a continuously inhomogeneous imperfect dielectric. The and conductivity of the dielectric depend on the distance r from the capacitor center and are given by the expressions e(r) = 3eo b/r and o(r) — aob 2 /r (a < r < b), where a and b are radii of the inner and outer capacitor electrodes, and cto is a (positive) constant. The capacitor is connected to a time-invariant voltage V. Find: (a) the conductance of the spherical capacitor permittivity bound charge capacitor, (b) the free charge distribution of the capacitor, (c) the distribution of the dielectric, and (d) the total free charge in the capacitor. Solution (a) Because of spherical symmetry, the current density vector given by Eq. (3.46). From the continuity equation, J(r) that the electric field intensity vector in the dielectric E= J r ^Trj with / 13 Note same that the voltages of the elements in Fig. 3.18(b) are = VR 2 /(R\ + r 47r<ro b in the entire dielectric in the dielectric is of the the same form as in Eq. (3.48), so (Fig. 3.19) = 4na(r)r2 being the current intensity of the capacitor. turns out to be the is is 2 = Er, (3.143) We note that the electric field intensity E = const). The voltage between the ( determined by the resistors: V\ = VR\/(R\ + Ri) - resistive voltage divider. In the case of the same system with perfect dielectric layers [Figs. 2.25(a) and 2.26(a)], which is an electrostatic system, the voltages are determined by the capacitors: V\ = VCi/{C\ -I- C2 ) and Vj = VC\/(C\ -I- C2 ) - capacitive voltage divider. Rl) and a Section 3.1 Analysis of Capacitors with Imperfect 1 Inhomogeneous / Figure 3.19 Spherical capacitor with a continuously inhomogeneous dielectric with losses; for Example 3.10. electrodes and the capacitor conductance amount b V= — Edr = E(b -a) [ to 14 G Ja (b) The electric flux density vector in the dielectric 1 4 na0 b 2 = (3.144) =V ~b^ given by is D = <•(/•)£?, where E = V/(b — a). By means density inside the dielectric volume charge of Eqs. (3.62) and (1.171), the free is E The (3.145) 3 e 0 bV d (3146) on the surfaces of the inner and outer capacitor free surface charge density electrodes are Psa = D(a + = ) e(a + )E = a(b - and ps t, = —D(b - ) a) 3g V — s{b )E = — 0 b — a' (3.147) respectively. (c) From Eq. (2.59), the polarization vector in the dielectric is given P= Using Eq. [g(r) bound volume charge (2.19), the - e 0 ]Er. (3.148) density over the ldr ?„1 „ by volume of the dielectric is — 2 r) b (30 £oU(3 l (b (3.149) - a) while Eq. (2.23) gives us the bound surface charge density on the surfaces of the dielectric: Ppsa (d) The — total free ‘ r Q= f P{& +. charge ) _ — - a)V a(b-a) eo(3 b and pp sft = 7 ) = 2e 0 V (3.150) b-a Q in the capacitor is zero. To prove that, we write b p(r) 471^ dr j P(b v +psa Sa + psb Sb = \2neobV + llnsoabV 12tt£o b b “r 2 —a V - 0 , dv (3.151) 14 Note that, since e(r)/a(r) capacitance C is found in ^ const, G Example 2.19. is not proportional to C, i.e., Eq. (3.101) is not satisfied. The Dielectrics 155 156 Chapter 3 Steady Electric Currents where dv is adopted to be the volume of a spherical shell of radius r and thickness dr [see 2 Fig. 1.9 and Eq. (1.33)], Sa = 4jra (surface area of the inner electrode), and Sb = 4nb 2 (surface area of the outer electrode). Problems: 3.10—3.16; Conceptual Questions (on Companion Website): 3.22 and MATLAB Exercises (on Companion Website). 3.23: ANALYSIS OF LOSSY TRANSMISSION LINES WITH 3.12 STEADY CURRENTS In this section, we analyze transmission lines with losses in a time-invariant (dc) regime. Consider, as an example, a coaxial cable with imperfect conductors of (finite) conductivity oc and an imperfect dielectric of (nonzero) conductivity 04. radius of the inner conductor of the cable is a, The the inner and outer radii of the outer conductor are b and c, respectively (a < b < c), and the length of the cable is /. The cable is fed by a time-invariant (dc) voltage generator of electromotive force £ and internal resistance R g as shown in Fig. 3.20. A load of resistance R l is connected to , the other end of the cable. Let I designate the current flowing The same current returns through cable is imperfect, there the cable conductors. is between symmetry of the structure, this current is and its density, J, depends only on the radial a leakage (stray) current through the dielectric, Due to cylindrical radial with respect to the cable axis distance r from the axis. through the inner conductor of the cable. the outer conductor. Since the dielectric in the The current leakage, in turn, causes a continual decrease meaning that I is a function of the coordi- of the current intensity / along the cable, nate z along the cable, I cylindrical closed surface — I(z). Applying the continuity equation, Eq. S of radius r and length A z (Fig. 3.20) gives I(z + Az) — a/ [currents I(z in + Az current leakage in a imperfect dielectric Figure 3.20 Coaxial cable with dc regime. 0 (3.152) rd Az) and /^Az leave 5, while /(z) enters it], where A 1 is the change A z, and I'd is the leakage current per unit length (p.u.l.) A/m). Hence, (^d)p.u.l.> transmission line with an imperfect conductors and = current I along a distance of the cable (in dielectric in a I{z) +72flrr (3.40), to a (3.153) / / Section 3.1 2 157 Analysis of Lossy Transmission Lines with Steady Currents and J A. = (3.154) 2rtr Using Ohm’s law the dielectric is Eq. in local form, computed (3.18), the electric field intensity vector in as —= E= I' (3.155) r, 2no&r cr(j so that the voltage between the inner and outer conductor of the cable amounts to V ^ln». a 27rad Edr = -S' (3.156) By definition, the conductance per unit length of a transmission conductance for one meter (unit length) of the line divided by 1 m, is , G' line, i.e., the /j = Gpul = ^. (3.157) , conductance per unit length of a transmission line (unit: also referred to as the p.u.l. leakage conductance. It is cable, from Eq. The unit is S/m. For a coaxial S/m) (3.156), 27T<7d = G' (3.158) In (b/a) conductance p.u.l. of a coaxial cable The drop of the current / along a distance AG = is Az equals A I = A GV, where G'Az (3.159) the conductance through the dielectric at the length Az. As Az approaches zero, the expression becomes the derivative of and (3.157), we have on the left-hand side of Eq. (3.153) combining Eqs. (3.153) / with respect to z. Therefore, d = -G'V. (3.160) dz This is transmission line with a lossy dielectric a first-order differential equation in z for the current and voltage of a trans- mission line with a lossy dielectric. along the line is It tells us that the rate of change of the current proportional to the negative of the line voltage, with the conduc- tance per unit length of the line as the proportionality constant. In the case of a transmission line with a perfect dielectric (no losses in the dielectric), G' that dl dz = 0, = 0, so yielding I Because of losses varies along the cable, in the i.e., — conductors V= V(z). const. (l/crc (3.161) / From Eq. 0), the voltage between conductors transmission line with a perfect dielectric (3.85), the total resistance (for time- invariant currents) of the cable turns out to be R = R 1+ R 2 = -L + -1-, (3.162) where R\ and R 2 are the resistances of the inner and outer conductor, respectively, and 5] and S 2 are cross-sectional surface areas of the conductors. The resistance per unit length of the line is given by R' = R p.u.l. R 1 (3.163) resistance per unit length of a transmission line (unit: £2/ m) < 158 Chapter 3 Steady Electric Currents (the unit dc resistance p.u.l. is £2/m), so that of a R' { 1 = The resistance along the length Az + \a ttoc coaxial cable 1 1 of the line AR = (3.164) ci~^bi is R'Az, (3.165) and the voltage drop across that resistance - V(z) + V(z = A RI. Az) (3.166) This equation can be rewritten as V(z + Az) - V(z) = -R'AzI, (3.167) AV which, as Az approaches zero, becomes transmission line with lossy (3.168) conductors We see that the rate of change of the voltage along a transmission line with lossy conductors is proportional to the negative of the line current, with the resistance per unit length of the line being the proportionality constant. For a transmission = conductors (PEC), R' line with perfect electric V transmission line with perfect 0 and dK/ dz = 0, so that - const. (3.169) conductors Eqs. (3.167), (3.153), and (3.157) of length Az can be represented by resistance R'Az and a shunt (parallel) resistor of transmission line can be replaced by in Fig. 3.21. us that each section of a transmission line tell a circuit cell consisting of a series resistor of Thus, a transmission line conductance G' Az, and the entire many cascaded is equal small cells, as indicated said to be a circuit with distributed (parameters per unit length), and Eqs. (3.160) and (3.168) are Kirchhoff’s laws and Ohm’s law for the cells of the circuit. ters in fact paramebased on Eqs. (3.160) and (3.168) are called transmission-line equations or telegrapher’s equations for time-invariant currents and voltages on transmission transmission line with both R! ^ 0 and G' lines. For a these two equations are coupled dif- 0, two unknowns: /(z) and V(z). By taking the derivative with respect to z of one equation and substituting its right-hand side by the corresponding expression from the other equation, we can eliminate one unknown (current ferential equations in l(z-Az) o A,A/\ R'Az /(z+Az) /(z) —+ >-VVv —+ rvA t R'Az V(z-Az) V(z+Az.) V{z) G'Az o < G'Az. — > c> Figure 3.21 Circuit-theory representation of a transmission line with losses in conductors and dielectric in a dc regime. r 1 R'Az z+Az z-Az Az —L G'Az ^ 1 Section 3.12 159 Analysis of Lossy Transmission Lines with Steady Currents or voltage) and obtain the following second-order differential equations with only voltage and only current as unknowns: —Vt -R'G'V = d 2 d 1 2 dz z — and 0 * dzl - R'G'I = 0 (R' 0, G' # 0). (3.170) Their solutions are exponential functions for the voltage and current along the R! If = Eq. (3.160) is and R' 0,1 line. V= const along the line [Eq. (3.169)], and the solution to / 0, a linear function for the current along the line. Conversely, if G' = 0 0 and G' = const along the line [Eq. (3.161)], and the solution to Eq. (3.168) is = 0, a linear function for the voltage along the line. Finally, if both G' —0 and R' both current and voltage do not vary along the line. Note that using the duality relationship in Eq. (3.101), the leakage conductance per unit length of a transmission line with homogeneous imperfect dielectric can be found from the capacitance per unit length of the line as (3.171) where e is duality of G 1 and C the permittivity of the dielectric [for a coaxial cable, see Eqs. (3.158) and Note also that R' ^ 1/G' in Eqs. (3.170). The power of Joule’s losses per unit length of a transmission line, P), can be obtained as the power of Joule’s losses in one cell in Fig. 3.21, divided by A z. The unit is W/m. Specifically, in conductors of the line (series resistor in Fig. 3.21), (2.123)]. ( A RI2 = Az pj) c = R'l 2 (3.172) . joule's or line ohmic losses p.u.l. in conductors (unit: W/m) In the dielectric (shunt resistor in Fig. 3.21), A GV2 Az «)d The total power of Joule’s 3.1 (3.173) = R'l2 + G'V2 is hence (3.174) . Transmission Line with Perfect Conductors and Imperfect Dielectric A transmission line of length and conductance per unit length G is connected to an ideal voltage generator of time-invariant emf £. The other end of the line is terminated in a load of resistance Pl- The losses in the conductors of the line can be neglected. Find: (a) the 1 l distribution of current along the line conductors, (b) the total line, and (c) the power of Joule’s losses in the power of the generator. Solution (a) With reference to Fig. 3.22, voltage along the line is 15 since there are 15 no losses in the line conductors, R' = 0, the [Eq. (3.169)] V(z) Although resembles a two-wire line, a =£ (0 <z<l). pair of parallel horizontal lines in Fig. 3.22 (3.175) is a symbolic repre- sentation of an arbitrary two-conductor transmission line, with conductors of completely arbitrary cross section and generally inhomogeneous dielectric. Joule's (ohmic) losses p.u.l. in line dielectric (unit: losses per unit length of the line P'j Example G'V2 W/m) . 160 Chapter 3 Steady Electric Currents m Figure 3.22 Transmission m j line with lossless conductors and lossy dielectric in a for Example 3.1 dc regime; 1 Eq. (3.160) becomes 2L = -<*, (3.176) az and by integrating it, we obtain — -G'£z + In I(z) (0 <z<l). the condition /(/) = V//?l — £/Rl (the line current through the load), the integration constant is found to be /0 (3.177) From at z = = G' £1 / equals the current + £/Rl [/o = 7(0) is the current of the generator], so that = I(z) G'£(l ~ R -z) + (R' = 0, a£ (3.178) 0). L (b) Based on Eq. (3.173), the P'j As = power of Joule’s (Pj) = G'V 2 = G’£ 2 = d fj does not vary along the the losses in the load, losses per unit length of the line line, comes out Pi= the total const (0 < power of Joule’s z < is given by (3.179) /). losses in the line, including to be ^l + C= v line G/+ ( ^)^ 2 (3 ' ' 180) ’ load (c) Eq. (3.121) tells us that the power of the generator is P =£1(0) =e(g'£1+ (3.181) { Note that Pi which, of course, is in =P 3 (3.182) , agreement with the principle of conservation of energy and power; namely, the power generated by the generator and dissipated is delivered to the rest of the circuit, to heat in the lossy dielectric of the transmission line and in the resistive load. Example 3.12 Thin Two-Wire Line with Losses Determine the resistance and conductance per unit length of a thin symmetrical two-wire line with lossy conductors and a lossy dielectric. The conductivities of both conductors and of the dielectric are erc and oj, respectively. The conductor radii are a and the distance between their axes is d (d a). The dc resistance two-wire p.u.l. of resistance per unit length of the line R' a thin = [see also Eq. (3.164)]. twice that of a single wire: 1 2 ac na 2 line is (3.183) — Analysis of Lossy Transmission Lines with Steady Currents Section 3.1 2 161 Figure 3.23 Cross section of a thin symmetrical two-wire line with imperfect dielectric coatings immersed conducting Example for From — — C' — n<Td Conductors of a thin two-wire o \ The of conductivity (3.1 ' . is is a, Conducting Liquid the thickness of coatings » a). The line d (d in a coated with thin coaxial layers of imperfect dielectric line are radius of each wire distance between wire axes 84) immersed is is also a , and the a2 in a liquid of conductivity . the conductance per unit length of this line? The evaluation of the (leakage) conductance per unit length of the line, whose is shown in Fig. 3.23, can be performed completely in parallel to the evaluation of the capacitance per unit length of the line in Fig. 2.31. Here, instead of Q' we have I'd - the Solution cross section current that leaks from conductor ductor 2 of the conductors, line, i.e., per unit of its 1 of the line through the coatings and the liquid to con- length. The leakage current densities the current density due to the leakage from conductor leakage into conductor 2, due to the individual and that due to the 1 M in Fig. 3.23, are h = id_- evaluated at the point _ V _k_ Ji 2.71 and T' 2n(d x — x) , (3.185) The total current density along the x-axis between the conductors is J = J\ + J2 The electric field intensity is E — J/o\ in the coatings and E = J/a2 in the liquid. The voltage between the conductors, V, is calculated as in Eq. (2.182), and the conductance per unit length of the line comes out to be respectively. . G' Note — V =7z ( In 2 + \ a\ — a2 that this expression for G' can be obtained also ln^-^) 2 a) (3.186) . by substituting the permittivities in the expression for C' in Eq. (2.183) by the corresponding conductivities. Namely, the inhomogeneity in terms of o in the system in Fig. 3.23 in terms of e in the system in Fig. 2.31, and a is of the same form as the inhomogeneity between the conductance and sort of duality capacitance of the two systems can be exploited. Example 3.14 Planar Line with Two Imperfect Dielectric Layers shows a planar transmission line consisting of two parallel metallic strips of width and a two-layer dielectric between the strips. The strips are perfectly conducting, whereas both dielectric layers are imperfect, with conductivities o\ and a2 The thicknesses of layers are d\ and d2 and the length of the line is /. The line is connected at one end to an ideal voltage generator of time-invariant emf £. The other end of the line is open. Calculate (a) the conductance per unit length of the line and (b) the current along the strips. Fig. 3.24 w . , Solution (a) An application of the continuity equation to a rectangular closed surface Fig. 3.24, much like that in S portrayed in Eq. (3.152), yields JwAz = I(z) - I(z conductance two-wire line Two-Wire Line with Imperfect Coatings Example 3.13 is In (d/ a) e is 3.1 3. Eqs. (3.171) and (2.141), the conductance per unit length of the line G' What in a liquid; + A z) = I' d Az (3.187) p.u.t. of a thin 162 Chapter 3 Steady Electric Currents Figure 3.24 Planar transmission line with a two-layer imperfect dielectric; for Example 3.14. ( I'd is the leakage current per unit length of the line), from which, the current density in both dielectric layers is /' /= A w (3.188) Having in mind Eqs. (3.133) and (3.157), the voltage between the conductance per unit length of the line are obtained as 1/ = —di + a1 (b) —d 2 = rs2 The current along W the line ( d A + ?l) V CTl is — w (— + 02 ) V<Tl given by Eq. (3.178) with I(z) = G'£(l w£{l -z) = d\/o\ — and the 1 . o2 'j ) °o (the line - strips is (3.189) open): z) +d2 /o2 (3.190) ’ Problems 3.17-3.22; Conceptual Questions (on Companion Website): 3.24-3.26; : MATLAB Exercises (on Companion Website). 3.1 3 GROUNDING ELECTRODES Consider a metallic body (electrode) of arbitrary shape buried under the flat surface body may represent a grounding electrode for some electrical or 16 electronic device (residing on the earth’s surface). Assume that the conductivity of the earth is the same, cr, in the entire lower half-space. Let a time-invariant current of intensity / flow from the electrode into the earth (this current is supplied to the electrode through a thin insulated wire), as shown in Fig. 3.25(a). As the conductivity of the electrode is much larger than a, Eq. (3.57) tells us that the current lines leaving the electrode are perpendicular to its surface. From Eq. (3.58), on the of the earth. This other side, the current lines (parallel) to the l6 Grounding boundary in the earth near to its boundary with air are tangential surface. electrodes, in general, are used for draining static electricity induced on parts of a device, for equalizing the electric potential of a device housing with the neighboring conducting objects, for reducing the so-called conduction interference (undesired currents induced “regular" currents in in a device, interfering with the device and with neighboring devices), for routing (together with lightning rods) the charge flow in lightning to the ground, etc. Section 3.1 3 Grounding Electrodes 163 Figure 3.25 Grounding electrode with a steady current (a) and the equivalent system with the earth-air boundary removed (b) (a) Consider now the system shown in Fig. 3.25(b), obtained by taking an image of the lower half-space of Fig. 3.25(a) in the earth-air boundary plane. Namely, the grounding electrode in the lower half-space is the same as in the original system and a symmetrical electrode (image electrode) with respect to the boundary plane is introduced in the upper half-space, while the entire space around the electrodes is filled with the medium of conductivity a. The current of the image electrode is the same as that of the original By symmetry, image). one (/), and it flows also out of the electrode (positive the total current density vector (due to both electrodes) at an arbitrary point of the plane of symmetry in the system in izontal component only. This to the plane of symmetry, in the system means and in Fig. 3.25(a). Fig. 3.25(b) has a hor- that the resultant current lines are tangential plane corresponds to the ground-air interface conclude that the distributions of current in the this We lower half-space in the systems in Figs. 3.25(a) and (b) are the same. This is the soimage theory (or theorem) for steady currents, which states that an arbitrary called steady-current source (e.g., a grounding electrode with a steady current flowing into the ground) situated in a conducting half-space near a flat boundary surface with a nonconducting medium (e.g., earth-air interface) can be replaced by a new system of sources in an infinite conducting medium. The new system consists of the original source configuration itself and its positive image in the boundary plane. The image electrode (i.e., the charge on its surface) actually represents an equivalent for the surface charge on the boundary between two half-spaces in the original system. There is no direct way to determine this charge and take its contribution to the total electric field into account in the original system, and that is why the equivalent (homogenized) system, with the boundary removed, is generally much simpler for analysis. After the current density vector, J, in the earth for a given current I of a grounddetermined by analyzing the equivalent system in Fig. 3.25(b), all other quantities of interest for a specific application can be found - in the original system in Fig. 3.25(a). The electric field intensity vector in the ground and on its surface is E = i/a. An important parameter of a grounding electrode is its grounding ing electrode is (b) - image theory steady currents. for 164 Chapter 3 Steady Electric Currents resistance, defined as R gr — grounding resistance ^gr (3.191) 1 ’ where Egr is the potential of the electrode with respect infinity. According to Eq. (1.74), this potential is given by to the reference point at infinity potential of a grounding V,gr -I E • (3.192) dl. electrode electrode (reference point at infinity) By integrating E along the field lines on the earth’s surface between two points that are a distance equal to an average person’s step apart, we get the so-called voltage of a step, Estep, around the grounding electrode. This voltage corresponds to the potential difference between the feet of a person walking in the direction along the field lines on the earth’s surface. above a grounding electrode is current intensity of the electrode even fatal The maximum voltage of a step in the region a parameter important for applications may where cause exceedingly large values of a large Estep , with consequences. Example 3.15 Hemispherical Grounding Electrode A hemispherical grounding electrode of radius a = m is buried in 1 the earth of conductivity — 10 -3 S/m, with its base up, as shown in Fig. 3.26(a). The current of the electrode is I = 1000 A. Find (a) the grounding resistance of the electrode and (b) the voltage of a step between points on the earth’s surface that are r i=2m and r2 = r\ + b away from the electrode center, where b = 0.75 m (average person's step). a Solution (a) Using the image theory for steady currents, Fig. 3.25, we get the equivalent system in Fig. 3.26(b), which consists of two hemispherical electrodes pressed onto each other, forming thus a spherical electrode with current 2/ flowing into a homogeneous medium of conductivity a. Due to spherical symmetry, the current density vector, J, in the medium is radial. Consequently, J in the original system is also radial and of the form given by Eq. (3.46). Applying the continuity equation to a spherical surface S of radius r shown Figure 3.26 (a) Hemispherical grounding electrode and (b) equivalent spherical electrode in a homogeneous medium; Example 3.1 5. for (a) (b) Section 3.13 165 Grounding Electrodes positioned around the electrode gives J(r) = ~ (a<r<oo). (3.193) 2tt^ 4tt7^ that the same result is obtained by applying the continuity equation to a hemisphersurface of radius r in the original system - in Fig. 3.26(a). Note ical Combining Eqs. (3.18) and (3.193), the electric field intensity in the earth £ (r = ) With the use of Eqs. (3.192) and the reference point at infinity and V* = (b) r°° L The voltage of a Eir step is - = Anam a (3-194) 2 (3.191), the potential of the electrode with respect to its grounding resistance come out to be _ v*_ — R gr — A / =^Ta 1 = 159 Q. grounding resistance of a hemispherical electrode as large as lb = 21.7 kV = r2 — r\). (b 2nar\r2 Two (3.195) 2Jioa Edr = Example 3.16 is (3.196) voltage of a step; b Hemispherical Grounding Electrodes emf £ = 100 V, two hemispherical grounding electrodes. The radius of each electrode is a = 2 m, the conductivity of the earth is a — 10 mS/m, and the distance between the electrode centers is d = 100 m. Compute the current in the circuit. The electric circuit shown in Fig. 3.27 consists of an ideal voltage generator of a wire of negligible resistance, and Solution Since d^> a, the potentials with respect to the reference point at infinity of the individual electrodes can be evaluated independently isolated grounding electrode. fore 2 7? gr is , The equivalent where the expression for /? gr is from each other, that is, as for a single resistance seen by the generator equals there- given in Eqs. (3.195). As the resistance of the wire negligible, the current in the circuit is 1 Example 3.17 = = naa£ = 6.28 A. Hemispherical Grounding Electrode (3.197) in a Two-Layer Earth Conductivity of a layer of earth around a hemispherical grounding electrode shown in Fig. 3.28 is o\ = 5 x 10 -3 S/m and that of the rest of the earth a2 = 10 -3 S/m. The of the electrode and the boundary surface between the earth layers are a b = 2 m, respectively. The current is average person 's step intensity of the electrode is I = = lm radii and 100 A. Determine (a) the grounding resistance of the electrode and (b) the power of Joule’s losses in the earth. + £ / Figure 3.27 cr Two hemispherical grounding electrodes in a dc circuit; for Example 3.1 6. an 166 Chapter 3 Steady Electric Currents Figure 3.28 Hemispherical grounding electrode in an earth v two concentric Example 3.1 7. consisting of layers; for Solution (a) Because of symmetry, the current flow both layers of earth in is radial. Applying the we obtain the same expression for the current density J in the earth as in Eq. (3.193). The electric field intensity vector in the earth is Ei = i/a\ for a < r < b and E 2 = i /02 for b < r < 00 By means of continuity equation to the closed surface S shown in Fig. 3.28, . Eq. (3.192), the potential of the electrode with respect to the reference point at is given by rb V, & Edr = -F roo Ei dr + 1 ( Using Eq. (3.32), — ^ resistance, Eq. (3.191), 2nb (b) = Jb Ja from which the grounding £2 dr / and combining b-a \ cj\a j (3.198) ,o\ is j_ 95.5 (3.199) fi. 02 with Eq. (3.193), the it infinity power of Joule’s losses in the earth turns out to be Pi f JEdv (°° = where dv Fig. 3.28). power is the — -= Jv 2nr2 Ja volume of a hemispherical Noting that the Ej Edr, I / (3.200) J, r shell of radius r last integral in this of joule's losses in the E2nr2 dr = and thickness dr (portrayed in Tgr [Eq. (3.198)], we have equation equals = Tgr 7 = E gr 7 2 (3.201) , earth surrounding a and Ej grounding electrode = 955 Example 3.18 kW for given numerical data. Spherical 17 Grounding Electrode Deep in the Earth A spherical grounding electrode of radius a = 30 cm is buried in a homogeneous earth of cona — 10 -2 S/m such that its center is at a depth d = 6 m with respect to the ground ductivity surface, as and its maximum l7 shown in Fig. 3.29(a). There is a steady current flowing potential with respect to the reference point at infinity tangential component of the Although derived for a specific resistance Rot and current /. Note electric field intensity vector grounding electrode, that of Joule’s losses in the earth in Eq. (3.201) that this is l/gr is is in Fig. 3.28, through the electrode, = 15 kV. Calculate the on the earth’s surface. the expression for the power true for an arbitrary grounding electrode, with grounding actually an expression of Joule’s law, Eq. (3.77). 3 Section 3.1 3 Grounding Electrodes Figure 3.29 167 (a) Spherical grounding electrode buried deep in the earth and (b) equivalent system - by virtue of the image theory for steady currents; for Example 3.1 8. Solution Since d^> a, the influence of the boundary with air meaning of the electrode can be neglected, in the infinite that R gr homogeneous medium of conductivity on the grounding resistance can be evaluated as for an electrode a. Invoking the duality relationship in Eq. (3.102), we can further relate R gr to the capacitance (C) of the same metallic sphere situated in an infinite dielectric medium of permittivity s. Eq. (2.121) gives this capacitance for s = eo, so that f? gr £p — =26.5 oC 4 noa Hence, the current of the electrode is I — Vgx /R gx = (3.202) £2. spherical electrode buried deep 565.5 A. and the superposition principle, the resultant current density vector at a point in Fig. 3.29(b), the position of which is determined by a radial distance r from the projection of the electrode centers on the plane of symmetry (point O), can be obtained as Applying the image theory for steady currents (Fig. 3.25) M J where J originai and + Jimagei ^original 3.2 33 due to the current flow from Jimage are the current density vectors Making use of the vidual electrodes. — fact that d^> indi- these current densities can be evaluated a, independently from each other, which results in Jo 'original (cos a = The r/R and — J« = •'image — R= tangential yjr 2 + d2 J ^2 ^ 2/original COS a — (3.204) 2 R3 ). component of the the original system, Fig. 3.29(a), electric field intensity vector on the earth’s surface in is _ 1 From — ^ J Ir _ 2na(r2 a the condition dE t (r) + J2 ) (3.205) 3 /2 = 0, (3.206) dr we £ is maximum on maximum field is get that the field intensity centered at the point O. This t (Ft)niax — Ft(r ma x) Note that the maximum voltage of a is a circle of radius r max — /V 9nad2 step, taking = 96 V/m. = d/sfl = 4.24 m, (3.207) 0.75 m as an average person’s step, = 72V, (3.208) b = approximately (Fstep) max ^(F t grounding resistance of a )m ax b in the earth 168 Chapter Steady 3 Currents Electric which much is less than the voltage found because the electrode in Fig. 3.29(a) is in Eq. (3.196), for instance. This buried deep Problems'. 3.23-3.29; Conceptual Questions (on MATLAB Exercises (on is expected, in the earth. Companion Website): 3.27-3.30; Companion Website). Problems 3.1. Charge density 3.5. ductor. Consider a hollow steel-reinforced alu- minum wire conductor of length / = ity o coordinates, = does not, e in the region density / const, = (e/<r)Va of the conductor is a = 5 mm, the thickness = 1.1 MS/m) rein- of the stainless-steel (aste ei W. • forcement is b — a = 5 mm, and Various computations for a copper wire with conductor, whose overall radius steady current. For a copper wire with a steady aluminum Example density of the power power of ohmic material, (e) the total the wire, (f) wire room temperature, and (g) the wire resistance at the _1 temperature of 100°C (ac u = 0.0039 K ). A an with capacitor Parallel-plate dielectric. homogeneous imperfect has find (a) is the resis- the ratio of — 10 5 S/m and 02 =4 x 10 5 S/m. The voltis V = 50 V. age between the resistor terminals Find the electric a field intensity, current density, current intensity, and power of Joule’s losses in dielectric of permit- each of the two cuboids if they are connected as in (a) Fig. 3.30(a) and (b) Fig. 3.30(b). e invariant), is = 2 cm, is Resistor with two cuboidal parts. A resistor is formed from two rectangular cuboids of the same size, with sides a = 8 mm, b = 2 mm, and c — 4 mm. The cuboids are made out from dif- <j\ and conductivity o. The separation between plates is d and the plate area is S. If the voltage between the plates is V (timetivity c is MS/m). What ferent resistive materials, with conductivities imperfect capacitor parallel-plate 35 ends of the wire? 3.6. losses in the resistance of the wire at = the rest of the current intensities lA\/htee\ in th e two materials for a given dc voltage drop between the two the electric field intensity in the wire, (c) the of Joule’s losses in the (ctai tance of this wire, and what compute: (a) the number of electrons that pass through a cross section of the wire during one hour, (b) 3.1, voltage between the wire ends, (d) the volume 100 m. The radius of the cylindrical hole in the central part const. If the electric potential given by p is whereas permittiv- V, show that the volume charge is current described in 3.3. Hollow steel-reinforced aluminum wire con- conductivity of the material varies with spatial 3.2. terms of the conductivity in gradient. In a region with steady currents, the current distribution the in the dielectric, (b) the current through the capacitor terminals, (c) the conductance of the capacitor, (d) the in power of Joule’s losses the capacitor, (e) the free charge distribu- tion in the capacitor, and (f) the bound charge distribution in the capacitor. 3.4. Spherical capacitor half ing liquid. A filled with a conductFigure 3.30 spherical capacitor has a poorly conducting liquid dielectric occupying a half of the space between the electrodes. electrodes are a and b (a ductivity of the dielectric conductance of < is The this capacitor. rectangular cuboids parallel; for Problem connected made out from in (a) series and (b) 3.6. radii of and the cona. Determine the b), Two different resistive materials 3.7. Conductivity gradient along the resistor current. Fig. 3.31 shows a right-angled cuboidal resistor of dimensions a, b, and c, made out 169 Problems from a continuously inhomogeneous given by the following function coordinate: o(x) where cto is = cr0 /(l + 9x/a) layer near the inner electrode are s x \ -4 4 x 10 and ctj S/m, and those of is = of the x- <x < (0 and conductivity of the relative permittivity resistive The conductivity of the material material. The capacitor a constant. Calculate the resistance V= voltage of this resistor. 12 the and CT2 = 5x 1CT 6 S/m. connected to a time-invariant =7 other layer e r2 a ), = is 100 V. (a) Find the current distri- bution in the dielectric. Using this result and integration (field-theory approach), compute conductance of the capacitor, (c) the power of Joule’s losses in each of the layers, and (d) the free charge on each of the boundaries of radii a b and c. (b) the , 3 11 . . , Solution using equivalent circuit with ideal ele- ments. Repeat parts (b)-(d) of the previous problem but using the equivalent circuit with two ideal capacitors and two ideal resistors in Figure 3.31 Right-angled cuboidal resistor made out from inhomogeneous Problems 3.8. 3. Fig. 3.18(b) (circuit-theory that the conductivity of the inhomogeneous imperfect shows a parallel-plate capacitor with circular plates of radius a and a continuously inhomogeneous imperfect dielectric. The permittivity and conductivity of the function of the dielectric are the following functions of the 3 . 12 resistive material; for Assume material in Fig. 3.31 y-coordinate, a(y) y < b, cto = const), a is = cto[1 + 9sin(7ry/h)] and (0 < find the resistance of Continuously 3.9. Integrated-circuit conductivity with resistor An profile. exponential integrated-circuit by diffusing a layer of ptype impurity into an rc-type background material. As a consequence of the diffusion process, the concentration of impurity is not uniform (IC) resistor is built across the treated region, so that the obtained 3.32 Fig. and ct(z) = where cto is a constant the separation between the plates. The z -coordinate: e(z) — 2(1 T 3 z/d), 0 < z < ct0 / (1 and d the resistor. IC . dielectric. 7-3. 9. Conductivity gradient normal to the resistor current. approach). a continuously is capacitor is + 3z/d)so d, connected to a time-constant volt- age V. Find (a) the current distribution in the dielectric, (b) the (c) the conductance of the capacitor, power of Joule’s losses in the capacitor, (d) the free charge distribution in the capacitor, and (e) the bound charge distribution in the capacitor. can essentially be represented by resistor the structure in Fig. 3.31 with the conductivity varying as cr=a(y). In specific, a(y) exponentially decreases from air-resistor surface to 02 = o\ = 0.1 100 S/m at the S/m at the inter- face with the n-type background, which is con- sidered to be nonconducting, and the length, Figure 3.32 and width of the p-type region are a = 4 p m, b — 1 pm, and c — 2 p,m, respectively. Determine the resistance of the IC Parallel-plate capacitor thickness, with a continuously inhomogeneous resistor. 3.10. Spherical capacitor with layers. The two = 5 cm and c = 15 cm is of two concentric imperfect dielec- where the radius of the boundary surface between the layers is b — 10 cm. The tric layers, Problem 3.1 2. with two lossy dielectric dielectric of a spherical capacitor with electrode radii a composed lossy dielectric; for 3.13. Parallel-plate dielectrics. capacitor homogeneous dielectric The permittivities of the are sj and £ 2 Both parts tor with a piece-wise shown in Fig. 3.33. dielectric lossy Consider a parallel-plate capaci- parts . 1 70 Chapter 3 Steady Electric Currents or 2 The d and the are lossy, with conductivities o\ and separation between the plates is cable - is / of the cross-sectional areas of the dielectric parts are 100 m. The relative permittivities are — = eri conductivities the and Si The voltage between the electrodes is V. By employing a field-theory approach (based on determining the current and field = layers are = 4 and e T 2 o\ = 10 -12 8, and S/m and 12 5 x 10~ distribution in the dielectric), calculate (a) the S/m. The cable is connected at an ideal voltage generator of timeinvariant emf £ = 30 V, and the other end of the cable is terminated in a load of resistance current through the capacitor terminals, (b) the R]_ conductance of the capacitor, (c) the power of Joule’s losses in each of the dielectric parts, and (d) the free charge distribution in the capacitor. in the dielectric, (b) the Si (72 . one end = to Compute 1 (a) the current density conductance per unit length of the cable, (c) the current intensity along the conductors, (d) the current density in each of the conductors, (e) the power of Joule’s losses in the cable and in the load, (f) the power of the generator, and (g) the free surface charge Figure 3.33 density on the boundary between the dielectric Parallel-plate layers. capacitor with two imperfect £ dielectric parts; for Problem c + d Rl b 3.1 3. 1ft. 3.14. Equivalent circuit with ideal capacitors and ei,cr\ Repeat the previous problem but employing a circuit-theory approach, i.e., generating and solving an equivalent circuit with ideal capacitors and resistors. e2 resistors. , o- 2 / Figure 3.34 Coaxial cable with two coaxial imperfect 3.15. Current distribution through circular plates. Find the distribution of dielectric layers in a the upper and lower plate of the parallelplate capacitor with imperfect dielectric fect dielectric. conductors Current distribution through a thin spherical Consider the spherical capacitor Example 3.3, and assume that the inner electrode is hollow, with a very thin wall. With this assumption, determine the distribution of current through the inner electrode. Represent this current using the surface current density vector. two lossy dielectric layers. shows a longitudinal cross section of a coaxial cable with two coaxial layers of imperFig. 3.34 dielectric. conducting, c = the tric 10 The conductors and their mm, and d = 12 radii are perfectly are mm. The a = 2 3 mm, radius of boundary surface between the dieleclayers is b = 7 mm and the length of the A coaxial cable with perfect filled is with an inhomogeneous = £q(1 + /a 2 ) charges 3.19. in the dielectric. Coaxial cable partly tric. 3.17. Coaxial cable with fect 3.1 7. and ad = cro/(l 4- r2 /a 2 ) (a < r < b), where a and b are the radius of the inner cable conductor and the inner radius of the outer conductor, and oq is a constant. The voltage between the cable conductors is V (dc). Calculate (a) the conductance per unit length of the cable and (b) the density of volume free r vector. with imperfect dielectric from Problem imperfect dielectric of parameters e be described using the surface current density electrode. for an inhomogeneous imper- 3.18. Coaxial cable with from Problem 3.12. Assume that the plates are thin enough so that the current through them can 3.16. dc regime; through current is A filled with a lossy dielec- coaxial cable with perfect conductors partly filled with an imperfect dielectric of conductivity < 7 , as shown ductor radii are a and in Fig. 3.35. b, the The con- cable length is /, and the length of the air-filled part of the cable is c. The cable is fed by an ideal dc voltage generator of emf £, and the other end of the cable is open. Find the expression for the current intensity through the cable conductors. 171 Problems the c and strips, (c) volume free charge in the dielectric. <j b da l filled with a condc regime; for Problem 3.19. Figure 3.35 Coaxial cable partly ducting dielectric 3.20. in a Coaxial cable with a poorly conducting spacer. Shown in Fig. 3.36 air-filled is line with a continuously inhomogeneous imperfect dielectric in a dc regime; a cross section of an coaxial cable with a spacer the conductors. Figure 3.37 Planar The spacer is between made 3.23. Grounding electrode earth. out of for in Problem 3.22. an inhomogeneous Conductivity of the earth around a hemispherical grounding electrode shown in an imperfect dielectric of conductivity a and its cross section is defined by an angle a. The conductor radii are a and b. What is the conductance per unit length of this cable? Fig. 3.38 can be described as the following func- tion of the radial coordinate (a < r < oo), trode and where a cto is is a constant. of the electrode o(r) = o^sfajr The current Find is I. r. the radius of the elec(a) the intensity grounding resistance of the electrode, (b) the total of Joule’s losses in the earth, and (c) the power power of Joule’s losses in a layer with thickness a around the electrode Figure 3.36 Cross (a < r < 2d). section of a coaxial cable with a poorly conducting spacer between conductors; Problem 3.20. Figure 3.38 for Hemispherical grounding electrode 3.21. Planar line with imperfect conductors. If the planar transmission line in Fig. 3.24 in a are not perfectly conducting, but have a finite earth; for conductivity ac find the dc resistance per unit Problem 3.23. strips of the , length of the each of the 3.22. Planar line dielectric. line. Assume strips is with that the thickness of t. 3.24. Grounding electrode in a two-sector earth. The earth around a hemispherical grounding electrode of radius a = 1 m can be represented as two sectors with conductivities 10~ 3 S/m and 02 = 8 x 10~ 3 S/m, as cti = 2 x shown in Fig. 3.39. The current intensity of the electrode amounts to 1 = 50 A. Compute (a) the grounding resistance of the electrode and (b) the power of Joule’s losses in each of the two earth sectors. 3.25. Two an inhomogeneous lossy Consider a planar (two-strip) trans- mission line shown in Fig. 3.37. The strips are perfectly conducting, very thin, w wide, and d apart from each other. The dielectric between the strips is imperfect and continuously inhomogeneous, with parameters e(x) = (4 + 3x/d)£Q and a(x) = cro/(l + 9 x/d), 0 < x < d, where cro is a constant. The line is connected to an ideal dc voltage generator of emf £, and is open-circuited at the other end. Calculate the distributions of (a) volume cur/ continuously inhomogeneous long, rent in the dielectric, (b) surface current over short-circuited grounding electrodes. Two grounding electrodes of radii a are galvanically connected to each other by a wire of negligible resistance, as identical hemispherical 172 Chapter 3 Steady Electric Currents 3.27. Two deep spherical grounding electrodes. Consider two identical spherical grounding homogeneous Both the distance between the sphere centers and their depth with respect to the ground surface equal d, where d^> a. What is the resistance between the two electrodes? metallic electrodes of radii a in a earth of conductivity a. Figure 3.39 Hemispherical grounding electrode in a two-sector earth; for shown earth is trode Fig. in Problem 3.24. The conductivity of 3.40. the 3.28. Cylindrical earth. o and the distance between the elecDetermine the is d ( d ~S> a). radius a centers grounding resistance of such a system A 10 electrodes. d 20 cm and -2 = S/m such — / = 5 m is buried 4 m 200 = depth below the ground surface, as depicted that in Fig. 3.42. If a I length homogeneous earth of conductivity a in a of = grounding electrode deep in the grounding electrode of cylindrical its axis is at a steady current of intensity A flows through the electrode, find the maximum voltage of a step on the earth’s sur- — m as an face. Adopt b step. Neglect end effects due to the 0.75 average person’s finite length of the cylinder. Two Figure 3.40 electrodes; for 3.26 short-circuited hemispherical Two grounding The grounding Problem 3.25. electrodes in a layered earth. electric circuit shown in Fig. 3.41 consists of an ideal current generator of current intensity / = 100 A, a wire of negligible resistance, and two identical hemispherical — electrodes of radii a 2 m. grounding The conductivity Figure 3.42 Cylindrical grounding electrode buried deep of the earth near the electrodes, within the radius b = 4 m, is o\ = = 10~ 2 S/m. The conduc-3 elsewhere is 02 10 S/m. The distance between the electrode centers is d = 80 m. What is the voltage between the electrodes? tivity Figure 3.41 Two hemispherical grounding electrodes layered earth; for Problem 3.26. in a in the earth; for Problem 3.28. 3.29 Two deep parallel cylindrical electrodes. Two grounding electrodes as the one in Fig. 3.42 are placed parallel to each other (at the same depth with respect to the earth’s surface) so that the distance between their axes is D = 6 m. If they are connected together in a circuit with an ideal current generator as in Fig. 3.41 (with I — 200 A), compute (for the numerical data given in the previous problem) the voltage between the electrodes. identical cylindrical 4 Magnetostatic Field in Free Space [ Introduction: n our studies of steady electric currents in preceding chapter, we introduced and discussed physical laws and mathematical techniques for determining the distribution of currents for given geometry, material properties, and excitation I the from the experimental law of the magbetween two point charges that move free space (a vacuum or air) - Coulomb’s law Starting netic force in for the magnetic field, we shall derive the Biot- Savart law, which, in turn, will serve as a starting of various structures. We now introduce a series of new phenomena associated with steady electric cur- point for the derivation of Ampere’s law. Both which are essentially the consequence of a new simple experimental fact - that conductors with netic field rents, currents exert forces on one another. These forces are called magnetic forces, and the field due to one current conductor in which the other conductor is and which causes the force on it is called the magnetic field. Any motion of electric charges and any electric current are followed by the magnetic field. The magnetic field due to steady electric currents is termed the steady (static) magnetic field or magnetostatic field. The theory of the magnetosituated static field, the magnetostatics, restricted to a vac- uum and nonmagnetic media chapter. is the subject of this The magnetic materials in the following chapter. will be discussed laws represent a means for evaluating the mag- due to given steady-current distributions, and we shall apply them to many theoretically and practically important configurations that do not include magnetic materials. The differential form of Ampere’s law will introduce a new differential operator, the curl. We shall derive the law of conservation of magnetic flux (Gauss’ law for the magnetic field), and complete the full set of Maxwell’s equations for the magnetostatic field in nonmagnetic media. The magnetic vector potential will be introduced as a counterpart of the electric scalar potential and the magnetic dipole as the equivalent of the electric dipole. Examples involving evaluation of magnetic forces and torques will also be presented. 173 174 Chapter 4 Magnetostatic Field in Free Space MAGNETIC FORCE AND MAGNETIC FLUX DENSITY VECTOR 4.1 The magnetic around moving fundamental property is that there is a force acting on any electric charge placed in the space, provided that the charge is moving. That wires carrying electric currents produce magnetic fields was first discovered by Oersted in 1820. To quantitatively describe this field, we introduce a vector quantity called the magnetic flux density vector, B. It is defined through the force on a small probe point charge Q p moving at a velocity v in the field, which equals the cross product of vectors Q \ and B, p field is a special physical state existing in a space electric charges and, thus, electric currents in conductors. Its 1 definition of B (unit: Fm T) This force is = Qp \ x called the magnetic force. B The (<3 p unit for note that the magnetic flux density vector, which induction vector, is -> is (4.1) 0). B is tesla (abbreviated T). We also referred to as the magnetic defined analogously to the electric field intensity vector, E, in electrostatics [Eq. (1.23)]. The basis for determining the magnetic flux density vector due to steady curan experimental law describing the magnetic force between two point charges that move in a vacuum. This law represents the equivalent of Coulomb’s law, Eq. (1.1), of electrostatics, and is an underpinning for the entire magnetostatic theory. With reference to Fig. 4.1, it states that the magnetic force on a point charge Q 2 that moves at a velocity \2 in the magnetic field due to a point charge Q\ moving with a velocity vj in a vacuum (or air) is given by rents is magnetic force Fml2 = Ho Q2V2 X (Q1V1 X the permeability of a permeability of a vacuum vacuum is the is (4.2) same as for Coulomb’s law in Fig. 1.1, and /xo is (free space), /xo (H ) R2 47r where the notation of vectors R 12 = 47t x 10 7 H/m (4.3) henry, the unit for inductance, which will be studied in a later chapter). mentioned, is the result of experiments, but it can also be derived from Coulomb’s law using the special theory of relativity. Note that it is sometimes referred to as Ampere’s law of force. Although of the similar form, the magnetic force law in Eq. (4.2) is mathematically more complicated than Coulomb’s law, because two vector cross products are implied in the equation. We first need to determine the vector resulting from the Eq. (4.2), as cross product Q\\\ x R[2, and then the cross product of (?2 V 2 and that vector [and, 2 of course, multiply the result by hq/(At: R )]. Let us apply this rule to the situation shown Figure 4.1 Magnetic force between two moving point in a vacuum, given by Eq. (4.2). in Fig. 4.2, mal to the where vectors Q\v\ and 02 v 2 are parallel to each other and northem. We see that the force between the charges is attractive line joining charges 'The cross product (also referred to as the vector product) of vectors a and b. a x b, is a vector whose is given by |a x b| = |a| |b| sin or, where a is the angle between the two vectors in the product. It is perpendicular to the plane defined by the vectors a and b, and its direction (orientation) is determined by the right-hand rule when the first vector (a) is rotated by the shortest route toward the second vector (b). In this rule, the direction of rotation is defined by the fingers of the right hand when the thumb points magnitude in the direction of the cross product. 175 Magnetic Force and Magnetic Flux Density Vector Section 4.1 HISTORICAL ASIDE Hans Christian Oersted (1777-1851), physicist, a was a professor and chemthe at istry University of Copenhagen. As of a classroom demon- part a wire with a current is also deflected in a magnetic field (by the magnetic force), thus laying the foundation for the construction of the electric motor. His epoch-making discovery of the magnetic effects of electric currents was announced to the French Academy of Sciences on September 4, 1820. It immediately set off an explosion of research activity by many bril- he discovered, apparently by accident, that an electric current flowing through a wire produces a magnetic field around the wire. Namely, while preparing an evening lecture on electricity and magnetism in April of 1820, in which he wanted to demonstration, a wire connected Ampere (1775-1836), Biot (1774-1862), and Savart (1791-1841). Oersted’s experiment was the between the terminals of first demonstration of a con- a voltaic source (battery), Oersted noticed that nection between electricity and magnetism, and the magnetic needle of a compass sitting next considered the foundation of the modern study to the wire spun off of the north position every of electromagnetism. In 1829, Oersted time the battery was in use. the was generally believed at that time that electricity and magnetism were not related. Therefore, he was extremely surprised and excited with what he if minds, including liant strate the heating effects of the electric current in continued experimenting during that which eventually led him to the conclusion that an electric current creates a magnetic field -and electromagnetism was born. He also discovered that, not only is a magnetic needle deflected by an electric current, but that lecture, physics of He saw. Danish the vectors are in the opposite directions same It direction [Fig. 4.2(a)], This [Fig. 4.2(b)]. is first is became director of the Polytechnic Institute in Copenhagen, now the Technical University of Denmark. (Portrait: AIP Emilio Segre Visual Archives, Brittle Books and repulsive Collection) they are in if formally just opposite to Coulomb’s law, where like charges repel and unlike attract each other. Combining Eqs. (4.1) and (4.2), and assuming that the second charge in Fig. 4.1 is a probe charge ( Q 2 = Q ), we can identify the expression for the magnetic flux p density vector of a point charge Q moving with a velocity v: fio 4 rt Qx x R R2 where R is the distance from the charge and R from the source point toward the field point, perpendicular to R and P0 Qx R 12 ml2 B due charge the unit vector along is as shown in Fig. 4.3. perpendicular to the plane of the vectors R to a moving point in free space directed Vector Q\ and B R. is Its P0 ,V 2 Vl (4.4) f Qx <22 _ Q\\x x R12 - Rli ® R 1 X F ml2 Figure 4.2 Two charges moving parallel to each other in the same direction (a) and in R 12 opposite directions (b) (a) (b) vacuum. in a 1 76 Chapter 4 Magnetostatic Field in Free comes from the orientation B Figure 4.3 Magnetic flux difference density vector due to a point in a definition of the cross product of vectors, i.e., it is vacuum. The the direction of the field vector with respect to the source. E, due to field intensity vector, magnetic flux density vector, B, other words, the lines of E Q is radial with respect to due to Qy is circular with Q electric whereas the (Fig. 1.7), respect to <2v (Fig. 4.3). In are radials starting at Q, while the lines of B are circles centered on the line containing the vector Q\. Conceptual Questions (on Companion Website): 4.1 and 4.2. HISTORICAL ASIDE Nikola ant American was inventor, for- of He made in (ac) and had about Croatia (then in the Habsburg Empire of Austria), He area of polyphase known U.S. alternating patents currents engineer 250 patents issued in the U.S. and other countries. Tesla was born in Smiljan near Gospic, Lika, in Serbian family. the and polyphase systems, including ac generators, alternating current practical in motors, and transformers, as well as principles no complete mal education. obtained most of his best brilli- a electrical with (1856- Tesla an 1943), in a studied mechanical engi- power distribution using alternating currents, 1887 and 1888, and delivered his famous lecture “A New System of Alternate Current Motors and Transformers” before the American Institute Engineers (now IEEE) on May June of 1888, George Westinghouse (1846-1914), head of the “Westinghouse Electric of Electrical 16, 1888. In Company” in Pittsburgh, bought the first seven of Tesla’s patents for polyphase systems. This marked the beginning of the five-year “war of Polytechnic School in Graz from currents” to decide whether Edison’s existing dc 1875 to 1878 and then natural philosophy at the University of Prague until 1880, but never gradu- systems or newly proposed Tesla- Westinghouse ac systems would be the chosen technology for the ated due to a shortage of funds after his father’s global electric death. While working as electrical engineer for in he came up with an idea of the principle of the rotating magnetic field and polyphase alternating currents producing it in 1882 and constructed the world’s first induc- nant transformer neering at the companies tion (ac) hours. Budapest and Paris, motor He four cents for in in 1883, both during after-work emigrated to America in 1884, with in his pocket, Thomas Edison and worked (1847-1931). He get Edison interested in the induction for a year failed Company” in New York to motor and alternating current, and later established his “Tesla Electric is determined by the right-hand rule in rotating Q\ by the shortest route to R. Note that the magnetic flux density is proportional to the product Q\, which can be regarded as a measure of sources of the magnetostatic field (the measure of sources of the electrostatic field is a charge Q ). Comparing Eq. (4.4) to the expression for the electric field intensity vector due to a point charge, Eq. (1.24), we observe the same dependence on the amount of sources (£9|v| and Q ) and on distance R. The constant no corresponds to the constant 1/eo- The only formal Q charge moving Space own City. Tesla power distribution of the future America. Tesla invented transformer) erating in known 1891, which alternating an air-core reso- as a Tesla coil (or Tesla currents was capable of genof extremely high frequencies for that time (up to a hundred of thou- sands of cycles per second or 100 notation). The coil kHz in today’s could also build up tremen- dously high voltages and cause spectacular spark discharges. Tesla’s new electric lamps based on a high-frequency power supply were the forerunners of our fluorescent tubes and neon signs. In a series of spectacular lectures-demonstrations in America Section 4.2 and Europe from 1891 to 1893 aimed at promoting alternating current and high-frequency technology, he was able to power electric lamps without wires, either through air or by allowing high-frequency electricity to flow through his body, produce artificial lightning flashes of different shapes, and even shoot large lightning bolts from his coils to the audience without harm. In early 1893, he proposed a complete radio system with a transmitter and receiver in the form of Tesla coils tuned to resonate at the same frequency as a basis for wireless communications. The “war of currents” ended on the evening of May 1, 1893, with the opening of the Chicago World Exposition, spectacularly illuminated by a hundred thousand lamps with alternating current from Tesla’s generators. Only couple of months later, Westinghouse was awarded Tesla’s is, less telegraphy. In 1898, Tesla world’s first it demonstrated the wireless remote control (of a boat model) at Madison Square Garden. He continued his investigations of the wireless transmission of power in 1899 and 1900 in Colorado Springs, where he built a huge, 200-kW radio transmitter consisting of a 142-foot metal mast with a spherical top and the world’s largest Tesla coil 51 feet in diameter. Tesla also patented a bladeless disk turbine, so-called Tesla turbine, in 1913, as well as many other devices and systems in various netic flux density, in 1960. (Portrait: © 16, first science and engineering. Nikola Tesla Museum, Belgrade, Serbia) 1896. Tesla filed his basic radio LAW we have many point charges moving is, (/L is £v) = Nin dv Qv d = Nv dv Q\ d = J dv, in charges in dv is computed (4.5) dv We the current density vector, defined by Eq. (3.3). permeability everywhere 2 is /xq. Hence, the magnetic assume that the due to the flux density vector as dB = /xq (J An dv) x R (4.6) R2 which is known as the Biot-Savart law. We see that the product J dv represents a macroscopic volume elemental source of the magnetic field, and we call it accordingly the volume current element. By integrating Eq. (4.6), we obtain the expression 2 We shall see in the next chapter that most materials we encounter in science and engineering are nonmagnetic materials, that is, their permeability is practically that of a vacuum. Examples are commonly used metallic conductors such as copper, aluminum, silver, gold, etc., most frequently encountered natural environments (air, water, and ground), biological tissues, and practically all insulators and semiconductors. areas of ac hydro- BIOT-SAVART where J in fact, was Guglielmo Marconi (1874-1937) who put it into practical and commercial use, and was awarded Nobel Prize in Physics in 1909 for the development of wirethe invention of radio, although in free space, the total magnetic flux density by the principle of superposition, a vector sum of the magnetic flux density vectors due to individual charges. In a current, a vast number of elementary free charges moves through a conductor in an organized manner with the macroscopic average velocity v d (drift velocity). The sum of products Q\ for all charges in an elementary volume dv is given by If vector 177 power plant on Niagara Falls, based on design, which was put into operation on November 4.2 patent applications in 1897, and this Law The SI unit for the magthe tesla, was named in his honor the contract to build the world’s electric Biot-Savart Some materials, such as much greater than /xq. permeability iron, steel, cobalt, nickel, ferrites, etc., on the other hand, have Biot-Savart law 178 Chapter 4 Magnetostatic Field in Free Space Figure 4.4 Magnetic flux density vector due to three characteristic current distributions - volume current (b), and line current for the resultant entire volume surface current magnetic flux density vector due to a current distribution an in v [Fig. 4.4(a)]: Ho Biot-Savart law for (a), (c). C (J dv) x volume R 4 tv Jv current R (4.7) 2 HISTORICAL ASIDE experimental physics and acoustics, and a Jean Baptiste Biot (1774-1862), a French physicist, was a student of Lagrange (1736-1813) at the Ecole Polytechnique and a junior professor of mechanics under sponsorship of Laplace (1749-1827) at the College de France. Biot’s most important contributions to science are in the theory of polarized light and effects on member of the Paris Academy. His most important work was in vibration and sci- and particularly in the physics of the violin. Biot and Savart showed experimentally in October of 1820, soon after Oersted’s discovery of magnetic effects entific acoustics, of electric currents, that the magnetic field pro- it duced by a current by organic substances. Felix Savart (1791-1841), another French physicist, was also a professor at the College de France, where he taught in a long, straight wire is inversely proportional to the distance from the wire. S [see Eqs. (3.12) and dS instead of /V dv in Eq. yielding J s dS as a surface (3.13)], (4.5), v s current element, where J s is the surface current density vector. The magnetic flux In the case of a surface current flowing over a surface we have N density vector due to a surface current distribution A(o Biot-Savart law for surface f (J s dS) x current Finally, for a line current along a line e.g., /, thus [Fig. 4.4(b)] R (4.8) R2 4 n Js is a current of intensity / flowing through a (generally curvilinear) thin wire of length / and cross-sectional area 5, dv = S d/, J = I/S [Eq. (3.5)], and ./ dv = JS d/ = / d /, where d / is an elemental segment along /. We conclude that the line current element equals / dl, where dl is oriented in the reference direction of the current flow. For curved lines, dl is tangential to the line. The magnetic flux density vector due to a line current is hence [Fig. 4.4(c)] Wi B_ Biot-Savart law for line f / dl x 47T Ji current In summary, we note that Eqs. (4.7)-(4.9) form: integral of /^(current element) x R R2 (4.9) have the same generic mathematical R/(4nR 2 ), with the following three current Section 4.3 179 Magnetic Flux Density Vector due to Given Current Distributions Figure 4.5 Part of a planar current loop P in the C and same a field point plane. elements as different source functions: Jdv — « J s dS > — < > /dl, (4.10) volume, surface, and line A m) point in current elements (unit: and they all represent one physical law - the Biot-Savart law. In addition, the general form of the Biot-Savart law for line currents, Eq. (4.9), can considerably be simplified in the case of a planar current contour (contour lying in one plane) and a field point P (at which the magnetic flux density vector is calculated) in the same plane, shown in Fig. 4.5. Here, the vector B is normal to the plane of the contour (the plane of drawing), and its magnitude is given by S= Because (R dO is nt—*~ x R| = = d/sina d/cos/3 = MN = Rdd (4.12) the length of an arc representing a differentially small portion of a circle of R centered at the point P), we have B= where 0 (4.11) (Fig. 4.5) |dl radius x R| /|dl X Mo 4 is the angle between contour plane. The reference the reference direction of B R ™ 4n Jc and an B R arbitrarily corresponds to the reference flow of - field; thumb - current). Note that the integral expression in Eq. (4.13) implies only scalar integration and the function that needs to be integrated is much simpler than that in Eq. (4.9). We shall, therefore, use the simplified form of the Biot-Savart law whenever possible, i.e., when dealing with planar current contours and calculating the magnetic flux density vector in the same plane. Eq. (4.13) can also be used in cases when a planar part of a nonplanar contour and a field point are in one plane. (fingers Conceptual Questions (on Companion Website): 4.3 for a loop and one plane adopted reference axis in the I, and related to the direction of I by the right-hand rule rise of 0 is that 4.3. MAGNETIC FLUX DENSITY VECTOR DUE TO GIVEN CURRENT DISTRIBUTIONS Eqs. (4.7)-(4.9), with Eq. (4.13) added, are general means for evaluating (by superposition and integration) the magnetic flux density vector, B, due to field z . 180 Chapter 4 Magnetostatic Field in j Free Space given current distributions Eqs. ( 1 .37)— (1 .39) are the space (or any nonmagnetic medium), just as in free means for evaluating the electric field intensity vector, E, due to given charge distributions in free space. In this section, we shall consider various characteristic examples of the application of the Biot-Savart law. The examples include current distributions that are theoretically and practically important on one hand and for which the integrals can be evaluated analytically on the other. Most of the magnetostatic structures we shall analyze have their electrostatic counterparts in Section 1.5, so that many solution strategies introduced and developed in that section apply also here. Example Magnetic 4.1 Current Loop Field of a Circular Consider a circular loop of radius a carrying a steady current of intensity / in the magnetic flux density vector along the axis of the loop normal to its plane. We Solution field computation Example in 1.6. contribution to the magnetic flux density vector at a point current element / dl at a point P' on the contour yj 2 [X()l = dB R= With reference P on dl x R (4.14) 47tR 2 + a 2 (z is the coordinate of the point P). As = 1 (unit vector), |dl x R| = dl. Hence, the plane; for Example its and R are mutually permagnitude of the vector dB dl equals dB = along the axis of a circular current loop normal to to Fig. 4.6, the the loop axis (e-axis) of a is pendicular and |R| Figure 4.6 Evaluation of the magnetic flux density vector Calculate subdivide the loop (contour C) into elemental segments and apply Eq. (4.9), analogously to the electric where air. The total magnetic Mo 1 d/ 47tR- (4.15) flux density vector at the point P is obtained as 4.1 B = dB (4.16) Because of symmetry, the radial (horizontal) components of the vector cel out in the integral (vertical) (much 1.11, in like in Fig. dB in Fig. 4.6 can- the electrical case), and only the axial components d Bz = dflcos(90° -cr) = dfisina = dB^ = contribute to the final result. Consequently, B= which yields [see Eq. B due to ( 1 <j> d Bz i M() la B = 2(z loop Magnetic / in 2 2 + a2 )3/2 z Field of a Finite Straight Consider a straight conductor of length current of intensity Jc -43) a circular current Example 4.2 (4.18) dl. 4n R} / (4.19) - Wire Conductor representing a part of a wire contour with a steady free space. Find the expression for the B field at an arbitrary point in space due to this current conductor. Solution The conductor and an arbitrary field point (P) always determine one plane, which that we can use the simplified form of the Biot-Savart law in Eq. (4.13). From Fig. 4.7, means / cos 0 The = d/R, so that the magnetic flux density solution to this integral Mo f 4n J, 7d0 is given by MO R 181 Magnetic Flux Density Vector due to Given Current Distributions Section 4.3 r e2 cos 9 dO. (4.20) 4nd Je=ei is B= And — (sin 02 (4.21) sin0i), B due to a straight wire conductor of length finite 9\ and 02 are the angles defining the starting and ending point of the conductor, respectively, and d is the perpendicular distance from the conductor to the point P. The expression in Eq. (4.21) can be combined for computing the magnetic field due to any structure assembled from straight line segments with steady current. In addition, it can be used for the evaluation of the contribution to the total field of straight segments contained where in structures that also include curvilinear segments. By taking 9\ = —n/2 and 02 = n/2 and in Fig. 4.7 in Eq. (4.21), we obtain the expres- sion for the magnetic flux density due to an infinitely long straight wire conductor carrying a current 1 in free space. With a notation cylindrical coordinate d = r, standing for the radial coordinate in the r system whose z-axis coincides with the axis of the wire conductor, this expression reads MQ/ B (4.22) 2 nr' Example 4.3 Magnetic Field of a B due to an infinite wire conductor Square Current Loop A square loop of side length a in free space carries a steady current of intensity Obtain the I. expression for the magnetic flux density vector at the loop center. Solution square The magnetic sides, in Fig. 4.8. flux density at the loop center B i, is given by Eq. By means (4.21) with d = a/2, 6\ due to the current along = —n/ 4, and 02 = 7t/4, as one of the can be seen of the superposition principle, the magnitude of the total magnetic flux density vector amounts to (4.23) na with respect to the reference direction of B indicated in Fig. 4.8. Figure 4.7 Evaluation Example 4.4 Magnetic Field of a of the magnetic field of a Loop with a Semicircular Part finite straight shows a wire contour consisting of a semicircle (of radius a) and a straight line (of length 2a). The contour is situated in air and lies in the xy-plane of the Cartesian coordinate system. If the contour carries a steady current of intensity I, calculate the magnetic flux Fig. 4.9(a) wire conductor; for Example 4.2. ® _L density vector at an arbitrary point along the z-axis. i at a point P on the z-axis due element / dl that belongs to the semicircular part of the contour [Fig. 4.9(a)]. This vector is given by Eq. (4.14), and we need to break it up into components suitable for integration, which are x-, y-, and z-components in this case, and this is very similar to the decomposition of the vector dE in Fig. 1.12 and Eqs. (1.46) and (1.47). So, we first decompose Solution Let dB' denote the magnetic flux density vector to a current dB' dB' in the = plane of the triangle dB], + dB'z z, d B'h — APOP' [Fig. 4.9(b)]: dfi'cosa, cos a = R d B'z = dB' sin a, sin a = R (4.24) where dB' is given y-components [Fig. in dB'h = Eq. (4.15). We then represent the horizontal vector dBj, by its x- and Figure 4.8 Evaluation of the magnetic Example dB'x x field at the center of a square current loop; for 4.9(c)]: + dB' y y, d B'x = dB'h coscj), d B'v = d5],sin^. (4.25) 4.3. - 182 Chapter 4 Magnetostatic Space Field in Free Figure 4.9 Evaluation of the magnetic field contour with due to a current a semicircular a linear part; for Example and 4.4. Finally, we integrate, as in Eqs. (1.48)-(1.50) in the electrical case, the three Cartesian com- ponents of dB' along the semicircle, with respect to the azimuthal angle 0 {-n/2 < 0 < n/2) as the integration variable [note that d/ = ad0 in Eq. (4.15)]: Iaz tiQlaz ^° B- = /hr' dB- = ' / 4 fx 0 Iaz An R? t* f 12 ... Tiei-n/2 —R Mo laz T 5 27T = cos 0 d(p B = n/2 f . La**** Hence, the resultant magnetic comes out to be = B‘ = f , °- MO la B = due n/2 A f d0 = MO la 2 (4.26) to the semicircular part of the contour t’I+ 2 2 R? y ' I, flux density vector ’> I, noIa , dS < dB ’> , = 4^5 Ln 4/? 3 (4.27) Z ) The magnetic the other hand, is flux density vector at the point P due to the linear part of the contour, on determined from Eq. (4.21) with d = z, 0\ = —a, and 9j = a: MO^™ B = Zsinaf— x) = • M()/a , The total B field B = along the z-axis B' +B = Mo la Example 4.5 Magnetic 2 — x+ n a • 3 1 „ . z R— y/ z 2 + a2 . (4.29) 2 7TZ Field of a Finite Solenoid shows a solenoid Fig. 4.10(a) (4.28) is ' 2 7? - x. 2nzR 47tz (cylindrical coil) consisting of wound uniformly and densely N turns of an insulated thin nonmagnetic support with a circular cross section of radius a. The length of the solenoid is / and the current through the wire is /. The medium is air. Find the expression for the magnetic flux density vector along wire the solenoid Solution in one layer on a cylindrical wound in a spiral axis. Because of the coil being closely and the wire being thin, the current flowing through the turns of the solenoid can be regarded as a thin cylindrical current sheet with surface current density [Eq. (3.13)] Js The current over an elemental = NI (4.30) ~T‘ length dz along the solenoid, d/ =7 S dz = shown in Fig. 4.10(b), Nldz (4.31) I 2 Figure 4.10 (a) Uniformly and densely wound 183 Magnetic Flux Density Vector due to Given Current Distributions Section 4.3 solenoidal coil with a steady current and (b) evaluation of the magnetic flux density vector along the solenoid axis; for Example 4.5. can be viewed as the current of an equivalent circular current loop of radius a and wire diameter dz. From Eq. (4.19), the magnetic flux density vector of this loop at an arbitrary P point at the solenoid axis [Fig. 4.10(b)] is HO d/a dB = To find the total field B at the point P, we 2 „ (4.32) integrate dB to sum the contributions of all equivalent loops along the solenoid (superposition principle): B f MoATa 2 [ 12 dz Zl In order to solve this integral, however, we note „ „ r^. ~ d from the point P) that the relationship in Eq. (1.55), with substituted here by a, exists between the length coordinate z (measured and the angular coordinate 0 of the position of the equivalent loop along the solenoid Multiplying it by a/R = cos 0 yields axis. 2 a dz ~w = cos# With this, the integral in Eq. (4.33) B is (4.34) reduced to a simple form: rv r<h HqNI cos 6 dO / 21 which d6. J9 =9 (4.35) z, \ results in NI B = Ho (sin #2 — sin#i) (4.36) z. 21 This expression is B along the axis of a finite solenoid valid for an arbitrary field point along the solenoid axis, both inside and outside the solenoid, with the position of the point being defined by angles 0\ and 6j. Note, however, that the position of the point P can also be defined by coordinates z\ and z\ = a tan 0\ and zi = a tan 02In the case of an infinitely long solenoid, 0\ in Fig. 4.10(b) and in Eq. 1 -* -oo) and 62 = n/2 ( Z2 where 00) (4.36), so that B= where N = —n/2 (z\ zj, ixqN'I, (4.37) = N // is the number of wire turns per unit length of the solenoid. The magnetic field uniform, with flux density given in Eq. (4.37), within an infinite solenoid even at points off of its axis, as we shall show in a later example. is B inside an infinite solenoid Chapter 4 Magnetostatic Space Field in Free Magnetic Example 4.6 An Field of an Infinitely Long Conductor Strip long conductor in the form of a thin strip of width a carries a steady current of The permeability everywhere is mo- Determine the expression for the magnetic infinitely intensity /. flux density vector at Solution shown This is an arbitrary point in space. a two-dimensional problem, and we solve in the cross-sectional it plane Eq. (3.82) tells us that the current I must be uniformly distributed in the cross section of the conductor. Because of the conductor being thin, however, its current can in Fig. 4.11. be regarded as a surface current. with density Fig. 3.3, = Js (4.38) a By virtue of width d / = we subdivide of the superposition principle, and each such dy, strip the conductor into elemental strips can be considered as an infinitely long line current of intensity =J d/ s dl = I dy (4.39) a The magnetic flux density vector due to individual line currents is circular with respect line, and, from Eq. (4.22), its magnitude at an arbitrary point P in space (Fig. 4.11) is dB = Mo d / R= ' 2 jiR where d dB into the perpendicular distance of is its x- + d2 P from the plane of (4.40) . the strip. We decompose vector and y-components, d Bx = dBy = dBcos#, dBsin#, and integrate them over the width of the elemental 2 y)y to the strip (4.41) conductor to sum the contributions of all strips, Mof 2 na f yz Jy=yi sin#dy R mo / f y2 2na Jy] ' cos9dy (4.42) R We now use the relationship in Eq. (1.55), where z isy here, to change the integration variable from y to 0. Multiplying its left-hand side by RcosO and right-hand side by d which by the fact that cos# = d/R in Fig. 4.11, we obtain , is justified cos#dv ~R- = A6 Figure 4.11 Evaluation of the an magnetic infinitely field long due to strip conductor (cross-sectional view); for Example 4.6. - (4.43) I Section 4.4 Formulation of Ampere's Law 185 which, multiplied by sin0/cos#(= tan 0) gives, in turn, —R~ = sin 6 Based on these two By = r Ho I bl 2na C Jg e2 dy sin 6 , t d e. relations, the integrals in Eqs. (4.42) are simple to solve: sin# n-nT cos 0 2na ms u. B cos 02 i nJ "i fit 2na ' where the use is made of the substitution u = cos 0 ( dn Hence, the total magnetic flux density vector due to the B with R\ and R2 (4.44) cos 9 2na xln^ + R\ r r ez Jg = «2 (e2 _ fll) 2na , = - sin 6 d 6) ( 4.45) in the first integration. conductor turns out to be strip -<9 i) y (6>2 P from being the distances of the point d(, l (4.46) B due to a B due to an strip conductor the starting and ending point, respectively, of the line representing the cross section of the strip conductor, in Fig. 4.11. Note R\ = R2 that if we let a 00 while keeping I /a = const, we get an infinitely wide and long planar current sheet with a uniform surface current density Js infinitely = — 7r/2, (infinite), 6\ and O2 = = I /a. Then, n/2, with which Eq. (4.46) becomes (4.47) infinite current sheet We conclude that the magnetic field due to an infinite planar sheet of current each side of the sheet, with field lines parallel to the sheet. 4. 1-4. 9; Problems'. is uniform at Conceptual Questions (on Companion Website): 4.4-4. 6; MATLAB Exercises (on Companion Website). 4.4 We FORMULATION OF AMPERE saw in Chapter 1 that the electric field S LAW due tributions in free space can be determined to highly symmetrical charge dis- much more easily using Gauss’ law, Eq. (1.133), than by direct application of Coulomb’s law and the superposition principle, i.e., Eqs. (1.37)-(1.39). In magnetostatics, Eqs. (4.7)-(4.9) (the Biot-Savart law) provide general solution procedures analogous to those in Eqs. (1.37)-(1.39), whereas the law that helps us evaluate the magnetic field due to highly symmetrical more easily is known as Ampere’s law (also called Ampere’s circuital law or Ampere’s work law). It states that the line integral (circulation) of the magnetic flux density vector around any contour (C) in a vacuum current distributions in free space (free space) we mark is equal to ho times the total current enclosed by that contour, which Figure 4.12 Arbitrary contour in a magnetostatic field - for the formulation of Ampere's law. as Ip, £ The reference B dl = ho direction of the current flow is c (4.48) related to the reference direction of means of the right-hand rule: the current is in the direction defined by the thumb of the right hand when the other fingers point in the direction of the contour, as shown in Fig. 4.12. This law may be derived from the Biot-Savart law (we recall that Gauss’ law is derived from Coulomb’s law), and the derivation is carried out in Section 4.10. For the present, we accept Ampere’s law temporarily as the contour by another law capable of experimental proof. Eq. (4.48) represents Maxwell’s second Ampere's law 186 Chapter 4 Magnetostatic Field in Free Space HISTORICAL ASIDE Ampere Andre-Marie French mathematician and phys(1775-1836), icist, a was a professor of mathematics, physics, to the French Academy of Sciences his discov- ery of magnetic forces between wires carrying He electric currents. two that same discovered experimentally parallel wires carrying currents in the direction attracted each other, whereas the and chemistry in Bourg and Paris, and inspector wires with currents flowing in opposite directions general of the science of the magnetic field due to electric cur- national under Napoleon. Although with university system little formal Ampere education, acquired the knowledge of mathematics and sciences for that time by reading many books and articles, with a great support and guidance from his father. As a teenager, he was already deriving his own mathematical theories and writing papers on some geometrical problems. However, his life was soon to be shattered when his father was sent to the guillotine by the Jacobins in Lyon in 1793 during the French Revolution. The effect on Ampere of his father’s death was devastating, and he did not return to his studies of mathematics for almost two years. After several years of tutoring mathematics in Lyon, he was appointed professor of physics and chemistry at Bourg Ecole Centrale in 1802, and then professor of mathematics at the Ecole Polytechnique in Paris in 1809. His mathematics research included a wide best possible variety of topics in probability, calculus, analytic geometry, and partial differential equations. He chemelectricity and contributed also to the theory of light, and, most importantly, to magnetism. Only a few weeks after hearing of Oersted’s experimental results, Ampere was ready, before the end of September of 1820, to report istry, equation for repelled each other. His experiments founded the rents. He around described the magnetic force circling a current-carrying wire in a way which is now known as the right-hand rule. On November 6, 1820, Ampere gave a talk on his circuital law of addition of magnetic forces - a basis of a general equation that He was the we first now Ampere’s (circuital) call law. to describe current as the flow of along a wire, analogous to the surge of water through a pipe. He predicted theoretically that a wire helix with current would behave as it were a bar permanent magnet, and called such electricity a helix a solenoid. manent magnets Ampere also explained per- as a collection of tiny electric currents circling eternally within them, and in this he was three-quarters of a century ahead of A comprehensive presentation of his and magnetism, including both description of experiments and mathematical derivations, appeared in his most important publication “Memoir on the Mathematical Theory of Electrodynamic Phenomena, Uniquely Deduced from Experience” in 1826. The same year. Ampere was appointed to a chair at the Universite de France, which he held until his death. In his honor, the intensity of electric current is measured time. his findings in electricity in amperes. (Portrait: Edgar Fahs Smith Collection, University of Pennsylvania Libraries) static fields in free space (as we know, there is a total of four Maxwell’s field). By expressing the current in terms equations for the general electromagnetic of the volume current density, J, Ampere’s law becomes B Ampere's law for volume (f) Jc current • dl = yU() 1 J dS, (4.49) Is where S is a surface of arbitrary shape spanned over (bounded by) the contour C and oriented in accordance to the right-hand rule with respect to the orientation of C (Fig. 4.12). Section 4.5 Figure 4.13 Closed path seven What is 187 C and 4.7. Algebraic Total Enclosed Current the circulation of the magnetic flux density vector along the contour The medium Law line currents in air; for Example Example 4.7 Applications of Ampere's C in Fig. 4.13? is air. According to the right-hand rule given with respect to the reference direction C indicated in Fig. 4.13, the reference orientation of the surface S spanned over C, i.e., the reference direction of the vector dS in Eq. (4.49), is from S upward. From Ampere’s law, the line integral of the magnetic flux density vector along C equals mo times the algebraic sum of all current intensities passing through S. We note that current I\ pierces S once in the positive direction (direction in agreement with the orientation of S ), I2 passes it four times, I4 and I-j once each but in the negative direction, I3 once in the positive and once in the negative direction, Is twice in the positive and once in the negative direction, while h does not traverse S at all. Hence, Solution of the contour £ What is • dl = + 4/2 -U+I5- h) mo (h (4.50) very important, the same result for the algebraic total current, and thus for the circu- lation of B, and B is obtained for any other surface we imagine with the contour C as the perimeter totally arbitrary shape. Example 4.8 Fig. 4.14 Contour inside a Conductor shows the cross section of a very long cylindrical copper conductor that carries a steady current of intensity 7 (7 > 0). Is the circulation of the magnetic flux density vector along the contour C less than, equal to, or greater than mo 7? Solution From Ampere’s law, < ) j since the total current appearing B • dl < (4.51) mo?> on the right-hand Figure 4.14 Contour inside side of Eq. (4.49) equals exactly that portion of the total current of the conductor (7) enclosed by the contour. Conceptual Questions (on Companion Website): 4.7-4.10. 4.5 APPLICATIONS OF AMPERE'S devoted to the application of Ampere’s law in evaluating the due to given steady current distributions in nonmagnetic media. the case with Gauss’ law, the use of Ampere’s law will also require This section magnetic As is LAW is field a conductor with a steady Example 4.8. current; for 188 Chapter 4 Magnetostatic Field in Free Space symmetry of the problem to determine which field components are present in the structure and which spatial variables the present components depend on. Eq. (4.48), although always true, enables us to analytically solve for the field only due to highly symmetrical current distributions. careful consideration of the Namely, as the unknown quantity to be determined (B) appears inside the intewe can use Ampere’s law to obtain a solution only in cases in which we are able to bring the magnetic flux density, B outside the integral sign, and solve for it. These cases involve current distributions for which we are able to adopt a closed path C, called the Amperian contour, that satisfies two requirements: (1) B is everywhere either tangential or normal to C and (2) B — const along sections of C where B is tangential. Along the portion of the path where B is normal to it (if such portion exists), the dot product B dl in Eq. (4.48) becomes zero. Along the remaining part of the contour, B dl becomes B dl, and the second requirement (constancy) then permits us to remove B from the integral sign in Eq. (4.48). The integration we are left with is usually trivial and consists of finding the length of that portion of the path to which B is gral in Eq. (4.48), , • • tangential. In this discussion, Gauss’ law, discussed we in notice and exploit the parallelism with the application of Section 1.13. All similarities and differences in the mathe- matical formalism are direct consequences of the mathematical form of the fields E and B due Gauss’ law, we to the elementary sources Q and Q\, Eqs. (1.24) and With (4.4). integrate over a closed surface (Gaussian surface) on the left-hand Ampere’s law implies integration along a closed path component contributing to the integral is a comthe Gaussian surface and tangential to the Amperian contour, side of the equation, while (Amperian contour). The ponent normal to respectively. On field the right-hand side of the equation, Gauss’ law involves finding the total charge enclosed by the surface, whereas the application of Ampere’s law involves finding the total current enclosed by the contour. For sources expressed by volume charge and current densities [Eqs. (1.135) and (4.49)], this means volume integration over the volume enclosed by the Gaussian surface and surface integration over the surface bounded by the Amperian contour, respectively. Example 4.9 Magnetic Field of a Cylindrical Conductor (a) dB'+dB" An infinitely sity /. long cylindrical copper conductor of radius a carries a steady current of inten- The conductor is situated in air. Find the magnetic flux density vector inside and outside the conductor. Solution current is Fig. 4.15(a) shows a cross section of the conductor. uniformly distributed in the cross section, so that its According to Eq. density (3.82), the is ( Figure 4.15 Cross section of a cylindrical conductor with a steady current /: (a) application of Ampere's law and (b) proof that the magnetic field lines are circles; for Example 4.9. - Because of symmetry, the lines of the magnetic field due to the conductor current are centered at the conductor axis. To show this, consider the direction of the magnetic flux density vector, B. at an arbitrary point P in space, either inside or outside the conductor. Let the distance of the point from the conductor axis be r, and let dB' and dB" represent the fields at P due to two symmetrical current elements denoted as J' dv and J" dv and shown in Fig. 4.15(b). In accordance to the Biot-Savart law, Eq. (4.6), these two elementary field vectors are such that their sum dB' + dB" is tangential to the circular contour C of radius r centered at the conductor axis. The same is true for any other pair of symmetrical current elements, which can also be in a plane that does not contain the point P, and all current elements constituting the current / in the conductor can be grouped in such symmetrical circles (b) 4 52 ) Applications of Ampere's Section 4.5 pairs. We at the point P is tangential to the contour C. const along C, i.e., the magnitude of B conclude that the resultant vector In addition, symmetry only on the radial coordinate with the conductor B= also implies that axis. r B 189 Law depends of the cylindrical coordinate system whose z-axis coincides Hence, we can write B= B(r) (4.53) <j>, where is the circular unit vector in the system. Based on the preceding discussion, it is now obvious that the contour <j> C in Fig. 4.15 satisfies both requirements for the Amperian contour for our problem. Along C, dl so that B dl = B dl [Fig. 4.15(a)]. The circulation of B along C thus turns out to be = d/ <j>, • B dl = (b B(r) dl = B(r) dl (b = B(r ) l = B(r) 2 nr, whereas the current enclosed by < r < oo, (4.54) C amounts to Ic Jnr2 = for r for r Eqs. (4.54) and (4.55) 0 Jc Jc c combined in B= Ampere’s < > a (4.55) a law, Eq. (4.48), finally give noIr/(2na 2 ) for r <a for r > J HQl/(2nr) B due a Note that the magnetic field outside the conductor (for r > (e.g., a thin wire with current I) and the a thick cylindrical current a) is identical to that of a line current of intensity I placed along the conductor axis, Eq. (4.22). This current to conductor with a steady | means that the line (thick) conductor of Fig. 4.15 are equivalent sources with respect to the region outside the conductor. Cylindrical Conductor with an Excentric Cavity A very long cylindrical nonmagnetic conductor of radius b has an excentric cylindrical cavity of radius a along its entire length, as shown in Fig. 4.16. The axis of the cavity is offset from + d < b). The medium in the cavity and outside the the axis of the conductor by a vector d {a conductor is air. Compute the magnetic flux density vector in the cavity, assuming a steady current of density J flowing through the conductor. Solution The current density in the cavity is zero, and it can be considered to result from two currents of the same density (J ) flowing in opposite directions. Accordingly, we can represent the current distribution in the hollow conductor, including the cavity, as a sum of current distributions of a cylindrical conductor of radius b carrying a current of density J without and another one of radius a with a current of the same density but flowing in the opposite direction, as depicted in Fig. 4.16. By the superposition principle, the magnetic flux density vector of the original (resultant) current distribution, B, can be obtained as a cavity B= Bi+B 2 , (4.57) Figure 4.16 Cylindrical conductor with an excentric cylindrical cavity, viewed as a superposition of two solid conductors with currents of the same density but opposite directions; for Example 4.1 0. 1 . 190 Chapter 4 Magnetostatic Field in Free Space where Bi and B 2 are the magnetic flux density vectors due to the partial current distributions and J 2 in Fig. 4.16. These vectors, according to Eq. (4.56), can be written in the form 3 Jt Ft = Bi qJ x ri where and rj Mo(— J) X B2 = and 2 (4.58) are the position vectors of the field point (P) with respect to the r2 conductor axis (Oi) and cavity axis (0 2 ), respectively. equations, we r2 2 combining the preceding Finally, get - r2 x (n MqJ B= MoJ x d ) (4.59) 2 2 and B = noJd/2. We conclude that the magnetic field inside the cavity is uniform, as the above expression for the resultant magnetic flux density vector does not depend on the position of the point P. The field lines are parallel to the plane of the conductor cross section and at right angles to the vector d. Example Magnetic 4.1 Cable Field of a Coaxial Conductors of a coaxial cable are made from copper, and its dielectric is air. The radius of the inner conductor is a, whereas the inner and outer radii of the outer conductor are b and c, respectively (a < < b c). A steady current of intensity / is established in the cable. Find the magnetic flux density vector everywhere. Solution Referring to Fig. 4.17, current densities in the cable conductors are and Figure 4.17 Evaluation of the magnetic B in 4.1 I n(c 2 Due to symmetry, the magnetic flux density vector, B, axis. Following a similar procedure as C of radius contour for all values of r, < r (0 r Example in < 00 ). The 4.9, circulation of — (4.60) b2 ) circular with respect to the cable is we apply Ampere’s law to the circular B along C is the same as in Eq. (4.54), while the expression for the total current enclosed by field of a coaxial cable; for Example h= four different characteristic positions of the contour. between the conductors is identical to that found in C is different for The field inside the inner conductor and Example 4.9 for a single conductor. 1 _ the dielectric of a coaxial HoJ\ r _ MO Ir (0 2 na 2 2 cable < r < MO-f B= a). {a < r < b). (4.61) 2nr For the contour positioned inside the outer conductor, the enclosed net current equals the entire current of the inner conductor minus that portion of the current of the outer conductor which is enclosed by the contour, so that B— Mo U ~ h^fr2 ~ b ] 2 )] = ^"^2 (b ^ )r < r < c). lc Finally, if the radius r is larger than the outer radius of the outer conductor, B— - /) = So, the external magnetic field (for r > c) 2nr mo(^ 0 is (c < r < 00 ). zero. This, we (4.63) see, results from currents of equal intensities and opposite directions in cable conductors (making zero total enclosed current in Ampere’s law), much as the external electric field of a charged coaxial cable is zero because of charges of equal magnitudes and opposite polarities per unit length of cable 3 Note B= that the magnetic flux density inside the solid cylindrical conductor in Fig. 4.15 can be written as n<)Jr/2, and, in vector form, respect to the conductor axis. B = /xoJ x r/2, where r is the position vector of the field point with r Law Applications of Ampere's Section 4.5 191 conductors (making zero total enclosed charge in Gauss’ law). [Such (equal positive and negative) charges and currents are standard in operations of all two-conductor transmission a very important property of a coaxial cable; the electrostatic and magnetostatic lines.] This fields of the cable are concentrated exclusively inside the cable, and is decoupled (shielded) with respect to the Magnetic Example 4.12 Shown The in Fig. 4.18(a) is interior, and vice exterior its is perfectly versa. Field of a Toroidal Coil a toroidal (“doughnut”-shaped) coil with a rectangular cross section. N turns of wire that are uniformly and densely wound along the length of coil consists of 7. The inner and outer radii of the toroid are and its height is h. The medium inside and outside the coil is nonmagnetic. Calculate the magnetic flux density vector inside and outside the coil. the toroid and carry a steady current of intensity a and b (a < b), Solution Fig. 4.18(b) portrays a horizontal cut of the toroid in Fig. 4.18(a). Because of symmetry, the lines of the magnetic field density vector, B, due to the current in the coil are circles centered at the toroid axis [z-axis in Fig. 4.18(b)], and the magnitude of B depends only on the distance r from the Eq. (4.53). axis. means This B that the vector Applying Ampere’s law to a circular contour of radius of the form given in is < r (0 r < oo), we obtain the following equations for three characteristic locations of the contour: = B(r) 2nr 0 B(r) B= Hence, (0 2n < < r = fio(NI — NI) = B(r) a), 2nr = hqNI (b 0 < r < < (a r < b ), (b) (4.64) oo). of the 0 outside the toroid, while inside B(r) it that this result (inside a thick toroidal coil). (4.65) only rectangular). If b ) /2 to the toroid is its is thin, mean radius. i.e., b - contours (0 we can assume a <^a, _ 27rc is c«c inside it, where c = (a + fipNI 2jtc = that < r < oo); for Example 4.12. flux density H-qNI / showing the winding and (b) cross-sectional view showing Amperian This implies that the magnetic field inside the toroid can be considered be uniform, with the where field of three-dimensional view = valid for an arbitrary shape of the toroid (vertical) cross section (not is magnetic a toroidal coil: (a) 2jxr Note Figure 4.18 Evaluation MO N'l, (4.66) B inside a thin toroidal coil l the length of the toroid and N' = N // is the number of wire turns per unit length. We realize that the final expression for B in Eq. (4.66) is the same as that in Eq. (4.37), found for an infinite solenoid. Namely, we can visualize an infinitely long solenoid as a toroid with an infinite radius. Magnetic Example 4.13 An infinitely long air-filled Field inside an Infinite Solenoid from Ampere's Law solenoid has N' turns of wire per unit of its length. A steady current of intensity / flows through the winding. Using Ampere’s law, prove that the magnetic flux density inside the solenoid Solution Knowing that the vector solenoid axis) and that magnitude of B it is B is given by the expression in Eq. (4.37). inside an infinite solenoid zero outside the solenoid, in the solenoid from Ampere’s law. it is now is axial (parallel to the very simple to evaluate the The Amperian contour is a rectangle The line integral of B along the Figure 4.19 Evaluation of positioned partly inside the solenoid as shown in Fig. 4.19. the magnetic edge of the rectangle that is inside the solenoid equals Bl, because B is parallel to that edge and does not vary along it (the structure is infinitely long). The line integral of B along the remaining three edges of the rectangle is zero, because B is perpendicular to an vertical (axial) infinite field inside solenoid using Ampere's law; for Example 4.1 3. 192 Chapter 4 Magnetostatic Space Field in Free portions of horizontal (perpendicular to the axis) edges that are inside the solenoid and zero outside it. The net current enclosed by the contour, on the other hand, equals the ber of turns over the length which /, is num- N'l times / (the current through each turn). Hence, is , Ampere's law gives = Bl hqN'II. (4.67) the same as in Eq. (4.37). Note that we did not in any way restrict the location of the left edge of the contour C in Fig. 4.19 to be on the solenoid axis or at a specific distance from it, which means that Eq. (4.37) is valid across the entire cross section of the solenoid, and not only on the solenoid axis (in Example 4.5, the magnetic field is evaluated along the solenoid axis only). In other words, the magnetic field is uniform (the same) throughout the entire volume enclosed by the solenoid winding (while zero outside it). We also note that the cross section of the solenoid in this discussion and calculation is not in any way restricted to be of circular shape, which implies that the magnetic field is the same for arbitrary shape of the i.e., solenoid cross section (provided that the solenoid Magnetic Example 4.14 Charged Cylinder Field of a Rotating A very long cylinder of radius a p. The cylinder uniformly is very long). is its volume by a charge of density an angular velocity w. The permeability uniformly charged over rotates about axis with its mo everywhere. Find the magnetic flux density vector inside and outside the cylinder, assuming that the charge distribution of the cylinder remains the same during the rotation. is Solution As the charges of the cylinder rotate with it, moves and the current density vector in the cylinder are circles =w x at that point centered We r = wn = pv(r) note that this current density, which Because of the currents series of can be found from Eq. (3.28). The current lines many very in = pwr is (0 zero, whereas contour B is C shown r < and the current density a). (4.69) electric field, the cylinder being circular, the cylinder can be visualized as a axial inside the cylinder. in Fig. can be regarded 3.17(a)], long coaxial solenoids of radii varying from r 4.20(b) results in (i.e., Ampere’s = outside 0 to all r — a. That is why the equivalent solenoids) is Eq. (4.49), applied to a rectangular [left-hand side of the equation is the same as in law, Eq. (4.67)] rci r(l B(r)l rotating charged cylinder: = equivalent impressed current density and (b) Amperian contour; for Example 4.1 4. < independent of the Jj [Fig. magnetic flux density vector, B, outside the cylinder (a) (4.68) j>, at the cylinder axis [Fig. 4.20(a)], as an impressed electric current density. Figure 4.20 Evaluation of the magnetic field of a the cylinder axis a function of r given by J(r) (b) away from r is is v is they form a volume electric current. M that Referring to Fig. 4.20(a), the velocity at which a point where dS is Mo •/(/) / V=r / dr' = mo pwl magnitude of B (4.70) dS the surface area of a thin strip of length variable). Hence, the r dr, / Jr "TT" 1 / and width d / ( r is the integration away from the cylinder axis comes at points that are r out to be B(r) Example 4.15 Using Ampere’s Magnetic law, = m)pw(a 2 - r2 )/! for r 0 for r Field of an Infinite prove that the magnetic Eq. (4.47). a (4.71) a Current Sheet from Ampere's Law flux density vector current sheet with a uniform surface current density 7 S in < > in free space due to an is infinite planar given by the expression Section 4.6 Solution Because of symmetry, the magnetic not vary in directions parallel to the sheet. C We from which the same result as in Eq. (4.47) = is B does 193 Ampere's Law of B BO along each of the two edges of the contour Bl and the enclosed current 2BI the sheet and Form apply Ampere’s law to a rectangular contour B portrayed in Fig. 4.21. The line integral of that are parallel to the sheet equals field lines are parallel to Differential jU 0 is Js l [Eq. 7s /, Pi Js (3.13)], so that (4.72) c obtained. l Figure 4.21 Evaluation Problems 4.10-4.19; Conceptual Questions (on Companion Website): 4.11-4.15; of the MATLAB Exercises (on Companion Website). an : magnetic infinite field of current sheet using Ampere's law; for DIFFERENTIAL 4.6 In electrostatics, we (1.166)] starting from FORM OF AMPERE S Example 4.1 5. LAW derived the differential form of Gauss’ law [Eq. (1.163) or its integral form [Eq. (1.135)]. Again, an analogous concept and transformation exist in magnetostatics. In this section, we shall utilize the inteform of Ampere’s law, Eq. (4.49), to derive its differential equivalent. What we expect to obtain is a spatial differential relationship between the magnetic flux density vector, B (field), and the current density vector, J (source), at a point in gral space. We consider first the one-dimensional case, and assume that J has only a z-component in the Cartesian coordinate system which is a function of the coordinate x only, that is, J = Jz (x) z (1-D current distribution). Then, by symmetry, the only present component of B is By (like in Example 4.15), i.e., B = B y {x) y. We apply Ampere’s law [Eq. (4.49)] to a narrow rectangular contour C lying in the xy-plane, with edges parallel to the x- and y-axes, as shown in Fig. 4.22. The dimension of the contour in the x-direction is dx and the length of the edges parallel to the y-axis is /. The magnetic field is constant along both edges with length / ( By does not vary with y), so that essentially no integration is needed on the left-hand side of Eq. (4.49), i.e., the integral along each edge reduces to B I (1 has the same direction as dl). No integration is needed on the right-hand side of Eq. (4.49) either, because dx is differentially small and we can take Jz (x) as constant over the surface spanned over C. Finally, as B is tangential to both edges with length l, B and dl are directed in opposite directions along the left edge, and in the same direction along the right edge, we can B v (x) ,B ,(x+dx) } T(6 BO © C 26Figure 4.22 For the derivation of the one- dimensional Ampere's law in differential form. write —B y {x) l + By (x + dx) / = iiqJz {x) /dx. (4.73) Noting that d By = By (x + dx) - By (x), (4.74) Eq. (4.73) becomes a differential equation, (4.75) is the one-dimensional Ampere’s law in differential form. We observe the analogy with the 1-D Gauss’ law in Eq. (1.158). We now generalize Eq. (4.75) to an arbitrary three-dimensional current distri- This The current density vector has now all three Cartesian components and them are functions of all three coordinates, bution. of J = Jx (x, y, z) X + Jy(x, y, z) y + Jz (x, y, z) z. all (4.76) 1-D differential Ampere's law 194 Chapter 4 Magnetostatic Field in Free Space C in Eq. (4.49) must, therefore, be differentially small in both dimensions. Additionally, one contour is not enough; we need three small con- The Amperian contour each of the three current density components, tours, oriented perpendicularly to respectively. All edges of the contours being differentially small, the line integral along each of them can be approximated by taking a constant value of the component tangential to the edge and multiplying field by plus or minus the edge length. This gives us the result for each pair of integrals along parallel edges within each contour that has exactly the same form as in the 1-D case [Eqs. (4.73)— (4.75)]. For the contour that lies entirely in the plane normal to the jc-component of J, shown in Fig. 4.23, we have <j) B dl it = -B z (x, y, z) d z + B y (x, y, z) dy + B z (x, y + -By (x,y, z + dz)dy dy, z) d z = noJx (x,y,z)dydz, (4 .77) which, divided by dy dz, results in §c B dl dy dz • _ B z(x, y + dy, Z) - B z (x, y, z) dy B y (x, y, z + dz) - By (x,y, z) dz dB z 8B, dy dz = Mo Jx - ( 478 ) For contours that are oriented perpendicularly to each of the remaining two coordinate axes, y- and z-axes, analogous procedures lead to equations dBx dBz dx dz = 3 Bv 8 By and ixoJy dx — MO Jz> (4.79) 8y by the unit vector x, and Eqs. (4.79) by y and and summing the three equations, we obtain the following equation with the vector J [Eq. (4.76)], multiplied by /xq, on the right-hand side: respectively. Multiplying Eq. (4.78) z, respectively, l Ampere's law (dB z dBy\ V 9y dz ) in differential form fdB x „ x+ ("97 This partial differential equation form B for field We dBr dBx dx dy z (PDE) in spatial Figure 4.23 For the derivation Ampere's law in differential form for an arbitrary current distribution. (4.80) It relates the rate of change of coordinates to the local current density vector, see that only those variations of individual of MoJ- represents Ampere’s law in differential an arbitrary steady current distribution. components — components J. that are in directions Section 4.7 Bx perpendicular to the direction of the component (change of and not along x, etc.) contribute in this relationship, which dependences in the differential Gauss’ law [Eq. (1.163)]. 4.7 is along y and 195 Curl z, just opposite to the CURL The expression on the left-hand side of Eq. (4.80) the so-called curl of a vector is 4 analogous to the divergence of a vector field (E) as used to express the differential form of Gauss’ law [Eq. (1.166)]. In addition, we notice that applying formally the formula for the cross product of two vectors function (B), written as curl B, system 5 to V x B, where the del operator obtain exactly curlB. Hence, in the Cartesian coordinate Eq. (1.100), we curl VxB=| B= 1 dB v \ '3/?, dy . (d bx + dz ) dz y — - 3 B z^ \ dx ) is given by A y dBx' *(Swhich can also be written in the curl curI in Cartesian coordinates Y- dy form of a determinant, B=V B= x AAA dx 3y (4.82) 3z Bx By B form is a convenient way for memorizing the expression for x B in the Cartesian coordinate system). Note that the curl is an operation that is performed on a vector, and the result is also a vector. Finally, the differential Ampere’s law can now be written in a short form as (using the determinant V curl B= VxB = /xqJ. (4.83) Ampere's law using curl notation In nonrectangular coordinate systems, the differentially small of its edges for instance, from becomes Amperian con- and with different expressions for lengths different current density components in each of the systems. For tour in Fig. 4.23 curvilinear Fig. 1.10, the contour perpendicular to the radial vector component Jr r in spherical coordinates is a curvilinear quadrilateral of edge lengths r d 0 and rsin# d <p. Carrying out similar derivations as in Eqs. (4.77)-(4.80) for this contour and contours oriented perpendicularly to other unit vectors the expression for the curl in cylindrical coordinates is in Figs. 1.25 and 1.26, obtained to be (4.84) 4 Note that 5 For vectors given by their Cartesian components, rot B is also used for curl B, “rot” being a short for “rotation” (or “rotational”). a x b = (ax = (ay b z x + ay y + a z z) - a z by ) x + x (bx x ( a z bx - + by y + b z z) ax b z ) y + (ax by - ay bx ) z. curl in cylindrical coordinates , 196 Chapter 4 Magnetostatic Field in Free Space and that for the spherical coordinate system curl — (sm an dO 1 B= V x B= 3 1 1 - + i _ dr 30 r [_sin# dBe \ • OBtt,) 07 v rsin0 curl in spherical coordinates f 30 3 — 3r l + rR 4> ; e 1 ’ , . „ ( 3 rBf) Br ' ) dO * (4.85) same way as with V E in Section 1.15, we use the notation V x B in cylindriand spherical coordinate systems not to refer to it as the actual cross product of vectors V and B, which is valid only in rectangular coordinates [Eq. (4.81)], but to merely symbolize the curl operation. With this, we also emphasize the fact that relations derived employing such vector formalism in the Cartesian coordinate system In the • cal curl B hold true (are identities) generally, cal quantities and relations coordinate systems (properties of physi- in all between them are the facts that are independent of the choice of coordinate system). Figure 4.24 For the definition of the curl of a vector field in Eq. (4.87). (4.81) and (4.78), we find that the jc-component of curlB, that can be expressed as the net circulation (line integral) of B along the incremental contour C of Fig. 4.23 divided by the area of the surface spanned over Combining Eqs. is, x • curl B, the contour, —r —B —— • x • curl B= <f dl ay az We can now (4.86) . formally proclaim the Cartesian x-axis to be an arbitrary linear axis and the contour (direction) in space C small contour bounding a surface AS, as shaped and write to be an arbitrarily shown in Fig. 4.24, Sr B dl — —-— differentially • n-curlB= alternative definition of the lim AS—>0 curl where n is (4.87) AS 1 AS (AS = ASn) and determined an equivalent mathematical definition of the the unit vector normal to the surface using the right-hand rule. Eq. (4.87) curl of a vector. is enables us to obtain a component of the curl of It direction at a given point by computing the circulation of B B along a desired along a contour in the plane perpendicular to that direction as the contour shrinks to zero about the point. The circulation of B about C, per unit area, appearing on the right-hand side of Eq. (4.87) equals n curl B = |curl B| cosa (|n| = 1). Therefore, for the orientation of AS defined by a = 0, we get the maximum in the circulation (cos a = 1), and • Eq. (4.87) becomes physical meaning |curlB| of the curl ( lim §c B • dl AS \AS->0 (a = 0). (4.88) max This means that (1) the magnitude of curl B equals the maximum (as the direction of the surface element AS = ASn is varied) net circulation of B per unit area with the area of the surface element tending to zero and (2) the direction of curl B is in the direction that gives the maximum value for the magnitude of the net circulation per unit area (a = 0). From it, we may regard the curl of a vector field (not only the magnetostatic field) physically as a measure of those local sources of the field field components with respect to that measure of sources that produce radial field components with respect to the point [Eq. (1.172)]. Ampere’s law tells us that the sources producing locally circular field components in the case of the magnetostatic at a point in point. We space which produce circular recall that the divergence is a > Section 4.7 field in free Curl space are elemental currents of intensities J AS. Quantitatively, the curl represents the surface density of such sources, and in our case this density concept of the curl of a vector field is used in numerous applications, in is J. many The areas of science and engineering. Replacing q,oJ by Eq. (4.49), leads to V x B [from Eq. (4.83)] in the integral form of Ampere’s law, £ Bd,= /s (V x B) • dS. (4.89) Although obtained here specifically for the magnetostatic field in free space, this equation is an identity, holding for any vector field (for which the appropriate partial derivatives exist). It is widely used in electromagnetics and other areas of science and engineering, and is known as Stokes’ theorem. In words, it states that the net circulation of a vector field along an arbitrary contour is the same as the net flux of its curl through any surface bounded by the contour, where the reference orientations of the contour and the surface are interconnected by the right-hand rule. We notice the parallelism with the divergence theorem, Eq. (1.173), which applies to a closed surface and relates the flux of a vector field with a volume integral of its divergence. To prove Stokes’ theorem (for an arbitrary vector field B), imagine the surface S subdivided into a large number of differentially small patches AS/ (A Si — 0) which are bounded by infinitesimal contours Q, as depicted in Fig. 4.25. By applying the definition of the curl in Eq. (4.87) to one of these patches, we can write £ B dl = ASi n curl B = (V x B) • AS/n = (V x B) • AS/. (4.90) A 5, comprising S and sum all the on the left-hand side of the resulting equation is the line integral of B along the overall contour C bounding the surface S, since the terms arising from the sides of small contours shared by any two patches cancel out during the summation, as can be seen in Fig. 4.25, and the only boundaries for which the Now, let results. us determine this circulation for every What we get cancelation does not occur are those forming the contour C. Hence, l B dl lim AS/-».0 V (V x B) • AS/. Figure 4.25 (4.91) Open surface 5 subdivided into differentially small patches - for the proof of Stokes' theorem, Eq. (4.89). Stokes' theorem 197 198 Chapter 4 Magnetostatic Field Free Space in summation on the right-hand side of this equation becomes an AS, becomes dS, and we have fs (W x B) dS, thus proving the theorem. By applying Stokes’ theorem to Eq. (1.75) or simply by analogy with the differential form of Ampere's law, Eq. (4.83), we arrive to the differential form of Maxwell's first equation for the electrostatic field: In the limit, the integral, curl of E • V in electrostatics We see that the electrostatic field 0. (4.92) and a curl-free or irrotational field, is property of any conservative vector E= x this is a with zero circulation along any field (the field closed path). Redo Example Solution This 4.9 but is Conductor Using Cylindrical Current Example 4.16 now Differential Ampere's Law using the differential form of Ampere’s law. completely analogous to the application of the differential Gauss’ law to problem with spherical symmetry in Example 1.22. For the interior of = J z and J = const), combining Eqs. (4.53), (4.84), and we now have in place of Eqs. (1.174) and (1.175) solve an electrostatic the conductor in Fig. 4.15 (where J (4.83), 7xB = with C\ = 0, ~ [rfi(r)] z = since there exists Fig. 4.15 [otherwise, this space (with J = — — ixqJ i no > rB(r) » B(r) = hqJ = J rdr y + = (0 < + r < a) C\ (4.93) , along the z-axis (for line current (of intensity Iq) constant would be C\ + Ci = r = 0) in surrounding /xq/o/( 2^)]. Similarly, in the 0), 9 [rB{r)\ = 0 — dr — >• rB(r) — = C2 B(r) > = Ci — < (a r < 00 ) / . (4.94) 2 goJa /2, from B(a~) — B(a + ). Both Eqs. (4.93) and (4.94) give the same corresponding results as in Eq. (4.56). where C2 = Companion Website): Companion Website). Problems'. 4.20-4.24; Conceptual Questions (on 4.17; MATLAB 4.8 LAW As vector B is Exercises (on OF CONSERVATION OF MAGNETIC FLUX called the magnetic flux density vector, called, accordingly, the magnetic flux (unit: magnetic flux. O Wb) and measured 4/ , is 4.16 and in It is -L webers (Wb), where its flux through a surface S is denoted as O, B dS, (4.95) Wb = T m 2 We • . recall that the electric flux, defined as the flux of the electric flux density vector, D, through a designated surface [Eq. (2.42)], and that the unit for The generalized Gauss' is C. law, Eq. (2.43), tells us that the net through a closed surface is equal to the analogy between electric and magnetic total outward electric flux charge enclosed by the surface. fields naturally Our imposes a question: what Section 4.8 Law of Conservation of 199 Magnetic Flux HISTORICAL ASIDE George Gabriel Stokes (1819-1903), a British mathematician and physicist, was a professor of mathematics at Cambridge University. Stokes was the oldest of the trio of Cambridge professors, James Clerk Maxwell (1831-1879) and Lord Sir dally contributed to the fame of the Cambridge was University’s Member of Parliament, and President of the Royal Society, the three offices that had only once before been held by one person, school of mathematical physics in the 19th century. Sir Isaac who Kelvin (1824-1907) being the other two, espe- Wilhelm Eduard Weber (1804-1891), a at German Newton (1642-1727). the coils). Furthermore, was used, it many or similar realizations, by in the same researchers and was a professor students over decades both to directly measure Gottingen University. the magnetic forces and torques due to given cur- physicist, Weber received his docdissertation toral from the University of Halle in 1826, with a topic on theory of the acoustic reed organ pipes, and was appointed professor of and to determine (indirectly measure) the (unknown) current from the measured torque (as a sort of ammeter). During his later years at Gottingen, Weber worked on a theoretical generalization and unification of laws describing forces between charges at rest (Coulomb’s law) and in motion (Ampere’s force law), as well rents Faraday’s law of electromagnetic induction physics at Gottingen in 1831. In collaboration as with Karl Friedrich Gauss (1777-1855), he built for time-varying currents (charges in accelerated the first practically useful telegraph (3 km long) in motion). He also contributed to understanding 1833, to connect his physics laboratory with Gauss’ the connection between light and electromagnetic astronomical observatory at Gottingen. They also phenomena and establishing the link between elec- mag- tromagnetism and optics, which was crucial for Maxwell’s development of electromagnetic field worked together on investigating terrestrial netism (earth’s magnetic is His main contributions were in the areas of viscous fluids, sound, light, spectroscopy, fluorescence, and X rays. During a period of time, he held the post of the Lucasian Chair of Mathematics at Cambridge, field). From 1836 to 1841, they organized a network of observation stations theory (that includes around the world to correlate measurements of terrestrial magnetism at different locations. In 1841, he developed the electrodynamometer, which could precisely measure the angular displacement of a coil with a current caused by another coil positioned perpendicularly to it and carrying the same current. Weber used this instrument for a final validation of Ampere’s previous conclusions about magnetic forces due to currents in wire loops (in with Gauss, in the early stage of the this, we consider first Working again jointly a significant impact development of a new coherent system of units to also include electromagnetic phenomena, which gradually evolved into the present International The SI includes the weber, named the magnetic flux density Brittle Books The answer dB System of Units (SI). unit of magnetic flux, in his honor. (Portrait: AIP Emilio Segre Visual Archives, the net magnetic flux through a closed surface equal to? To prove light). Weber provided is: Collection) zero. of a single current element J dv in free space, as depicted in Fig. 4.26. From the Biot-Savart law and the expression for dB in Eq. (4.6), we know that the lines of this field are circles centered on the straight line containing the current element (Fig. 4.26). We can imagine the entire space surrounding the element divided into thin closed tubes of uniform cross section formed by bunches of field lines, where one such tube is shown in Fig. 4.26. According to Eq. (4.6), again, the magnitude of the magnetic I 200 Chapter 4 Magnetostatic Field in Free Space flux density vector is constant along each and thus over the entire volume field line, of each tube (because the tubes are thin). Consequently, the magnetic flux through any tube is the same in magnitude at any cross section of the tube, regardless of whether that cross section is perpendicular to the tube axis or not. Imagine now an arbitrary closed surface S in the field, Fig. 4.26. Some of the elementary tubes pass through S, but always an even number of times. Therefore, the net outward flux of dB through all small surfaces representing intersections of thin tubes and the surface S equals zero. Since such intersections cover the entire surface S, we have dB dS = • i Our proof continues then by invoking of which the actual magnetic flux density Figure 4.26 For the derivation of the law of conservation of magnetic free space can be decomposed (4.96) 0. the superposition principle, by B means of an arbitrary current distribution in into elementary flux densities dB due to individual current elements making the current distribution, so that flux. B dB, -L (4.97) Vcur where v cur is the domain with currents (sources of the magnetic field). Applying the integration over v cur as an operator to Eq. (4.96), and interchanging the order of integral signs, we obtain m dB -dS = (4.98) 0. Vcur Finally, substituting Eq. (4.97), law of conservation of magnetic B dS = (4.99) 0, • i flux which completes our proof. This relation is known as the law of conservation of magnetic flux, and also as Maxwell’s fourth equation. For obvious reasons, it is sometimes referred to as Gauss’ law for the magnetic field. Together with Ampere’s law (Maxwell’s second equation), it forms a complete set of Maxwell’s equations for the magnetostatic field in free space (or any nonmagnetic medium). We notice that the law of conservation of magnetic flux has the identical form as the continuity equation for steady currents, Eq. (3.40). Consider next an arbitrary contour arbitrary shape that are both C bounded by and two open surfaces, 5] and S2 with the contour and oriented in the same , way - according to the right-hand rule with respect as shown in Fig. 4.27. The magnetic fluxes through Oi Figure 4.27 For the proof a contour is unique. [ B dS S2 (S = Si flux <^) which flux through a contour, Fig. 4.27 zero, is B dS = 0 — 0 2 1 all that the magnetic flux through any have a common (4.100) (4.101) , from the law of conservation of magnetic | that dS. through the closed surface S formed by Si and 0 = 02 meaning • 2 { The total outward U S2 ) is respectively. the surfaces are 02 = [ B Js and • Js that the magnetic flux through = to the orientation of the contour, flux. Hence, (4.102) , number contour bounding them is of surfaces of arbitrary shape the same, provided that all the r , Section 4.9 201 Magnetic Vector Potential same way. This enables us to link the flux to a contour bounded by the contour (and there is an infinite number surfaces are oriented in the rather than to a surface of such surfaces), and to use the term flux through a contour or flux linked by a is uniquely determined by the shape of a contour and by its orientation, and is the same for any surface spanned over the contour and oriented in accordance to the right-hand rule with respect to contour. In other words, the flux (through a contour) the orientation of the contour. By analogy with the differential form of the continuity equation for steady cur- rents, Eq. (3.41), the law of conservation of magnetic flux in differential form given by V B= We see that the B (4.103) 0. • is flux conservation law differential field is means in form another divergenceless (divergence-free) or solenoidal be local sources of radial magnetic field components with respect to a point, i.e., that there exist no positive or negative “magnetic charges” and no free north or south magnetic poles, which would correspond to electric charges, with density p [Eq. (1.166)]. Equivalently, the magnetic field lines close upon themselves, as there are no “magnetic charges” for the lines to begin and terminate on. We can now summarize the two Maxwell’s differential equations governing the vector This field. that there cannot electrostatic field in free space, VxE = 0 V-E = and — (4.104) conservative (4.105) solenoidal field field eo and those for the magnetostatic field in free space, V x B= which, in a condensed form, shows between the two • 0, the similarities and fundamental differences fields. Problems 4.25 and : all V B= and ixq J 4.26; Conceptual Questions (on Companion Website): MATLAB Exercises (on Companion Website). 4.18-4.23; MAGNETIC VECTOR POTENTIAL 4.9 In electrostatics, electric fields we introduced the electric scalar potential (E) to help us describe and evaluate the electric field intensity vector (E). potential due to a point charge free space is Q , as The an elementary source of the electric scalar electric field, in [Eq. (1.80)] V= 1 Q Atzsq R Following the analogy established in connection with Eq. (4.106) (4.4), which is the basis of the Biot-Savart law, the potential due to an elementary source of the magnetostatic field, Q\, is defined as A = A Mo Qy (4.107) s?i magnetic vector potential (unit: is called the magnetic vector potential, and its unit is T m. It is a whose direction is very simple to determine - the same as the direction of Qy, and whose magnitude is proportional to 1 /R (and not to 1/R 2 which is present in This quantity • vector , T m) • — 202 Chapter 4 Magnetostatic Field in Free Space B due the expression for to Q\). By the same reasoning as with obtaining the three versions of the Biot-Savart law for volume, surface, and line currents, Eqs. (4.7)(4.9), the corresponding integral expressions for the magnetic vector potential are: A due to volume current ( 4 108 ) A due to surface current ( 4 109 ) ( 4 110 ) A due to line current In general, the solutions for the magnetic vector potential due . . . to given current distributions are substantially simpler than the corresponding solutions for the magnetic flux density vector. Let us find the curl of A. By representing the expression in Eq. (4.107) in the spherical coordinate system shown in Fig. 4.28, in which Q is at the coordinate origin, so R = r (r being the radial spherical coordinate), and v is z-directed. A = A z = A cos 9 r — A sin# 0 = A r + AyO, A — r MoQv 4nr ( 4 111 ( 4 112 ) . ) where A = A cos 9 = HoQvcosd r (Afj, = 0). Aq 4nr HoQv sin 9 = —A sin 9 — — 4nr Using the expression for the curl in spherical coordinates, . Eq. (4.85), we have V x A= r (rAg) dr mo Qv 4^r Noting that we - 3A r 4> ~d9 d — (— sin dr cos 9 d . 6>) 2 <*> = — — /u 0 Qvsin9 ( 4 113 ) ( 4 114 ) ( 4 115 ) : ; 9 4nr 2 d9 <P . (Fig. 4.28) = Qv Q\ xr = Qv i x ? V x A= | z x r| <|> x r = Qv sin 9 <J>, . obtain Mo 47t Figure 4.28 Magnetic vector due to a point charge moving in free space, and its components in a spherical potential coordinate system. gv 2 r . which R= r. exactly the magnetic flux density vector due to Q\, Eq. (4.4) with is R= 203 Magnetic Vector Potential Section 4.9 r and Hence, B=V x A, (4.116) magnetic flux density from potential and the same is true for the magnetic vector potential dA due to an arbitrary ume, surface, or line current element, Eq. (4.10). Integrating the expression V x over a domain v with volume currents, we have J (V x dA) where the integration and =V dA = V x ) x A, vol- dA (4.117) differentiation (del) operators can readily interchange places because they are completely independent ~ the integration is performed with respect to the coordinates of the source point (point at which the current density vector is J), while the differentiation carried out with respect to the coordinates is of the field point (point at which the magnetic vector potential is A). Consequently, the magnetic flux density vector (B) and the magnetic vector potential (A) due to an arbitrary volume current distribution, Eq. (4.108), are related at an arbitrary point in space as in Eq. (4.116), and similar proofs can be carried out for the magnetic potential due to surface currents, Eq. (4.109), and line currents, Eq. (4.110). This relationship, in conjunction with Eqs. (4.108)-(4.110), provides an alternative produced by steady currents, where A is found from the potential by differentiation. Potentials, generally, are auxiliary quantities that are used to determine fields indirectly. Finally, we note that, contrary to V in electrostatics, A does not have any general method evaluated first for evaluating the B field by integration, and then B is simple physical interpretation in magnetostatics. On the other side, the divergence of the magnetostatic potential which A is always consequence of the steady current density J being a divergenceless vector, i.e., of the continuity equation in Eq. (3.41) or (3.40). Namely, in analogy to Eq. (4.117), we can apply the divergence operator to the integral expression in Eq. (4.108) and write zero, V • is, essentially, a A= (4.118) is the closed surface bounding v, and the use is made of the divergence theorem, Eq. (1.173), to convert the volume integral to a surface (flux) one. For the special case when v is a spherical domain centered at the field point (where the potential is being computed), 1/R can be brought out of the flux integral, as S in where S R and the integral, in turn, becomes zero, by virtue of the continuity equation, Eq. (3.40). In the general case, the arbitrary domain v can be subdivided into a stack of concentric spherical layers with that case is §s J dS = a spherical surface of radius , 0, each of them being a part of a full spherical shell of thickness d R centered at the with the current of density J flowing through that part and no current field point, The same conclusion about a zero flux integral based on the continuity equation can then be derived for the surface of each shell, and thus in the rest of the shell, outside v. V A= • 0, (4.119) divergence of currents which holds true for the magnetic potential due to surface and as well. line steady currents A due to steady V 204 Chapter 4 Magnetostatic Field in Free Space The magnetic expressed flux through an arbitrary contour, Eq. (4.95), can terms of the magnetic vector potential, in 4> jT_B-dS = now = y (VxA)-dS, be (4.120) which, by using Stokes’ theorem, Eq. (4.89), becomes magnetic flux O= from the £ potential Adi. (4.121) The orientation of the contour C and the surface S are in accordance to the righthand rule, as shown in Fig. 4.29. Eq. (4.121) represents a means for determining the magnetic flux by evaluating a contour integral (of A) rather than a surface integral (of B), which is very convenient in some computations and derivations. Note that this is yet same for another proof that the magnetic flux through a contour surfaces bounded by C ). is unique (the all Figure 4.29 For the evaluation of the magnetic through by integrating the magnetic flux a surface vector potential along Problems: 4.27; Conceptual Questions (on Companion Website): 4.24 and 4.25; MATLAB Exercises (on Companion Website). its boundary. PROOF OF AMPERE 4.10 We now ready are On A. to prove S LAW Ampere’s law by utilizing the magnetic vector potential, we have, by the left-hand side of the differential form of the law, Eq. (4.83), means of Eq. (4.116), VxB = Vx(Vx A). (4.122) Applying formally (symbolically) the vector identity for expanding the vector triple product, 6 a x (b x to the product V c) = b(a • — c) c(a • b), (4.1 23) x (V x A), we get the following identity for expanding the curl of the curl of an arbitrary vector field (A): V The div first x (V x A) term in this A=V A= • = V(V A) - (V V)A expansion • is V2A is x this, in it is 24) zero, because Eq. (4.122) becomes B=— 2 A. called the Laplacian of the vector function Eq. (1.100), can be written (4.1 • grad(div A), and in our case 0 [Eq. (4.119)). With V Vector = V(V A) - V 2 A. (4.125) A, which, having in mind Cartesian coordinates as 2 '2 2 \ ( 6 A cross (vector) product of a vector (a) with a cross product of two (other) vectors (b and c) the vector triple product. Note also that the vector triple product identity in Eq. (4.123) the “bac-cab" or “back-cab" identity, as the equation. word “bac-cab" can be read on is is called known as the right-hand side of the z Section 4.1 0 d 2 Ax ~dx 2r 'd + We 2 A x 3 2A x \ „ X+ + ~d^ l 2 ') A 7 3x 2 2 3 + + d 2 A 7 d + 2 A 7 \ 'd 2 Av dx + 2 2 Av 3 + 2 3z 3y 3y Law A, 2 „ z. (4.126) 3 z2 see that Cartesian components of the Laplacian of components of A, where the Eq. (2.94), and hence 7 the corresponding scalar field, 3 Proof of Ampere's A equal the Laplacian of operator latter is the Laplacian of a V 2 A = V 2 A* x + W 2 A y y + V 2 A Z z. (4.127) Laplacian of a vector in Cartesian coordinates From Eq. (4.108), A = Ax x + Ay y + A = z Jx dv Mo /*/*/* „ (Jx Mo _ z 4tt [ JVv f Jy dv x + Jy y + Jz z) dv R f „ Jz dv \ (4128) and we conclude that each Cartesian component of A is actually produced by the same component of the current density vector, namely, A x is produced by /*, and so on. Recalling the expression for the electric scalar potential due to a volume charge in free space, Eq. (1.82), we identify the duality between V due to p and A x due to/*: V= 1 f p dv Ax = , i / ^-=r 47T£ 0 Jv R Mo / Jx dv ^l-Z—. (4.129) R 4 n Jv By the same duality principle, in turn, there must be a differential equation for which has the same form as the differential equation for V, that is, Poisson’s equation [Eq. (2.93)]. Therefore, by changing the variables, we get Ax V2 E = - — = -mo/*. V 2A Z = ~p-qJz VL4* £o (4.130) Similarly, V 2 A y = — Mo/y and substituting all and these back in Eq. (4.127), V2 A = This is (4.131) , -MoJ- (4.132) a second-order vector partial differential equation (PDE) for the magnetic vector potential of a volume current in free space (or any nonmagnetic medium), usually referred to as the vector Poisson’s equation. in solving (analytically or We use it numerically) for can be used as a starting point we can A V Note It to a given current distribution J. here for deriving Ampere’s law. Namely, returning to Eq. (4.125), for the Laplacian of and obtain now substitute 7 A due x B = -V 2 A = moT (4.133) does not have equally simple counterparts in cylindrical and and 1.26). In these systems, the Laplacian of a vector func2 tion (A), V A, is computed, from Eq. (4.124), as V 2 A = V(V A) — V x (V x A) = gradfdiv A) — curl (curl A), namely, as the gradient of the divergence of A minus the curl of the curl of A, using the expressions for the gradient, divergence, and curl in Eqs. (1.105), (1.108), (1.170), (1.171), (4.84), and (4.85). that the expansion in Eq. (4.127) spherical coordinate systems (Figs. 1.25 vector Poisson 's equation 205 206 Chapter 4 Magnetostatic Field in Free which Space Eq. (4.83), and this completes our proof of the Ampere’s law in differtheorem, Eq. (4.89), which is derived from the mathematical is ential form. Stokes’ definition of the curl of a vector, gives Note its integral equivalent, Eq. (4.49). that the proof carried out in this section of the magnetic vector potential, which is essentially based on the concept defined by Eq. (4.107) and related to the magnetic flux density vector by Eq. (4.116). This latter relation is obtained from Eq. (4.4), we which is is the rudimentary version of the Biot-Savart law. This means that have, in fact, derived Ampere’s law from the Biot-Savart law. 4.1 MAGNETIC DIPOLE 1 A small loop with a steady current constitutes the magnetic equivalent of the elec- dipole of Fig. 1.28, and is referred to as a magnetic dipole. The reason for and what we mean by “small” will soon be evident. A magnetic dipole is characterized by its magnetic moment, defined as tric this m = 7S, magnetic dipole moment where / is (4.134) the current intensity of the loop and S = Sn is the loop surface area vector, oriented in accordance to the right-hand rule with respect to the reference direction of the current. Note that m analogous to the electric dipole moment, p, m A m is defined by Eq. (1.116). The unit for We would like is 2 • . magnetic vector potential and the magnetic flux density vector due to a magnetic dipole at large distances compared with the loop dimensions. To this end, we consider a rectangular loop with sides a and b, shown in Fig. 4.30. Let the axis of the loop coincide with the z-axis of a spherical coordinate system and the center of the loop be located at the origin of the system. Far away from the loop ( r a, b ), the loop is observed as being small and the magnetic vector potential of the loop can be evaluated as that due to four line current elements, given by / a, 7b, /(—a), and /(— b), using Eq. (4.110). Let us first calculate the potential at a point P due to the pair of parallel elements of length to find the expressions for the a. Fig. 4.30, (4.135) where r\ and r2 are the distances of the point P from the centers of the first and second element, respectively. Noting that the position vector of the first element with respect to the second element is d = — b, we can now use the same approximations z Figure 4.30 Magnetic dipole. X A,•aa A A : Section 4.11 207 an electric dipole, with which Eq. (4.135) becomes as in deriving Eq. (1.115) for i^o/a d • Mo^a(b r r) (4.136) 4 nr2 47T due to the other pair of elements, those of length b Similarly, the potential Magnetic Dipole , is given by ~ ^ oIh(a ~ 4 icr2 a Abb ' ?) (A" ' 1 J By superposition and using the formula for the vector triple product in Eq. (4.123), the total magnetic vector potential of the dipole turns out to be A = A aa + A bb = As the magnetic moment [b(a • of the loop r) - is (Fig. 4.30) a(b r)] • = £j^(a m = /S = lab z = /a A can be expressed in terms of m as _ ixq m x r (4.1 38) r. (4.139) b, (4.140) r2 471 We x x b) x moment and on the shape of the see that the magnetic potential depends on the dipole magnetic the position of the field point with respect to the dipole, and not mxr = msin0f be arbitrary (not only rectangular). Finally, since which directly comes from Eq. (4.114), Eq. (4.140) can be rewritten as loop, which can Horn sin 9 A= * (4.141) 4nr2 magnetic dipole potential We conclude that the magnetic vector potential due to the magnetic dipole has only a 0 -component in the spherical coordinate system, which is a function of coordinates r and The magnetic 9. flux density vector of the dipole formula for the curl in spherical coordinates, Eq. is now determined applying the (4.85), to the expression for A in Eq. (4.141), B=V x A= (sin 6 rsin# 3 9 ix 0 m 47rr 3 Comparing Eqs. (1.117) and 2 cos 9 r (4.142), 1 3 r dr (rA<p) 0 #) + sin 9 0)• we observe (4.142) that the E and B field of the electric and magnetic dipoles, respectively, are identical in form, so that the corresponding field lines have identical shape. large distances from the dipoles trated in Fig. 4.31. normalized We field lines However, this is true (relative to their dimensions), which only at is illus- see that, although identical far from the sources, the due to the two dipoles are fundamentally different close on the two charges forming the electric to them; the electric field lines terminate dipole, whereas the magnetic field lines close upon themselves through the current loop. We shall see in the next chapter that the concept of a magnetic dipole is fundamental for understanding the behavior of magnetic materials, much like the electric dipole was used in Chapter 2 in studying the electric field in the presence of dielectric materials. The field of a magnetic dipole, Eq. (4.142), is also used, again in magnetic dipole field 208 Chapter 4 Magnetostatic Field in Free Space Figure 4.31 Normalized electric field lines of an electric dipole (a) and magnetic field lines of a magnetic dipole (b). parallel to the electric field of an electric dipole, as an approximation for the and quasistatic (low-frequency) magnetic which is important in EMI considerations. field produced by an Problems 4.28-4.31; Conceptual Questions (on Companion Website): : MATLAB Exercises (on Companion Website). static electrical device, 4.26; Q Q The Lorentz Force and Section 4.12 4.12 THE LORENTZ FORCE AND HALL EFFECT From the definition of the electric field intensity vector, Eq. (1.23), electric force on a point charge we have Eq. that the Q situated in an electric field of intensity E is Fe = QE. Similarly, 209 Hall Effect (4.1) tells us that the (4.143) magnetic force on a point charge electric force on a particle Q moving at a velocity v in a magnetic field equals Fm = Qy X B, (4.144) magnetic force on a particle where B is the flux density vector of the field. Finally, the force on a moving charge due to both an electric and a magnetic field is obtained by superposition, F = Fe + Fm — This equation + tric is magnetic known = Q E + Qx x B. (4.145) as the Lorentz force equation or law. electromagnetic) force on the particle is The Lorentz force total (elec- called the Lorentz force. An and important manifestation of the motion of free charges in is the Hall effect, which we describe briefly here. We consider a conducting strip of width a situated in a uniform steady magnetic field of flux density vector B that is perpendicular to the strip, as interesting a material under the influence of the Lorentz force depicted in Fig. 4.32. Let a steady current of density J flow through the charges constituting the current can be positive or negative in a semiconductor), and shows both Fig. 4.32 cases. (e.g., Due strip. The free holes and electrons to the magnetic force given by Eq. (4.144), the charges move (deflect) across the strip in the direction perpendicular to both B and J, which results in a charge separation on the two sides of the strip. In Fig. 4.32(a), the free charges are positive velocity) is in the same direction as J [Eq. (3.11)], x v<j ( > 0), v = VH (a) vj (drift B is directed to the right, and move to the right. In Fig. 4.32(b), on the other hand, the free charges are negative ( < 0), Vd is in the direction opposite to J, Vd x B is directed to the left, and Fm is again directed to the right (because Q < 0); hence, so is Fm ; thus, the positive charges edge of the strip. Accumulated charges E) across the strip. This field, in turn, acts on the free charges with an electric force, Eq. (4.143), which is in the opposite direction to the magnetic force. In the equilibrium, the two forces are equal in magnitude, i.e., the Lorentz (total) force on the charges is zero, the negative charges produce an end up at the right electric field (of intensity Lh (b) F = Q(E + Vd x B) = (4.146) 0, from which, The voltage between the strip The is known as the Hall voltage, left in Fig. 4.32(a), the polarity of the Hall voltage Note that is v&Ba. this represents a p-type or n-type. tells method (4.148) and the is free charges and (b) negative free charges. edges amounts to direction of the Hall voltage drop charges [from right to (4.147) Vd-B. Eh = Ea = effect. in a material with (a) positive E= This voltage Figure 4.32 Hall effect effect itself is called the Hall different for positive and from and negative left to right in Fig. 4.32(b)]. So, us the sign of free charge carriers in a material. for determining whether a given semiconductor Hall voltage 210 Chapter 4 Magnetostatic Field Free Space in HISTORICAL ASIDE Hendrik Antoon Lorentz (1853-1928), a Dutch mathematician and physicist, a professor of mathematical physics the at (of space and time coordinates), which describe time dilation and length contraction for a body moving at velocities close to the speed of light and represent the foundation of Einstein’s (1879- He University of Leiden, was 1955) special theory of the winner of the the Lorentz force law for a parti- cle in fields. Nobel Prize He further Maxwell’s in 1902 Physics. developed electromag- and proposed the electron theory according to which oscillating electrons within atoms constitute miniature equivalent Hertzian radiators that emit light. In 1904, he netic theory of light, Edwin Herbert Hall (1855-1938), an American was a profesHarvard University. physicist, sor at He received his doctor- from Johns Hopkins University under Professor Henry Augusate in physics tus Rowland (1848-1901), who was one of the world’s most brilliant physicist of the last quarter of the 19th century. introduced his famous Lorentz transformations As a part of his dissertation work. Hall pursued the question as to whether the resistance of a current-carrying conductor was affected by the presence of an external magnetic field. In experiments with guidance from Professor Example 4.17 Hall relativity. formulated moving charged the presence of electric and magnetic Lorentz is also the author of the so-called Lorentz- Lorenz formula, jointly with Danish physicist Ludwig Lorenz (1829-1891), who discovered it independently. The formula provides a mathematical relationship between the index of refraction (of light) and density of a medium. (Portrait: AIP Lande Emilio Segre Visual Archives, Rowland in 1879, Collection) he used a conductor mounted on of a strip of a gold leaf in the form a glass plate and placed it between the poles of an electromagnet such that the magnetic field lines were perpendicular to the current flow in the strip. What he observed was the development of a significant transverse electric field in the conductor and the associated voltage across the strip as the result of the applied magnetic ings in the Magnet on famous field. article Hall published his find- “On a Electric Currents” in New Action of the American Journal November of 1879, and this phenomenon soon came to be known as the “Hall of Mathematics in effect.” He was Harvard 10, appointed professor of physics at Como, Italy -Sept. in 1895. (Portrait: “Voltiana,” 1927 issue, courtesy AIP Emilio Segre Visual Archives) Element for Measuring the Magnetic Flux Density A Hall element for measuring the magnetic flux density is in the form of a strip with width a and thickness d, shown in Fig. 4.33. The concentration of free charge carriers in the strip is /V v and the charge of each carrier is Q. The strip carries a steady current of intensity /, and the magnetic flux density vector is perpendicular to the strip. A voltmeter shows a voltage V \2 between the strip edges. What is the algebraic intensity of the magnetic flux density a Figure 4.33 Hall element for vector (B) with respect to the reference direction indicated in Fig. 4.33? Solution From Eqs. (3.11) and (3.5), the current density in the element can be written as measuring the magnetic flux density; for Example 4.1 7. J = Nv Qv d = — , (4.1 49) Section 4.1 3 = which yields v d Fig. 4.33, namely, Evaluation of Magnetic Forces 211 I/(Nv adQ). The Hall voltage is given in Eq. (4.148), and Eh = V 21 in opposite to the measured voltage V 12 Hence, the magnetic flux density it is . turns out to be = B= 50) Lorentz Force due to a Rotating Charged Contour Example 4.18 A (4.! vd a vd a uniformly charged circular contour of radius a and total charge Q rotates in free space uniform angular velocity w = w z, as shown in Fig. 4.34. A charged particle q moves with a uniform velocity v = v y along a path that belongs to the plane z — a and is parallel to the y-axis. Find the Forentz force on the particle at an instant when it is at about axis with a its the point P(0, The Solution with z 0, a), above the center of the contour. 2^/2 E= The time period for 16nsQa 2 z in Fig. 4.34 is given by Eq. (1.44) ^ = one rotation of the contour _ T= (full P electric field intensity vector at the point = a, Ez. (4.151) is 2n — w (4.152) z Fe 1 angle divided by the angular velocity). Noting that the total charge of the contour, Q, passes any given reference point on the contour during the time T, we conclude that is equivalent to a line current along the contour of intensity the rotating charged contour [Eq. (3.4)] , _QD _QDu;W T From Eq. (4.19), the (4.153) ' magnetic flux density vector due to B /r 0 /V2 „ z (4.145), the this current at the point P (z noQwV2 —- z„ = B z.„ =— Forentz force on the charge q comes out to be qQV2 F = q(E + vyxB) = qEz + qvB x = 2 = a) is (4.154) I6na 8a Using Eq. 2n (eojUQVH'a x+ (Fig. 4.34) z). (4.155) 16^-eofl Problems'. 4.32. EVALUATION OF MAGNETIC FORCES 4.13 Eq. (4.144) gives the Lorentz magnetic force (i.e., the magnetic component of the total Lorentz force) on a single point charge moving in a magnetic field. If we have many charges constituting a current in some domain, principle as in Eq. (4.5) element J dv is B is utilize the superposition given by dFm where we and conclude that the magnetic force on a volume current = (Jdv)xB, (4.156) the flux density vector of the external magnetic Eq. (4.10), the magnetic force on a surface current element dF m = (J s dS) x B, field. Then, from is (4.157) TV Q — Figure 4.34 Lorentz force on a charged particle moving above a rotating charged contour; for Example 4.1 8. 212 Chapter 4 Magnetostatic Field in Free Space and that on a line current element dF m = /dlxB. (4.158) Integration of Eqs. (4.156)-(4.158) leads to the following integral formulations for the total magnetic force for volume, surface, and line current distributions: magnetic force on volume (4.159) current magnetic force on surface (4.160) current magnetic force on line current (4.161) In the last integral, I can be taken out of the integral sign because it is always constant along the line (continuity equation for steady currents). homogeneous conductor of arbitrary cross section with we have J = const [Eq. (3.82)] x B = const and Eq. (4.159) becomes In the case of a straight a steady current placed in a uniform magnetic field, and B = const, so that J Fm = where v and / x B) (J are the surface area of its l dv = (Jv) x B= (J57) xB = volume and length of the conductor, cross section, the direction of the vector direction of the current flow along the conductor, and intensity of the conductor, / F m on a straight conductor a uniform magnetic (751) |1| = /. x B, (4.162) 5 is the same as the respectively, is I the Introducing the current = 75, we obtain Fm in = 71 x B (4.163) field Example 4.19 Two Force Between parallel, very long Two Long and thin wires flowing in the same direction. Parallel Wires with Current in air carry currents of intensities The distance between the wire axes d. is I\ and /2 , both Find the magnetic forces on wires per unit of their length. Solution shows the cross section of the two wires. From Eq. (4.22), the magnetic due to the current in the first wire (/i), assuming that it is infinitely long, of the second wire is Fig. 4.35 flux density vector at the axis B, doh 2nd « (4.164) y we can assume uniform and given Since the wires are thin (as compared to the distance between their axes), that the magnetic field across the entire cross section of the second wire Figure 4.35 Evaluation of the force between two long parallel current-carrying wires; for Example 4.19. is I Section 4.1 3 Evaluation of Magnetic Forces by Eq. (4.164), so that Eq. (4.163) can be used. Hence, the force on the part of the second conductor that is / long equals Mo hhl Fm 2 — 12(H) x B — 2nd * Mo hhl x y z 1 „ (4.165) 2nd and the force per unit of its length dohh Fm2 Ki = ~T The per-unit-length force on the We same conductor first F' is ml see that the magnetic force between the wires and - (4.166) 2nd = — F^ 2 . attractive is if the currents are in the and repulsive if they are in opposite directions (/ 1/2 < 0). So, just in contrary to charges Q 1 and Q 2 and Coulomb’s law, “like” currents (in parallel wires) attract and “unlike” currents repel each other (also see direction, i.e., if both I\ I2 are either positive or negative, Fig. 4.2). Force on a Loop near a Long Wire Example 4.20 A long straight wire carries a steady current of intensity lies in the same plane as the wire, with sides (of length a) perpendicular. of the loop two I. A rectangular conducting loop sides (of length b ) parallel to the wire The distance between The loop carries a steady current of the same intensity, and the shown in Fig. 4.36. Determine the net magnetic force on the loop. is c. currents are and two the wire and the closer parallel side directions of Solution The magnetic flux density vector due to the current in the long wire at any point in is normal to the plane, and at a distance x from the wire its magnitude is the plane of the loop = ^~. 2nx B(x) The forces on each given in Fig. 4.36. in magnitude and with opposite it is forces on the and sides 1 and their directions are directions, — Fm4 (4.168) . 3 are also in opposing directions, but their from the long wire different because of their different distances v Fm 1 = (4.163), obvious that the forces on the sides 2 and 4 are equal Fm2 The from Eq. side of the loop are obtained From symmetry, (4.167) = tutu \ IbB(c) do — 2 (x = c and x magnitudes are = c + a), b (4.169) (repulsive), 2nc 2 Fjn 3 = IbB(c Mo I b 2n(c + a) + a) (4.170) (attractive). y © j (D — L c / F ml B |fVl 1 a Fm2 (D Fm3 b x © Figure 4.36 Evaluation of the magnetic force on a rectangular current loop near a long wire with current; for Example 4.20. 213 . 214 Chapter 4 Magnetostatic Field in Free Space The total force on the loop, assuming that it is rigid (i.e., the loop maintains under the influence of the magnetic forces on its sides), is hence given by its shape even Fm — F m T F m2 + Fm3 + F m 4 — F m T F m 3 i l^pl i 2 b /I 2n \c \ 1 _ c . _ + aj 2 mo I ab 2nc(c + a) . (4.171) is pushed away from the long wire. Note that if the polarity of either (but not both) of the two currents (in the wire and the loop) were reversed, the loop would be pulled Thus, the loop toward the long wire. Force on a Loop in a Uniform Magnetic Field Example 4.21 Prove that the net magnetic force on a contour of arbitrary shape with a steady current uniform magnetic field is zero. Solution For a uniform magnetic Fm Figure 4.37 In it is =/® B= dl x const, so that Eq. (4.161) B= / (b dl becomes x B. (4.172) view of the head-to-tail rule for vector addition, field, in a For any closed path, obvious that dl = (4.173) 0, the integral of dl along a closed path for is always zero; which is evident from Fig. 4.37, and hence Fm = 0. Example 4.21 Torque on a Current Loop Example 4.22 A in a Uniform Magnetic Field is situated in a uniform steady magnetic field of flux There is a steady current of intensity I in the loop. The angle between the plane of the loop and the plane normal to the vector B is 6. The loop is mounted such that it is free to rotate about the axis O-O' which is perpendicular to the plane of the drawing. Find (a) the net force and (b) the net torque on the loop. rigid square loop of side length a density B, as shown in Fig. 4.38(a). Solution (a) From Eq. (4.172), there is no net force on the loop, magnetic forces on the loop sides (b) The is i.e., the vector sum of individual zero. on the sides of the loop that are normal to the axis O-O' tend to stretch the do not produce torques on it. The torques (moments) of the forces on the sides and 2 of the loop (sides parallel to the axis O-O') calculated with respect to the center forces loop, but 1 of the loop are Ti = rj x Fm i Figure 4.38 Evaluation of the torque in a uniform magnetic field, (b) field: (a) and T of T2 = x F m2 , magnetic forces on a current loop position of the loop relative to the magnetic magnetic forces on loop sides producing relationship with the magnetic r2 moment m and (c) the Example 4.22. a torque, of the loop; for (4.174) Problems where Fm i and Fm 2 are the magnetic forces on the respectively, sides, and ri and r2 215 are the position vectors of the centers of sides with respect to the loop center [Fig. 4.38(b)]. Since Fm i+Fm2 = the resultant torque on the loop T= Ti +T2 = ri x Fm i (4.175) 0, given by is +r2 x Fm2 = - (ri r2 ) x Fm i = ri 2 x Fm i. (4.1 76) r X2 = ri - r 2 joins the point of application of F m2 to that of F m i and is independent of the choice of origin of the two vectors iq and r2 Therefore, the torque is also independent of the choice of origin, i.e., it is the same when calculated about any reference point, provided that the total force on the loop is zero. Using Eq. (4.163), the magnitudes of the forces Fm i and Fm2 are The vector . Fm i=Fm2 = IaB (4.177) , which, substituted in Eqs. (4.174), gives the magnitudes of the corresponding torque vectors: m = T2 = T\ (|ri | = a/2). „ |ri x Fm i| = a „ -F m \ smd = la 2 B sin 6 .. , (4.1 Hence, the magnitude of the resultant torque vector 78) T is T = 2TX = Ia2 Bsm9, (4.179) and its direction is shown in Fig. 4.38(b). Of course, the same is obtained from Eq. (4.176). Noting that the angle between the unit normal n on the loop surface oriented in accordance to the right-hand rule with respect to the reference direction of the loop current and the vector B is also 9 [Fig. 4.38(c)], we conclude that the resultant torque of magnetic forces on the loop can be compactly expressed as T= m x B, (4.180) torque on a current loop uniform magnetic m where is the magnetic moment in a field of the loop, given by Eq. 4.134, so that T = |m x B| = mBsm.6 = 2 Ia Bsind. (4.181) a general expression for the torque on a current loop of arbitrary shape uniform magnetic field. We see that the torque on the loop always tend to turn the loop so as to align the vectors and B. In other words, it tends to align the magnetic field produced by the loop current (which coincides with the direction of - see Figs. 4.6 and 4.8) with the applied (external) magnetic field that is causing the torque. Finally, the magnetic field across a small current loop, i.e., a magnetic dipole (Fig. 4.30), can always be considered as locally uniform, which means that Eq. (4.180) gives us the torque on a magnetic dipole (with any shape and the magnetic moment m) in any (generally nonuniform) magnetic field (with the local flux density B). We observe the analogy with the torque on an electric dipole, T = p x E, in Eq. (2.3). Eq. (4.180) and is size in a m m Problems 4.33-4.39; Conceptual Questions (on Companion Website): 4.27-4.30; : MATLAB Exercises (on Companion Website). Problems 4.1. Rectangular current loop. Consider a rectangular loop with sides a and b in air. If the loop carries a steady current of intensity /, find the magnetic flux density vector at an arbitrary point along the axis of the loop perpendicular to its plane. 216 4.2. Chapter 4 Magnetostatic Triangular current loop. Field in Free A loop Space form in the 4.6. Solenoids with different length-to-diameter of a triangle representing a half of a square ratios. of side a carries a steady current of intensity core and /, as shown in Fig. 4.39. The medium Calculate the magnetic flux density vector at P located a point of the solenoid vertex of the at the fourth of the P (or last) wire turn, first < oo) for the solenoid radius equal to (a) a = 25 cm and = 2 cm, respectively. (b) a Helmholtz coils. Fig. 4.41 Figure 4.39 Triangular very short coils, current loop; for Problem 4.2. wire, in Current loop with circular and linear parts. shows a wire contour composed from air. a half of a circle of radius a and a half of a is d situated in air carries a steady current of intensity I. shows two each with N identical circular turns of The distance between the d, is and the wire turn centers radii are a. The coils carry steady currents of equal intensities, /, and are oriented in the same way. When Fig. 4.40 square of side 2a. The contour < z of the coils and and sketch the function B(z) along the solenoid axis (— oo 4.7. 4.3. = is / field at the solenoid center, at the center of the 250th (or 750th) wire turn, and at the center square. a = 1 A. The length 50 cm. Calculate the B carrying a steady current I air. is Consider a solenoid with a nonmagnetic N = 1000 tightly wound turns of wire — a, the magnetic field near the center of (midway between the two the structure Find coils) approximately uniform, and the structure is referred to as Helmholtz coils, (a) For an arbiis the magnetic flux density vector (a) at the central point the point O and (b) at a point la apart from O along the line perpendicular to the d (d trary distance a), find the expression for the magnetic flux density B(z) at an arbitrary plane of the contour. point along the axis of the coils (z-axis), assuming that B due to each of the coils equals N times the flux density of a single turn of wire (circular current loop, Fig. 4.6) at the respec- a and three linear Problem 4.3. current distribution. There is z b with a hole of radius a (a < = d/2 (and any verify that d d a = 2 that 0 at d, relative to a), (c) £/dz 2 = a (note that even Then C when 3 d 5/dz 3 = 0 at C in this 0 at the point case). surface current over a circular surface of radius lies in = the center of the structure (point C), so for parts; for The surface d5/dz contour composed from semicircle 4.4. Circular surface Show Figure 4.40 Current tive coil location, (b) b) in free space. the plane z = d 0 of a cylin- drical coordinate system, with the coordinate origin coinciding with the surface center. surface current density vector (a < r < b ), where 7s o is is Js The = J o(a/r)ty s Compute a constant. the magnetic flux density vector along the z-axis. 4.5. Magnetic field of a rotating charged disk. A ciris uniformly charged over its surface by a charge of density p s The disk uniformly rotates in air about its axis (perpen- Figure 4.41 cular disk of radius a Helmholtz coils; for Problem 4.7. . dicular to the disk) with an angular velocity 4.8. Spherical coil. Consider a w. Find the magnetic flux density vector at N an arbitrary point along the axis of rotation. formly and densely Assume netic sphere of radius a. that the charge distribution over the disk remains the same during the rotation. coil consisting of turns of an insulated thin wire the wire is /. in wound uni- one layer on a nonmag- The current through The surrounding medium is air. 217 Problems Find the magnetic flux density vector sphere center. 4.9. at the calculate the everywhere Two parallel strips with opposite currents. Two 4.12. (0 magnetic flux density vector r < 00 ). < An Rotating cylinder with a surface charge. long conducting nonmagnetic cylin- parallel, very long identical strip conductors of infinitely width 2 a carry currents of the same intensity 7 der of radius a The distance between the strips is 2 a as well. The cross section of the structure is shown in Fig. 4.42. The permeability everywhere is p-o- Compute the B field at surface with a charge density p s The cylinder rotates in air about its axis with a uniform and opposite directions. the center of the cross section (point O). is uniformly charged over angular velocity w. Find the cylinder and outside 4.13. B field inside the it. Rotating nonuniformly charged hollow cylinder. An infinitely long hollow cylinder of inner radius a and outer radius b (a Figure 4.42 Cross section of two its . < b) in air is charged with a volume charge density p(r) = Por/a (a < r < b), where po is a constant and r the radial distance from the cylinder axis. parallel strip conductors with currents The same magnitude and opposite directions; for Problem 4.9. of the cylinder uniformly rotates about its axis with an angular velocity w. Assuming that the charge distribution of the cylinder does not change during the rotation, determine the 4.10. Magnetic ductor. of a hollow cylindrical con- field A through an infinitely long hollow cylindrical copper conductor of radii a and b (a < b ). The cross section of the conductor is shown in Fig. 4.43. magnetic flux density vector everywhere. steady current of intensity 7 flows The medium the conductor is 4.14. Two parallel infinite planar current sheets. Two parallel infinite planar current sheets in air have the same uniform surface current density 7S The distance between the sheets is d. Find the magnetic flux density vector everywhere if the currents of the sheets run in (a) the same direction and (b) opposite directions. . and outside nonmagnetic. Find the magin the hole netic flux density vector everywhere. 4.15. Magnetic field inside a thin plate with current. A thin copper plate of length b, width a and , thickness Figure 4.43 Cross section of intensity a hollow cylindrical Magnetic field same « a) carries a steady current of shown in Fig. 4.44. The Neglecting the end plate effects, is use to find the distribution of the magnetic flux density vector inside the plate [note that this is mathematically completely analogous to the application of a similar integral equation in Eq. (6.146) and Fig. 6.25(b)], of a triaxial cable. The cross cylindrical conductors, looks the as air. Ampere’s law section of a triaxial cable, having three coaxial 7, situated in conductor with a steady current; for Problem 4.1 0. 4.11. d (d as the cross cut of the system of three concentric spherical conductors in Fig. 1.56. The radius of the inner conductor of the cable 1 mm, is conductor are b = 2 mm and c = 2.5 mm, = 5 mm those of the outer conductor d e = 5.5 tric, all mm. The = and and cable conductors and dielec- as well as the surrounding Figure 4.44 Thin conducting plate with medium, are nonmagnetic. Assuming that steady cur- rents of intensities I\ /3 a the inner and outer radii of the middle = —1 A — 2 A, I 2 = —1 a A, and steady current; for Problem 4.15. flow through the inner, middle, and outer conductor, respectively, with respect to the same reference all given direction, 4.16. Magnetic tor. field in a leaky parallel-plate capaci- Calculate the magnetic flux density vector 218 Chapter 4 in the 4.17. Magnetostatic Field in Free Space imperfect dielectric of the parallel-plate 10 cm and Magnetic vector field in a from Example leaky spherical capacitor. der from Problem 3.17 Magnetic field 4.27. Magnetic (Fig. 3.34). around a grounding electrode. Find the magnetic flux density vector on the surface of the ground for the grounding electrode from Example 3.15 [Fig. 3.26(a)]. B makes Law Magnetic field in a leaky coaxial cable. Determine the magnetic flux density vector in height h Current distribution from field distribution. Using the differential Ampere’s law, show that the magnetic field given by Eqs. (4.53) and from vector flux (a) 4.22. 4.23. is A = 2 r2 T m (r in m). <\> this region, (b) Obtain the magnetic flux through a circular contour plane z = origin, (c) circulation of 4.11 but employing Ampere’s in differential form. charged Rotating 4.28. Potential cylinder by A differential m is in radius that lies in the centered at the coordinate Check the by evaluating the results A along the contour. and magnetic Am 1 0 and Redo Example field due to a magnetic dipole. dipole with moment a m= Ampere’s law. Redo Example 4.14 but with the use of the differential form of Ampere’s law. 400 /x ical coordinate system. Calculate the magnetic Thin plate with current, differential Ampere’s law. Redo Problem 4.15 but applying the differential Ampere’s law [adopt the vectorcomponent notation as in Fig. 4.22 and Eq. (4.75), and use the analogy with the application of the differential Gauss’ law in potential Ampere’s law B= by in in differential and is Magnetic through a cylindrical surface. Calculate the outward magnetic flux through the flux lateral m, surface of a cylinder of radius at the follow- 0, 0), (b) (1 (1 magnetic dipole 4.30. magnetic dipole. far away from it. Rectangular current loop as a magnetic dipole. Check the result for the B field due to a rectangular loop obtained in Problem 4.1 by sides, of length 2 in (b). B flux density giving the magnetic flux density vector of a m, parallel to the x- and y-axes. (c) Confirm Ampere’s law in integral form and Stokes’ theorem by evaluating the net circulation of B along the contour defined and » given 3 the xy-plane, with the center at the coordinate and located at the origin of a spher- Refer to the circular current loop in Fig. 4.6, a, Eqs. (4.19) and and show that for |z| (4.142) become the same, the latter equation integral form. current enclosed by a square contour lying in origin A is m, ;r/2, n/2), (c) (1 m, n, 0), m, 7r/4, 0), (e) (10 m, jr/4, 0), and (f) (100 m, 7r/4, 0). The dipole dimensions are much smaller than 1 m. (1 (d) m). The density, z ing points defined by spherical coordinates: (a) — l) x + 2x y + xy z] mT {x, y, z medium is air. (a) Find the current (b) From the result in (a), find the total 2 [4 (z 2 4.29. Circular current loop as a 1.23]. In a certain region, the magnetic field 4.25. potential. In a cer- magnetic vector potential is Example 4.24. cylin- Find the magnetic flux density vector in Coaxial cable using differential Ampere’s law. law uniform of conservation of magnetic flux. For the magnetic field defined in Problem 4.24, confirm the law of conservation of magnetic flux in differential and integral forms by evaluating (a) the divergence of B and (b) the outward flux of B through the surface of a cube, respectively. Let the cube be centered at the coordinate origin, with edges parallel to coordinate axes and 2 m long. cal coordinate system: infinitely long cylinder of radius a in free space. 4.21. in a S = 1T. The given as the following function in a cylindri- produced by a uniform volume current of density given by Eq. (4.52) along an (4.56) cm 20 an angle of 60° with the tain region, the 4.20. = field of flux density axis. 4.26. 3.3 (Fig. 3.6). the imperfect dielectric of the coaxial cable 4.19. — magnetic Find the magnetic flux density vector in the imperfect dielectric of the spherical capacitor 4.18. a capacitor from Problem 3.12 (Fig. 3.32). comparing it with the corresponding dipole- field expression, away from 4.31. from Eq. (4.142), at points far the loop. Dipole equivalent to a surface current distribuConsider the circular nonuniform surface current distribution from Problem 4.4, and a), show that far away along the z-axis (|z| tion. » 219 Problems this 4.32. current distribution can be replaced by 4.35. Wire-strip transmission line. Fig. 4.46 shows a an equivalent magnetic dipole located at the cross section of a two-conductor transmission coordinate origin. Find the moment, m, of the line consisting of a equivalent dipole. of the strip Lorentz force due to a rotating charged disk. Refer to the rotating charged disk from Problem 4.5, and assume that a charged particle Q moves with a uniform velocity v along a path parallel to the plane of the disk. Find the Lorentz force on the particle at an instant when it is at the point that belongs to the axis the conductors of disk rotation and is at a distance a is wire and a strip. The width 2 a and the separation between is The a. dielectric air. is If the current I runs along the line, calculate the magnetic force on the of its strip conductor per unit a section of a length. Figure 4.46 Cross from the transmission line disk center. consisting of a wire 4.33. Forces between three parallel wires with current. Three parallel very long and thin wires — in air carry currents of intensities I\ I2 to 1 A, — —1 A, and I3 = 2 A, all given with respect the same reference direction. The distance between any two of the wires is and a strip; for Problem 4.35. 4.36. A d=lm, which is made from two identical strips of width a carry steady currents of intensities I and —I given with respect to the same tor, so their cross section constitutes an equilateral triangle (with sides d ). (a) , Determine the The direction and magnitude of the per-unit-length magnetic force on the wire with current I3 (b) Redo the problem if I2 = 1 A. reference direction. Force on a wire due to a semicylindrical conductor. very long aluminum conductor in portrayed in Fig. 4.47. The materials and the the form of a half of a thin cylindrical shell of per-unit-length magnetic force A radius b 40 situated in mm air. and thickness d — 0.5 mm is a. cross section of such a transmission line is ambient medium are nonmagnetic. Find the A = distance of the wire from both ends of the corner conductor . 4.34. Line composed of a wire and a corner strip wire and a 90° corner conducconductor. on the wire. is Another very long aluminum conductor, in the form of a very long wire = 1 mm, of radius a is positioned along the The two conductors carry steady currents of the same magnitude I = 100 A and opposite directions, as shown in Fig. 4.45. Find the magnetic force on the wire Figure 4.47 Cross axis of the semicylinder. section of a transmission line consisting of a and wire a 90° corner strip conductor; for Problem conductor per unit length. 4.36. 4.37. 1 Coaxial cable with off-centered cavity. in Fig. 4.48 is Shown a cross section of a coaxial cable in which the cylindrical cavity of radius b representing the inner surface of the outer conductor is off-centered by a vector d with respect to the common axis of the inner con- ductor and the outer surface of the outer conductor. The other two radii are a and Figure 4.45 System composed of a wire conductor along the axis of c, and the relationship b + d < c is satisfied. The permeability everywhere is hq. If a steady curestablished in the cable, a thin semicylindrical rent of intensity I conductor; for Problem determine the magnetic force on the inner conductor per unit of its length. 4.34. is i i i 220 Chapter 4 Magnetostatic Field in Free Space Figure 4.48 Coaxial cable with off-centered inner Figure 4.49 surface of the outer Triangular current conductor; for loop in a uniform magnetic field; for Problem 4.38. Problem 4.37. 4.39. 4.38. Torque between two magnetic Force and torque on a triangular current loop. small current loops are A space. rigid loop in the form of an equilateral tri- is situated in a uniform magnetic field of flux density B. The steady field lines magnetic are parallel to the plane of the loop and are perpendicular to one of its angle of side length a sides, as shown in Fig. 4.49. If a steady current m; The = mi loop has a magnetic first and is centered ing locations of of its magnetic moment mx, and the net torque on the (c) on each of the the net force and (d) loop. at the origin , (e) m = 0.1 Am 2 0) (O) of = m2 : 1 directions (a) P(0, a, 0) and m 2 = my, (c) P(0, P(0, 0, a) and m 2 = and (d) P(0, a, a) and a P and center its loop sides, as well as established in the loop, find (a) moment torque on the second loop, for the follow- the force and (b) the torque is free in the Cartesian coordinate system. Obtain the m 2 = m z, (b) P(0, a 0, a) and m 2 =mz, of intensity I Two dipoles. positioned and ni 2 = mz. Take 10 m. I I I Magnetostatic Field Media in Material Introduction: O ur study of magnetostatics has so far been restricted to the magnetic field due to steady vacuum and other nonmagwe shall introduce phenomena associated with the mag- electric currents in a netic media. In this chapter, and discuss netostatic field in the presence of magnetic materials. Many of the basic concepts, physical laws, and mathematical techniques constituting the analysis of materials in the magnetic field are entirely analogous to the corresponding concepts, laws, and techniques in electrostatics, which makes our discussions in this chapter much easier. The most important difference, however, with respect to the analysis of dielectric materials linear behavior of the is the inherent non- most important class of magnetic materials, called ferromagnetics. This is a class of materials with striking magnetic properties (many orders of magnitude stronger than in other materials), with iron as a typical example. We shall start with a qualitative characterization of microscopic magnetic phenomena in substances and describe the behavior of different types of magnetic materials based on the classical atomic model. By analogy with the polarization vector in electrostatics, the magnetization vector will be used to describe the magnetized state of a material on a macroscopic scale. Magnetization volume and surface current density vectors will be defined as macroscopic equivalents to a vast collection of tiny electric currents that are micro- scopic sources of the magnetization of a magnetic These current densities will enable us due to magnetized materials using free-space formulas and techniques from the preceding chapter. We shall derive and discuss Maxwell’s equations and boundary conditions material. to evaluate the magnetic field for magnetostatic systems that include arbitrary media. The concept of permeability of a material will allow for additional macroscopic character- ization of magnetic materials. Finally, a section on magnetic cores circuits (consisting of of different ferromagnetic shapes with current-carrying 222 Chapter 5 windings) will Magnetostatic Field in Material Media represent a culmination of the theory netic materials. be applied to perform the analysis of such which essentially resembles the dc analysis will of the magnetostatic field in the presence of mag- circuits, Most of the work of the chapter of nonlinear electric circuits. MAGNETIZATION VECTOR 5.1 According to the elementary atomic model of matter, all materials are composed of atoms, each with a central fixed positively charged nucleus and a number of negatively charged electrons circulating around the nucleus in various orbits. Both these orbital motions and the inherent spins of the electrons about their own axes can be represented by small current loops, i.e., magnetic dipoles. These tiny currents are referred to as Ampere’s currents. The magnetic moment of each elementary loop is given by Eq. (4.134). In the absence of an external magnetic field, the equivalent magnetic dipoles have random orientations with respect to one another, resulting in no net magnetic moment. With an applied field, however, the equivalent current loops experience torques, which lead to a net alignment of microscopic magnetic dipole moments with the external magnetic field [see Eq. (4.180)] and a net magnetic moment in the material on a macroscopic scale. The process of inducing macroscopic magnetic moments by an external magnetic field is called the magnetization of the material. This process is practically instantaneous, and the material in the new magnetostatic state is said to be magnetized or in the magnetized state. When magnetized (by an external field), a material is a source of its own magnetic field, and the total field at an arbitrary point in space (inside or outside the material) is a sum of the external (primary) field and the field due to the magnetized material (secondary field). In the analysis, we can replace the material by a collection of microscopic Ampere’s current loops (magnetic dipoles) residing in a vacuum, as the rest of the material does not produce any field. Then, the secondary field can, in principle, be determined using the expression for the magnetic field due to a magnetic dipole in free space, Eq. (4.142), and superposition. Instead of analyzing every single atom and all microscopic magnetic dipole moments, however, we rather introduce a macroscopic quantity termed the magnetization vector to describe the magnetized state of a material and the resulting field. Analogously to the definition of the polarization vector (P) for electrically 1 polarized dielectric materials, Eq. (2.7), the magnetization vector, M, is defined as the density of the equivalent elementary magnetic moments in a magnetic material at a magnetization vector given point: (Vm). /in , av (5.1) jyj (unit: dv A/m) We note that M dv represents the dipole moment of a magnetic dipole that alent to an elementary volume dv Ampere’s currents) within ' A it. The in the material, i.e., to unit for the magnetization vector third source of equivalent microscopic magnetic dipole moments is equiv- the dipoles (microscopic all in atoms is is A/m. nuclear spin, but it provides a negligible contribution to the macroscopic magnetic properties of materials. Note, however, that nuclear spin represents the basis of magnetic resonance imaging (MRI), used in medicine, and also in many areas of science and engineering. Section 5.2 223 Behavior and Classification of Magnetic Materials In any magnetic material, the magnetization vector at a point is a function of the magnetic flux density vector at that point, M = M(B), (5.2) and this relationship is a characteristic of individual materials. It is entirely analogous to the relationship between the polarization vector and the electric field intensity vector, Eq. (2.9), in electrostatics. 5.2 BEHAVIOR AND CLASSIFICATION OF MAGNETIC MATERIALS A thorough understanding and precise quantitative characterization of microscopic magnetic phenomena in materials require a full quantum mechanical treatment. Here, however, we describe qualitatively the behavior of different types of magnetic materials based on the classical atomic model. Generally, materials can be classified according to their magnetic behavior into diamagnetic, paramagnetic, ferromagnetic, antiferromagnetic, ferrimagnetic, and superparamagnetic materials. In diamagnetic materials, the magnetic moments of electrons orbiting about their nuclei are dominant compared to the magnetic moments attributed to electron spin. In order to describe diamagnetic behavior, which is present to a greater or lesser extent with all magnetic materials, we consider first a model of an atom with a single electron that circulates about the nucleus along an orbit of radius a with a uniform angular velocity wo = wq z, as shown in Fig. 5.1(a). In the absence of an external magnetic field, the outward centrifugal force on the electron, given by Fc f = m e vn o = m e wla, (5.3) a where mt and vo are the mass and velocity of the electron (vo = woa), respectively, balanced by the centripetal (attractive) electric (Coulomb) force, nucleus and electron. Thus, the balance equation reads is Ft = m e Wo«. Fe , between the (5.4) The orbiting electron is equivalent to a small current loop (magnetic dipole), where the current I of the loop is given by Eq. (4.153) with Q = — e and is Figure 5.1 Atom with a single electron orbiting about the nucleus, an applied magnetic opposite to it (c). field, whose in the absence of an external magnetic field (a), direction coincides with the direction of the electron angular velocity vector (b) or and with is m 224 Chapter 5 Magnetostatic Field in Material Media directed oppositely to the direction of electron travel (because the electron charge is From Eq. negative). (4.134), the magnetic moment of the dipole — m = lna 2 {- z) = () is 2 In the presence of an external magnetic field, there (5.5) is an additional force on the electron - the magnetic force, Eq. (4.144), = Qv Em B e xt = — X B ex X (5-6) t) with Bext denoting the magnetic flux density vector of the applied field. For the situation in Fig. 5.1(b), where the electron angular velocity vector is in the same B ex = new balance equation direction as the applied field (inward), so that the the magnetic force z, t is Fe + ewaB ex = m t wa, w — wq + Aw. i (5.7) is compensated Combining Eqs. Here, the force unbalance created by the magnetic force increase Aw centripetal is (Aw > of the orbital angular velocity 0). for by an and (5.4) (5.7) gives m e (w 2 - = ewB ext Wg) Since the perturbation of the electron velocity strongest applied magnetic fields, 9 9 w — Wq = we can (w + is (5.8) . small, i.e.. Aw « wq, even for the write — wq)(w 2 wAw. wq) ( Substituting this into Eq. (5.8), the increase of the angular velocity Aw = new m (5.10) ewa 2 „ _ z momentum L= where O (|r| r is —- ea,2 - w. (5.11) 2 of the electron, mer x v = Xw, (5.12) the instantaneous position vector of the electron with respect to the origin The increase of the dipole the electron, in Eq. (5.10), — L = — —2 m 2m e moment due is Xw. to the increase of the orbiting velocity of given by called the induced magnetic antiparallel (i.e., in the (5.13) e Am = — — XAw — - ( —— ^ 2me \2 e ) is is = a) andX = m e a 2 is the rotational inertia of the electron, Eq. (5.11) becomes m= and moment ‘ = 2 Introducing the angular to be e equivalent magnetic dipole m . ^®ext 2 In place of Eq. (5.5), the comes out 5 9) moment (5.14) XBext- of the electron. We opposite direction) to the applied magnetic note that For an electron whose orbital angular velocity vector and the opposed, as in Fig. 5.1(c), the force F m on and the force unbalance is compensated new magnetic dipole moment is smaller moment Am is again in the —z the electron is in the Am B ex field B ex field is t- t are outward direction, by a reduced velocity, i.e.. Aw < 0. The magnitude than m so that the induced for in direction, that () , is, still antiparallel to the applied Section 5.2 Behavior and Classification of Magnetic Materials Am and given by the same expression in Eq. (5.14). This same expression for obtained also in the case of an arbitrary mutual position of the electron orbit and is electrons, the applied magnetic field, as well as for the model of an atom with field, N travel along orbits that are arbitrarily oriented with respect to the applied which magnetic field, where X should be replaced by the average rotational inertia of electrons in the atom, X — Xav With no external magnetic all . random field applied, the orientations, so that the net magnetic in the material are zero. ferential induced magnetic dipoles of the atoms have moment and the magnetization vector With an applied field, however, each atom acquires a difgiven by Eq. (5.14), with X = Xav and all these moments moment , are antiparallel to the magnetic flux density vector, B, in the material. the net magnetic moment of the atoms opposes the applied diamagnetic material placed in a magnetic field. field will orient itself As a result, Thus, a rod of a perpendicular to the applied field, i.e., across the field lines. The term diamagnetism originates from the Greek word “dia” meaning “across.” If a diamagnetic specimen is brought near either pole of a strong bar magnet, it will be repelled by the magnet. From Eqs. (5.1) and (5.14), the magnetization vector is -~- B, M = iV Am = — Nwe 4 mg (5.15) v Av being the concentration (number per unit volume) of atoms in the material. We note that this equation represents a special form of Eq. customary to write M= where Xm is — Mo (5.2). In addition, B, it is (5.16) a dimensionless proportionality constant given by /xoAv e^Xav (5.17) 4 nil It is rials medium, Xm = 0.) We vacuum, which is see that the diamagnetic mate- are linear magnetic media, because the relationship M(B) in Eq. (5.16) is a linear one. Typical diamagnetic materials are bismuth, silver, lead, copper, gold, silicon, germanium, graphite, sulfur, hydrogen, helium, sodium chloride, and water. Substituting typical values for the quantities involved in the expression in Eq. (5.17) -4 -6 indicates that Xm of diamagnetic materials is of order — 10 (water) to — 10 (bis- We see that the magnetic susceptibility of diamagnetic materials is negative and very small. The macroscopic magnetic effect in diamagnetic substances is very weak and negligible in most practical situations. In general, the diamagnetic effect is present in all materials when placed in a magnetic field, because it arises from an interaction of orbiting electrons in atoms with the external field. However, in other types of magnetic materials, the diamagnetic reaction is completely masked by other effects. In paramagnetic materials, the atoms have a small permanent net magnetic dipole moment associated with them, due almost entirely to spin magnetic dipole muth). moments of electrons. In the absence of an external magnetic orientation of the atoms produces an average magnetic finite volume. When an external susceptibility of diamagnetic materials called the magnetic susceptibility of the material. (For a the only truly nonmagnetic magnetic field is applied, field, moment however, there is the random of a zero in a a small torque given by Eq. (4.180) on each atomic moment, which tends to align the moment in the direction of the applied field. This alignment acts to increase the value of 225 226 Chapter 5 Magnetostatic Field in Material Media the magnetic flux density vector, B, within the material over the external value. However, the external field also causes a diamagnetic effect of the orbiting elecmagnetic materials exhibit diamagnetic behavior), which counteracts the increase in B. The alignment process is also impeded by the forces of random thermal vibrations, and the resulting increase in B is quite small. The overall macroscopic effect is equivalent to that of a small positive magnetization. The mag- trons (all netization vector, M, is in the same direction as the vector B. Eq. (5.16), then, tells us that the magnetic susceptibility of paramagnetic materials small. itself When a rod of a paramagnetic material along the field lines, Greek means “along”). is is positive placed in a magnetic field, and very it orients and hence the term paramagnetism (the word “para” paramagnetic substance is brought be attracted to it. Typical paramagnetic materials are palladium, aluminum, and oxygen, and typical values of Xm -6 are of orders 10 (oxygen) to 10~ 3 (palladium). Note that air is a paramagnetic in Additionally, near a pole of a strong bar magnet, medium as well. Obviously, the it if a will paramagnetic effect is also very agnetic materials can be treated as nonmagnetic media (x m = weak and param- 0) in most practical applications. The remaining four classes of magnetic materials (ferromagnetic, antiferromag- and superparamagnetic) all have strong atomic magnetic dipole moments caused primarily by uncompensated electron spin moments. The interaction of adjacent atoms leads to an alignment of the atomic moments in either an aiding (parallel) or opposing (antiparallel) manner. The macroscopic magnetization of ferromagnetic materials can be many orders of magnitude larger than that of paramagnetic materials. Interatomic forces in a ferromagnetic sample cause the atomic moments to line up in a parallel fashion over regions containing large number of atoms (e.g., 10 15 atoms). These regions, called the magnetic domains (or Weiss' domains), range in size from a few microns to several centimeters in each dimension. With no external field applied, the domain moments vary in direction from domain to domain, as indicated in Fig. 5.2 for a polynetic, ferrimagnetic, Figure 5.2 Magnetic domains in a polycrystalline ferromagnetic specimen with no external magnetic field applied. crystalline ferromagnetic specimen. Due to overall vector cancelations, the material whole has no net magnetization. Upon the application of an external magnetic field, however, the volumes of the domains that have moments aligned or nearly aligned with the applied field grow at the expense of their neighbors, and the magnetic flux density vector of the secondary field (that due to the material) increases greatly over that of the external field alone. For weak applied fields, the movements of magnetic domain walls are reversible, i.e., the domains go back to their initial as a is turned off. Above a certain (not large) field magnitude, however, this process becomes irreversible. Additionally, the domains start rotating toward the direction of the applied field, so that completely random domain orientations are no longer attained after the external field has been removed. In other words, there is a residual or remanent net magnetization of the material that states after the field remains after the complete removal of the primary field. This means that the magnetization of the material lags behind the field producing it, and also that the magnetization state of the material at an instant of time is a function not only of the magnetic field, i.e., its flux density vector, at that instant, but also of the magnetic history of the material. This phenomenon is called hysteresis (which is derived from a Greek word meaning “to lag”). As the applied field becomes even much stronger, a total alignment of all the domain moments with point, the ferromagnetic material is the applied field occurs. At this said to be saturated. In the state of saturation, further increase of the external magnetic flux density vector increase of the magnetization vector. no longer causes an Section 5.3 Magnetization Volume and Surface Current Densities 227 Typical ferromagnetic materials are iron, nickel, and cobalt, and their alloys. The name ferromagnetic comes from the Latin word for iron, “ferrum.” Ferromagnetics are the most important magnetic materials in engineering. They are widely used in cores of inductors and transformers, and in electric motors, generators, electromagnets, relays, and other devices that use magnetic forces and torques, as well as in magnetic heads and tracks of computer hard disks and other magnetic storage (recording) devices. Note that above a certain (very high) temperature, called the Curie temperature, the thermal vibrations of atoms completely prevent the coupling (parallel alignment) of atomic magnetic moments, so that ferromagnetic materials lose all their ferromagnetic characteristics and revert to paramagnetic materials. The Curie temperature for iron is Illllllllltlll (a) 770° C. In antiferromagnetic materials, the forces between adjacent atoms cause the atomic magnetic moments to line up in opposite directions so that the net magnetic moment of a specimen is zero, even in the presence of an applied magnetic field. as well as many oxides, sulfides, and chlorides, Examples are manganese oxide (Mn02), ferrous (FeS), and cobalt chloride (C 0 CI2 ). Antiferromagnetism is not of practical sulfide this class of materials. importance. aligned antiparallel, but the moment A of a specimen. therefore occurs, although moments moments of adjacent atoms are also are not equal, so there is a net magnetic large response to an externally applied magnetic field it is substantially less than in ferromagnetic substances. comparison of the atomic magnetic dipole moment structure for ferromagnetic, antiferromagnetic, and ferrimagnetic materials. The most important subclass of ferrimagnetics are the ferrites, which have much lower Fig. 5.3 depicts schematically a electric conductivity (a) 7 compared with 10 S/m material why when than the ferromagnetics (for instance, 10 for iron). Low many for high-frequency transformers, shifters, in spite -4 to 10 2 S/m, conductivity limits induced currents in the alternating (ac) fields are applied (so-called the ferrites are used in eddy currents), and this applications at high frequencies, such as cores AM, short-wave, and FM antennas, and phase of their weaker magnetic effects (compared to ferromagnetics). The reduced eddy currents lead to lower Joule’s (ohmic) losses in the material examples of ferrite substances are iron ferrite (FesCL), nickel ferrite (NiFe 2 C> 4 ), and cobalt ferrite (CoFe 2 C> 4 ). Ferrimagnetism also disappears at temperatures above a critical value - the Curie temperature. Finally, superparamagnetic materials are composed of ferromagnetic particles suspended in a nonmagnetic (dielectric) matrix. Each particle contains many magnetic domains, but the domain walls cannot penetrate the matrix material and there is no coupling with adjacent particles. Thin dielectric (plastic) tapes with suspended ferromagnetic particles can store large amounts of information in magnetic form because the particles are independent from each other and it is possible to change the state of magnetization along the tape abruptly in very small distances. Such superparamagnetic tapes are widely used as audio, video, and data recording tapes. (core). Typical MAGNETIZATION VOLUME AND SURFACE CURRENT DENSITIES In this section, we shall obtain the expressions for evaluation of the macroscopic volume and surface current densities equivalent to microscopic Ampere’s currents in a magnetized body - from a given distribution of the distribution of |i|i|i|i|i|i|i (c) In ferrimagnetic materials, the magnetic 5.3 (b) Chromium and manganese, belong to is Illllllllltlll Figure 5.3 Schematic structures of atomic magnetic dipole moments for (a) ferromagnetic, (b) antiferromagnetic, and (c) ferrimagnetic materials. Chapter 5 Magnetostatic Field in Material Media magnetization vector, M. The vector M, in turn, moments is obtained by averaging the Ampere’s current The equivalent macroscopic current is called the magnetization current, and the corresponding volume and surface current densities are denoted by J m and J m s, respectively. The expressions for these current densities are similar to the expressions for calculating the volume and surface bound (polarmicroscopic magnetic dipoles (magnetic of microscopic loops) in the magnetic material. and p ps ) from the polarization vector (P) in a polarized be used later for free-space evaluations of the magnetic flux density vector (B) due to magnetized bodies. Let us first find the intensity of the total magnetization current Im c enclosed by an arbitrary imaginary contour C situated (totally or partly) inside a magnetized magnetic body, as depicted in Fig. 5.4. We note that Im c is actually the current that passes through any surface S bounded by C. Let the magnetic moments of small Ampere’s current loops in a vacuum that constitute the magnetization current be expressed as ization) charge densities (p p dielectric body. They will (5.18) where S m = |S m is | the surface area of the loop. the loops that pass through S It is obvious in Fig. 5.4 that all do not pass through S 7 m c Only loops that pierce S twice, as well as those that at all, contribute with zero net current intensity to only once, that the loops that encircle C, contribute actually to the total current is, intensity flowing through S. To evaluate /m c the loops that are strung along C in the general case, (like pearls the contribution of such loops as either / or . on a —/ string). In (note that /, we therefore count doing that, we count generally, differs from loop to loop), by inspecting whether the direction of the loop current traversing 5 agreement or disagreement with the reference orientation of S. Consider an element d / of C and the case when the angle fi between the vector (or the average of elementary dipole moment vectors m near d/) and vector dl, which is oriented in accordance to the orientation of C, is less than 90°, as shown in Fig. 5.5(a). Note that centers of loops that are positioned close to dl and encircle it are inside an oblique cylinder with bases S m and height in is M dh so that the total number = d/cos^5, (5.19) of these loops equals the concentration of loops (magnetic dv = S m d h. The refcontour C by means of erence orientation of the surface S is related to that of the normal vector ii on S in the right-hand rule, so the reference direction of the unit dipoles), 7V v [see Eq. (2.6)], times the Figure 5.4 Contour C in a magnetized magnetic body. volume of the cylinder, C in Figure 5.5 Element of the contour and dl: (a) 0 < p < 90° and (b) 90 ° Fig. 5.4, in 229 Magnetization Volume and Surface Current Densities Section 5.3 two cases with regards between to the angle p M <p< 180°. from S upward. We then realize that all the loop currents encircling dl pass through S in the positive direction (direction in agreement with the orientation of S ) and contribute to the total current with the intensity / (we assume that all the loops near dl, which is differentially small, have the same moments and currents). Hence, the corresponding contribution to the magnetization current intensity through S is given by Fig. 5.5(a) is = Ny Sm dl cos p I dlm when p > In the case and are taken all 90°). = dl cos(7r — P) = d/(— cos/3), (5.21) the loop currents encircling dl pierce S in the negative direction as —I, dlm = NvS m dl (—cos /l) (-/) (90° <p < which is the same result as in Eq. (5.20). = y m, where Eq. (5.1) can be interpreted here as and hence, for an arbitrary p (0 < p < 180°), we have M N d/m [note that the (5.20) 90°, Fig. 5.5(b), dh and, because <p < (0 boundary = Nw m dl cos p = case, p = yVv m • dl = M 180°), m • is (5.22) given by Eq. (5.18), dl 90° (the contour element dl (5.23) is tangential to the S m of the loops) and d/m = 0, is also properly included in this formula]. by integrating the result for d/m along the entire contour C, we get surfaces Finally, /mC = £M-dI. (5.24) total magnetization current enclosed by a contour This is an integral equation similar in form to Ampere’s law, Eq. (4.48). It tells us that the circulation (line integral) of the magnetization vector along an arbitrary contour in a magnetostatic system that includes magnetic materials tion of the current flow right-hand rule. is is equal by that contour. The reference direcrelated to the reference orientation of the contour by the to the total magnetization current enclosed C 230 Chapter 5 Magnetostatic Field in Material Eq. (5.24) is Media true for any contour C. Let us apply an elementary surface AS it to the contour C enclosing inside a magnetic material: (An)through §c M AS AS • dl (AS -> AS (5.25) 0), where both sides of the equation are divided by AS. The expression on the left-hand above equation represents the component of the magnetization volume current density vector normal to AS, side of the . » n (n — • — (An)through AS .. (5.z6) AS the unit vector normal to the surface AS), while the expression on the right- is hand is, by definition [Eq. n curl M. Hence, side of the equation M along n, that is, n Jm • and, since this is component of (4.87)], the the curl of • true for any n = n curl M, (5.27) • and thus for all components of the vector J m , it implies that magnetization volume Jm -- curl M = V x M. (5.28) current density vector This is material If form of the integral relationship in Eq. (5.24). It tells us that volume current density vector, in A/m 2 at an arbitrary point in a a differential the magnetization is , equal to the curl of the magnetization vector at that point. M = const inside the magnetic material (uniformly magnetized material), M are zero, and from Eq. (5.28), all spatial derivatives of M — const no volume magnetization Jm — 0. (5.29) current in a uniformly magnetized material Physically, the currents of adjacent Ampere’s current loops that flow in opposite directions cancel everywhere in the interior of a uniformly magnetized material, and there is no net volume current. If magnetization current exists only M ^ const, however, if then volume macroscopic the magnetization vector varies throughout the volume of the material (nonuniformly magnetized material) nonzero, otherwise J m = in a way that its curl is 0. On the surface of a magnetized magnetic body, there always exists surface macroscopic magnetization current (there are parts of small Ampere’s current loops pressed onto the surface that cannot be compensated by oppositely flowing currents and the of neighboring loops). The only exception are parts of the surface where microscopic magnetic dipoles are normal to the surface, that is, the surfaces S m of the Ampere’s current loops are laying in the surface of the body. To determine the magnetization macroscopic surface current density vector, J ms (in A/m), equivalent to the microscopic currents, we apply Eq. (5.24) to a narrow rectangular elementary contour, with length A / and height Ah (Ah -> 0), shown in Fig. 5.6. The contour being differentially small, the magnetic moments m of Ampere’s current loops near the contour are all parallel to each other. The resultant of all the corresponding Ampere’s currents, and thus the vector Jm S is perpendicular to the local magnetizalies in the plane tion vector, M. The contour C in Fig. 5.6 is positioned such that of the contour and Jms is perpendicular to that plane. There is no magnetization (M = 0) in free space (a vacuum, air, or any other nonmagnetic medium) surroundAl in Eq. (5.24) is reduced to ing the body, so that the circulation of vector M , M M M • along the lower side of C. By the definition of the surface current density vector . Magnetization Volume and Surface Current Densities Section 5.3 231 Figure 5.6 Elementary contour used for deriving the boundary condition for the vector M on the surface of a magnetic body. [see Eq. (3.13)], the total current enclosed Eq. (5.24), equals /ms A/ where a the angle that is M = • Al = = MA/sina, MA/cos/i (5.30) M makes with the normal on the surface directed from the magnetic body outward (a or, in a by C, appearing on the left-hand side of Jms Al, and we have = 90° — Hence, /$). Jms = M sin a, (5.31) Jms = Mx (5.32) vector form, n, magnetization surface current density vector; n outward with n standing for the outward normal unit vector on the surface. This ary condition for the vector is the bound- M on the surface of a magnetic body, connecting the normal on a magnetic body surface magnetization vector in the body near the surface and the magnetization surface current density vector M contributes to Jms on the surface. Note that only the tangential component of . The expressions in Eqs. (5.28) and (5.32) can be used for determining the disvolume and surface magnetization current densities, J m and J ms of a magnetized body, assuming that the state of magnetization of the body is described by a given distribution of the magnetization vector, M, inside the body. We can tribution of , then, considering these macroscopic currents to reside in a vacuum, calculate the magnetic flux density vector, B, due to the magnetized body (and any other related quantity of interest) using the appropriate free-space equations (e.g., various forms of the Biot-Savart law and Ampere’s law) and solution techniques suitable to specific geometries and current distributions. Example Nonuniformly Magnetized Ferromagnetic Cube 5.1 The magnetization vector in a ferromagnetic cube shown in Fig. 5.7 M(x, where Mq is a constant. y) = M ~ a 0 The surrounding medium is given by (5.33) z, 1 air. is Find the distribution of magnetization currents of the cube. Solution inside the From cube Eqs. (5.28) and (4.81), the magnetization is 3M Jm 3y Z A x dM z dx „ y M o (xx volume current density vector -y y). Figure 5.7 Ferromagnetic (5.34) cube with magnetization Mpc, y); for Example 5.1 232 Chapter 5 Magnetostatic Field Material in Media The magnetization surface current density vector Jmsi = x x = on the front side a sin = = \ (5.35) a Uniformly Magnetized Ferromagnetic Disk thin ferromagnetic disk of radius a its M(*, a~) x y M [in uniformly magnetized throughout bases and = (5.32), respectively, and J ms = 0 on the remaining = 0 on the back side and left-hand side, terms of Eq. (5.31), 0 on the top and bottom sides of the cube]. Example 5.2 A J ms2 by means of Eq. and right-hand side of the cube, four sides of the cube whereas and y a is, magnitude is and thickness d (d a) is situated in volume. The magnetization vector its M. Determine (a) the distribution of is air. The disk is normal to disk magnetization currents of the disk and (b) the magnetic flux density vector at an arbitrary point along the disk axis normal to the bases. Solution Eq. (5.29) (a) tells us that there According to Eq. (5.32) and no magnetization volume current is Fig. 5.8, a inside the disk. magnetization surface current flows circumfer- entially along the side disk surface, with density = J ms Mx fii = Mi x r = 0, because while on the upper and lower disk bases, J ms and = (5.36) M<j>. M is collinear with both n 2 thin, the circumferential sheet of magnetization current 113. Figure 5.8 Magnetization surface current on a (b) Since the ferromagnetic disk uniformly magnetized ferromagnetic current intensity Example disk; for its is surface can be replaced by an equivalent circular current loop with radius a and over Assuming is that the loop is given by Eq. (4.19), that in a = -Ans d — Md. (5.37) vacuum, the magnetic flux density vector along the z-axis An 5.2. is. IxoMda 2 2(z 2 +«2) Example 5.3 A Cylindrical Bar (5.38) 3/2 Magnet magnet of radius a and length / is permanently magnetized with a uniform The magnetization vector, M, is parallel to the bar axis. The medium around air. Compute the magnetic flux density vector at the center of the magnet. cylindrical bar magnetization. the magnet is the magnet is the same as that and the vector B both inside and outside the magnet can be found as that due to a cylindrical current sheet of radius a and length / in a vacuum with the surface current density given in Eq. (5.36). This sheet, on the other hand, is equivalent to the solenoid in Fig. 4.10, so that the flux density along the magnet axis is given by Eq. (4.36) with NI/l substituted by/ms = [see Eq. (4.30)]. Specifically, B at the center of the magnet equals Solution The magnetization surface current density over over the magnetized disk in Fig. 5.8, M R = ^° l y/l2 Nonuniformly Magnetized Fig. 5.9(a) shows an tization given by —M (5.39) + 4a 2 Infinitely Long Cylinder infinitely long cylinder of radius a in air, M = A/o(l - 2 r /a 2 ) i, where Mq is having a nonuniform magne- a constant. Find (a) the distribution of 233 Magnetization Volume and Surface Current Densities Section 5.3 magnetization currents of the cylinder and (b) the magnetic flux density vector inside and outside the cylinder. Solution (a) Applying the formula for the curl in cylindrical coordinates, the streamlines of the magnetization centered at the cylinder axis, as volume current Eq. (4.84), we obtain that inside the cylinder are circles indicated in Fig. 5.9(b), with the following current density vector: L - V x M = - dM dr (b) 2M5“ — - 0 z r (5.40) 4>- There is no magnetization surface current over the surface of the M(a~) = 0 cylinder, because in the cylinder close to the surface. We note that the vector J m in Eq. (5.40) is of exactly the same form as the current density vector J inside the rotating charged cylinder in Fig. 4.20 computed in Eq. (4.69). The only Eq. (4.69) vs. 2Mo/a 2 in Eq. (5.40). Consequently, the magnetic flux density vectors in the two systems are also of the same difference are the multiplicative constants, pw in form, the only difference being those two multiplicative constants. [The magnetic field was obtained by visualizing the current and applying Ampere’s law, Eq. (4.49).] By substituting, therefore, pw in Eq. (4.71) by 2Mo/a 2 the flux density due to the magnetized cylinder in Fig. 5.9 turns out to be due to the rotating charged cylinder in Fig. 4.20 distribution J as a series of coaxial infinitely long solenoids , B= inside the cylinder (0 < r < p.qMq and a) B= 1 ~ [ J 0 outside i = mqM (5.41) (b) it. Figure 5.9 Magnetization vector (a) and A ferromagnetic sphere of radius a M. The sphere magnetization volume Uniformly Magnetized Ferromagnetic Sphere Example 5.5 is surrounded by vector at an arbitrary point is air. currents (b) inside a uniformly magnetized, and the magnetization vector is Calculate (a) the magnetization surface current density on the sphere surface and (b) the magnetic flux density vector at the sphere center. Solution (a) at the sphere center and the z-axis par- in Fig. 5.10, M = Mz and the magnetization For a spherical coordinate system with the origin allel to the magnetization vector, as shown surface current density vector over the sphere surface at a point 9 P defined by an angle is J ms = M x n = Msin Z(M, n) = Msin0, 0 < 9 < n. (5.42) been removed, we apply the superposition principle due to the spherical sheet of current described by the function in analogous to the computation of the electric field due to a uniformly (b) After the ferromagnetic material has to find the B field Eq. (5.42), which is 2.7. We subdivide the sphere surface into thin rings ad9, which is depicted in Fig. 5.10. Each such ring can be viewed as an equivalent circular wire loop with the same current. The current of the ring containing polarized dielectric sphere in Fig. of width d/r the point = P equals d/m Using Eq. (4.19), the magnetic = /ms d/ = Masin9 d9. (5.43) r flux density vector of this ring at the sphere center is nonuniformly magnetized infinitely long ferromagnetic cylinder; for Example 5.4. 1 i 234 Chapter 5 Magnetostatic Field in Media Material Figure 5.10 Ferromagnetic sphere with a uniform I magnetization; for Example = asm# with a r 5.5. standing for the ring radius. Hence, the resultant flux density comes out to be PoM B „ 2 where the /* tt sin jf Problems'. MATLAB integral in 0 5. 3 p #d0 = 1-5.6; J is tt sin / 2 3 Q d6 = - /zqM, (5.45) J0=O evaluated as [see also Eq. (2.32)] , / . ^1 — cos 2 9j sin0 d6 = (—cos 0-1 — £os^ 0 \ Conceptual Questions (on Companion Website): Exercises (on Companion (5.46) 5. 1-5.3; Website). I ! GENERALIZED AMPERE'S 5.4 We now i I consider a general magnetostatic system where, lent magnetization currents surfaces, LAW we have conduction all addition to equivatheir currents (free currents) flowing through conductors (including conducting magnetic materials). vector, B. in (bound currents) inside magnetic bodies and over these currents act as if As sources they were in a of the magnetic flux density vacuum, and Ampere’s law, I Eq. (4.48), becomes Ampere's law for with conductors B a system • dl = po ( Ic + /me). (5.47) and magnetic materials with Ic and /m c standing for the total conduction current and the total magnetiza- by an arbitrary contour C. Dividing this equation by po, moving Im c to the left-hand side of the equation, then substituting it by the circulation of the magnetization vector, M, from Eq. (5.24), and finally joining the tion current, respectively, enclosed two integrals along C into a single integral, we • get the equivalent integral equation: dl = Ic , (5.48) I which generalized Ampere's law is conveniently written as l H dl = lc . (5.49) I Section 5.4 This equation is referred to as the generalized Ampere’s law. Generalized Ampere's Law 235 The new quantity on the left-hand side of the equation, H= B M (5.50) Ho is (unit: called the magnetic field intensity vector to the electric flux density vector, The generalized Ampere’s law and is easier to use than the D, magnetic and field intensity vector A/m) measured in A/m. It is analogous which is defined by Eq. (2.41). is in electrostatics, valid for magnetostatic fields in arbitrary media, is form in Eq. (5.47) because it has only free (true) is that currents on the right-hand side of the integral equation. The most general representation of conduction current volume current density, J, which yields X *3 II JC by means of the f J dS, (5.51) • Js generalized Ampere's law in terms of the volume current where S is a surface of arbitrary shape bounded by the contour C (orientations of C and S are in accordance to the right-hand rule). Since this integral relation holds for any choice of C, applying Stokes’ theorem, Eq. (4.89), results in the following density differential relation: V x H = J, (5.52) generalized differential Ampere's law namely, the differential form of the generalized Ampere’s law. Eqs. (5.51) and (5.52) represent, respectively, the integral and differential Maxwell’s second equation for the magnetostatic field in an arbitrary Toroidal Coil with a Ferromagnetic Core Example 5.6 A uniform /. If field intensity Solution N turns of wire and dense winding with netic core of length magnetic Due medium. there is l (4.66)]. placed over a thin toroidal ferromag- a steady current of intensity I through the winding, find the vector in the core. to symmetry, magnetic field lines in the core are circular as in the air-filled toroid in Fig. 4.18. Moreover, the field in the core Eq. is ferromagnetic Applying generalized Ampere’s along the toroid axis, as shown H= — is uniform, because the toroid law, Eq. (5.49), to a circular in Fig. 5.11, gives HI (thin toroid with = contour is thin [see C of length NI, from which, an arbitrary core). (5.53) Figure 5.11 Evaluation of Note that this result holds true for a core (including nonlinear and made from an arbitrary magnetic material inhomogeneous media). the magnetic vector a Prove that the Closed Surface flux of the in a magnetic Uniformly Magnetized Material field intensity vector through a closed surface situated inside a uniformly magnetized ferromagnetic material equals zero. Solution According to our experience so far with vector calculus and different forms of Maxwell’s equations, the flux of a vector through a closed surface in integral notation corresponds, in differential notation, to the divergence of that same vector (with similar cor- respondence between the circulation along a contour and the curl). Let us therefore consider the divergence of the magnetic field intensity vector, H, in the material. Since the material is uniformly magnetized, the divergence of the magnetization vector, M, at an arbitrary point with ferromagnetic core; for Example Example 5.7 field intensity in a toroidal coil 5.6. 236 Chapter 5 Magnetostatic Field in Material in the material zero is on the other hand, field Media (M = const). The divergence of the magnetic flux density vector, B, always zero [Eq. (4.103)]. Consequently, the definition of the magnetic intensity vector, Eq. (5.50), gives is V-H = — V-B — V-M = (5.54) 0, Mo i.e., the divergence of H is also zero in the material, which in integral notation [see Eq. (1.173)] reads HdS = 0, j) for any closed surface S (5.55) situated inside the material. Problems 5.7-5.11; Conceptual Questions (on Companion Website): : MATLAB Exercises (on Companion 5. 4-5.7; Website). PERMEABILITY OF MAGNETIC MATERIALS 5.5 We now introduce the concept of permeability for macroscopic characterization of magnetic materials, which is analogous to the permittivity concept in electrostatics. Substituting Eq. (5.2) into Eq. (5.50), we obtain that the magnetic field intensity vector (H) at a point in any magnetic material is a function of the magnetic flux density vector (B) at that point, or vice versa, constitutive equation of B = B(H). an (5.56) arbitrary (nonlinear) This magnetic material the general constitutive equation for characterization of magnetic materials, is For linear magnetic materials, the magnetizaand hence also to H [see Eq. (5.50)], parallel to Eq. (2.46) in electrostatics. tion vector, which is M, is linearly proportional to B, customarily written as M = X mH. Here, Xm is (5.57) the magnetic susceptibility of the material, that is, the same dimension- less quantity defined in Eq. (5.16). Note, however, that Eq. (5.16) holds only for diamagnetic and paramagnetic materials, where, as B ^ mqH, while Eq. (5.57) holds for B = m0 where p T — + Xm 1 is ( all we shall see later linear magnetic media. in this section, Thus we have H + M) = mo( + Xm)H = MoMrH, 1 a dimensionless proportionality constant called the relative permeability of the material, which e r in electrostatics. In practice, pT is entirely analogous to the relative permittivity (or Xm) for a given material can be experimentally. In analogy to Eq. (2.50) in electrostatics, we determined also introduce the permeability (or absolute permeability) of the material. permeability (unit: M = H/m) (5.59) /ZrAO), where the value of the permeability of a vacuum, po, is given in Eq. (4.3), with which, (5.60) constitutive equation of a linear magnetic material The unit for media, constitutive equation for free space pr = p 1 is henry per meter (H/m). For free space and other nonmagnetic and B = Mo H. (5.61) Section 5.5 For diamagnetic materials (see Section than the permeability of a vacuum 5.2), (e.g., /z r = the value of pr is Permeability of Magnetic Materials 237 slightly smaller 0.999833 for bismuth, a substance which shows diamagnetism more strongly than most materials), whereas p, of paramagnetic materials is slightly greater than p-o (e.g., p, r = 1.0008 for palladium, one of the strongest paramagnetic materials). Because p, differs only insignificantly from li o, it is very common to assume p, = p-o for diamagnetic and paramagnetic materials, as well as for antiferromagnetic substances, in most practical applications. Thus, these three classes of materials are commonly said to be nonmagnetic. For ferromagnetic, ferrimagnetic, and superparamagnetic materials, on the other hand, p is much larger than /xq. These materials, especially ferromagnetic ones, often exhibit permanent magnetization, highly nonlinear behavior, and hys, teresis effects, as the function we shall discuss in more B(H ) in Eq. (5.56) is detail in a later section. In ferromagnetics, in general nonlinear and has multiple branches. The magnetization properties of the material depend on the applied magnetic field intensity, H and also on the history of magnetization of the material, i.e., on its pre, vious states. In other words, the value of not unique, but is a function of p, for a ferromagnetic material generally is H and the previous history of the material. Typically, maximum value of p r is around 250 for cobalt, 600 for nickel, and 5000 for iron 0.4% impurity), whereas it is as high as about 200,000 for purified iron (0.04% impurity) and 1,000,000 for supermalloy (79.5% Ni, 15% Fe, 5% Mo, 0.5% Mn). In many applications involving ferromagnetics, we assume the (with (5.62) perfect magnetic conductor (PMC) and such media are customarily referred to as perfect magnetic conductors (PMC). As we shall see in the next section, the magnetic flux density vector, B, in a nonmagnetic medium near the surface of a ferromagnetic (or PMC) body is always normal to the surface, the same as for the electric field intensity vector, E, near the surface of a perfect electric conductor (PEC), with a -> oo. Shown in Table 5.1 are values of the relative permeability of an illustrative set of selected materials, Table 5.1. Relative permeability of selected materials Material Mr Material Mr Bismuth 0.999833 Titanium 1.00018 Gold Mercury 0.99996 Platinum 1.0003 0.999968 Palladium 1.0008 Silver 0.9999736 Manganese 1.001 Lead 0.9999831 Cast iron 150 Copper 0.9999906 Cobalt 250 Water 0.9999912 Nickel 600 Paraffin 0.99999942 Nickel-zinc ferrite (Ni-Zn-Fe 203 ) Wood 0.9999995 Manganese-zinc Vacuum 1 Steel Air 1.00000037 Iron (0.4% impurity) Beryllium 1.0000007 Silicon iron Oxygen Magnesium 1.000002 Permalloy (78.5% Ni, 21.5% Fe) 1.000012 Aluminum 1.00002 Mu-metal (75% Ni, 14% Fe, 5% Cu, Iron (purified - 0.04% impurity) Tungsten 1.00008 Supermalloy (79.5% Ni, ferrite 650 (Mn-Zn-Fe 203 ) 1200 2000 (4% 5000 7000 Si) 7 x 10 4% Mo, 2% Cr) 15% Fe, 5% Mo, 0.5% Mn) 10 4 5 5 2 x 10 10 6 238 Chapter 5 Magnetostatic Field in Material Media of considerable theoretical and/or practical interest - in terms of their magnetic properties. A magnetic material homogeneous when said to be is do not change from point to point material is inhomogeneous [e.g., p = its magnetic properties region being considered. Otherwise, the in the p(x,y, z) in the region]. Finally, some magNamely, an applied magnetic field netic materials, such as ferrites, are anisotropic. (vector B) in one direction can produce magnetization becomes direction in a material. Accordingly, Eq. (5.60) ~ BX~ = By - permeability tensor of an anisotropic magnetic |/i| LB Z ] material where [yu] is ~ ftxx Mjry l^xz Mz* Mzy Mzz the permeability tensor. _ vector M) in another a matrix equation. Hx ~ Hy Pyz l^yx fLyy _ (i.e., (5.63) - LtfzJ analogous to the permittivity tensor, It is [e], defined by Eq. (2.52). However, Eq. (5.50) remains valid (it represents the definition of the field), although B, H, and are no longer parallel at a point. The M H anisotropy of ferrites is used in a number of microwave devices, including some types of gyrators, directional couplers, and isolators. and homogeneous magnetic material, we can Ampere’s law and bring 1 /p (because it constant) outside the integral sign in the integral form of the law, Eq. (5.49), or Note substitute is that in a linear, isotropic, H by B/p in the generalized outside the operator (curl) sign in <j) respectively. B • dl = differential form, Eq. (5.52), yielding its VxB = pi, and pic (5.64) We realize that these equations are identical to the corresponding free- space laws, Eqs. (4.48) and (4.83), except for po being substituted by p. In the same way, replacing pg with p in Eq. (4.7) gives the version of the Biot-Savart law for a volume conduction current in a homogeneous magnetic medium 4it x f (J dv) p of permeability p: R (5.65) & l H = B Ip. Additionally, Eq. (4.132) and the concept of magnetic permeability imply that the Laplacian of the magnetic vector potential of a volume current in a with homogeneous magnetic medium satisfies the V2A = Finally, following differential equation: -//J. (5.66) dv f J (5.67) Eq. (4.108) becomes A= p inLir From Eq. (5.50), we find that the magnetization vector in a linear and isotropic magnetic material can be expressed in terms of the magnetic field intensity vector as M= — -H= (— If the material vector, J m at , = J H = (/* - 1)H. (5.68) r / homogeneous, then the magnetization volume current density a point in the material can be obtained directly from the conduction is also volume current density Jm 1 V Mo Mo vector, J, at that point: VxM = Vx l(p T — 1 )H = 1 (p T — 1 ) V X H= (p r — 1 )J. (5.69) We see that there cannot be magnetization volume current (J m neous linear magnetic medium with no free current (J = 0). Problems'. 5.12-5.15; Conceptual Questions (on 5.6 From 239 Maxwell's Equations and Boundary Conditions for the Magnetostatic Field Section 5.6 = 0) in a Companion Website): homoge- 5.8. MAXWELL'S EQUATIONS AND BOUNDARY CONDITIONS FOR THE MAGNETOSTATIC FIELD the Biot-Savart law for the magnetic flux density dB of a single conduction current element J dv in free space and the superposition principle, we proved in Section 4.8 the law of conservation of magnetic flux, Eq. (4.99), for the magnetostatic field in free space. We can repeat here that same proof for a magnetization current element, J m dv (or J ms dS), and get the same result, which means that the net magnetic flux (the flux of the vector B) due to any distribution of conduction and magnetization currents (which, equivalently, reside in a vacuum) through a closed we conclude surface always equals zero. In other words, that the law of conserva- tion of magnetic flux (Maxwell’s fourth equation) holds for structures that include arbitrary magnetic materials. We now write down the of Maxwell’s equations governing the mag- full set netostatic field in an arbitrary medium, together with the associated constitutive equation: fc H-dl = /s J-dS Maxwell's second equation, static field $s B dS = 0 B = B(H) [B • The corresponding (5.70) = fiH] Maxwell's fourth equation constitutive equation for B differential Maxwell’s equations are: V x H=J V B= and • (5.71) 0. In addition to integral and differential Maxwell’s equations for any field, bound- Maxwell's differential equations, magnetostatic field ary conditions always represent the third form of field equations, and they are derived from the respective integral equations (differential equations apply only Let us derive here the boundary conditions for the magnetostatic field on the boundary surface between two arbitrary media. Let J s be the density vector of a surface conduction (free) current that may exist on the boundary. We apply the generalized Ampere’s law in integral form, Eq. (5.49), to a narrow rectangular elementary contour C positioned such that J s is normal to the plane of the contour, as depicted in Fig. 5.12. Having in mind Eqs. (2.79) and (5.30), we obtain at a point). <j> H • dl = Hn A/ - H A = Ic = J 2t l s Al, (5.72) which yields H\t ~ H2 = Js- (5.73) t In vector form [also see Eq. (2.84)], n x Hi — n x H2 = J s , (5.74) where n is the normal unit vector on the surface, directed from region 2 to region If no surface conduction current exists on the boundary, Eq. (5.73) becomes H\t — H2 t (J s =0 ), 1. (5.75) boundary condition for directed from region 2 region 1 H to t ; n 240 Chapter 5 Magnetostatic Field in Material Media Figure 5.12 Deriving the boundary condition for tangential components of vector H on the boundary surface between two arbitrary magnetic media. that is, the tangential component of H is continuous across the boundary free of conduction current. Noting, on the other hand, that the law of conservation of magnetic flux, Eq. (4.99), and the continuity equation for steady currents, Eq. (3.40), have exactly the same form, we conclude that the corresponding boundary conditions must have exactly the same form as well. Therefore, from the boundary condition for normal components of the vector J in the steady current field, Eq. (3.55), we directly write the boundary condition for normal components of the vector B in the magnetostatic boundary condition for field: n B] Bn • It tells us that At an Bn is interface — n • B2 = holds that or B\ n = B2 n (5.76) - always continuous across a boundary. between two linear magnetic with no surface conduction current (J s field lines 0 is = 0), media of permeabilities /xj and \x 2 the law of refraction of the magnetic entirely analogous to the corresponding laws in electrostatics, field, Eq. (3.56). With aq and 012 denoting the angles and region 2 make with the normal to the interface, as Eq. (2.87), and steady current that field lines in region shown in Fig. 5.13, 1 we have tanoq /j.\ tan [i 2 law of refraction of magnetic (5.77) field lines 012 field lines are bent farther away from the normal in the medium with the higher permeability. Bending of field lines is basically due to unequal magnetization surface currents on the two sides of the interface. This relationship indicates that magnetic Figure 5.13 Refraction of magnetic field lines magnetic-magnetic at a interface. Section 5.7 Image Theory for the Magnetic 241 Field Note that if region 2 is filled with a ferromagnetic material and region 1 with a nonmagnetic material, then q. 2 pi and p-i/p.2 ~ 0, and Eq. (5.77) results in » oq^O (|n 2 »/ri) (5.78) any a 2 (except for a 2 = 90°, i.e., for field lines in region 2 parallel to the intermeans that magnetic field lines in a nonmagnetic medium near the interface with a ferromagnetic medium (or PMC) are always normal to the interface. Finally, using Eq. (5.32) and adding up the magnetization surface current density vectors that accumulate on the two sides of a magnetic-magnetic interface, analogously to deriving Eq. (2.89) in electrostatics, we arrive to the following boundary condition for the tangential components of the magnetization vector: for face). This n x where n total is directed from Mi - n medium 2 to M x 2 medium = J ms (5.79) , 1 (Figs. 5.12 and 5.13) and J ms is the magnetization surface current density vector at the interface. Problems'. 5.16; Conceptual Questions (on Companion Website): 5.9-5.12; MATLAB Exercises (on Companion Website). 5.7 IMAGE THEORY FOR THE MAGNETIC FIELD Magnetostatic systems often include current conductors in the presence of large flat ferromagnetic bodies. described in Section 1.21, By utilizing image theory, similarly to the procedure we can remove the ferromagnetic body from the system and replace it by an image of the original current distribution. The equivalent problem is then much simpler to solve because it consists of a known current distribution image) in free space. Consider a straight current conductor in a nonmagnetic half-space in the vicinity of an infinite planar interface with a ferromagnetic (or PMC) half-space. Let the conductor be parallel to the interface. Eq. (5.78) tells us that the ferromagnetic material in the lower half-space, i.e., the induced magnetization current in the material, influences the resultant magnetic flux density vector, B, such that it has no tangential component at the upper side of the interface, as shown in Fig. 5.14(a). This condition remains unaltered, however, if we replace the ferromagnetic block by another conductor parallel to the interface that is positioned symmetrically with respect to the original conductor and carries a current of the same intensity and the same direction, Fig. 5.14(b). We conclude thus that, as far as the magnetic field in the upper half-space is concerned, systems in Fig. 5.14(a) and Fig. 5.14(b) are equivalent. This is an example of the image theory (theorem) for the magnetic field. The theory is not restricted to conductors parallel to the material interface only. It states that an arbitrary current configuration above an infinite ferromagnetic (or PMC) plane can be replaced by a new current configuration in free space consisting from the original current configuration itself and its positive image in the ferromagnetic plane. The equivalence is with respect to the magnetic field above the ferromagnetic plane, the component of that field due to the induced magnetization current in the ferromagnetic material being equal to the field of the image. As another example, Fig. 5.15 shows the image of a current conductor consisting of three segments with different orientations with respect to the ferromagnetic interface in Fig. 5.15(a), where it is a simple matter to conclude that the vector B in the plane of symmetry in Fig. 5.15(b) has no component tangential to the plane. (original plus boundary condition for M t 242 Chapter 5 Magnetostatic Field in Media Material ® © / / i MO Mo ' ; (a) Figure 5.14 X I ® i i i i (b) (a) Straight PMC) current conductor parallel to the interface of a image theory, the influence on the magnetic field in the upper half-space can be represented by a positive image of the original ferromagnetic (or half-space, (b) By of the ferromagnetic material M» Mo current. Example 5.8 A Force on a Conductor above a Ferromagnetic Plane very long and thin wire (a) parallel to surface, its the wire per unit of its By image Solution wire system in air, as of is length its and is situated in air at a height h carries a current of intensity Here, B theory, the system wire-ferromagnetic shown in Fig. 5.16. The force due i.e., (b) Figure 5.15 Image theory magnetic field due to a current conductor with arbitrarily oriented segments above a ferromagnetic (or for the PMC) plane: (a) original system with the material interface and (b) equivalent free-space system. An infinitely is I\ = /2 = / (original) and d conductor per unit 2 h, that is, (5.80) gj. along the upper conductor due to the lower PMC) We see block always attracts a current conductor running Strip Conductor between Two PMC Planes long thin strip conductor of width a carries a steady current of intensity PMC planes, as shown in Fig. 5.17(a). the space between the PMC planes, assuming that placed between two parallel flux density vector in Solution — it. Example 5.9 strip equivalent to a symmetric two- to the magnetization current in the ferromagnetic material. that a large ferromagnetic (or parallel to is on the upper then obtained using Eq. (4.166) with the magnetic flux density is above a ferromagnetic half-space, Determine the magnetic force on length. n, = IB = conductor, /. By /. The Find the magnetic it is multiple applications of the image theory for the magnetic air-filled. field, we obtain the equivalent system in Fig. 5.17(b), which represents an infinite planar current sheet with surface current density 7S = I /a in free thus uniform, with flux density B= space. ijlqJs/2 The magnetic = ^1 /(2a) field between the PMC planes and Eq. (4.47)]. [see Fig. 4.21 is B B / 243 Magnetization Curves and Hysteresis Section 5.8 i <•>- h -F' A m Mo M0 M»M0 M0 Figure 5.16 Evaluation of the h force on a horizontal straight wire with a current above a ferromagnetic plane, using 0 / Uniformly Magnetized Hemisphere on a Example 5.10 A PMC uniformly magnetized ferromagnetic hemisphere of radius a in The magnetization vector (/r r oo) plane. shown in Fig. 5.18. Calculate the is M, and We first is for Example 5.8. Plane air is lying on a PMC perpendicular to the plane, as magnetic flux density and center of the bottom surface of the hemisphere (point Solution it image theory; field intensity vectors at the O in the figure). evaluate the distribution of equivalent magnetization currents of the hemisphere. The magnetization volume current density vector inside the material and the magnetization surface current density vector on the bottom surface of the hemisphere are both zero. There is, however, a hemispherical sheet of magnetization current over the upper surface of the hemisphere given by Eq. (5.42), ory for the magnetic field, where now 0 < 8 < we then supplement this sheet, 8 Using the image the- Till. considered to be in a vacuum, with another hemispherical sheet below the plane of symmetry, which is also described by symmetric with respect which corresponds to a positive image of the current, exactly as required by the the function in Eq. (5.42), but with to 6 = jr/2, n /2 < 0 <n (note that sin# is image theory). We thus obtain the full spherical sheet of magnetization current in Fig. 5.10, and conclude that the magnetized hemisphere on the PMC plane is equivalent to the magnetized sphere of Fig. 5.10, in free space. Hence, the magnetic flux density vector at the sphere center, as well as that at the point O in Fig. 5.18, is given by Eq. (5.45). Finally, by means of Eq. (5.50), the magnetic field intensity vector at the point O is MO 8/ a 8 PMC 8 H= — -M= ^M-M = — 3 8 PMC (5.81) 3 (b) (a) Problems 5.17-5.19; Conceptual Questions (on Companion Website): : 5.13; MATLAB Exercises (on Companion Website). Figure 5.17 (a) Strip conductor between two parallel PMC planes and (b) equivalent infinite planar 5.8 MAGNETIZATION CURVES AND HYSTERESIS current sheet for In this section, we consider in more detail the B-H relationship, Eq. (5.56), for ferromagnetic materials. This relationship, being nonlinear in general, is usually given as a graph showing B (ordinate) as a function of (abscissa). curve representing H A Figure 5.18 Ferromagnetic hemisphere with a uniform magnetization lying on a plane; for Example 5.10. PMC Example in 5.9. free space; Chapter 5 Magnetostatic Field in Material Media Figure 5.19 Simple apparatus for measurement of magnetization curves. the function 77(77) on such a diagram is called a magnetization curve and is obtained by measurement on a given material specimen. Fig. 5.19 shows the simplest apparatus for measurement of magnetization curves, where a uniform winding is placed over a thin toroidal core (ring) cut from a ferromagnetic sample we want to measure. If a current 7 is established in the toroid (primary coil), the magnetic field intensity 77 in the core is given by Eq. (5.53), where N is the number of wire turns of the coil and I is the mean length of the toroid. By varying 7, therefore, we directly vary 77 ( 77 is proportional to 7). The magnetic flux density B in the core is then measured, as a response to 77, by a ballistic galvanometer (BG) connected to another winding (secondary coil) that is placed over the core. Namely, the galvanometer measures the charge that passes through the secondary circuit, as a consequence of the change of magnetic flux through the secondary coil. This charge, as we shall see in the next chapter, is proportional to the flux change, so that the galvanometer actually serves as a fluxmeter. Thus, by changing, step by primary coil and field 77 in the core, and measuring the corresponding values of the flux density B, we obtain, point by point, the magnetization curve of the material. step, the current 7 in the Shown in Fig. 5.20 is a typical initial sample, where the material Figure 5.20 Typical initial magnetization curve for a ferromagnetic material. 0 is magnetization curve for a ferromagnetic completely demagnetized and both B and 77 are zero d Magnetization Curves and Hysteresis Section 5.8 is applied. As we begin to apply a current in the primary circuit in magnetic flux density also rises, but not linearly. Moreover, the value of rapidly at first and then more slowly. The first part of the initial magnetization before a field Fig. 5.19, the B rises curve (roughly up to the point P in Fig. 5.20) represents the region of easy (steep) magnetization. In the upper section of the curve, the increase in magnetization due domains in the material not already parallel to H and the curve tends to become flat. This is the region of hard (flat) magnetization. Very strong magnetic fields are usually required to reach the state of saturation, where all the moments of magnetic domains in the material are parallel to H and the magnetization curve flattens off completely. The permeability at any point on the magnetization curve is given by to gradual rotations of magnetic is more difficult, B = H where B is (5.82) jj, the ordinate of the point (in T) and 77 relative permeability is on the curve with the then /z r = is the abscissa (in A/m). n/no- The maximum permeability largest ratio of B to 77, i.e., at the point of The at the point is tangency with the straight line of steepest slope that passes through the origin and intersects the magnetization curve (Fig. 5.20). Note that the maximum the steepest slope of the magnetization curve, because slope of the curve B/ d 77), but equals ( the ratio n \i is does not correspond to not proportional to the B/H. Having reached saturation, let us now turn to Fig. 5.21, where we continue our experiment - by reducing 7 and 77. As we do so, the effects of hysteresis begin to show, and we do not retrace the initial-magnetization curve. Hysteresis means that B lags behind 77, so that the magnetization curves for increasing and decreasing the applied field are not the same. Even after 77 becomes zero, B does not go to zero, but to a value B = B r termed the remanent (residual) magnetic flux density. Note that the existence of a remanent flux density in a ferromagnetic material makes permanent magnets possible. We then reverse 77 (by reversing the polarity of the , battery in Fig. 5.19), and increase zero at 77 = — 77c , where 77c is it the negative direction further, the material negative polarity. It B comes in the negative direction, so that the so-called coercive force. As 77 to increased in is becomes magnetized even more, with passes through the stages of easy and hard magnetization, the magnetization curve flattens off, and negative saturation is reached. The end of the curve on the left-hand side of the diagram is for the field 77 = — 77m after which , Figure 5.21 Typical hysteresis loop for a ferromagnetic material. 245 246 Chapter 5 Magnetostatic Media Field in Material Figure 5.22 Demagnetization sample by H; of the normal of a ferromagnetic reversals of the applied field definition magnetization curve. we start gives B reducing H. The next intercept of the curve with the B = -B x (negative remanent flux density). At this point, axis (for we H — 0) reverse the bat- H in the positive direction. This makes the flux density zero at a positive field H = H (coercive force). With a further increase in H, the material reaches positive saturation (at H = Hm and a full and continue - by increasing tery polarity again c ), cycle in the curve during BH diagram is completed. The loop traced out by the magnetization this cycle is referred to as the hysteresis loop. Having carried our ferromagnetic specimen to saturation magnetization curve in Fig. 5.21, we now move on to Fig. 5.22, the applied field H , but over successively smaller ranges ( ±H at both ends of the to continue to cycle is brought to smaller and smaller amplitudes on each reversal). We obtain thus a series of hysteresis loops that decrease in size, and the residual B for H = 0 eventually becomes zero, that is, the material is left in a demagnetized state. Such a process is used for demagnetization of objects that have a residual magnetization (i.e., remanent flux density) under conditions of zero applied magnetic field. In practice, the process can be carried out by inserting the object to be demagnetized inside a coil with a lowfrequency ac current, and then gradually reducing the current amplitude in the coil or slowly removing the object from the coil with the current amplitude constant. The curve connecting the tips of the hysteresis loops in Fig. 5.22 known is another charac- normal magnetization curve. For a particular ferromagnetic material, the normal and initial magnetization curves are teristic of ferromagnetic materials as the very similar. H row Ferromagnetic materials having small coercive forces c and therefore nar(thin) hysteresis loops (with small loop areas), as illustrated in Fig. 5.23, are referred to as soft ferromagnetics. As we enclosed by the hysteresis loop proportional to energy loss per unit volume is shall see in a later chapter, the area of the material in one cycle of field variation. This is so-called hysteresis loss, which corresponds to the energy lost in the form of heat in overcoming the friction encountered during the movements of magnetic domain walls and rotations of the domains in a ferromagnetic material. Soft ferromagnetic materials have a large magnetization for a very small applied field and exhibit low hysteresis losses. Additionally, they have very large values of initial permeability [/u. in Eq. (5.82) for a very small H], For example, supermalloy, a typical soft ferromagnetic, has an initial relative permeability on the order of 10 5 B r = 0.6 T, and c = 0-4 A/m. Soft ferromagnetic materials are used for building transformers and ac machines , H Section 5.9 Magnetic Circuits - Basic Assumptions Figure 5.23 Hysteresis loops for soft and hard ferromagnetic materials. (motors and generators), where the material is permanently exposed to alternating magnetization. Hard ferromagnetic materials, on the other hand, have large coercive forces and hence broad (fat) hysteresis loops (Fig. 5.23). They have small initial perc meabilities, and are used for building permanent magnets and dc machines. In these applications, fields do not change frequently, and hysteresis losses, therefore, H are not significant, in spite of large hysteresis loop areas. of permanent-magnet materials teristic are permanently magnetized even with is a high remanent no magnetic The field applied), important that their coercive force be large, so that the material ily essential charac- B flux density but may it (they r is also not be eas- demagnetized. Alnico, an aluminum-nickel-cobalt alloy with a small amount of is a typical hard ferromagnetic, having an initial [x x around 4, B r around 1 T, copper, and H c on the order of 50,000 A/m. Conceptual Questions (on Companion Website): 5.14-5.16. MAGNETIC CIRCUITS THE ANALYSIS 5.9 - BASIC ASSUMPTIONS FOR Magnetic circuit in general is a collection of bodies and media that form a way along which the magnetic field lines close upon themselves, i.e., it is a circuit of the magnetic flux flow. The name arises from the similarity to electric circuits. In practical applications, including transformers, generators, motors, relays, magnetic recording devices, etc., cores of various shapes, that wound about windings magnetic may or may circuits are not have formed from ferromagnetic air gaps, parts of the cores. Fig. 5.24 with current-carrying shows a typical magnetic circuit. The analysis of magnetic circuits with steady currents in the windings (dc netic circuits) in arbitrary is media, Eqs. (5.70). Given, however, great complexity of the rigorous magnetic circuits, we introduce here a set of approximations conjunction with Maxwell’s equations, make the analysis much simpler analysis of practical that will, in mag- based, of course, on Maxwell’s equations for the magnetostatic field and yet accurate enough for engineering applications. for the Analysis 247 1 248 Chapter 5 Magnetostatic Field in Material Media Figure 5.24 Typical magnetic circuit. Firstly, we assume that the magnetic flux is concentrated exclusively inside the magnetic circuit, i.e., in the branches of the ferromagnetic core and air gaps. This is never exactly true. For a toroidal ferromagnetic core with a uniform winding shown in Fig. 5.25(a), however, the flux is restricted to the interior of the toroid as a consequence of the geometrical symmetry of the structure, provided that the wire turns are very tightly wound over the core. age flux between the wire turns. In addition, In reality, there in circuits is always some leak- containing coils placed only over parts of the core, some flux lines bridge the space between the core sections through field Boundary conditions as indicated in Fig. 5.25(b). air, for the magnetostatic applied to the interface between the ferromagnetic and air in Fig. 5.25(b) tell us that magnitudes of the magnetic flux density vector at points a and b in the figure are related as so J3 a we conclude % /z r £b- As /z r is very large for ferromagnetic materials, that practically the entire magnetic flux is Ba B b, restricted to the ferro- magnetic core, and the larger the q, r the more accurate this assumption. Note that in electric circuits this is always true for the current flow because the conductivity of air is zero, whereas the permeability of air is not. Secondly, we assume that air gaps in the magnetic circuit are narrow enough so that the fringing flux near the gap edges can be neglected. For more precise analysis, formulas for an effective length and cross-sectional area of the gap may be used to incorporate the fringing effects into the basic equations for the 6 6 (b) Figure 5.25 Toroidal ferromagnetic core with a (a) uniform winding and (b) concentrated winding. we assume circuit. uniform throughout the volume of each branch of the circuit. Then, in applying Maxwell’s equations, every cross section of a branch is assumed to have the same area, and the path of every flux line along the branch is assumed to be of the same length, equal to the mean length of that part of the circuit. This is true only for thin magnetic circuits. In many applications involving thick cores, however, the error due to thin magnetic circuit approximations is acceptable, which is illustrated in the following examples. Finally, Example Assume 5.1 that the magnetic field is Thick Toroid with a Linear Ferromagnetic Core that the toroidal coil in Fig. 4.18 is wound about a core made of a linear ferromag- Determine the magnetic flux through the core, (b) Find the error made in the flux computation if the magnetic field in the core is assumed to be uniform, specifically for b — a = 0.1a (thin toroid) and b - a = a (thick toroid). netic material of permeability /u. (a) Solution (a) The magnetic From Eq. // inside a thick toroidal coii field intensity vector, H, is = - the same as in the air-filled toroid in Fig. 4.18. (4.65), H(r) — 2nr , a < r < b. (5.83) . Magnetic Section 5.9 Circuits - Basic Assumptions for the Analysis Figure 5.26 Evaluation of the magnetic flux through a cross section of a thick toroidal coil with a linear ferromagnetic core; for The magnetic flux through the core is — where dS (b) Under is f B(r) 5.1 1 then obtained by integrating the flux density B(r) /xH(r) over a cross section of the toroid, as <t> Example = dS shown /xNIh 2n r = in Fig. 5.26, b b —— In-, dr fjiNIh 2n r Ja (5.84) a the surface area of a thin strip of length h and width dr. the assumption of a uniform field distribution in the core, the path of every flux assumed to be of the same length, equal to the mean length of the toroid, / = n(a + b ), and the magnetic field intensity is given by Eq. (5.53). The approximate flux is then obtained by multiplying the constant flux density B = ixH by the surface area S of line is the core cross section, = BS = $ approx FF The associated (iH(b -a)h tc relative error in the flux l^approx ~ computation 2 (b/a Q| <5<j> (1 For b b — a —a= = a case), the ( 0.1 a, that b/a = 2), is, 5<t> = is a computed - (5.85) +b as 1) (5.86) + b/a) In (b/a) b/a = 1.1, the error turns out to be 5<j> = 0.075%, while for 3.8%. We see that even for a quite thick toroid (the latter = approximate expression in Eq. (5.85) is reasonably accurate. Thick Toroid with a Nonlinear Ferromagnetic Core Example 5.12 be wound around a core made from a nonlinear ferroshows the cross section of such a structure, with a = 2 cm, 6 = 4 cm, 6 = 1 cm, N = 200, and / = 1 A. Assume that the initial magnetization curve of the material can be approximated by the piece-wise linear curve shown in Fig. 5.27(b). Calculate the magnetic flux through the core (a) rigorously and (b) assuming that the field in Let the toroidal coil of Fig. 4.18 magnetic material. the core is Fig. 5.27(a) uniform. B B / 0 Figure 5.27 Analysis of a thick toroidal for initial Example coil 1000 with a nonlinear ferromagnetic core: magnetization curve of the material, and 5.1 2. -) Hk = A/m (b) (a) idealized H Ilk — (c) distribution of a 1 c i- b (c) (a) cross section of the structure, (b) the magnetic flux density along the radial axis; n : 250 Chapter 5 Magnetostatic Field in Material Media Solution (a) From Eq. (5.83), the tfmin minimum and maximum magnetic = = H(b) respectively. Since the field value tization curve in Fig. 5.27(b) which is A/m and Hk = 1000 796 H max = H(a) = A/m, satisfies the where Ua is the initial Ma = ^ no = permeability of the material. = Hk H(c ) we in the linear is regime and above (5.88) Hk is in saturation [ B(r ) = Sm ], in the linear regime. In this latter case, is = Ma H(r), B(r) (5.87) condition which H(r ) > that the part of the core for whereas the remaining part of the core A/m, Hmax* ^min < Hk < we conclude 1592 core are representing the limit on the magne- up to which the material (“knee” value), in saturation field intensities in the 0.001 From H/m, (5.89) the condition (5.90) , obtain the radial distance c that represents the boundary between the two parts of the core: C= [see Fig. 5.27(a)]. Sketched 2^ =3 2Cm (5 9,) ' in Fig. 5.27(c) is ' the distribution of the magnetic flux density, B(r), along the radial axis. The magnetic through the core flux is (see Fig. 5.26) b f <&= B(r) d S = B m h(c - a) + , Ja b / —dr Jc r fi*NIh — f 2n . * saturation linear regime = Bm h (b) If we assume where 1061 NI a) (c is 5.10 r mea Hk 5.20; = - 191.4 c n Wb. , = is (a + b)/2 = (5.92) 3 cm. That implies that the entire core = B m (b — a)h — relative to the result in Eq. (5.92) given that the core Exercises (on ln is, H = H(rmean — ) is in saturation. The hence ^approx Problems - which, being greater than magnetic flux through the core The error b , a uniform field distribution in the core, then the field intensity every- in the core equals that for r A/m, + 2nH tt k is <5<p 200 = gWb. (5.93) 4.5%, and this is very reasonable both thick and nonlinear. Conceptual Questions (on Companion Website): 5.17; MATLAB Companion Website). KIRCHHOFF S LAWS FOR MAGNETIC CIRCUITS With the assumptions made (in the previous section) that the field is restricted to the branches of the magnetic circuit (flux leakage and fringing are negligible) and is uniform in every branch, we now specialize general Maxwell’s equations for the magnetostatic field to obtain the laws analogous to Kirchhoffs laws in the electric circuit theory. Thus, applying the law of conservation of magnetic flux, Eq. (4.99), to , l , Section 5.10 Figure 5.28 A Kirchhoff's Laws for Magnetic 251 Circuits closed surface S about a node and a closed path C along the axes of branches of a magnetic circuit - for the formulation of Kirchhoff's laws for a closed surface as shown magnetic S placed about a node (junction of branches) circuits. in a magnetic circuit, in Fig. 5.28, yields M B dS = • £ where Si and Bi (i (i = m 1,2, ... M , ) J2BiSi = 0, (5.94) Kirchhoff's "current" magnetic M) 1,2, 0 law for law for circuits are cross-sectional areas of the branches in the junction are the magnetic flux densities in these branches. Applying, on the other hand, the generalized Ampere’s law, Eq. (5.49), to a contour C placed along a closed path of flux lines in the circuit (Fig. 5.28) gives P £ = H • dl = Ic Q E^lE^ j =l (5.95) P) standing for the lengths of the branches along the path P) for the magnetic field intensities in the branches, whereas 2, Q) are the numbers of wire turns and current intensities, respectively, in the coils that exist along the path. The product Nkh, expressed in ampere-turns, is termed a magnetomotive force (mmf), in analogy to an electromo- with lj (j 1, 2, . and Hj (j = 1, 2, Nk and Ik (k = 1, . tive force . . . , . , . . . (emf) in electric circuits. Eqs. (5.94) and (5.95) are referred to as Kirchhoff’s laws for magnetic cuits. In addition to these circuital laws, we need cir- the “element laws” describing individual parts of the circuit, analogous to current-voltage characteristics for elements (e.g., resistors) tionships B= in the analysis of electric circuits. These laws are the relaB(H), i.e., the magnetization curves, for the branches of the circuit, including air gaps (where B = hqH). Because of the nonlinear nature of the ferromagnetic portions of the magnetic circuit, and nonlinearity of magnetization curves, the analysis of magnetic circuits often resembles the analysis of nonlinear electric circuits which contain diodes and other elements with nonlinear current-voltage characteristics. This is, at the same time, the most important difference between the analysis of magnetic circuits and the electric circuit theory, which primarily deals with linear electric circuits. where the ferromagnetic materials in the circuit can be considered however, an equivalent electric circuit with linear resistors and timeinvariant voltage generators can be introduced. By solving the equivalent circuit using some of the standard circuit-theory techniques, we obtain thus all required In cases as linear, Kirchhoff's "voltage" magnetic k= circuits 252 Chapter 5 Magnetostatic Field in Material Media quantities for the original magnetic circuit. As an note that the mag- illustration, netic flux through a simple magnetic circuit in Fig. 5.29(a) can be expressed, having in mind Eq. (5.53), as O = BS = ix HS = NI /jlNIS (5.96) H' / where reluctance (unit: H 1 (5.97) J is the so-called reluctance of the core. It is defined generally as the ratio of the mag- netomotive force (ampere-turns) to the flux. The SI unit for reluctance is ampere _1 per weber ( A/Wb) or inverse henry (H ). We note that the final expression for 0 in Eq. (5.96) is analogous to the expression for the current in a simple electric circuit in which an ideal voltage source of emf NI is connected with a resistor of resistance IZ, as indicated in Fig. 5.29(b). Eq. (5.97), moreover, has the same form as the expres- sion for the resistance of a resistor with uniform cross section and conductivity a in Eq. (3.85). The concept of reluctance can be used for the analysis of arbitrary linear magnetic circuits, where the equivalent electric circuit is obtained by replacing the individual parts of the core and the air gaps by resistors with resistances calculated from Eq. (5.97) and representing the coils by voltage generators with electromotive magnetomotive forces NI. forces equal to the O Simple Nonlinear Magnetic Circuit with an Air Gap Example 5.13 Consider a magnetic an air R= gap shown 50 £2. emf is Iq coil has — 4 mm, N= ferromagnetic core with a is is £ = coil and 1000 turns of wire with the total resistance circuit is the cross-sectional area of the toroid of the generator in the coil circuit curve of the material and The The length of the ferromagnetic portion of the (width) of the gap the circuit consisting of a thin toroidal in Fig. 5.30(a). 200 V. The idealized given in Fig. 5.30(b). Find the magnetic = / is S 1 = m, the thickness 2 So = 5 cm and initial , magnetization field intensities in the core in the air gap. Figure 5.29 Reluctance concept: a simple linear magnetic circuit (a) and the Solution We adopt standard approximations for the analysis of magnetic circuits and neglect the field nonuniformity in the ferromagnetic core, as well as the flux leakage from H) and Bq Hq) equivalent electric the core and fringing around the air gap edges (Section 5.9). Let (B. circuit (b). ignate the magnetic flux density and field intensity in the core and the gap, respectively, as Figure 5.30 Analysis of a simple nonlinear magnetic circuit with an (a) circuit air gap: geometry, with flux density and field intensity vectors and in the circuit, (b) idealized initial magnetization curve of the material, with the load line and operating point for the circuit; for Example 5.1 3. ( , des- Section 5.10 indicated in Fig. 5.30(a). From Kirchhoff's Laws for Magnetic Circuits 253 Kirchhoff’s laws for magnetic circuits, Eqs. (5.94) and (5.95), we have — BS = B 0 S0 B = £0 » (5.98) and HI where I = £/R = 4 A is + HqIq = AT, the current intensity of the (5.99) coil. The above equations, combined with the constitutive equation for the air gap, Bo = (5.100) between the result in the following relationship + B— = HI and flux density field intensity in the core: NI. (5.101) Mo With the numerical data substituted, H + 3183B = 4000 and (H in A/m; BinT), equation of the load line for the magnetic circuit (the locus of this represents the possible combinations of values B and all H for this given circuit configuration and excitation, not yet taking into account the characteristics of the core material). in the (5.102) BH chart [Fig. 5.30(b)], we conclude that its Upon plotting this line intersection with the magnetization curve of the core material belongs to the second part of the curve (hard magnetization section - also see Fig. 5.20). This intersection (point P) represents the operating point for the circuit, i.e., it determines the actual position of the ferromagnetic core on the magnetization curve and excitation (NI). To for the given circuit configuration ordinate of the point P, we and find numerically the abscissa composed of the load solve the system of equations line equation, Eq. (5.102), and the equation of the line describing the hard magnetization segment of the magnetization curve, read from Fig. 5.30(b), (5.103) + 2.5 x 10" 5 (//- 1000), and what we get is B = 0.44 T and H = 2600 A/m. By means of Eq. (5.98), Bo = 0.44 T as well, and Eq. (5.100) then gives Hq = 350 kA/m (field intensity in the air gap). We see that B= Hq 0.4 This can also be concluded without actually solving the circuit, by introducing the -4 P, m = B/tt = 1.7 x 10 H/m, into Eq. (5.98) and tt. permeability of the material at the point expressing the magnetic field intensity in the gap as tt0 = — tt = (5.104) 135.3tt. M0 (Note that /r here is a function of 77, not a constant.) It air gaps, in general, that in the is typical for the magnetic field intensities in the gaps are magnetic much circuits with larger than those ferromagnetic material. Example 5.14 Simple Linear Magnetic Circuit with an Air Gap n o Assume that the core of the circuit of Fig. 5.30(a) material of relative permeability Solution Combining Eq. 4> where the reluctances [Eq. 72 = —— = q, r = 1000, and m {jrs + 7kY (5.97)] of the core B= flux =HAAAAArn through the core. iiH we can write now , NI = 7Z (5.105) + 72o and the gap are Figure 5.31 Equivalent l 1.6 x 106 HT 1 and 72 0 MrMOO and the equivalent electric -4 the core amounts to 4> = 5 x 10 Wb. respectively, a linear ferromagnetic compute the magnetic (5.101) with the relationship = BS = made from is = = 6.37 x 10 6 H -1 (5.1 , 06) MoSo circuit is shown in Fig. 5.31. electric circuit for the magnetic Hence, the flux through circuit of Fig. 5.30(a), assuming that the ferromagnetic material is linear; for Example 5.14. Chapter 5 Magnetostatic Field in Material Media Figure 5.32 Analysis of a nonlinear magnetic circuit with three branches: (a) circuit geometry, with the adopted independent node and closed paths in the circuit, and (b) idealized initial magnetization curve of the material, with computed operating points for the branches; for Example 5.15. Nonlinear Magnetic Circuit with Three Branches Example 5.15 Dimensions of the magnetic 5i = £2 = ^3 = 1 circuit cm 2 The mmf . shown in the first in Fig. 5.32(a) are branch of the = l\ circuit is l3 = N1 = 2lj = 20 cm and 400 ampere-turns. The initial magnetization curve of the core can be linearized as in Fig. 5.32(b). Calculate the magnetic flux densities and field intensities in the three branches of the circuit. Solution Let us orient the branches of the circuit as in Fig. 5.32(a). There is a total of two nodes and three closed paths in the circuit. Thus, in complete analogy with the analysis of electric circuits, Kirchhoff’s “current” and “voltage” laws for magnetic circuits, Eqs. (5.94) and (5.95), are to be applied to one node (independent node) and two closed paths (independent closed paths), respectively. We choose the node ]\f\ and closed paths C\ and C2 in Fig. 5.32(a). The corresponding equations -B 1 S1 are: + B 2 S2 + B 3 S3 = —» 0 H + H2 = NI -H + H = l l1 Depending on whether 3 l3 (5.107) , , (5.108) 0. (5.109) l2 2 l2 Bi=B 2 + B 3 the operating points for the individual branches of the circuit belong to the linear part or to the saturation part of the magnetization curve in Fig. 5.32(b), we may have B,= li a Hi initial them is A/m for Hi > 900 A/m is /r a = 0.9/900 H/m = 0.001 H/m. So, there combinations for the magnetization stages of the branches and only one true. Suppose, in all T permeability of the material exist a total of eight of Hi < 900 (5.110) 0.9 where the for first, that none of the magnetizations becomes in the branches is in saturation ( B= [x a H three branches). Eq. (5.107) then Hi =H 2 + H3 , (5.111) H and the solution of the system with Eqs. (5.111), (5.108), and (5.109) is H\ = 1500 A/m, 2 = 1000 A/m, and 3 = 500 A/m. This is contradictory to the assumption of linearity in all three branches, since both H\ and 2 appear to be larger than 900 A/m. We conclude that our initial guess is not correct, and the circuit has therefore to be solved again, with another assumption, i.e., that some of the branches are in saturation. To reduce the number of additional trials (out of remaining seven combinations) to a minimum, we note first that the adopted reference directions of the magnetic flux density vectors in Fig. 5.32(a) likely reflect the actual flux flow in the branches, meaning that B 1, B 2 and B 3 are all positive. Eq. (5.107) then tells us that B\ must be larger than both B 2 and 63 individually, so it is logical to expect that the branch with the coil would first reach saturation ( B\ — 0.9 T). The other two branches, however, must remain in the linear regime (B 2 = a H2 and B 3 — /x a H3 ), because B 2 = 0.9 T would imply B 3 = 0 and vice versa, which H H , )JL is impossible. With this, Kirchhoff's = H2 — 600 A/m, Eq. (5.107) becomes H2 + H3 = 900 A/m, and the solution of the Laws Section 5.1 0 new system of three equations (5.112) is H\ 1700 A/m, H and 3 = 300 A/m. Obviously, these values are consistent with the new assumption made about the magnetization stages in the branches, and this is the true solution for the circuit. The actual positions of the operating points for the branches on the magnetization curve are indicated in Fig. 5.32(b). The remaining two flux densities in the circuit are B 2 = 0.6 T and B3 = 0.3 T. Reverse Problem Example 5.16 For the magnetic circuit shown the Magnetic Circuit Analysis in in Fig. 5.33, = l\ l3 = 20 cm, l2 = l' 2 + 1% = 10 cm, Iq = 1 mm, S3 = 1 cm S2 = 2 cm and N2 I2 = 1000 A turns. The core is made from the material whose initial magnetization curve can be approximated analytically by the function 5i = 2 2 , , 1 B= If 500 C f_T + H’ H ~° the magnetic flux in the central branch of the circuit downward reference Solution This is is $2 = (5.113) )- 200 n Wb with respect to the direction, find N\I\. a reverse problem in the magnetic (magnetic flux in one or more branches of the or 5inT (HinA/m; more magnetomotive forces) that produces circuit analysis: for a given circuit), find the it. unknown response excitation (one In the analysis of magnetic circuits with nonlinear magnetization curves (given either graphically or analytically), reverse problems are generally much simpler to solve than direct problems, because in the latter case to simultaneously solve the full set of equations written in we have accordance to Kirchhoff’s circuital laws together with nonlinear material characteristics for individual parts of the circuit. In reverse problems, on the other hand, we with the given start field quantities for branches and then applying the circuital and material equations one at a time one or more we solve for other field quantities, one by one, and for the required magnetomotive forces. With reference to the notation in Fig. 5.33, the flux densities in the of the second (central) branch of the circuit B2 From the magnetization curve, i.e., ( B 2 ) and in =B 0 = ferromagnetic section the air gap (Bo) are ^ = 1T. (5.114) by solving the magnetization equation, Eq. (5.113), for the corresponding field intensity in the material, h2 = 500B 2 1.5 - B2 = 1 kA/m, (5.115) Figure 5.33 Magnetic circuit with a nonlinear magnetization curve given analytically; for Example 5.1 6. for Magnetic Circuits 256 Chapter 5 Magnetostatic Field in Material whereas Media in the gap. //o ~= = 795.8 kA/m. (5.116) MO Kirchhoff’s “voltage” law for the closed path N H„ = 2 I2 ~ H2 l2 C2 now in Fig. 5.33 H0 = 521 ~ lo 3 yields A A/m, , (5.117) h with the corresponding flux density in the third branch given by B3 = I.5//3 500 0.765 T. + //3 (5.118) The flux density in the first branch is next obtained using Kirchhoffs “current” law node M\ in Fig. 5.33, and from it the associated H value, B = 1 —— = B 2 S2 — A 3 S 3 ----- 1.235 T Hi = we apply Kirchhoffs “voltage” law to the N\I\ = H\l\ l in a ±_ = 2.33 kA/m. Magnetic (5.119) -fli closed path C\ and obtain the + H2 2 + H0 lo = Demagnetization Example 5.17 500 Ai 1.5 5i Finally, — for the 1362 A turns. mmf we seek: (5.1 20) Circuit The ferromagnetic core of the circuit shown in Fig. 5.34(a) has the cross-sectional area 5 = 1 cm 2 mean length / = 20 cm, and air-gap thickness /o = 1 mm. The number of wire turns of the coil is N = 1000, the resistance of the winding is R = 20 £2, and the emf of the generator is £ = 20 V. The core does not have residual magnetization and the switch K is in , the off position (open). By turning the switch on, the mmf is applied to the circuit and the magnetic flux through it rises following the initial magnetization curve of the material. This curve can be considered as linear, as shown in Fig. 5.34(b), where the initial permeability is = B m in the stationary fj a = 0.001 H/m. The magnetic flux density in the core becomes B . state. Figure 5.34 Magnetization and demagnetization in a magnetic circuit with an air gap: (a) circuit initial geometry, (b) idealized magnetization and demagnetization curves of the material, (c) magnetic flux density and field intensity vectors in the circuit, and operating points for the two stationary Example 5.1 7. in (d) circuit states; for The switch is then turned off (opened) and a new stationary state established in the Section 5.10 circuit. Kirchhoff's Laws for Magnetic 257 Circuits The demagnetization curve for the material can be approximated by two straight-line [Fig. 5.34(b)], where the coercive force is given by H c = Hm What is the magnetic segments . gap field intensity in the in the new state? Solution After the switch K is turned on (closed), the current in the winding is [Fig. 5.34(c)] = £/R = 1 A, and the relationship between the flux density B and field intensity in the core is the same as in Eq. (5.101). Moreover, as H I B = p a H, we obtain [see also Eq. (5.105)] B= Nl(— + — ) \Ma and 1 (5.121) maximum this is the = 1 T, (5.122) MO/ B = Bm — value for the flux density, T. Hence, 1 Hm = 5 m //r = a kA/m. After the switch K is opened, Eq. (5.101) HI becomes + B— = (5.123) 0, Mo and this represents the equation of the load line for the magnetic circuit in this state. We suppose first that the intersection of this line with the demagnetization curve in Fig. 5.34(b) belongs to the horizontal segment of the curve, i.e., that B = B m = 1 T. From Eq. (5.123), = — 4 kA/m, which is impossible because < —Hc ( c = m = 1 kA/m). We conclude thus that the operating point for the circuit belongs to the vertical segment of the demagnetization curve, as indicated in Fig. 5.34(d). The magnetic field intensity in the core is therefore H H H = —H = —1 kA/m. c B= 0.25 T. For the magnetic = Eq. (5.123) then gives the corresponding flux density in the core, field intensity in the gap is Hq = 5/mo = 200 kA/m. The magnetic Example 5.18 N\ Complex Linear Magnetic shown circuit made can be considered Solution 5.35(a), in Fig. h —h= as linear, with relative permeability parts of the core all Circuit 10 cm, =5 cm, I3 Iq = 1 mm, N2 = 500, and 7i = I2 = 1 A. The ferromagnetic material from which the core 1000, area of H H Fig. 5.35(b) S is = 1 cm 2 . Find the magnetic shows the equivalent = ju. r 1000. The is cross-sectional field intensity in the air gap. electric circuit for the problem, with reluctances, Eq. (5.97), (a) 72i = 72 2 = 272. 3 = — = 8 X 10 5 H -1 , Ko = figure also 8 x 10 6 H _1 . 24) (5.1 M o5 MrMCP The = !R2 22.1 shows the adopted loops for the loop analysis of the (electric) circuit. The corresponding loop equations are + 72-3 + 72o)4>i + 1Z\ 2 = N\I\, 72i 4*1 + (72i + 722) ^2 — 7Vi/i + A^2?2, (72i and their solution for the flux 722 4>i 72q72i The magnetic through the central branch of the M (5.125) <t> — IZ 1 N2 H I] + 72o72-2 + 72i72.2 + 72-i723 + 722723 field intensity in the = (5.126) circuit 2.84 x 1(T 5 Wb. (5.127) Figure 5.35 Analysis of a gap comes out to be linear magnetic circuit with three branches: (a) circuit H0 = = 226 kA/m. M0 5 (5.128) geometry and analysis; for Problems 5.21-5.27; Conceptual Questions (on Companion Website); 5.18-5.21; : MATLAB Exercises (on Companion Website). (b) equivalent electric circuit for loop Example 5.1 8. 258 Chapter 5 Magnetostatic Field in Material Media MAXWELL'S EQUATIONS FOR THE TIME-INVARIANT ELECTROMAGNETIC FIELD 5.1 1 Maxwell’s equations for the magnetostatic field, Eqs. (5.70), represent a general mathematical model for determining the magnetic field (H) from a distribution of steady electric currents (J), which considered to be known. The distribution of is currents, however, for the geometry, material properties, and excitation of a given structure can be obtained from Eqs. (3.59), which also yield the solution for the electric field (E) in the system. In addition, Eqs. (3.60) provide a means for evalu- ating the associated distribution of charges (p) in the system from the electric field distribution. All these equations together represent a full set of Maxwell’s equations that govern both the electric and the magnetic field due to steady electric currents. These two fields, moreover, can be considered as components of a more complex field - the electromagnetic field - produced by steady currents. The combined field is called the time-invariant electromagnetic field, Maxwell’s equations Maxwell's first form for in differential equation, and we summarize here the four this field: V x E=0 V xH=J V D=p static field Maxwell's second equation, static field (5.129) • Maxwell's third equation V-B = 0 Maxwell's fourth equation What very is important, the and magnetic electric constituting fields the and independent from each other, and can be analyzed separately, as we have done in Chapters 3 (electric part) and 5 (magnetic part). This is not the case, however, with the electromagnetic time-invariant field are we unrelated entirely Under same form whereas the first two equations have additional terms on the equations, which are responsible for the coupling between the time-varying electromagnetic field, as shall see in the following chapters. nonstatic conditions, the third and fourth Maxwell’s equations retain the as in Eqs. (5.129), right-hand side of and magnetic fields that change in time. In chapters that follow, the electric and magnetic fields will always be treated together, as related integral parts of the electric time-varying electromagnetic field. Continuity Equation from Ampere's Law Starting from Maxwell’s second differential equation for the time-invariant electromagnetic field derive the corresponding form of the continuity equation. Solution Let us take the divergence of both sides of the differential form of Maxwell’s second equation (generalized Ampere’s law) for V The second-order vector any vector 2 field, 2 The divergence of is if A =V • (5.130) J. derivative on the left-hand side of this equation the curl of an arbitrary vector function (A) V obvious [Eq. (4.103)]. (V x H) is equal to zero for so the right-hand side of the equation must be zero as well, operator. This • static fields, in Eqs. (5.129): is • (V x A) = it for A representing any vector a (b x c) = (a always zero, or, in • J = 0, and terms of the del 0. the magnetic vector potential, because then To prove is V field, x b) B=V x A [Eq. (4.1 16)] and V B= • 0 however, we apply formally the identity c L 259 Problems the differential form of the continuity equation under the static assumption, this is exactly We Eq. (3.41). see now that Eq. (3.41) is actually included implicitly in the set of Maxwell’s equations in Eqs. (5.129). Problems : 5.28; MATLAB Exercises (on Companion Website). Problems 5.1. Nonuniformly magnetized parallelepiped. rectangular ferromagnetic parallelepiped uated in the air in A is sit- octant of the Cartesian first coordinate system (x,y,z >0), with one vertex at the coordinate origin, lengths a, b, and y, and c, z, respectively. + Figure 5.36 Uniformly The magnetization vector magnetized square given by M(x, y, z) = sin(7rz/c)y + sin(^x:/fl) z], in the parallelepiped Mo[sin(7ry/h) x and the edges, of parallel to coordinate axes x, ferromagnetic is plate; for Problem where Mo is a constant. Compute (a) the volume magnetization current density vector in the parallelepiped and (b) the surface magnetization current density vector over 5.2. Hollow drical cylindrical bar magnet. its sides. < b ), and permanently magnetized with a uniform magnetization and situated in air. The magnetization vector, of magnitude M, is parallel to the bar axis. Find (a) the distribution of magnetization currents of the magnet and (b) the magnetic flux density vector along the length /, is axis. 5.3. M= assume that the magnetization vector is Mo y (Mq — const). Neglecting end effects, center (point O). 5.5. Nonuniformly magnetized ferromagnetic disk. A thin ferromagnetic disk of radius a and thickness d (d <$C a) in air has a nonuniform mag= Mo(r/a) 2 z (Fig. 5.37), netization, given by where Mo is a constant. Calculate (a) the dis- M tribution of magnetization currents of the disk and (b) the magnetic flux density vector along the z-axis. Uniformly magnetized square ferromagnetic plate. A uniformly magnetized square ferromagnetic plate of side length a and thickness d (d <^a) is situated in air. With reference to the coordinate system in Fig. 5.36, the magnetization vector in the plate is given by = Mo z, where Mo is a constant. Determine the magnetic flux density vector at an arbitrary Figure 5.37 Nonuniformly magnetized thin ferromagnetic M point along the z-axis. 5.4. (in disk; for Problem 5.6. Infinite cylinder Magnetization parallel to plate faces. Consider the square ferromagnetic plate in Fig. 5.36, and An infinitely product V (V x V (V • • (cross product of a vector with itself the expression for the curl (of be zero. is with circular magnetization. long ferromagnetic cylinder of change the A), and obtain x A) = (V x V) • A=0 always zero). Alternatively, the divergence, using Eq. (1.167), of any vector) in the 5.5. radius a in air has a nonuniform magnetization. a scalar triple product, the cyclic permutation of the order of the three vectors does not result) to the scalar triple find the magnetic flux density vector at the plate A hollow cylin- bar magnet of radii a and b (a 5.3. Cartesian coordinate system in Eq. (4.81) turns out to 260 Chapter 5 Magnetostatic Field in Material In a cylindrical coordinate system r < a), where Mo is a constant. Find (a) volume magnetization current density vec- 5.13. Thin toroidal coil A with with a linear ferromagnetic N zation current density vector on the cylinder uniformly and densely over a thin toroidal surface, (c) the magnetic flux density vector in ferromagnetic core of permeability the cylinder, and (d) the magnetic flux density steady current of intensity / vector outside the cylinder. the Find the magnetic field intensity vector, H, (a) along the axis of the magnetized disk from Example 5.2, netic field intensity vector, (b) the magnetic Magnetic field intensity vector. magnet coil turns of wire is fx. If a established in the circulation of (a) the mag- coil, find and flux density vector, (c) the vector through the core, along 5.14. Solenoidal wound is coil magnetization its mean length. with a linear ferromagnetic from Example 5.3, and (c) inside and outside the nonuniformly magnetized infinitely long cylinder from Example 5.4. core. Total (conduction plus magnetization) current ber of wire turns per unit length of the core in a tion The magnetic is flux density vector, B, ferromagnetic material of spatial There is a uniform and dense solenoidal winding wound over a very long cylindrical ferromagnetic core of permeability [x. The num- a is coordinates, known func- Prove that (a) ume J current density in the material, J to t + Jm , can be obtained as J to (b) Specifically, compute J to t t =V for B = B(x, y, z) = {[(2x + z)/y + 2y) z] T (x, y, z in m). nates: (x ] x + (2 /y) y + face magnetization current density vector over the surface of the core. 5.15. volume current density vector, J m = in air. dl = situated vector, (b) the magnetic flux density vector, and magnetization vector inside and outside the cylinder, as well as (d) the distribution of magnetization currents of the cylinder. 5.16. boundary Magnetic-magnetic Assume that the plane z = (z > 0) and medium 2 1 = tive 0. permeabilities ix T \ = conditions. 0 separates (z < 0), 600 and medium with rela/x T 2 = 250, The magnetic field intensity vector + in medium 1 near the boundary (for z = 0 ) is Hi = (5 x — 3 y + 2 z) A/m. Calculate the magrespectively. magnetic The magnetization vector, current density vector, J, flux density vector. M, and conduction are known at every netic field intensity vector in point of a magnetic body. Find the expression for the circulation of the magnetic vector along a closed path C boundary flux density magnetization and conduction current. Consider an imaginary closed path C inside a homogeneous magnetic material of relative (for z = and Js 5.17. (b) there = 3y A/m is H2 medium 2 near the no conducon the boundary (J s = 0) 0~), tion current exists situated entirely inside the body. 5.12. Total /x r is a uniform conduction current of distribution of (a) the magnetic field intensity (c) the const), in a current-free (J to t 5.11. Circulation of the is density J flowing along the cylinder. Find the 0) region, prove the following vector identity: §c There . Closed path in a uniform field. Considering an arbitrary contour in a magnetic field with no spatial variation of the magnetic flux density vector (B Ferromagnetic cylinder with a conduction current. A very long ferromagnetic cylinder of radius o and relative permeability Constant flux density vector in a magnetic region. In a certain magnetic region, the magnetic flux density vector does not vary with spatial coordinates. The conduction volume current density vector is J. Find the magnetization steady current of intensity / flows density vector in the core, as well as (e) the sur- x B/ixq. given as the following function of Cartesian coordi2 A N'. through the winding. Determine (a) the magnetic field intensity vector, (b) the magnetic flux density vector, (c) the magnetization vector, and (d) the volume magnetization current the total (conduction plus magnetization) vol- 5.10. conduction current total core. density. 5.9. The Ic What is the total magnetization current /m c enclosed by C? is tor in the cylinder, (b) the surface magneti- (b) at the center of the cylindrical bar 5.8. /x r . C enclosed by < the 5.7. permeability 2 -axis M = Mo(r/a)<\> coincides with the cylinder axis, (0 whose Media , if (a) a surface current of density on the boundary. Force on a conductor above an infinite PMC corner. Two PMC (/x T -> 00 ) half-planes are connected together at an angle of 90° with . 261 Problems respect to each other, as A shown in Fig. 5.38. /J-T—^OO very long and thin wire in air runs parallel from each to the half-planes, at a distance h of them. If a steady current of intensity I M is established in the wire, find the per-unit-length magnetic force on it. h r ’ Figure 5.40 Uniformly a Figure 5.38 Cross magnetized cylinder between two 0 section of a system consisting of a parallel Mr-— oo and an Problem 5.18. corner; for 5.21. 5.1 7. Uniformly magnetized hollow disk on a PMC plane. A uniformly magnetized hollow ferromagnetic disk surrounded by air is lying on a PMC plane. The magnetization vector is M, and it is normal to the plane, as shown in Fig. 5.39. The disk radii are a and b, and thickness is d (d a, b). Obtain the magnetic flux density « vector along the z-axis for z > planes; infinitely large PMC 90° PMC Problem 5.19. for current-carrying wire 0. Simple nonlinear magnetic circuit. Fig. 5.41 shows a magnetic circuit consisting of a thin magnetic core of length / = 40 cm and 2 cross-sectional area S = 2.25 cm and an air gap of thickness Iq = 0.25 mm. The winding has TV = 800 turns of wire with a steady current of intensity I = 1 A. The core is made from a nonlinear ferromagnetic material whose initial magnetization curve can be linearized in parts as in Fig. 5.27(b). Find the magnetic field intensities in the core and in the air gap. S Figure 5.39 Hollow ferromagnetic disk with a uniform magnetization lying on a PMC plane; for Problem 5.18. Figure 5.41 Simple nonlinear magnetic circuit with an 5.19, Magnetized cylinder between two A ferromagnetic cylinder of radius a and height h placed between two parallel PMC planes, 5.22. is as portrayed in Fig. 5.40. The The medium around is uniformly magnetized throughout its volume, with the = Mq z, where magnetization vector given as Mo is a constant. Find the magnetic flux density the cylinder is air. cylinder 5.20. gap; for Problem 5.21 PMC Complex nonlinear magnetic circuit. Dimensions of the magnetic circuit shown in Fig. 5.42(a) are l\ = / 3 = 2/2 = 20 cm and S\ = S2 = S3 = 2 cm 2 The magnetomotive forces in the circuit are N\Ii = 100 ampere-turns and . M vector in the space between the air PMC planes. TV2/2 = 300 ampere-turns. The idealized magnetization curve of the core planes, Fig. 5.42(b). Compute both inside and outside the cylinder. sities Magnetic flux branches of the Example 5.12 but for the piece-wise linear through a thick toroid. Repeat approximation of the initial magnetization curve of the core material given in Fig. 5.30(b). 5.23. and Magnetic is field intensities in each of the three circuit. circuit 5.42(a), in the magnetic flux den- with a zero flux branch. Referring to the magnetic Fig. initial shown let l\ = I3 = 30 cm, I2 in one circuit in - 10 cm, H 262 Chapter 5 Magnetostatic Field l\,S\ in / 3, Material Media 53 Figure 5.43 Nonlinear magnetic circuit with (a) three branches, two gaps, and air two mmf's; for Problem 5.24. B[ T] k 5.25. 1 1 1 1 — //[A/m] the material whose initial magnetization curve can be approximated analytically by the func> 0. tion B = arctan(///250) T ( in A/m), 1 1000 0 Reverse problem with a nonlinear magnetic circuit. Refer to the magnetic circuit shown in Fig. 5.43 and described in the previous problem, and assume that the core is made out from H (b) Under Figure 5.42 Nonlinear magnetic three branches and geometry and two mmf's: (b) idealized circuit these circumstances, find the in the third with through the (a) circuit branch such that the magnetic flux first branch (branch with the given = mmf 7Vi/i) is initial magnetization curve of the material; for mmf N2 I2 upward reference i 125 ^uWb with respect to the direction. Problem 5.22. I 5.26. Si=S3 = 10 cm 2 S2 = 20 cm 2 Ni = 1500, and N2 = 1000. Let the idealized initial magne, , tization curve of the core be that in Fig. 5.27(b). If the current of the winding in the second branch of the current I\ circuit is I2 = 0.5 of the winding in the A, find the first branch such that the magnetic flux in that branch 5.24. Remanent flux in a circuit with zero mmf. Repeat Example 5.17 but for / = 50 cm and is /o 5.27. = mm. 0.5 Linear magnetic circuit with three branches. Assuming that the ferromagnetic material out of which the core of the magnetic circuit from Problem 5.24 linear, is made can be considered with relative permeability /x r = as 1000, N2 I = 500 A turns, (a) find the reluc- zero. and that Nonlinear magnetic circuit with two air gaps. For the magnetic circuit shown in Fig. 5.43, l[ = l'[ = / = /3 = 12 cm, l2 = 5 cm, Iq = 0.6 mm, 3 and N\I\ = 1100 A turns. The cross-sectional area of each of the branches is S — 2.5 cm 2 The initial magnetization curve of the core can be approximately represented as in Fig. 5.27(b). Determine the mmf N2 I2 of the second coil such that the magnetic flux through that coil is tances of the individual parts of the core and 2 the air gaps and generate an equivalent electric circuit for the problem, (b) electric circuit in (a), find the By solving the magnetic field intensities in each of the air gaps. . zero. 5.28. Continuity equation from Ampere’s law in from Maxwell’s second form for the time-invariant integral form. Starting equation in integral electromagnetic integral field, derive the time-invariant form of the continuity equation. 1 Slowly Time-Varying Electromagnetic Field Introduction: W e now introduce time variation of electric and magnetic fields into our electromagnetic model. The new field is the time-varying electromagnetic field, which is caused by time-varying charges and currents. In the new model, all the quantities, in general, change in both space and time. As opposed to static fields, the electric and magnetic fields constituting the time-varying electromagnetic field are coupled to each other and cannot be analyzed separately. Namely, a magnetic field that changes in time produces (induces) an electric field and thus an electric current, and this phenomenon is known as electromagnetic induction. Additionally, a changing electric field induces a magnetic field. As we shall see in a mutual induction of time-varying and magnetic fields is the basis of time of electromagnetic waves and of electromagnetic radiation. The time-retardation concept is one of the most important phenomena in electromagnetics. cally tells us that there is a time lag It basi- between a change of the field sources, i.e., of time-varying charges and currents, and the associated change of the fields, so that the values of field intensities at a distance from the sources depend on the values of charge and current densities at an earlier time. In other words, it takes some time for the effect of a change of charges and currents to be “felt” The time lag equals the time needed for electromagnetic disturbances to propaat distant field points. gate over the corresponding distance. We shall see later chapter, this in a later chapter that the velocity of propagation electric of electromagnetic disturbances in a =3 vacuum or air x 10 8 m/s (speed of light retardation in electromagnetic systems (lagging in (free space) equals c$ time of time-varying electric and magnetic fields behind their sources), and as such of propagation and other electromagnetic waves in free space). Hence, the time lag in free space is r = R/cq, where 263 264 R Chapter 6 Slowly Time-Varying Electromagnetic Field the distance between the source and field is (observation) points. If the time r for that all combinations of source and domain of field points in a interest is much than the time of change of the sources shorter the [e.g., period of change of time-harmonic (steady-state sinusoidal) charges / and currents, T =\/f, where the frequency of the sources], the retardation is new effect in the system can be neglected. With ignating the maximum dimension interest (containing all sources of the and D des- domain of all field points of interest for the analysis), which most often is the under consideration, so that R < D, and r the corresponding (maximum) time lag in the domain, we thus have r = D/co <£ T = 1// as the condition under which the retardation is insignificant. This means that the system size and the rate of change of charges and currents are such that electromagnetic disturbances propagate over the entire system (or the useful part of the system) before the sources have changed significantly. We refer to such charges and currents as slowly time-varying sources and the corresponding electromagnetic slowly time-varying indicating that fields, is not present under the static assumption electromagnetic induction, and the fundamen- is governing law of electromagnetics describing new phenomenon is Faraday’s law of electromagnetic induction. However, this law is probatal this most important basic experimental fact of It is the underpinning of all of the dynamic-field practical applications, from electric motors and generators, through propagation of electromagnetic waves, to antennas and wireless communication. bly the electromagnetics. We entire system fields as feature of the quasistatic electromagnetic field shall start the study of the slowly time- varying electromagnetic induced electric by introducing the field field intensity vector due to a single point charge in accelerated motion as an experi- mental postulate, and then generalize this concept to the evaluation of the induced electric field intensity vector due to any spatial distribution of slowly time-varying currents. The component due the system, with the will Coulomb electric field to time-varying excess charge in same form as in electrostatics, be discussed and added to the field equations. The concept of the induced electromotive force slow when compared to the velocity of travel of electromagnetic disturbances (waves). In time-harmonic electro- will magnetics, a slow time variation corresponds to the tions for the slowly time-varying electromagnetic the rate of their change in time low frequency, and the retardation is field negligible is is model which the time in On induced electric field due to motion of conductors in static magnetic fields will be introduced as an time-varying time-harmonic) field field. (e.g., the other high-frequency cannot be analyzed without taking into account the travel time of electromag- from one point in the system to another. In this chapter (and the following one), we our attention to slow time variations and low frequencies of sources and fields. restrict The slowly time-varying (low-frequency) has time-invariant the This is why tromagnetic ties field being it is field. many formal (static) elec- similarities with electromagnetic field. also called the quasistatic elecIn be completed, along with the associated version of the continuity equation. In parallel, the side, the rapidly tromagnetic field will called accordingly the low-frequency electromagnetic netic disturbances be introduced and Faraday’s law of electromagnetic induction derived using the magnetic vector potential. The full set of Maxwell’s equa- addition to now time-dependent, 6.1 all the quanti- the only essentially field and discussed in the con- All new concepts and equations will be applied whole variety of quasistatic electromagnetic systems. Examples will include systems based on transformer induction (stationary conductors in changing magnetic fields) and those involving motional induction (moving conductors in static magnetic fields), as well as structures in which both types of electromagnetic induction are present at the same time (moving conductors in changing magnetic fields). to the analysis of a INDUCED ELECTRIC FIELD INTENSITY VECTOR We know from field, impressed electric text of Faraday’s law of electromagnetic induction. Chapter 1 that a point charge Q in free space is a source of an electric predicted by Coulomb’s law and described by Eq. (1.24). the Biot-Savart law (Chapter 4) tells On the other side, us that there will also be a magnetic field, given Section 6.1 by Eq. (4.4), if this charge moves with some velocity v in space. Induced Electric Field Intensity Vector 265 We now introduce whenever the velocity deceleration) a = dv/d t of the a third field, which will exist in the space around the charge v changes in time, charge is i.e., whenever the acceleration not zero. This new an field is (or electric field in its nature. It is called the 2 induced electric field and its intensity vector Eind(0 is dv G Jt_ — 4tt which is an experimental result as a time-varying field, that (4.107), we conclude is, Ei nd well. is Of (6.1) point charge in an ’ accelerated motion is V/m. This is Comparing Eqs. (6.1) and course, the unit for Ei ncj a function of time. 1 any instant actually equal to the negative of the time rate of is change of the magnetic vector potential, A, Eind(0 this R that the induced electric field intensity vector at of time and any point of space Combining then given by — at that point, 3A (6.2) IT' induced electric field intensity vector (unit: temporal differential relationship between A V/m) and Ej nc with J, and A in i the spatial integral relationship between the current density vector, Eq. (4.108), we obtain the following integral (more precisely, integro-differential) expression for the induced electric field intensity vector due to an arbitrary volume current distribution: ind _ “ _mo 4 7t _9_ f Jdv dtjv R Here, the time derivative operator can be directly to the current density vector _ _mo f ~ 4 tv Jv (3J/3Q dv (6.3) R Ei n d due to volume current moved inside the integral sign and applied because the differentiation with respect to time spatial integration over the volume v are entirely independent operations and can be performed in an arbitrary order. Similarly, Eqs. (4.109) and (4.110) yield the corresponding expressions for a surface current of density J s and line current of and intensity v? f (3J s /3r) dS 4 tt Js R Mo _ _M0 Ejnd(0 - Eind(0 Mo [ f ( di/dr) 4tc Ji (6.5) Ejnd due to line current dl R 'While using the notation Ej n d(0 for the induced electric field vector to emphasize its time dependence, and similarly designating other time-varying field quantities (field intensity vectors, flux density vectors, potentials, charge and current densities, etc.) that will be introduced in this chapter, we always keep mind that all these quantities, in general, are functions of spatial coordinates as well, e.g., Ei„d = Eind (x,y,z,t). 2 When dealing with time-varying spatially distributed quantities, respect to time (e.g., we use the partial derivative with 3 A/3 1) rather than the ordinary (total) derivative (dA/dr) to emphasize their multivariable character. 3 As mentioned in Chapter 3, for circuit-theory quantities (e.g., we use lowercase notation (e.g., and v) to distinguish in time, i constant (dc) regime, which are capitalized (/ and V). to surface current Ei n d We conclude from Eqs. (6.3)-(6.5) that an induced electric field will exist in a system whenever time-varying electric currents (representing accelerated or decelerated motion of electric charges) exist in the conductors. Such currents are said to induce in due (6.4) current intensity and voltage) that vary it from the same quantities in the time- x . 266 Chapter 6 Slowly Time-Varying Electromagnetic Field the field. On the other side, it will be zero time at all instants of constants with respect to time (time-invariant currents) at means Eqs. (6.3)-(6.5) represent a general when J s and / are J, , points of the system. all for evaluating (analytically or numerically) the electric field due to any spatial distribution of slowly time-varying we currents in free space. In a later chapter, in these shall introduce certain corrections expressions to extend their validity to rapidly time-varying current dis- wave propagation tributions. Essentially, these corrections introduce computation of Example due to rapidly time-varying fields Induced 6.1 effects in the currents. Electric Field of a Straight Wire Conductor 1 5 Find the expression for the electric by a slowly time-varying current wire structure in We Solution field intensity vector at an arbitrary point in space induced segment of length in a straight i(t) / representing a part of a air. use the integral expression in Eq. (6.5) and refer to Fig. 6.1. Let P' and P field point positioned arbitrarily denote, respectively, a source point along the segment and a The coordinate defining the position of the point in space. and *2 (*2 —x\ = /) P' is x, x\ < x < * 2 where , x\ are the coordinates of the starting and ending point of the segment, Note that both x\ and X 2 can be either positive or negative, as well as zero, depending on the actual position of the point P with respect to the segment. Finally, let d respectively. / Figure 6.1 Evaluation of the induced electric field due mark induced electric (Fig. 6.1), the to a finite straight wire conductor with d/ /no a slowly dt 4tt time-varying current; for Example finite straight f dl J, = is /x o di the segment. „ C *2 An d t * JX[ y/x 2 + d2 and given by dx (6.6) ' 2 jx + d2 where di/dt can be brought outside the integral sign because the current intensity i(t) does not change along the wire (given that the current is slowly time-varying). 4 The solution of 6.1 this integral Ejnd - field intensity R ~ As R vector at the point P P from the perpendicular distance of the point = dxx dl I is (6.7) wire conductor We note that the expression electric field due in Eq. (6.7) can be combined for computing the induced any number of straight wire segments with a slowly to structures containing time-varying current. Example 6.2 A Induced Electric Field of source of electromagnetic interference (EMI) can be approximated by a square current contour of side length a = current whose intensity, i(t), is cm 5 in free space, as a pulse function electric field intensity vector at the point Solution the is 4 an EMI Source (Square Contour) We note domain of M. for electromagnetic disturbances to propagate over interest (from source points at the on the order of As we needed that the time shown in Fig. 6.2(a). The contour carries a shown in Fig. 6.2(b). Compute the induced r = q/cq = 0.167 ns (cq = contour to the 3 x 10 8 field point m/s). Since this time M) is in Fig. 6.2(a) much shorter shall see in a later chapter, the intensity of a rapidly time-varying current in a wire conductor, in general, changes along the conductor. 5 Note that 2 f Ax/Jx + easily verified 2 cl =ln(x+ by differentiation. \l 1 + d2 j + C (C being the integration constant), which can be 1 I Induced Section 6.1 Electric Field Intensity 267 Vector all M r[ns] all L (a) X Eind [V/m] Figure 6.2 Evaluation of the induced electric field near square current contour: 333- Ejndi Ejnd (a) 1 o E Ejnd2 M © ,111 Ejnd4 y 0 5 10 f[ns] and -333 (d) By means (rise or fall) of the current intensity i(t) At in Fig. 6.2(b), = 5 ns, the of the superposition principle, the total induced electric field intensity vector [Fig. 6.2(c)] Ejnd = + Ej n(j2 + Ej n d3 + Ej n d4> Ej nlji (6.8) where the field intensity vectors due to individual sides of the contour, Ej n di, Ej n d 4 are obtained from Eq. (6.7). Due to mutual antisymmetry of the second and fourth current segments with respect to the point M, E; nd 2 = — Ej n d 4 The distances d in the expressions for computing E; n di and E; n d 3 equal all and 3a/2, respectively, whereas xi = —a /I and *2 = a /I are the same in both expressions. Hence, the resultant Ej nd at the point comes out to be . . . , , . M Eind — = Ej n di i T Ej n d3 — 7 —1.11 x 10 In , 4 7Z dt — xV/m + Vi i + 72 + VTo — i + vTo l In (di/df in A/s), (6.9) dt Ejnd Fig. 6.2(b), dr/d/ is nonzero only during the rise and fall intervals when di/dt = Ai/At = ±15 A/(5 ns) = ±3 x 10 9 A/s, which = ±333 V/m. The function Ej n d(0 is plotted in Fig. 6.2(d). We see that very current pulse, of yields strong pulses of the induced electric field in the form of “spikes” are generated in the vicinity of the contour. This field thus may cause a very strong undesirable interference (EMI) into the operation of neighboring circuits in the system. Example 6.3 Induced Electric Field of a Circular Wire Segment (Arc) Consider a wire conductor in the form of an arc representing a part of a wire contour with a slowly time-varying current of intensity a and angle a, as shown in Fig. 6.3(a). intensity vector at the arc center (point Solution the same, (d) resultant field intensity Example current can be considered as slowly time-varying and the system as quasistatic. From to as a function of time; for (c) the due individual sides of the contour, ® given by of the problem, contour current intensity intensity vectors Ejnd3 is geometry (b) as a function of time, (c) field 15 1 than the time of change i(t) in free space. The arc is defined by a [see Fig. 6.3(a)], O) due we can its radius Find the expression for the induced electric bring field to this current conductor. Since the distance of the field point from the source point in this case R= a it is always outside the integral sign in Eq. (6.5), which 6.2. - 268 Chapter 6 Slowly Time-Varying Electromagnetic Field Figure 6.3 (a) Electric field induced by a slowly time- varying current along a circular wire segment and (b) application of the head-to-tail rule for vector addition to solve the integral of dl along the segment; for Example 6.3. leads to MO d / — Ejnd dl, ( A rca dr 6 10 ) . M and N are the starting and ending points of the conductor, respectively. Using the where we observe from head-to-tail rule for vector addition, dl Fig. 6.3(b) that =MN ( 6 11 . ) (note that this result holds true not only for a path in the form of an arc but for an arbitrarily M and N). Finally, as the distance between shaped path, planar or nonplanar, between points = 2asin(a/2), points and N equals M MN r Emd Ejnd - circular wire segment Example 6.4 = —Ana — Mo A dt 7 - — 2n « - Mo sin- MN = dr di . x. ( 2 dr 6 12 ) . Current Contour of Complex Shape Consider the contour consisting of two semicircular and two linear parts in Fig. 1.51, and assume that it carries a time-harmonic current of intensity i(t) = cos(9 x 10 6 r) A (r in s) with respect to the clockwise reference direction, as well as that a = 10 cm and b = 20 cm. Under these circumstances, determine the induced electric field intensity vector at the contour center (point O). Solution Since the period of change of the current intensity i(t), T = 2n/u) = 0.7 gs (where a> = 9 x 106 rad/s is the angular frequency), is much longer than the time z = b/co = 0.667 ns needed for electromagnetic disturbances to propagate from source points at the larger semicircle to the field point O in Fig. 1.51, the time-harmonic current in the contour can be considered as a low-frequency (slowly time-varying) current. Referring to Fig. 6.4 and employing the superposition principle, the resultant induced electric field intensity vector c Ejndl that is, — Ejndi Mo d / -i QR An a d/ , is Mo 2n — , + Ej n d3 = 0. Ej n d2 given by Eq. - Eind 3 — M0 dt"i An dt ~z (6.7), = (6.12), -J^ Anb SP — (-x), = 2n (6.13) at on the other hand, + y/b 2 + 0 a + ya 2 + 0 b In From Eq. dt df From Eq. = (6.8). „ X — — An = - M0 b dz dt In a „ x. (6.14) Section 6.2 269 Slowly Time-Varying Electric and Magnetic Fields Figure 6.4 Evaluation of the induced electric field a current due to contour with two and two linear Example 6.4. semicircular parts; for Due to symmetry, Ej n d 4 Ej nc i = Ej ncj 2 so that the total Ej nc equals = 2Ej n d2 = — rr^-r 27r at Problems : 6. 1-6.9; i , In - x= 1.25sin(9 x 10 6 f) x V/m (6.15) (fins). a Conceptual Questions (on Companion Website): 6. 1-6.8; MATLAB Exercises (on Companion Website). 6.2 SLOWLY TIME-VARYING ELECTRIC AND MAGNETIC FIELDS The slowly time-varying electric field, in general, is composed of two components: the induced electric field, given by Eqs. (6.3)-(6.5), and the field - the field field intensity due to excess charge, vector is which we denote here as Coulomb E q The . electric total electric thus E(0 = Ein d (f) + E*(0. (6.16) total electric field plus The Coulomb field component has the same form due to excess volume charge 1 E q (t) 47T£o R2 Jv induced by now time-dependent. For in free space f PC) dv = fields as in electrostatics, given Eqs. (1.37)-(1.39), however, the charge densities are instance, the field Coulomb ^ is obtained as (6.17) Equivalently, E qit) = -W(t), (6.18) electric field due to excess charge where the electric scalar potential V(t) is expressed as f Pit) 4 Tteo Jv [see Eqs. (1.101) and (1.82)]. R dv (6.19) slowly time-varying electric potential 270 Chapter 6 Slowly Time-Varying Electromagnetic Field As E q (t) the spatial distribution of the field electrostatic field, we can has the properties of the all write E At) [Eq. (1.75)] or V E x = dl • (t) = (l 0 ( 0 6 20 ) . (6.21) [Eq. (4.92)], which are the mathematical expressions of the conservative character of the field E q (t). and In accordance to Eqs. (1.88) (1.90), the time-varying potential M and N in space difference (voltage) between points given by is /•N vmn(0 The tial due field = ^m(0 - = E^fr) / • (6.22) dl. JM to excess time-varying charge also obeys Gauss’ law. In differen- notation [Eq. (1.166)], V E q (t) = However, • Ej nc i (6.23) ' so and (4.119) leads a combination of Eqs. (6.2) v Pit) — 3A\ = -V to 3 = A) (V = (6.24) 0, 3t 31 where the time derivative operator can be brought outside the divergence operator (which implies spatial differentiation) because these two operations are entirely independent from one another and can be performed in an arbitrary order. Hence, the divergence of the vector E(r) in Eq. (6.16) equals V-E(f) =V • +V Ei nd (0 • E,(f) = P{t) (6.25) So which means that Gauss’ law holds for the total electric field as well. In the case of dielectric materials in the time-varying electric ization vector [Eq. (2.7)] and bound charge densities [Eqs. the material are time-dependent. Eq. (2.44), is field, the polar- and (2.23)] in (2.19) The generalized Gauss’ law in integral form, written as <j) D(l) dS • = J pit) dv, (6.26) where D(r) is the total electric flux density vector in the material, given by Eqs. (2.41) and (2.47) with E representing the total electric field intensity vector. On the other side, the slowly time-varying magnetic field has the same form as the magnetostatic field. The magnetic flux density vector in free space can thus be obtained using the time-varying version of Eqs. (4.7)-(4.9). For instance, the due to a volume current distribution in free space is given by [J(r) P-0 Bit) field B(/) R ‘ 4tt The dv] x field l (6.27) can also be obtained indirectly, via the magnetic vector potential, as [Eq. (4.116)] slowly time-varying magnetic B(f) =V x A it). (6.28) field where slowly time-varying magnetic potential A(l) _ .lit) /fo f 4?r Jv dv R with analogous expressions for surface and line currents. (6.29) Section 6.3 Faraday's Law of Electromagnetic Induction In magnetic materials, the magnetization vector [Eq. (5.1)] and magnetization current densities [Eqs. (5.28) and (5.32)] are now time-dependent. The integral form of the generalized Ampere’s law, Eq. (5.51), can be rewritten as <j> H(f) • dl = J J(r) • dS, (6.30) that of the law of conservation of magnetic flux, Eq. (4.99), as and <j> B(f) • dS = (6.31) 0. Problems'. 6.10. 6.3 FARADAY'S We now LAW OF ELECTROMAGNETIC INDUCTION introduce Faraday’s law of electromagnetic induction, as the most impor- and the explicit change in time. Following Oersted’s discovery in 1820 that electric currents produced magnetic fields, Michael Faraday was convinced that the reverse was also possible - that a magnetic field could produce an electric current. In 1831, Faraday set up an apparatus consisting of an iron toroidal core (ring), like the one in Fig. 5.19, with two coils wound on it. The primary coil was connected through a switch to a battery (voltaic cell) and the secondary coil was short-circuited by a wire running above a compass, as sketched in Fig. 6.5. Thus, any electric current in the secondary coil would, by means of its magnetic field, deflect the compass needle. Upon closing the switch, Faraday observed a momentary deflection of the needle, indicating a brief surge of current induced in the secondary coil. The same happened when the switch was opened, terminating the current in the primary coil, but the needle deflection was opposite in polarity with respect to the previous one. In steady states, however, i.e., once the current in the primary coil reached its final value (equal to the battery voltage divided by the resistance of the primary circuit or to zero), there was no current in the secondary coil and the compass needle was at its zero position. Faraday realized that a current was produced (induced) in the secondary coil by a changing magnetic field in the iron core (the field was changed from zero to a final steady value, corresponding to a steady current intensity established in the primary circuit, and then back to zero when the current was terminated). This extraordinary discovery led to the formulation of the law of electromagnetic induction, named after tant governing law of the slowly time-varying electromagnetic field relation between the electric and magnetic fields that Faraday. The mathematical statement of Faraday’s law of electromagnetic induction describes the time variation of the magnetic flux through an arbitrary contour as Figure 6.5 Sketch of the apparatus used 1 in Faraday's 831 experiment that led to the discovery of electromagnetic induction. 272 Chapter 6 Slowly Time-Varying Electromagnetic Field HISTORICAL ASIDE when member permanent magnet was moved in and coil, a current was induced in the coil. He then demonstrated that a continuous current could be generated by rotating a copper disk between the poles of a large permanent magnet and taking leads off the rim and the center of of the Royal Society at the disk. This invention, referred to as Faraday’s Michael Faraday (17911867), an English physicist and chemist, selfeducated from books he was binding to earn a living, became a He was age 34. unfamiliar with that a out of a wire mostly wheel (see mathe- tric Fig. 6.35), was the first dynamo (elec- generator). Faraday used his concept of lines matics, but at the same was an enor- of force to explain the principle of electromag- time netic induction observed in his experiments. experimen- talist explained that an electric current was induced in a conductor only when magnetic lines of force cut of the greatest scientists ever. across he mously gifted and imaginative thinker, and certainly one Upon a recommendation by Sir Humphry Davy (1778-1829), the discoverer of six chemical elements, Faraday was appointed Chemical Assistant at the Royal Institution on March 1, 1813. After repeating Oersted’s (1777-1851) 1820 experiment, he demonstrated in September of 1821 that the magnetic field around a straight wire with an electric current was circular, and, in the same set of experiments, went a large step further - by making a current-carrying wire suspended above a permanent magnet circle around the magnet, he invented the first electric motor. Generalizing from the patterns formed by iron filings around magnets, he introduced the concept of electric and magnetic field lines (he called them lines of force) as a new approach to studying electricity and magnetism. Faraday was elected to the Royal Society in 1824 and was made Director of the Laboratory at the Royal Institution in 1825. In 1826, he started Christmas Lectures for children at the Institution, which not only continue, but now and it, this He could be either because the lines of a changing magnetic field expanded and col- lapsed in space cutting thus across the conductor (transformer induction) or because the conductor moved induction). across the static field lines (motional He that the realized magnitude of the induced current was dependent on the num- ber of lines of force cut by the conductor in unit time, which is a true equivalent to a more mathematical formulation of what is now known as Faraday’s law of electromagnetic induction. It was the discovery of electromagnetic induction in 1831, more than any other, that allowed electricity to be turned, during the remainder of the 19th century, from a scientific curiosity into a powerful technology. In the 1830s, Faraday also studied the relationships between the amount of material deposited on electrodes of an electrolytic cell, the amount of electricity passed through the and chemical properties of different elements, and formulated fundamental principles of electrocell, chemistry (Faraday’s laws of electrolysis). He also over the discovered that light could be affected by a mag- world. In his famous August 29, 1831 experiment, netic force - he demonstrated in 1845 that a strong he wound two magnetic are televised to giant audiences coils of all wire on the same iron ring and discovered that a current in one coil if changed in time induced a current in the other coil. He concluded that an electric current could be produced by a time-varying magnetic field and thus discovered electromagnetic induction. Faraday’s induction ring was the world’s first electric transformer. In subsequent experiments in the fall of 1831, he attempted to create a current using a permanent magnet. He discovered (see Fig. 6.5) field could rotate the plane of polar- - which later became Faraday magneto-optical effect. In his famous lecture “Thoughts on Ray-vibrations” at the Royal Institution in April of 1846, he suggested that the propagation of light through space ization of polarized light known as the consisted of vibrations of lines of force, which, intuitively, was not far from Maxwell’s (1831— 1879) explanation, given much later (in 1865) and based on rigorous mathematical derivations, Section 6.3 light was an oscillatory electromagnetic disturbance - electromagnetic wave. In a series of that studies from 1846 Faraday evolved his model based to 1850, global theoretical electromagnetic on the “force - the field” in tension filling the space ies, field of lines of force around charged bod- current-carrying conductors, and permanent magnets, and thus established the field theory of electromagnetism. These concepts of electric and magnetic force fields were put into a mathematical Faraday's Law of Electromagnetic Induction form a generation made it later 273 by Maxwell, who himself very clear in his texts that the basic ideas for his classical electromagnetic field equations came from Faraday. Despite his epochal achievements and far-reaching contributions to humanity, Faraday remained a modest and humble person throughout his life. We honor Faraday also by using farad (F) as the unit for directly scientific capacitance. Edgar Fahs Smith (Portrait: Collection, University of Pennsylvania Libraries) the cause of the induced electromotive force along the contour. In developing the general electromagnetic model, it is usually taken as an experimentally based postu- However, because we started our study of electromagnetic induction by taking the mathematical expression for the induced electric field intensity vector due to a point charge in accelerated motion as an experimental postulate, we are now able to actually derive Faraday’s law from the facts that we already know about the induced late. electric field. In a region with free charge carriers electric intensity vector, field [Eq. (4.143)], where line integral of Ej nd a conducting wire), the induced (e.g., in E, nci, acts on the carriers by the force <2Ej nd Q is the charge of a carrier (e.g., a free electron). Therefore, the M and N in space represents along a line joining any two points the electromotive force (emf) induced in the line: /N e ind The induced emf is measured = in volts M Eind ' (6.32) dl. J/ induced electromotive force (emf), in volts and defined in the same way as the emf of a voltage generator in Eq. (3.112). In fact, the line can be replaced by an equivalent voltage generator whose emf is Cj n d, as shown in Fig. 6.6. For a closed line (contour) that does not change or move use Eq. (6.2) and write e ind We now recall = j> E ind • dl =£ ^ • dl = £ in time, Fig. 6.7, A we (6.33) dl. that the circulation of the magnetic vector potential along a contour equals the magnetic flux through the contour, Eq. (4.121), so that (6.34) Faraday's law of electromagnetic induction This equation is known as Faraday’s law of electromagnetic induction. the most important experimental pillars of electromagnetics. shows, It is first one of of all, one another under nonstatic conditions. More specifically, it states that a magnetic field that changes with time produces (induces) an electric field and an electromotive force, as well as an electric current in conducting media [by virtue of Ohm’s law in local form, Eq. (3.18)]. that the electric and magnetic It fields are related to 274 Chapter 6 Slowly Time-Varying Electromagnetic Field it quantifies the induced emf in an arbitrary contour as being equal to the negative of the time rate of change of the magnetic flux through the contour, i.e., Finally, through a surface of arbitrary shape spanned over the contour and oriented in accordance to the right-hand rule with respect to the orientation of the contour. This rule tells us that the flux is in the direction defined by the thumb of the right hand when the other fingers point in the direction of the emf, as indicated in Fig. 6.7. Expressing the magnetic flux using the flux density vector [Eq. (4.95)] leads to M £E oN ind -dl —^jfB-dS, (6.35) where the reference directions of dl and dS are interconnected by the right-hand rule: fingers - dl, thumb - dS (see Fig. 6.7). Since the circulation of the field intensity vector due to excess charge is zero [Eq. (6.20)], we have that ^ind lE d,= £ Ejnd ‘ which means that Faraday’s law can be expressed am (6.36) dl, in terms of the total electric intensity vector as well. In addition, the time derivative Figure 6.6 Induced emf in two points in a line joining = space. on the right-hand field side of Eq. (6.35) can be moved inside the surface integral, provided that the surface S does not change or move in time. Combining these two conclusions, we obtain the following version of Faraday’s law of electromagnetic induction in integral form: Faraday's law in integral form (6.37) We note the formal similarity between this equation and the generalized Ampere’s law in integral form, Eq. (6.30), where — 3B/3 1 in Eq. (6.37) stands for J in Eq. (6.30) in the flux integrals, while the electric and magnetic field intensity vectors, E and H, appear in the corresponding line integrals on the left-hand side of equations. The contour C in Fig. 6.7 can be an imaginary (nonmaterial) contour, i.e., it does not need to be a conducting wire loop for Faraday’s law of electromagnetic induction to be true. The electric field and emf are induced by a magnetic field that changes with time regardless of whether or not conducting wires are present. However, in the case when C does represent a conducting wire contour, there is a current of intensity induced current find in the wire, as shown in Fig. 6.8, and of the equivalent closed emf ejnd [this, basically, — ^ind (6.38) ~R~ where R is the total resistance of the contour circuit including the ideal voltage generator with comes from the version of Kirchhoff’s voltage law in Eq. (3.118)]. This current is called the induced current. We can say thus that, in general, time-varying magnetic fields, B (t), induce electric currents in conducting media, which also change with time. On the other hand, if the conducting wire loop is not closed (e.g., there is a small gap in the loop), there is no current flowing through it, 6 and the loop behaves like an open-circuited generator with emf n d, as in Fig. 6.6. The voltage across the gap equals the induced emf, i.e., vnm( 0 = ^ind(0 [ see Eq. (3.115)]. air Figure 6.7 Arbitrary contour in magnetic a time-varying field <?j - for the formulation of Faraday's law of electromagnetic 6 induction. current can exist even As opposed to dc and slowly time-varying cases, as we shall see in a later chapter, a rapidly time-varying in open-ended wire conductors that do not form closed current circuits. Section 6.3 Faraday's Law of Electromagnetic Induction 275 HISTORICAL ASIDE As Heinrich Friedrich Emil Tartu), Estonia, then a part of Russian Empire. Lenz geophysical scientist, he traveled around the world Russian a a (1804-1865), was physicist, of professor physics at the University of St. Petersburg. of the entists, He was one three great sci- together with Faraday (1791-1867) and Henry (1797-1878), who independently from each other investigated electromagnetic induction at about the same time at three remote places on the Lenz was born and educated in Dorpat (now and made extremely accurate measurements of the salinity, temperature, and specific in the 1820s gravity of sea waters. He also studied electricity and magnetism, and discovered, in 1833, that the conductors increases with a rise in temperature [see Eq. (3.22)]. In 1834, he discovered that an induced current always proresistivity of metallic duces effects that oppose cause. This its became to be known as Lenz’s law. From 1840 to 1863, Lenz was the Dean of Mathematics and Physics at the University of St. Petersburg. globe. emf in the contour is magnetic flux through the contour that caused the emf in the first place. This fact, contained in Faraday’s law, is an experimental result referred to as Lenz’s law. To illustrate it, assume that at an instant t the flux in Fig. 6.8 increases in time, i.e., d<h/ dt > 0. From Eq. (6.34), the emf £j n d(0 at that instant is negative, and so is the induced current intensity ij n d(0 in Eq. 6.38. The induced current produces a secondary magnetic field, whose reference direction is determined by another application of the right-hand rule: fingers current, thumb - field [for example, see Fig. 4.31(b)]. Hence, given the reference direction of /; nd in Fig. 6.8, the reference direction of the secondary magnetic field, and its flux, will be the same as that of the primary (original) magnetic field B(t) and flux 0(0- So, with respect to that reference direction, the secondary flux is negative, because q n d(0 is negative at the time instant considered. We conclude thus that the magnetic field due to the induced current opposes the change (increase in this case) in the primary magnetic field, which caused the induced emf and current in the first place, and this is the statement of Lenz’s law. Generally, Lenz’s law represents a rule for (quickly) determining the actual direction of an induced current in a loop (circuit), without fully applying Faraday’s law. This direction is always such that the magnetic field due to the induced current opposes (tends to cancel) the change in the magnetic flux that induces the current. 7 By virtue of Stokes’ theorem, Eq. (4.89), or simply by analogy with the differential form of the generalized Ampere’s law, Eq. (5.52), we obtain the differential equivalent of Eq. (6.37), namely, Faraday’s law of electromagnetic induction in The minus sign in Eq. (6.34) indicates that the induced in a direction that opposes the change in the Figure 6.8 Induced current in a conducting wire loop situated in a time-varying magnetic field. differential form: V x 9B E=- (6.39) Hi' 7 The magnetic field much weaker than due to the induced slowly time-varying current form in typical thin-wire circuits the primary magnetic field inducing the current. Therefore, ble with respect to the primary field electromagnetic induction. and falls far Faraday's law it is is usually negligi- short of fully canceling the flux change causing the in differential 276 Chapter 6 Slowly Time-Varying Electromagnetic Field In words, the curl of the time-varying electric field intensity vector existing at any point of space and any instant of time equals the negative of the time rate of change of the magnetic flux density vector at that point. Note that the differential form of Faraday’s law can also be derived by taking the curl of both sides of Eq. (6.2) and using Eq. (6.28). This results in /3A\ ^ = V x Ei„ d = -V x (— ) which is (6.40) the version of the law with the induced electric field intensity vector. Since the curl of the field intensity vector due to excess charge conclude that V x E= V x Ej nd zero [Eq. (6.21)], is |VxE? = Vx E which gives the version of the law with the Eq. (6.39). 6.4 3B --9 (V x A) = ind we (6.41) , total electric field intensity vector, MAXWELL S EQUATIONS FOR THE SLOWLY TIME VARYING ELECTROMAGNETIC FIELD Faraday’s law of electromagnetic induction, given by Eq. (6.37) or Eq. (6.39), represents Maxwell’s first equation for the time-varying electromagnetic field. It is essentially different magnetic from Maxwell’s first equation for the time-invariant electro- field in Eqs. (5.129). It tells us that a magnetic field changing with time The remaining three Maxwell’s equations are given by Eqs. (6.30), (6.26), and (6.31), and they retain the same form as in the time-invariant case. We now summarize the full set of Maxwell’s equations in differential form gives rise to an electric field. along with the constitutive equations for the slowly time-varying electromagnetic field in a linear isotropic Maxwell's first equation Maxwell's second equation, medium: V x E(f) V x = H(0 = J(f) quasistatic field Maxwell's third equation v Maxwell's fourth equation V B (t) • D(r) • = pit) - 0 constitutive equation for D D(t) = eE(t) constitutive equation for B B(0 = J (0 = constitutive equation for J crE(t) Because of the new term on the right-hand side of the first equation, the time- varying electric and magnetic fields are coupled together and cannot be analyzed opposed to time-invariant electric and magnetic fields, which are independent from each other. On the other hand, as we shall see in a later separately, as entirely chapter. Maxwell's equations for the rapidly time-varying electromagnetic field dif- from the corresponding equations for the slowly time-varying field only in the generalized Ampere’s law (second equation). Namely, an additional term exists on fer the right-hand side of this equation electric field rapidly second equation in in the general case, expressing the fact that an changing with time gives rise to a magnetic field. Therefore, the the system of Eqs. (6.42) represents the quasistatic version of 1 Section 6.5 Maxwell’s second equation and time-harmonic) fields. is true for slowly time-varying Computation (e.g., of Transformer Induction 277 low-frequency Eqs. (6.19) and (6.29) give the expressions for slowly-time varying electromagand (6.18) in Eq. (6.16) leads netic potentials in free space. Substituting Eqs. (6.2) to the following expression for the electric field intensity vector, E, in terms of the potentials: dA vm (6.43) electric field via potentials We see that both potentials are needed for E, whereas A alone suffices for the magnetic flux density vector, B, in Eq. (6.28). Eqs. (6.43) and (6.28) represent a for evaluating (by differentiation) the electromagnetic field (E, B) means from potentials (V, A), the evaluation of potentials being, in general, considerably simpler than the direct evaluation of field vectors. In addition to Maxwell’s equations for a given class of electromagnetic we always have in mind the associated version of fields, the continuity equation, which represents one of the fundamental principles of electromagnetics - the principle of conservation of charge, but can also be derived from Maxwell’s equations. Thus, by taking the divergence of both sides of the quasistatic version of the generalized Ampere’s law in Eqs. (6.42), as in Eq. (5.130) for the static case, we obtain that V which is • J(f) = (6.44) 0, the differential form of the continuity equation for slowly time-varying cur- We note that it has the same form as the continuity equation for time-invariant (steady) currents, Eq. (3.41). We also note that it can be regarded as the special rents. case of the general continuity equation for time-varying currents, Eq. (3.39), with dp/dt ^ 0. Namely, in the slowly time-varying field, the rate of the time-variation in excess charge slow enough to be neglected while evaluating the current continuity is balance. In integral notation, J(0 l dS = (6.45) 0, time-varying currents which is the same as Eq. (3.40) for steady currents. Eq. (6.45) implies that the slowly time-varying current intensity conductor, just as the steady current intensity I along a wire, cross section of the conductor chapter. is It also means - the fact that we have is i(t) the along a wire same in every already used throughout this that Kirchhoff’s circuital law for slowly time-varying currents given by X>(0 = 0, (6.46) k= where N is the number of conductors (branches in a circuit) meeting at a node. see that Kirchhoff’s current law for low-frequency ac circuits has the We same form as that for dc circuits [Eq. (3.42)]. 6.5 COMPUTATION OF TRANSFORMER INDUCTION devoted to the application of Faraday’s law of electromagnetic induce\ n q, and electric field intensity vector, Ej n d, in stationary contours due to given slowly time-varying current distributions and their magnetic fields. This kind of electromagnetic induction is called transformer This section continuity equation for slowly is tion in evaluating the induced emf, 278 Chapter 6 Slowly Time-Varying Electromagnetic Field the basis of current and voltage transformation by magnetic induction, because it is coupling between circuits. one Namely, enables time-varying currents and voltages it (primary circuit) to be transformed, by induction, to time-varying currents and voltages in another circuit (secondary circuit), where the transfer of in circuit energy between the circuits is actually performed by the magnetic field due to the currents in the primary circuit causing the induced electric field in the secondary circuit. The electromagnetic induction due to motion of conductors in magnetic fields will be introduced and studied in the next section. In some contour (e.g., applications, we flux emf induced in a we apply the version of Faraday’s law Eq. (6.34), where we only need to evaluate the are interested only in the total a wire loop). In such cases, of electromagnetic induction in magnetic , through the contour and compute a simple task to do. However, its time derivative, which to find the actual distribution of the is usually emf along a contour or to find the distribution of the induced electric field intensity vector in space, which is necessary in many applications, we have to employ the version of the law in Eq. (6.35). This equation, although always true, enables us to analytically solve for the field Ej nc only due to highly symmetrical primary current distributions. These are the cases where the vector Ej nc is tangential to some (or all) sections of the contour and has a constant magnitude along such sections, while being perpendicular to the remaining sections of the contour (if such sections exist). In other words, these are the cases in which we are able to bring the induced electric field intensity, find* outside the integral sign on the left-hand side of Eq. (6.35), and solve for it. Note that Eqs. (6.3)-(6.5), on the other hand, provide general solution procedures for computing Ej n d. Because Faraday’s law and Ampere’s law have the same mathematical form, there is a complete formal analogy in their application. In solving for Ej n d due to B j j in highly symmetrical situations, therefore, application of Ampere’s law in solving for whenever infinitely unit of its Electric Field of the solenoid The solenoid is air. is The magnetic Find the induced electric Solution H) due to J, discussed in Section 4.5, an Infinite Solenoid long solenoid with a circular cross section of radius a has N' turns of wire per length. wound about slowly time-varying current of intensity in the shall exploit the parallelism with the (or possible. Induced An we B field /(f) a ferromagnetic core of permeability flows through the winding. Because of symmetry, the dE- nd and P due to currents induced in the core can be neglected. vector inside and outside the solenoid. lines of the induced electric inside the solenoid, at a distance r dE" d be the field intensity vectors at this elements denoted as / dl' A field intensity solenoid winding are circles centered at the solenoid arbitrary point /x. The medium outside and / dl" and shown axis. field due To show to the current this, consider an from the solenoid axis (Fig. 6.9). Let point due to two symmetrical current in Fig. 6.9. In accordance to Eq. (6.5), these dE"nd is tangential to the circular contour C of radius r centered at the solenoid axis. The same is true for any other pair of symmetrical current elements, which can also be in a plane that does not contain the point P. As all current two vectors are such that their elements constituting the current we conclude sum dEj nd /(f) in the -F winding can be grouped that the resultant vector Ej ncj at the point P is in such symmetrical same conclusion holds for a point outside the solenoid. In addition, £j n d the magnitude of Ei n<j depends only on the radial coordinate the solenoid axis. Hence, system whose z-axis is Ejnd where <j> is — £j ntj(r the circular unit vector in the system. > 0 pairs, tangential to the contour C. This = const along C, r of the cylindrical i.e., coordinate (6.47) j r Section 6.5 Computation of Transformer Induction 279 Figure 6.9 Evaluation of the induced electric field due to a slowly time-varying current in an infinite solenoid (cross-sectional view); for Example 6.5. The magnetic field due to any spatial distribution of slowly time-varying currents has the same form as that due to the same spatial distribution of steady currents. Therefore, we can use the result of the analysis of a solenoid with a steady current of intensity I in the winding, performed Example in 4.13. This (field lines are parallel to the means that the magnetic field inside the solenoid H = H{t) i = N'i(t) while H=0 magnetic is axial solenoid axis) and uniform, given by (6.48) z, outside the solenoid. Note that this result does not take into account the field due to currents induced in the ferromagnetic core, which can be neglected. We now use Faraday’s law of electromagnetic induction, Eq. (6.35), and apply it to the contour C in Fig. 6.9 in a manner completely analogous to the application of Ampere’s law in Eqs. (4.54)-(4.56). The circulation of Ej nd along C equals Find 2nr d<f> = (6.49) ~df~' The magnetic flux through the contour <J> (there is no is = flux outside the solenoid), Bnr2 for r Bn a 2 for r where — /j-N'r d/ (r 2 < a) and a (6.50) a H B = iH solutions for the induced electric field intensity inside Find < > £ind = df and is given in Eq. (6.48). The and outside the solenoid come out to be i liN'a 2 di ( 2r > a). (6.51) df respectively. We note that the magnetic field inside the solenoid varies synchronously with the current winding [H(t) a /(f)], whereas the induced electric field varies synchronously with the in the time derivative of the current [E; n d(0 d/(f)/df]. Inhomogeneous Conducting Loop around Example 6.6 a Solenoid Consider the solenoid from the previous example, and assume that a wire loop of radius b (b > a) is placed coaxially around it, as shown in Fig. 6.10(a). The two halves of the loop made from different materials, with conductivities o\ and 02 The cross-sectional area of are both wire parts . is S. The magnetic field due to currents induced in the core and in the wire Find of an infinite solenoid + a 280 Chapter 6 a n Slowly Time-Varying Electromagnetic Field loop can be neglected. Calculate the voltage between the junctions of two wire parts (points M and N). Solution The system C contour Having shown in Fig. 6.10(b), R2 R\ and in Fig. 6.10(a), while the contour. can be analyzed from the circuit-theory point of view, in Fig. 6.10(a) using the equivalent circuit diagram mind the second expression in , a emf in the emf amounts to t di 2 (6.52) due field to the induced currents are of Eq. (3.85), the resistances are = Ri Employing Eq. the induced dr Electromotive forces generated by the magnetic By means is in Eqs. (6.51), this d/ neglected. ex d —dOT- = -nixN — = e ind where are resistances of the two wire parts constituting R2 = and (6.38), the current in the circuit, i.e., (6.53) the induced current in the loop, is given by ^ind ' ind ~~ (6.54) R + R2 ' { between points Finally, the voltage V MN Note = • 2 obtained from is — R2 R\ e ind ~ D R\l ind that in the case of a M and N — nix(a2 + R2 2{R\ Fig. 6.10(b) as o\)N' 2 dr (6.55) e nd ‘ 2 (cti ) homogeneous wire loop (p\ = a2 ), dr’ cr2 ) there is no voltage between different points along the loop. (b) Figure 6.10 composed (a) of parts with different conductivities placed around an infinite solenoid carrying a slowly time-varying current and (b) equivalent circuit diagram; for Example Example 6.7 Wire loop two Magnetic Refer to the system N = 1 and 1000 turns/m, ix = hq a in — due to Induced Current Field and assume that 2 20 cm, S = 1 o\ 6.10(a), Fig. 10 cm, = b mm Under (air-filled solenoid). Loop in a — i(t) , = 57 A (r MS/m, a2 = 15 2 cos lOOOr in s), MS/m, these circumstances, find the magnetic flux density vector at the loop center due to the induced current in the loop. Solution Let us first check whether the current in the solenoid 6.6. time-varying. Since the solenoid infinitely long, this is problem one, and the check of whether a low-frequency analysis is can be considered as slowly is actually a two-dimensional sufficiently accurate here should be performed in the cross section of the structure containing the wire loop. The maximum dimension of the structure relevant to the field computation in this cross section is 2b, and the corresponding time of propagation of electromagnetic disturbances is r = 2b /cq = 1.33 ns. This time is much shorter than the period of change of i(t), which equals T = 2-n/u> — 6.28 ms (w = 1000 rad/s). We conclude that this structure can indeed be analyzed as a low-frequency (quasistatic) problem. This performed in the From Eq. means that (6.54), the induced current intensity 6nd(l) IXqo\<j2 N' — (cri The magnetic flux density vector Eq. (4.19) for the field Let us finally compare the primary /(/) in due = solenoid B(/) S d/ the results of the analysis of the structure in the = loop is 1.5 sin 1000/ A. (6.56) d/ to this current at the center of the loop — , 0 and the contour radius 2 = 2b By means is obtained using b: (6.57) 4.7 sin 1000/ z fxT. field Bj n d(/) to the the solenoid winding. field inside the 2 + o2 )b point defined by z Bind (') current we can use previous example. primary magnetic field - that due to the of Eq. (6.48), the flux density vector of the is = nuN'i(t) z = 2.5 cos 1000/ z mT. (6.58) Computation Section 6.5 We see that |Bj n d(?)|/|B(?)| = 3 x 10 1.9 281 of Transformer Induction 1 , i.e., i I Bjnd (0 « |B(0|. 1 (6.59) due to the induced current in the loop is absolutely neglimagnetic field of the solenoid, which caused the induced current in also note that the waveforms of these two fields (sin 1000? and cos 1000?) In other words, the magnetic field gible with respect to the the first place. We are such that B; n d(?) tends to that actually created compensate for the change Of course, it. in B(?), opposes the action i.e., it accordance with Lenz’s law. this is in I Open-Circuited Coil around a Solenoid Example 6.8 i = 2 m and circular cross section of radius a = 10 cm has air-filled solenoid of length = 1750 turns of wire. There is a low-frequency time-harmonic current of intensity 6 rad/s. An open?(?) = Io sin a>f flowing through the winding, where 1$ = 10 A and co = 10 — 10 turns of wire is placed around the solenoid, as shown in circuited short coil with Nj An (a) / N\ Compute Fig. 6.11(a). the voltage between the terminals of the coil. » Solution The solenoid is very long (/ a), so that the end effects can be neglected while computing the magnetic field about its center. This means that the solenoid can be considered as infinitely long while computing the magnetic flux through the short coil in Fig. 6.11(a). As the coil consists of 2 wire turns, this flux is given by N = ^2 ^single turn. t < > (6.60) Figure 6.1 where Single the flux through a surface spanned over any of the turns. In other words, turn is the electromotive forces induced in individual turns in the coil e ind is N2 times that in Eq. (6.52) with n = = -^2 n HQN\N2a fl’l’single turn d? There no current is 2 ncL>iXQN\N2a lQ cos dt because it is cot. (6.61) l open-circuited, so that the voltage across the coil terminals is [Fig. 6.11(b)] v(?) = -e ind (?) = ncotioN\N2a 2 Io cos cot = 3.45 cos 10 ? kV (? An coil placed around a very long solenoid carrying a low-frequency current: 2 d? I in the coil, /jlq add in series, and hence the total emf and N' = N\/l, which results in all 1 open-circuited short in (6.62) s). three-dimensional view showing the windings and (b) cross-sectional view showing reference directions for the emf and voltage; for Example 6.8. (a) / Example 6.9 An infinitely Rectangular Contour near an Infinite Line Current long straight wire carries a slowly time-varying current of intensity ?'(?). A dx same plane with the wire, with two sides parallel to it, as shown in Fig. 6.12. The distance between the wire and the closer parallel side of the contour is c. Determine the emf induced in the contour. rectangular contour of side lengths a and b lies in C i the By = a X ; Mo i(t) 2tc x integrating this density across the flat surface (6.63) spanned over the contour, we obtain the magnetic flux through the contour: Figure 6.12 Evaluation of the emf in a contour in infinitely rc+a 3 >(?) = B(x, / Jx=c t) b dx = 2n dS c+a W(t)b l dx X — Mo i(t)b 2n , c +a In is similar to that in Fig. 5.26). rectangular the vicinity of an long wire with a slowly time-varying current; , (6.64) c where dS is the surface area of a thin strip of length b and width dx in Fig. 6.12, and the flux is determined with respect to the reference direction into the plane of drawing (note that this integration dS S Solution The magnetic flux density vector produced by the current in the wire at any point plane of the contour is perpendicular to the plane and at a distance x from the wire (Fig. 6.12) and an instant ?, its magnitude is [see Eq. (4.22)] ?) S 8 in the B(x, B (h c for Example 6.9. Chapter 6 Slowly Time-Varying Electromagnetic Field For the adopted direction of the direction for the induced emf — ^ind(f) Induced Emf Example 6.10 A coil with N = 400 core of length I = 40 dO Hob dr 17 is function sketched in Fig. 6.13(b) made is c is di (6.65) di wound uniformly and 5=1 cm cross-sectional area slowly time-varying current whose intensity, The core +a "’— with a Nonlinear Core in a Coil turns of wire cm and right-hand rule gives the clockwise reference flux, the contour, with respect to which, in the i(t), is densely about a thin toroidal 2 , as shown in Fig. 6.13(a). A a periodic alternating triangular-pulse established in the coil, where Iq — 0.1 A and T= 1 ms. of a nonlinear ferromagnetic material that exhibits hysteresis effects. In steady state, the operating point ( B H) , can approximately be represented as periodically circumscribes a hysteresis loop that in Fig. 6.13(c), where B m /Hm = hu = 0.001 H/m. The resistance of the coil can be neglected. Find the voltage across the coil terminals in the time interval 0 < Solution < t T. The magnetic the winding [H{t) cx /(f)] core varies synchronously with the current in field intensity in the and is given by H(t) = Ni(t) (6.66) I [see Eq. (5.53)]. From the core, B(t ), (for H = //m afterwards The function H(t) is plotted in Fig. 6.13(d), where the hysteresis loop in Fig. 6.13(c), first ), it is varies as a linear function of time then becomes time-invariant while Figure 6.13 Analysis of electromagnetic induction = v\ 2 ); for from H is Hm = NIq/1 — 100 A/m. that the magnetic flux density in B = -B m H = 0) to B = B m H = Hm to H = 0, (for reduced from when H(t) is reversed and and so on. The amplitude of again a linear (now decaying) function of time increased in the negative direction from material (note that v we conclude Example 6.10. H=0 in a coil to H = -H m with a core , made from a nonlinear ferromagnetic Section 6.6 the periodic trapezoidal-pulse waveform of B B m = PhHm = is 0.1 T, 283 Electromagnetic Induction due to Motion and B(t) is plotted in Fig. 6.13(e). The magnetic ^ind(O) is through the flux coil is = NB(t)S <t>(f) and the induced emf given by Faraday’s law of electromagnetic induction, Eq. (6.34). of the coil is negligible, the voltage across the coil terminals v{t) This function is = V 12 (f) = -Cind (0 = d4>(0 _ NS dB(t) dt amplitude is We (6.67) dt proportional to the slope of the function B(t) and proportional to the slope of B(t) in the NS(2B m )/(T/4) = 8 NSB m /T = see that the induced the time derivative of its in the coil, the resistance is a periodic alternating rectangular-pulse function with the It is As first is plotted in Fig. 6.13(f). same period ( T ). The pulse quarter of the period, that Vq is, — 32 V. emf and voltage of current, which the coil do not vary synchronously with a consequence of the nonlinearity and hysteresis is behavior of the core material. Problems'. 6. 1 Companion 1—6.17; Conceptual Questions (on Website): 6.9-6.16; MATLAB Exercises (on Companion Website). ELECTROMAGNETIC INDUCTION DUE TO MOTION 6.6 Consider a conductor moving with a velocity v in a netic field of the flux density B. The field exerts the mag- static (time-invariant) magnetic Lorentz force, Fm , given by Eq. (4.144), on each of the charge carriers in the conductor. This force “pushes” the carriers to move, and separates positive and negative excess charges We can formally divide Fm by the charge of a carrier and obtain quantity, although expressed in V/m, is not a true electric field intensity vector, because it is not produced by an excess charge [Eq. (6.18)] or by a time-varying current [Eq. (6.2)]. By definition given by Eq. (3.107), it represents an impressed electric field intensity vector, which we term here the induced electric field intensity vector due to motion, and write in the conductor. Fm IQ = vxB. This new Eind =V X B. ( 6 68 ) . induced electric field intensity vector due to motion (unit: This field generates an induced electromotive force, as given by Eq. (6.32). Hence, the emf along motion a line through a conductor in a time-invariant magnetic between points = / r dl Ei n d 7m This emf is referred to as the emf due For a moving contour (closed = N (v / 7m X B) dl. (6.69) to motional induction or simply motional emf. line). e ind — £ (v x B) • (6.70) dl. Note that the velocity of different parts of the contour need not be the same, including cases when some V/m) field is /N eind M and N (Fig. 6.6) due to parts are stationary while other In other words, the motion of the contour may move in arbitrary directions. include translation, rotation, and deformation (changing shape and size) of the contour in an arbitrary manner. When a contour moves and/or changes in a static magnetic field, the magnetic flux through the contour changes with time. It is possible to relate the total motional emf (unit: V) ) Chapter 6 Slowly Time-Varying Electromagnetic Field Figure 6.14 A contour moving a time-invariant in magnetic field. emf induced in the contour to the rate of change of the flux, i.e., to express the motional emf in terms of Faraday’s concept of changing flux through the contour, as in Eq. (6.34). To see this, consider the moving contour C in Fig. 6.14. Let v be the velocity of an element dl of the contour. In a time interval dr, this element moves a distance dp = vdr (see Fig. 6.14). Eq. (6.70) thus becomes Cjnd = <j) x B) (v = dl • (dp x B) <j) • dl. (6.71 Applying the identity (a x b) c = (c x a) b (scalar triple product is unaffected by permutation of the order of vectors) to the scalar triple product (dp x B) dl and noting that dl x dp equals the vector surface element dS shown in Fig. 6.14, we have • • cyclic • = —® last integral x dp) = — <c dr Jc B (6.72) represents the magnetic flux through the strip AS swept out by the dr This B dS. ej n d contour C contour in Fig. 6.14). • (dl Jc during the interval dr (tinted strip between the positions Let us mark this flux dO t h r0 ugh 1 and 2 of the a s, so that d^through A S — ^ind by (6.73) dr The strip AS represents the difference in area between the surface S the contour in position position 2 (at instant r 1 (at + instant r) and the surface = S' dO = is bounded by the contour in U AS. Designating the magnetic flux through S and increment in flux from r to r + dr equals this bounded by dr), as S and S' O' - (6.74) 5' by O and O', respectively, the O, (6.75) exactly the negative of the flux through the strip AS, — dO hrough as- Finally, substituting this equation in Eq. — dO/ dr, which is the same as in Eq. (6.34). We conclude t i.e., (6.73) leads to that the dO = ej nci = same form of Faraday’s law of electromagnetic induction holds for both transformer and motional emf in a contour. Moving A metallic bar of length a allel metallic rails, as Metallic Bar in a Static Magnetic Field — shown 2 in m slides without friction at a constant velocity Fig. 6.15. The bar is perpendicular to the over par- rails and the . Section 6.6 mechanical force acting on the bar Electromagnetic Induction due to Motion 285 Fme ch = 4 N. The whole system is situated in a uniform B = IT. The field lines are perpendicular to the is connected of the page. A load of resistance R = 5 is time-invariant magnetic field of flux density plane of the rails between the rails. and directed out The and in the rails, as well as the magnetic field due be neglected, (a) Find the velocity of the bar. (b) the load and discuss the energy balance in this system. losses in the bar to induced currents in the system, can Evaluate the power of Joule’s losses in Solution (a) The bar moves field and the emf is induced in it due to motional and the field is uniform (i.e, the same everywhere), the motional emf in Eq. (6.69) becomes in a static induction. Since the bar the expression for magnetic is straight = eind where the emf v is directed from point the velocity of the bar (which is and 1 are all (v x B) • (6.76) 1, M to point N along the bar (Fig. 6.15), 1 = MN, and emf in a straight conductor moving in a uniform magnetic field to be determined). Moreover, since the vectors v, B, is orthogonal with respect to each other and ejnd = |1| = a, we can write vBa. (6.77) M The bar and the rails constitute a conducting loop, that is, an electric circuit of the form shown in Fig. 6.8. Hence, there will be a time -invariant induced current in the loop of intensity v Ba find _ “ R _ ~ R t ind [see Eq. (6.38)], given with respect to the (6.78) same reference direction as the emf, where we The on the other hand, produces a secondary magnetic field, which is neglected in evaluating the emf in Eq. (6.76). Note that this field opposes the increasing flux of the primary field B through the loop, the area of which expands as the bar moves to the right, as yet another example of Lenz’s law. N neglect the resistance of the bar and the rails and the corresponding Joule’s losses. current 7j nci, We know presence of a magnetic that, in the From Eq. experiences a magnetic force. in Fig. 6.15 comes out (4.163), the field, a current-carrying conductor magnetic force on the metallic bar to be Fm — Jjndl x B — Im &aB. * This force opposes the motion of the bar and the generation of the again is in accordance with Lenz’s law. In other words, Fm is (6. emf in it, 9) which directed oppositely to the Fmec h on the bar. Furthermore, as the velocity of the bar is constant, these two forces must be exactly equal in magnitude (Newton’s second law), that is, mechanical force Fmech Hence, the velocity we seek emd (b) By = 10 now = F ™2g2 (v = const). (6.80) = m/s 5 (6.81) - give the values for the induced emf and current in the loop: V and /ind = 2 A. Joule’s law, Eq. (3.77), the power of Joule’s P} = the — is v Eqs. (6.77) and (6.78) y g2^2 = Em = Rlfnd losses in the load resistor = 20 W. is (6.82) Note that, on the other side, the power generated by the induced emf in the bar, power of the equivalent ideal voltage generator of emf Cj n d in Fig. 6.8, equals Find — ^ind^ind — 20 W i.e., (6.83) Figure 6.15 A moving uniform in a metallic bar time-invariant magnetic field (elementary electric generator); for Example 6.1 1 286 Chapter 6 Slowly Time-Varying Electromagnetic Field power used [see Eq. (3.121)]. Finally, the mechanical to move the bar at the velocity v is obtained as ^Anech We = ^mectA — — = Ph 20 W. (6.84) see that, as expected, ^rnech which is in Fig. 6.15 is a ^ind (6.85) compliance with the principle of conservation of energy. The system in simple example of an electric generator based on electromagnetic induction, where the applied mechanical energy is converted into the electric energy and delivered to the load. The agent by which the energy transfer is carried out is the induced emf in the moving bar, and the energy of this emf is ultimately dissipated to heat in the resistor. Rotating Wire Loop Example 6.12 A axis in a its Fig. 6.16(a). page. At an uniform time-invariant magnetic The vector B instant t = 0, field perpendicular to the plane of drawing and is the loop the plane of drawing. lies in co of flux density B, as depicted in The is directed out of the resistance of the loop is R. due to induced currents can be neglected. Calculate (a) the induced emf the loop and (b) the instantaneous and time-average mechanical power of loop rotation. The magnetic in Field rectangular wire loop of edge lengths a and b rotates with a constant angular velocity about CO Magnetic in a Static field Solution ^ind (a) B Referring to Fig. 6.16(b), the magnetic flux through the loop ® =B <t> C where 9 is =B — abB cos 9, abn (6.86) the angle between the plane of the loop at an instant perpendicular to the vector B. © S • is From t and the plane the definition of angular velocity, n T3 1-5 II 3 9o + cot (6.87) And Since co — const, the solution for 9 is = 9(t) a where i 9q = 0 (9 = 0 for = t 0). (6.88) , Thus, 4>(t) (a) = abB coscot, (6.89) which, substituted in Eq. (6.34), leads to the following expression for the induced motional emf in the loop: — = coabBsmcot. e jnd (r) (6.90) df (b) The current intensity in the loop amounts find© The magnetic Figure 6.16 A rotating uniform in a wire loop time-invariant magnetic field field due — to ej nd coabB R R to this current is neglected in the computation of the an instant t = 0 and Example 6.12. /; in From Fig. 4.38 on the loop is and Eq. (4.181), the instantaneous torque of magnetic forces acting given by Tm (t) - m ^abB sin 9 = coa 2 b2 B 2 sin i cot. (6.92) ~R (b) cross-sectional view at an arbitrary instant emf Eq. (6.90). (elementary ac generator): (a) top view at (6.91) sinoV. for Eq. (4.180) tells us that the direction of this torque is such that it opposes the rotation of the loop (Lenz’s law), i.e., it is opposite to the direction of an externally applied mechanical torque, T mec h, that rotates the loop. In order to sustain the rotation at a constant rate, R Tmech must be exactly equal in magnitude to T m (Newton’s second law Hence, the instantaneous mechanical power used to rotate the loop is P mech (0 — co — Tm CO T’jnechO) 287 Electromagnetic Induction due to Motion Section 6.6 angular form). in 2 2 a b2 B2 sin R 2 (6.93) cot. The same result can also be obtained from energy conservation, as the mechanical power of loop rotation equals the electric power of the circuit, i.e., the power of Joule’s losses dissipated in the loop [as in Eq. (6.85)]. This yields — Anech(0 — P](f) Ri- j co — 2 2 a b 2 B2 sin (6.94) cot. ~R Given that the time-average value of the function r 1 T T T= 2 n/co is -cos2 cot 1 . . L where T 1 Sm2 <**=7 [ JQ 1 2 sin cot is T T / ~ df = 2 f [Jof f . dt ~ . power of loop co 2 \ ’ emf and (6.95) 2 ) the period of time-harmonic variation of the loop, the time-average mechanical . COS20Jtdt Jo current in the rotation equals a2 b 2 B 2 (6.96) (^mech) ave 2. Note that the system in Fig. 6.16 represents a rotational version of the generator based on translational motion in a static magnetic field in Fig. 6.15. It illustrates the basic principle of an alternating current (ac) generator, where the loop is mechanically rotated in a static magnetic field at a constant rate and the emf and current are induced in the loop of a sinusoidal (time-harmonic) waveform. The angular frequency of this waveform equals the angular velocity intensity = c. dc Line Current Infinite from that the current in the straight wire conductor 7, and that the contour moves away from the wire in Fig. 6.17. x of the loop rotation. Moving Contour near an Example 6.13 Assume co At t = 0, E indl Fig. 6.12 is time-invariant, with at a constant velocity v, as shown the distance of the closer parallel side of the contour from the wire Determine the emf induced is in the contour. Solution The magnetic field produced by the current in the wire is time-invariant, and the system in Fig. 6.17 represents a motional induction version of the same geometry with transformer induction in Fig. 6.12. The magnetic field is nonuniform (magnetic flux density changes in space), and that is why the magnetic flux through the contour is time-varying. As the contour moves uniformly away from the wire, the distance of its closer parallel side from the wire increases linearly in the course of time, and is given by = c + vt. x(t) The magnetic (6.97) around the wire has the same spatial distribution as in Fig. 6.12, so that the magnetic flux through the contour has the same form as in Eq. (6.64), with i(t) substituted by I and c by x(t): flux density vector Mo lb <H0 = , x (6.34), the d4> x motional emf induced reference direction (Fig. 6.17), £md(0 +a In 2n From Eq. MO lb — — c , In 2n + a + vt c + vt (6.98) given with respect to the clockwise in the contour, is dO dx dO /xolabv dx dr dx 2n iiolabv 1 x(x + a) 2 n{c + vt){c + a + vt)' (6.99) The emf in the contour can also be computed using Eq. electric field intensity vector Ej n d E ind2 7 due to motion, given by Eq. (6.70). Note (6.68), is that the induced perpendicular to the b a Figure 6.17 Evaluation of the emf in a rectangular contour moving in the magnetic field due to an infinitely long wire with a steady current; for Example 6.1 3. 288 Chapter 6 Slowly Time-Varying Electromagnetic Field pair of contour edges of length a and hence there , Its magnitude along the and left £.ndi = vfli right = ^ 2.71 emf in the contour e ind which is the — same is (f Jc in Fig. 6.17 and Eind2 x equals = VB 2 = - ^ 2:t(x (6.100) + a) are the corresponding magnetic flux densities. Finally, the total obtained as E; nc dl • | - F) ncn£> Conducting shows a system for — £ind2^ = ( x V (6.1 01 + aj : \x 2n ) Eq. (6.99). result as in Example 6.14 Fig. 6.18 B2 where B[ and respectively, edges no emf in these edges. On the other hand, two contour edges, which are parallel to it. is Ej ni] does not change along each of the remaining Fluid Flow in a Static measurement of Magnetic Field fluid velocity that consists of a parallel-plate capacitor situated in a uniform time-invariant magnetic the capacitor plates and the voltage between the plates A liquid flows between measured by a voltmeter. The field. is The magnetic field lines are parallel The conductivity and permeability of the liquid are a and p. o, respectively. The capacitor plate area is S and the separation between plates is d. Fringing effects in the capacitor can be neglected. The flux density of the applied magnetic field is B. The internal velocity of the fluid can be considered to be uniform. to the plates. resistance of the voltmeter is negligible. Show is Ry, while the resistance of the interconnecting conductors that the velocity of the fluid is linearly proportional to the voltage V indicated by the voltmeter and find the proportionality constant. Solution We have a motion (flow) of the conducting liquid (with an unknown velocity v) magnetic field, and therefore an electric field due to motion is induced, given by Vectors v and B are mutually orthogonal (Fig. 6.18), and we can write in the static Eq. (6.68). E ind = vB. This field forces the charge carriers in the liquid to (6.102) move perpendicularly to the direction of the liquid flow, so that positive and negative excess charges are accumulated on the lower and upper capacitor These charges produce a Coulomb plates, respectively. to excess charge), which field E9 (field due perpendicular to the plates and practically uniform in the space is between them. Since the voltmeter current flowing through is not ideal, its i.e., its terminals. By current continues with the current of the /v internal resistance is not infinite, same = /liquid = JS, a steady is, (6.103) Figure 6.18 Measurement of based on motional electromagnetic induction; for Example 6.14. is intensity through the conducting liquid, that /V fluid velocity there the continuity equation for steady currents, this V Section 6.7 Total Electromagnetic Induction with J standing for the current density in the liquid. This current density, in turn, can be expressed in terms of the total electric J field in the liquid, where the scalar form of the equation Ei n d, and E q adopted in Fig. 6.18. The voltage terminals (M and — = ctE = cr(E g + Ejnd) is J > given by Eq. (6.16), as follows: = CT(£j nd - Eq ), (6.104) obtained for the reference directions of vectors that the voltmeter indicates equals the potential difference between if its N): V=VM -Vv, which, J, (6.105) evaluated using Eq. (6.22) along a (straight) path through the liquid, can also be expressed as -Vn = Eq d Vu (6.106) (the interconnecting conductors in the voltmeter circuit are ideal, Finally, Ohm’s law RySo (vB — V = Ryly = RyJS = Ry So (Ejnd — Eq) from which the solution see that, indeed, v —V (6.1 07) for the velocity of liquid flow turns out to be v We and thus equipotential). gives is = oSRy +d oSdRyB (6.108) V. where the proportionality constant and the conductivity of the fluid. linearly proportional to V, depends on the parameters of the system in Fig. 6.18 Problems 6.18-6.28; Conceptual Questions (on Companion Website): 6.17-6.22; : MATLAB Exercises (on Companion Website). TOTAL ELECTROMAGNETIC INDUCTION 6.7 Consider now the most general case of electromagnetic induction - that of a moving conductor in a time-varying magnetic field. This is the case where both sources of and conductor motion, Hence, the induced electromotive force in a contour that is moved and/or deformed in a magnetic field that itself varies with time is the sum of the transformer emf, Eq. (6.37), and the motional emf, Eq. (6.70). We thus write the induced electric field, namely, the magnetic field change act simultaneously. =- eind V* "" -dS ' + ^(v ^ V transformer emf and call e m & ^ x B) N" ' -dl, " ' ( 6 109 ) . ^ motional emf here the total (transformer plus motional) or complex (combined) emf in the contour. The motional emf term of Eq. (6.109) can be transformed as described by Eqs. (6.71)-(6.75) and Fig. 6.14, so that the total emf in the contour can be expressed as e ind = — d<i>/dr [same as in e ind This last expression Eq. — $ (6.34)], or Ej n d on the right-hand dl — B dS. d t fs (6.1 1 0) side of the equation represents the total derivative of the magnetic flux through the contour with respect to time, where total induction 289 . Chapter 6 Slowly Time-Varying Electromagnetic Field the change in flux with time due to a change These two parts of the partly is due change to a in the shape, orientation, partly in the magnetic field and and/or position of the contour. change correspond to the transformer and motional appears, however, that Eq. (6.34) is the most general form of Faraday’s law of electromagnetic induction, which includes both mechanisms by which the magnetic flux through a contour could change. These two mechanisms are the magnetic field variation and contour motion, and except for them, there are no other possibilities that may result in an induced emf in the emf terms flux of Eq. (6.109). 8 It contour. Moving Contour near Example 6.15 Refer to Fig. 6.17 a Time-Varying Line Current and assume that the current time-varying, with intensity wire conductor in the infinite Find the emf induced i(t). in the is slowly moving contour. Solution We now have a motion of the contour in a time-varying magnetic field, produced by the current in the wire, i.e., a combination of systems in Figs. 6.12 and 6.17. Therefore, the emf is induced in the contour due to combined (transformer plus motional) induction. Combining Eqs. (6.64) and (6.98), the magnetic flux through the contour is <*>(0 From Eq. = m(t)b 2n c , In + a + vt c + vt (6.110) [or Eq. (6.34)], the total (combined) Cjnd(0 d<t> — yu-o + a + vt c + vt — b c i ~dT 2tz — note that the emf; it Eq. (6.65) Eq. (6.99) is /zo i(t)abv 2n(c dr + vt)(c +a+ ( 6 112 ) . vt) motional emf in this expression represents the transformer part of the total and becomes the same as in The second term represents the motional part case of a stationary contour and becomes the same as in the case of a steady current in the wire it becomes zero in the case of a Example 6.16 Assume term the contour case of a stationary contour. in the of the total emf; in first becomes zero emf in di transformer emf We (6.111) in the steady current in the wire. Rotating Loop in a Time-Harmonic Magnetic Field is a low-frequency time-harmonic magnetic field Rosinwr, and obtain the emf induced in the rotating loop. Identify the parts of the emf corresponding to the transformer and motional induction. that the applied field in Fig. 6.16 with flux density B(t) Solution = we The magnetic Obviously, in Fig. 6.16. are now adding flux in <h(f) Hence, the emf induced in e in d(0 = abB(t) cos 0 the contour = a transformer induction component to the system Eq. (6.86) becomes dO at = abB(t) cos cot. (6.1 1 3) (6.1 1 4) is dB —ab—- cos cot + coabB{t) sin a>t df v transformer emf , motional emf 8 Note that the division of the induced emf between the transformer and motional parts depends on the chosen frame of reference. The particular division in Eq. (6.109) is given for the stationary frame of reference, attached to the field B, with respect to which the contour moves at the velocity v. In other words, it however, is is given as measured by a stationary observer (so-called laboratory observer). The total emf, unique and the same for any chosen frame of reference and any observer, including the one moving with the contour. 291 Total Electromagnetic Induction Section 6.7 For the given time-variation B(t), the terms corresponding to the transformer and motional induction appear to be = — COdbBo COS ^ind(transformer) respectively, = e ind (r) We and the total —coabBo (cos and cot (6.1 15) cot, emf 2 cot — sin 2 = cot) —coabBo coslcot emf (and see that the frequency of the induced frequency of the applied magnetic = coabBocos(2cot + n). current) in the contour is (6.1 1 6) twice the field. Stationary Loop Example 6.17 — (OClbBo sin Cind(motional) in a Rotating Magnetic Field R A rectangular loop of resistance R produced by two mutually perpendicular large coils with low-frequency time-harmonic currents. The field due to each coil can be considered to be uniform. The currents in the coils are of equal amplitudes and 90° out of phase, so that the magnetic flux densities they produce are given as Bi(t) = Bo cos cot and Bi{t) = So sinruf, respectively, and shown in Fig. 6.19(a). The sides of the loop are a and b long. Neglecting the magnetic field due to induced current in the loop, find the time-average torque of magnetic forces on the loop. Solution From situated in the magnetic field is magnitude of the resultant magnetic Fig. 6.19(b), the B(f) at = Bi(r) flux density vector, + 82(f), (6.117) an arbitrary instant of time equals |B(f)| i.e., it is = ^B\{t) + B\{t) = 7s 2 (cos 2 cot T sin 2 cot) = that the tip of the vector B(f) traces a circle Tm Such a vector belongs to a class of so-called circularly mark the angle between vectors B(f) and Bi(f) at (b) constant with respect to time. This of radius Bq (6.118) Bq, in the course of time. means polarized time-harmonic vectors. Let 6 time t, Fig. 6.19(b). The tangent of this angle is Figure 6.19 = tan 0(f) and hence the rate at a which it changes #2(0 Bq Bi{t) Bq cos cot in time is sin cot = tan cot and currents = cot [see Eq. 0(f) At in the coils. t = 0, B2 (6.119) , given by Eq. (6.87). This means that constant angular velocity, equal to the angular frequency densities and at = 0 and B= co B rotates of the individual magnetic flux Bi, which implies that 0(0) = 0 (6.88)]. and vector B changes in time (its magnitude is constant, but its direction changes), this is a system based on transformer induction. On the other hand, for the generation of emf it is irrelevant whether B rotates around a stationary loop or a loop rotates (at the same rate) in a static B. This latter case is exactly the system based on As the contour is motional induction in Fig. 6.19, stationary Fig. 6.16. Exploiting this equivalency, 9 the flux through the loop in emf, current, and instantaneous torque of magnetic forces on the loop are given by Eqs. (6.89), (6.90), (6.91), and (6.92), respectively, with the time-average torque coa direction of this torque which is in (6.95), is the same 2 b 2 Bl ( 2R 6 120 ) . as the direction of the field rotation [Fig. 6.19(b)], accordance with Lenz’s law. Namely, the induced current in the loop and the associated magnetic 9 B substituted by Bq. Using Eq. is (Lm ) ave The moment of the loop produce a torque (T m ) that tends to rotate the loop The possibility to approach the problem of a stationary contour in a rotating magnetic field induction case is the reason for which we analyze B2 it in the section devoted to as a motional total induction. A rectangular wire loop exposed to two time-harmonic magnetic equal amplitudes and 90° out of phase (a), fields of which superposed to each other represent a rotating magnetic field Example 6.1 7. (b); for 292 Chapter 6 Slowly Time-Varying Electromagnetic Field along with the applied rotating decrease the angle which 0], field in is contour (caused by the increase r~ Assume same that the loop generated the emf in 6) that Rotating Loop Example 6.18 Wo B to rotate n closer to [i.e., in Fig. 6.19(b) and thus opposition to the change in the magnetic flux through the in a the first place. Rotating Field - Asynchronous Motor exposed to the rotating magnetic direction with an angular velocity in < coo (too field from Fig. 6.19 also rotates in the as indicated in Fig. 6.20. This device to), represents an elementary asynchronous motor. Calculate (a) the time-average power of Joule’s losses dissipated in the loop, (b) the time-average torque of magnetic forces (0 loop, and on the motor. (c) the efficiency of the Solution PB (a) This in field asynchronous motor in the form of a rectangular wire loop rotating with an co o in angular frequency > wo (top view for Example co, at t on total (mixed) induction - the magnetic field called the asynchronous It is field. The changes (rotates) motor because the relative rate of rotation of the with respect to the rotating part of the motor (the loop in our case), called the rotor, equals Aco = co — coq, (6.121) a which rotating magnetic field of co a system based loop does not rotate in synchronism with the Figure 6.20 Elementary angular velocity is time and the loop moves (rotates). where — referred to as the slipping angular velocity of the asynchronous motor. is Consequently, this system can be replaced by either an equivalent system with a station- ary loop and a magnetic field rotating with a velocity Aco (transformer induction case, as 0); in Fig. 6.19) or 6.1 8. magnetic field (6.121), the an equivalent system with a loop rotating with a velocity Aco (motional induction case, as in magnetic flux through the loop 0(f) Eq. (6.94) then tells = From Fig. 6.16). in a static Eqs. (6.89), (6.118), and in Fig. 6.20 is ab |B| cos Acot — abBo cos(a> — coo)t. us that the time-average (6.122) power of Joule’s losses in the loop can be where = written as (Fj)ave (b) By means = k(co - coor, a2 b2 B2 (6.123) 2R of Eq. (6.92), the time-average torque of magnetic forces on the loop (Tm )ave where the k coefficient k is 1 k(a> — is (6.1 coq), that in Eq. (6.123). This torque has the same 24) direction as the slipping velocity of the motor. (c) The rate of the loop rotation in Fig. 6.20 is coo, so that Eq. (6.93) gives the following expression for the time-average mechanical power used to rotate the loop: (P mech)ave The efficiency of the motor is n where we neglect the Example 6.19 = Tm)ave ( too == k too(to — too). (6.1 25) given by = (P mech)ave toO (6.126) ’ (Pmech)ave T" (P]) ave losses in the stationary part of the Charge Flow due to a to motor (the stator). Magnetic Flux Change Consider a wire contour of resistance R situated in a magnetic field, as shown in Fig. 6.21. If this field is changed and/or the contour is moved in the field during an arbitrary interval of time so that the corresponding net change of the magnetic flux through the contour is A0, find the total charge flow Q in the contour during this process. Section 6.7 Since the magnetic flux of the contour, Solution emf which we can process, an for is induced in the varies in time during the considered $>, contour (due to the total induction d4> is i in the general case), write fiind where = -—r dr and e ind = Ri, the intensity of current in the contour (Fig. 6.21). (6.127) Combining these two equations = -Ri dr. d<J> d<2 = dr, i We we (6.128) Figure 6.21 Evaluation of the charge flow (3.4), the charge that flows through the contour during an elementary time dr is resulting in = -RdQ. d<l> tion, e ind Q gives From Eq. 293 Total Electromagnetic Induction (6.129) On then integrate both sides of the above equation. AO, from thus obtain the total change of flux, its the left-hand side of the equa- in a wire contour as a consequence of a change of the magnetic flux through the contour; for Example 6.1 9. starting value (4>i) to the ending value (O 2 ) in the process, dO = 4>2 4>i = AO. (6.130) Q equals the total charge flow, Q, 1 On the right-hand side of the equation, the integral of d and hence (6.131) charge flow due where the reference directions of the charge flow and the magnetic by the right-hand flux are interconnected to in a wire contour a magnetic flux change rule, as indicated in Fig. 6.21. Example 6.20 Fluxmeter Based on a Charge-Flow Measurement B A fluxmeter consists of a small coil (magnetic sonde) connected to a ballistic galvanometer, is S and the number of wire turns is N. and the galvanometer is R. The coil is placed in a uniform time-invariant magnetic field such that the magnetic field lines are perpendicular to the flat surface spanned over the coil cross section (Fig. 6.22). The coil is then removed from the field, and the charge flow indicated by the galvanometer is Q. What is the flux density of the magnetic field? as in Fig. 6.22. The The cross-sectional area of the coil total resistance of the coil Solution The magnetic flux of the coil while <Di it is in the field = NBS, whereas $2 = 0 after the coil is removed from the through the galvanometer is given by Q= A4> 4>2 = IT is (6.132) field. - $1 = R Using Eq. (6.131), the charge flow galvanometer (BG); for Example 6.20. a ballistic <J>i ~R which yields the following expression for the magnetic = NBS (6.133) ~1T' flux density of the field: (6.134) NS We see that B is linearly proportional to Q, where the proportionality constant is determined by the parameters (R, N, and S) of the fluxmeter in Fig. 6.22. Thus, by directly measuring the charge flow through its terminals, a ballistic galvanometer connected to a magnetic sonde can be used to indirectly measure the flux density of an unknown magnetic field. In some cases, like in this example, the measurement is based on motional electromagnetic induction (field is time-constant and a coil is either removed from the field or brought into it). In other applications, like in the apparatus for magnetization curves in Fig. 5.19, a coil is Figure 6.22 Fluxmeter consisting of a small coil stationary in a field that is measurement of from either established and ) 294 Chapter 6 Slowly Time-Varying Electromagnetic Field zero to its final measurement value is B Prove that the magnetic <1> = const Solution A R= ( B to zero, so that the Magnetic Flux through a Superconducting Contour Example 6.21 resistance, be measured) or reduced from some value (to based on transformer electromagnetic induction. through a superconducting wire contour cannot be changed. flux superconducting wire has zero 0, resistivity and zero [see Eq. (3.23)] total so that Eqs. (6.127) give — =0 d<t> through a (R = (6.135) 0), dr superconducting contour that is, <t> = const, which completes our proof. Having in mind that denotes the <t> can be explained as follows. (external or primary magnetic field) current is induced in the total existing flux through the contour, is changed and/or the contour contour whose magnetic field (secondary <t> moved its in the field, a completely cancels extreme form), such through the contour remains constant. Problems'. 6.29-6.33; Conceptual Questions (on MATLAB is field) the change of the magnetic flux through the contour (Lenz’s law in that the total flux this result the magnetic field in which a superconducting contour resides If Exercises (on Companion Companion Website): 6.23-6.25; Website). EDDY CURRENTS 6.8 Whenever induced in a conducting medium, electric current is same time-dependence as the field (we assume that the medium is linear in terms of its conductivity). An example is a conducting wire loop in a time-varying magnetic field (Fig. 6.8), where the intensity of the induced current in the wire is given by Eq. (6.38). This section is devoted to studying volume induced currents in solid conducting bodies, where many current contours are established throughout the volume of the body as a result of the induced electric field. These currents are perpendicular to the magnetic flux in the body and, since they flow like “eddies” (in water), we call them eddy currents. The eddy current density vector, Jeddy is related to the electric field intensity vector, E, through Ohm’s law electric field is also established, with the > in local form: density of eddy currents (6.136) where a is The vector E represents the actual (meais composed of the induced excess charge, E [see Eq. (6.16)]. The vec- the conductivity of the material. surable) electric field in the material, which, in general, electric field, Ej ncj, and the field due to tor Jeddy in Eq. (6.136), therefore, material, the corresponding component cr f/ the actual (total) current density vector components of which are a Ej n d and a E f/ . Note in the that the is usually identified as the induced current density vector However, as the accumulation of excess charge in the majority of Ej n d alone in the material. is practical systems with an induced electric field is also a result of electromagnetic components of the eddy current density vector can be said to be induced by the same cause that induces the field Ej n d and electromotive force in the system, i.e., to represent the induced current in the material. This cause, on the induction, both other hand, can be related to either transformer or motional induction, as well as to total (combined) induction. Section 6.8 As a consequence of eddy currents, according to Joule’s law, and this is electric power is lost to 295 Eddy Currents heat in the material, the principle of induction heating. In so-called induction furnaces, for instance, eddy currents are created on purpose to produce local heating in metal pieces and high enough temperatures to melt the metal. In ac machines and transformers, however, the power loss due to eddy currents induced in ferromagnetic 10 cores sity of the power of is undesirable. From Eq. (3.31), the Joule’s losses at a point in the material is volume den- proportional to the square of the density of eddy currents, /e ddy> at that point. Using Eq. (3.32), the total instantaneous power of Joule’s (ohmic) losses in the entire body is obtained as r I Pj= 2 dv, Jv (6.137) o By due to eddy currents where v denotes the volume of the body. Another important consequence of eddy currents produce. joule's (ohmic) losses is the magnetic field that they Lenz’s law, this field (secondary magnetic field) opposes the change in the primary magnetic flux inside the body, which caused the eddy currents in the place. first While the secondary magnetic circuits is practically often this the rial is field due to induced currents in thin-wire always negligible with respect to the primary magnetic field, not the case with volume eddy currents in solid bodies. The larger volume of the body and areas of eddy current contours (eddies) in the matethe larger the induced emf along the contours and the current intensities, as well as the magnetic field they produce. This effect is also usually not desirable. For instance, the magnetic field due to eddy currents in ferromagnetic cores of ac machines and transformers tends to cancel the primary magnetic flux in the core and thus considerably reduces the efficiency of the device. As an illustration, consider a large core of a rectangular parallelepipedal shape in a uniform time-varying primary magnetic field B, as depicted in Fig. 6.23(a). The secondary (induced) mag- (a) aj : netic field, Bi n d, is the strongest at the center of the cross section of the core, i.e., center of all eddy current contours, where all the fields due to these contours add up. Hence, the resultant (primary plus secondary) magnetic flux density is not uniformly distributed over the core cross section; it is the smallest at the center and at the the largest near the core surface. In other words, practically only the “skin” region below the surface of the core carries the the operation of the device. This magnetic phenomenon is flux and is effectively used for referred to as the skin effect in ferromagnetic cores. Note that the skin effect in current conductors at high frequencies is also a consequence of induced (eddy) currents and their magnetic field. To show this, consider a cylindrical conductor carrying a time-harmonic (ac) current of density J, Fig. 6.23(b). Using the analogy with the dc case in Fig. 4.15, the lines of the magnetic field B due to this current are circles centered at the conductor axis. This field induces an electric field Ej ncj, which is axial in the conductor (field lines are parallel to the conductor axis) and can be qualitatively analyzed by applying Faraday’s law of electromagnetic induction, Eq. (6.35), to the rectangular contour C shown in Fig. 6.23(b). The direction of vectors Ej nC and J e ddy is determined by Lenz’s law (i.e., by the minus sign in Faraday’s law). It is such that the secondary magnetic field, Bj ncj, opposes the primary field B. Hence, the eddy current density vector tends to cancel j 10 Ferromagnetic materials are electrically conducting, with large conductivities iron). (e.g., ape = 10 MS/m for (b) Figure 6.23 Illustration of the skin effect in a parallelepipedal ferromagnetic core with a time-varying magnetic (a) and field in a cylindrical conductor with a time-harmonic current (b). 296 Chapter 6 Slowly Time-Varying Electromagnetic Field the current density J inside the conductor volume, while adding to near the conductor surface. Therefore, the magnitude of the vector is its magnitude total current density small at the conductor axis and increases towards the conductor surface. The higher the frequency (i.e., law) the larger the induced skin effect. At very high the faster the time rate of change d/d f in Faraday’s emf in the contour C frequencies, the current and the more pronounced the is restricted to a very thin layer (“skin”) near the conductor surface, 11 practically on the surface itself, and can be considered therefore as a surface current and described using the surface current density vector, J s [see Eqs. (3.12) and (3.13)]. Examples in this section include evaluation of eddy current distributions and Joule’s losses in several characteristic systems based on each of the three types of electromagnetic induction (transformer, motional, and total induction). In cases, we skin effect in body would require on numerical field-computation techniques. the distribution of eddy currents in the analysis based Example 6.22 A Eddy Currents in a is a much more complex Thin Conducting Disk thin conducting disk of radius a, thickness 8 (8 lio all magnetic field due to eddy currents and the associated conducting bodies. Taking this field into account in the evaluation of shall neglect the positioned inside an infinitely long air-filled low-frequency time-harmonic current of intensity a), conductivity a, and permeability shown solenoid, as /(f) = /q cos cot in Fig. 6.24(a). A flows through the winding. The number of wire turns per unit of the solenoid length is N'. (a) Determine the distribution of eddy currents induced in the disk, neglecting the magnetic field that they produce, (b) Find the time-average power of Joule’s losses dissipated in the disk, (c) Evaluate the magnetic field due to eddy currents at the disk center. Solution (a) Eddy currents are induced in the disk due to transformer induction in this structure. induced electric field inside the solenoid Using Eq. (6.136) and referring to is given by the first The expression in Eqs. (6.51). Fig. 6.24(b), the distribution of eddy currents in the Figure 6.24 (a) Thin conducting disk inside an infinitely long solenoid with a low-frequency time-harmonic current and (b) evaluation of in the disk and magnetic field at the disk center; for Example 6.22. eddy currents their 11 For example, we shall see (b) in a later chapter that the thickness of the layer that carries most of the current in copper conductors at frequencies higher than about millimeter. I MHz is less than a fraction of a = 0 0 Section 6.8 disk is described by the following expression for the induced current density: 7eddy (C 0 — E\n& (f > 0 — liQcrN'r di 2 dr cofioaN'Ior = sin cur where r is the radial distance from the solenoid due to excess charge is practically zero. (b) From Eq. (6.137), the instantaneous (0 < axis power of Joule’s 27rrdr<5 = J/r= losses in the disk = is sin 3 2 r cot dr Jo . sin ? (6.139) cur, with dv being the volume of an elementary hollow disk of radius [Fig. 6.24(b)], electric field 4 no{con.pN'Ip) a 8 ness (height) 5 (6.138) a), ra na(a)/xoN' Ip) 8 dv 2 < and we assume that the 2 PjU) r 2 Having mind Eq. in (6.95), the r, width and dr, time-average of this thick- power amounts to na (cofipN'Ip) 2 a 4 8 > (6.140) (7 j)ave 16 (c) As 8 a, the elementary hollow disk of volume dv in Fig. 6.24(b) can be replaced by an equivalent circular current contour (wire) of radius r and current intensity d/eddy (r, 0 = ./eddy O', 0 S dr (6.141) which the current of density J e ddy flows is a The magnetic flux density the center of the disk is obtained using Eq. (4.19) for z = 0 and r (cross section of the hollow disk through small rectangle of side lengths 8 and dr, and surface area 8 dr). due to this current at substituting a: MO d/eddy( r 0 > d/?i n d(r By virtue of the is , (6.142) t) 2r superposition principle, the resultant magnetic field due to eddy currents given by — a co/aiaN'IpS /^ind(0 [ dB m<xir Jr = — , t) Comparing the amplitude primary field B(f) = (iqN'Iq cos f sin cot 4 Jo dr = —— co/iiaN'IpaS i ! sin cot. 4 Bj n do of the field B; n d(f) to the amplitude cot [Eq. (6.48)] inside the solenoid, BindO _ ~ we (6.143) Bo of the see that co/dpaaS (6.1 4 44) Hence, the magnetic field due to eddy currents in the disk is negligible with respect to the magnetic field due to primary currents in the solenoid only if nfunaaS where / tion is = « (6.145) i frequency of the currents, and whether or not this condidepends on the numerical values of the parameters of the structure in As an example, for / = 60 Hz (power frequency), a — 58 MS/m (copper), co/(2it) is the satisfied Fig. 6.24(a). and 8 — a/20, we obtain that only for disks with quite small radii (a 5 cm), Bj n do Bo, whereas eddy currents in larger disks produce magnetic fields that cannot be neglected with respect to primary fields (at least at the disk center, where the secondary magnetic maximum). Note that the magnetic field Bj n d in Eq. (6.143) is evaluated based on the distribution of eddy currents given by Eq. (6.138). However, in the case when the condition in Eq. (6.145) is not satisfied, this evaluation is not accurate enough and provides only qualfield is itative results, because the starting expression for the induced electric field in Eqs. (6.51) Eddy Currents 297 298 Chapter 6 Slowly Time-Varying Electromagnetic Field is obtained taking into account only the primary magnetic on the right-hand field side of Faraday’s law of electromagnetic induction. Example 6.23 Eddy Currents in a Thin Ferromagnetic Plate A thin conducting ferromagnetic plate of length b, width a in a uniform low-frequency time-harmonic magnetic The field lines are and thickness d(d <5C of flux density B(t) a) = is situated Bq cos cot. perpendicular to the plate cross section, as shown in Fig. 6.25(a). The conductivity of the plate by eddy currents, find total , field is a. Neglecting the end effects (a) the distribution of these currents and the magnetic field produced throughout the plate and (b) the time-average power of Joule’s losses associated with them. Solution (a) This By is another example of a structure with eddy currents due to transformer induction. neglecting the end effects (since d <& a), we assume that the current streamlines in the plate are straight and parallel to the plate surfaces, as indicated in Fig. 6.25(b). From Eq. (6.136), the same is true for the lines of the total electric field intensity vector, E, in the plate. By the same token, the magnitude E of this vector does not depend on the coordinate y in Fig. 6.25(b). Applying Faraday’s law of electromagnetic induction, Eq. (6.37), to the rectangular contour C shown in Fig. 6.25(b), we get 2E(x, t) l =— 2x1, —-<x<^-, 2 df (6.146) 2 which yields E{x,t) dB = —x -—= coBqx sin cot, (6.147) dr where we neglect the magnetic these currents is given by field •^eddyCr Figure 6.25 (a) Thin conducting ferromagnetic plate in a uniform low-frequency time-harmonic magnetic field, (b) distribution of eddy currents in the plate, (c) N insulated thin plates forming a core of an ac machine or transformer, and (d) a core with the same dimensions made of a single piece of material; for Example 6.23. 0 due to eddy currents in = oE(x, t) = cooB^x sin cot. the plate. The density of (6.148) 2 Section 6.8 (b) The total instantaneous the plate power of volume v of is Jldd^X ^ rd/2 ’ Pj(0 Joule’s losses dissipated throughout the 299 Eddy Currents -l abdx — a>,2 oabBQ9 sin"9 tot 2 x dx = to 2 oabd 3 Bl / Jx=-d/ dv - sin 2 art. 12 (6.149) where dv represents the volume of a differentially thin slab of thickness dx used volume integration. Finally, averaging in time [Eq. (6.95)] results in to 2 aabd3 Bq (6.150) (^*j)ave Note 24 machines and transformers are made of mutu- that ferromagnetic cores of ac ally insulated in the stacked thin plates, as portrayed in Fig. 6.25(c), rather than of a single shown With this, the areas of eddy current contours and so are the electromotive forces and current intensities along the contours. Consequently, both undesirable effects of eddy currents (Joule’s losses and secondary magnetic field) are reduced significantly as well. In specific, the time-average power of Joule’s losses in the laminated core in Fig. 6.25(c) can be obtained as N times the power given in Eq. (6.150) for a single thin plate, where N is the piece of material, in Fig. 6.25(d). in the core are considerably reduced, number of insulated thin plates. homogeneous core losses in the On the other hand, the time-average in Fig. 6.25(d) thin-plate expression in Eq. (6.150) for the plate thickness Nd. Hence, (^j)avel oc AW3 and power of Joule’s can roughly be estimated using the same (Ej) a vc2 o< (Nd) 3 , we can write (6.151) ohmic respectively, for these laminated core is two powers. We see that the reduction of Joule’s losses in the estimated to be as large as by roughly a factor of TV 2 as compared to losses in a laminated homogeneous core of an ac machine or core vs. transformer homogeneous core with the same dimensions. Also, the skin effect caused by the magnetic field due to eddy currents is much more pronounced in the core in Fig. 6.25(d), where the resultant magnetic flux is “pushed” to the “skin” region near the surface the of the core only, so that the entire interior of the core is practically flux-free. In the core in Fig. 6.25(c), on the other side, the “skin” regions are formed in each of the thin insulated plates, so that, although not entirely uniform over the cross section of the core, the resultant flux is much more core and the ferromagnetic material densely distributed throughout the volume of the is much more effectively used for the machine or transformer operation. Note also that, from Eq. (6.150), 2 (Pj)ave CX / (6.152) CT, dependence of ohmic due which means that the power loss due to eddy currents in the core increases very rapidly with an increase in the operating frequency of the device (f) and also that it can be reduced by using core materials that have low conductivity (u). That is why ferrites (which have high permeability but low conductivity) are used instead of ferromagnetics in some applications at high frequencies (e.g., for the cores of high-frequency transformers or multi turn loop antennas). Example 6.24 A Eddy Currents in a Rotating Strip very long, thin conducting strip of length about its /, axis at a constant angular velocity width to in a a, and thickness 8 (5 <sc a <SC /) rotates uniform time-invariant magnetic field shown in Fig. 6.26(a). At an instant t = 0, the strip is perpendicular The conductivity of the strip is a, and permeability ptQ. Determine (a) the eddy currents and (b) the instantaneous power of Joule’s losses in the strip. of flux density B, as to the vector B. distribution of Neglect the end effects and the magnetic field produced by eddy currents. to losses eddy currents on frequency and conductivity 300 Chapter 6 Slowly Time-Varying Electromagnetic Field Solution This (a) a structure with is induced electric field ®B due to currents shown the strip, / = Ejnd o-, /i 0 where 6 a t v x is in in Fig. The vector At a point P, perpendicular to the cross section of the strip. Ej n<j 6 26(b). the velocity of which is Note between the is = v ~\ <x< B = vB sin 9{— z) = —coxB sin cot z, the angle [see Eq. (6.88)]. Ej n d (a) = cot eddy currents generated because of motional induction. The given by Eq. (6.68), where we neglect the magnetic field in the strip is cox, (6.153) \' and the reference horizontal plane strip at time that, for the position of the strip in Fig. 6.26(b), the direction of > 0 and into the page for x < 0 (point P'). The density of given with respect to the reference direction out of the page in out of the page for x >s eddy currents in the strip, Fig. 6.26(b), is /eddy where we neglect the end near the ends of the lates = crFind = ojcjxB sin cot, (6.1 54) i.e., the electric field due to excess charge that accumuand causes the current streamlines to bend and close into effects, strip themselves near the ends. (b) The total instantaneous power of Joule’s losses in the strip comes out to be f — Pj(t) a/2 ^eddv / Jx=-a/2 where dv rotating in a and at dx co 77 oI8B 2 sin a f 2 cot a is 7 x2 dx / = volume of an elementary 2 oa i l8B 2 — sin 2 cot, (6.155) 12 J -a/2 the co strip of length /, thickness 8, and in a strip uniform top view at t = (b) cross-sectional an arbitrary time r; Rotating Cylinder Example 6.25 time-invariant magnetic field: (a) 18 7 2 = width dr. Figure 6.26 Evaluation of eddy currents = dv 0 view for Example 6.24. in a A very long conducting cylinder of length angular velocity density B and coo about field radius a, and conductivity a rotates while being exposed to a rotating magnetic its axis, at a constant field of flux (w > wo), as shown in Fig. 6.27(a). Neglect the end effects due to eddy currents, and calculate the total instantaneous power of angular frequency and the magnetic /, Rotating Magnetic Field co Joule’s losses in the cylinder. Solution This we can use is a system based by a cylinder rotating Rotating Figure 6.27 (a) cylinder rotating magnetic field and in a (b) equivalent system with the cylinder rotating magnetic Example 6.25. static field; for in a on total (transformer plus motional) induction. However, the concept of slipping velocity given in Eq. (6.121) and replace this system in a static magnetic field with the rate Aw = w — wq in the opposite Problems direction, as depicted in Fig. 6.27(b). In the equivalent system, induction only, the induced electric field point P found is is based on motional Eq. (6.153). At a in a similar fashion as in in Fig. 6.27(b), = Ej n d = vBsin6(—x) = — AcorB sind z v x (By) = — AtorB sin(Acot + 9q)z, where which 301 r is < < a, the radial distance from the cylinder axis (r = 0 r between the vector —n < |r|), v < jt , (6.156) = A cor is the velocity of 0 and the x-axis, which equals 6q at t = 0 and is given by Eq. (6.88) with A co as rotation rate. Note the opposite directions of Ei n d at points P and P' in the cylinder cross section, i.e., for 0 positive and negative, respectively, which tells us how the streamlines of eddy currents (J e ddy) close throughout the point, and 0 is the angle r the cylinder. The total instantaneous power of Joule’s losses dissipated in the cylinder is obtained by integration: Pj(t)= f crEfnd Jv dv= f f aEfnd (r,d)lrd6dr Jr=0Jd=—7x 'fl a = (A co) i 2 oB 2 f l / 71 As dr JO , f 7 = sm 2 6 dQ , n (co — coq ) 2 a a 4 IB 2 (6.157) , 4 J-7T where dS is an elementary patch in the cylinder cross section [Fig. 6.27(b)], the sides of which are dr and r d 0 long, and dv = / dS. We see that the dissipated power is constant with respect to time, which is a consequence of the cylindrical (rotational) symmetry of this problem. Problems 6.34-6.45; Conceptual Questions (on Companion Website): 6.26-6.30; : MATLAB Exercises (on Companion Website). Problems 6.1. Induced electric field of a circular current loop. Assuming in Fig. 4.6 is electric field above a square current vector at a point N placed at a height a above induced electric field intensity vector at an arbitrary point along the the vertex representing the junction of sides 1 z-axis (point P). coordinates of the point thus being x Induced and i(t), electric find the field of a triangular and 2 of the square contour cur- Fig. 4.39, calculate the electric field intensity vector at the point P induced by time-varying current of intensity 6.5. a slowly i(t) in the loop. Magnetic field of an EMI source (square contour). Consider the square current contour described in Example 6.2, and find the magnetic field intensity vector at the point Fig. 6.2(a). Compare M in the result with that for z in Fig. 6.2(c), the = y = a /2 = — a, induced by the pulse current i(t) in Fig. 6.2(b). rent loop. For the triangular current loop in 6.3. Induced contour. Determine the electric field intensity not steady but slowly time-varying, with intensity 6.2. 6.4. that the current in the circular loop Magnetic field of a current contour of complex shape. For the wire contour with semicircular and linear parts carrying a low-frequency timeharmonic current from Example 6.4, compute the magnetic field intensity vector at the point O in Fig. 6.4. 6.6. Induced electric field of a semicircular- rectangular loop. Find the induced electric field intensity vector at the point O in Fig. 4.40, the induced electric field intensity vector in assuming that a slowly time-varying current Fig. 6.2(d). of intensity i(t) flows along the loop. 2 302 6.7. Chapter 6 Slowly Time-Varying Electromagnetic Field Current contour with circular and straight segments. A current of intensity i(t) = sin(10 8 /) A (r in s) flows along a wire contour with two circular (quarter-circle and 3/4-circle) and two linear parts, , shown in Fig. 6.28, where a lengths b and c ( b c , field (a) = 3 cm > 2a), as is placed coaxially around shown in Fig. 6.29. it The magnetic due to induced currents can be neglected. What is the total induced Find the emf the edge in emf MN in the loop? of the loop, and b — 9 cm. (a) Verify that this is a lowfrequency current, and compute (b) the in- ^indMN» in the following two ways, respectively: (b) by integrating along the edge the induced duced electric field intensity vector and (c) the magnetic field intensity vector at the point O. electric field intensity vector in the due to the current solenoid winding [use the relationship Eq. (4.43) to solve the integral] and (c) by showing that Cj n dMN equals the induced emf in the triangle in Fig. 6.29, and then comin AOMN Figure 6.28 Wire contour with a puting this latter quarter-circle, from emf as a part of the total emf (a). and two 3/4-circle, linear parts carrying a N P low-frequency time-harmonic current; Problem 6.7. Figure 6.29 for Rectangular wire loop placed coaxially around 6.8. Induced electric field at the axis of a circular segment. Repeat Example 6.3 but for the field the solenoid for in Fig. 6.9; Problem 6.12. point at an arbitrary location (defined by the coordinate z) along the z-axis (normal to the plane of drawing) in Fig. 6.3(a). 6.9. Induced electric field composed of two made of a i(t). lengths, Determine the induced parts with the is same con- and S 2 (Si ^ S2), and different determined by angles a and 2n — a respectively, as shown in Fig. 6.30. an arbitrary electric field intensity vector at point along the z-axis assum- S\ carries a slowly time-varying current of intensity 6.6 but ductivity, a, but with different cross-sectional areas, it Repeat Example ing that the wire loop around the solenoid Example 4.4, semicircle and a straight line from Solenoid and a loop of wire with nonuniform cross section. above or below a semicir- cular loop. Consider the wire contour and assume that 6.13. M in Fig. 4.9(a). from current distribution and total electric field. In a domain v in free space, there is a slowly time-varying distribution of volume currents and charges. We know the current 6.10. Voltage density vector, J, at every point of v, as Figure 6.30 Structure Fig. well a loop of wire of as the (total) electric field intensity vector (due to currents and charges outside Find the voltage between any two points, 6.11. it. in v), E, at M and N, outside A in a quasistatic is placed quasistatic electromagnetic field, for magnetic vector potential, A, tric field What 6.12. is in a which the known at every are (a) the total elec- inside the wire between the ends, and (b) the voltage M and N, of the wire? Example 6.14. 6.1 3. Complex wire assembly inside a solenoid. Let the wire loop composed of two semicircular parts with conductivities a\ and 02 from Example 6.6 6.5, a rectangular wire loop of edge have radius a/ and be placed coaxially inside an air-filled solenoid, and let two additional linear pieces of wire, with con<73 and 04, be attached to it, at points M and N, as depicted in Fig. 6.31. Linear wire segments are positioned radially with respect to the solenoid axis, and their ends (points ductivities Rectangular wire loop around a solenoid. Consider the solenoid described in and assume that solenoid; for Problem straight metallic wire point of space. nonuniform cross section around the every point v. Voltage along a straight wire field. in 6.10(a) but with 303 Problems P and Q) are very close to one another (the gap between them Compute is much smaller than the voltage between points a). P and Q. Figure 6.31 Wire assembly of four parts with different conductivities placed inside an air-filled solenoid (the gap between points Q is P and very small); for Problem 6.14. 6.15. Emf in a rectangular loop due to a two- wire line current. A very long lossless thin two-wire line in air, with distance equal to 4a, is between axes of conductors generator of low-frequency time -harmonic current intensity ig (t) — 7go cos cot, while the other rectanguend of the line is short-circuited. A lar (b) (a) fed at one end by an ideal current wire loop of side lengths a and b in the plane of the line, such that its is Figure 6.32 Electromagnetic induction in a rectangular loop due to a time-varying current of a thin two-wire (a) loop between of the line; for line line: conductors and (b) loop on a side Problem 6.1 5. placed two sides are parallel to the line and the distance of one of the sides from one of the line conductors a. Neglecting end and propagation effects, Figure 6.33 is Magnetically coupled i.e., computing the magnetic field of the line as if it were infinitely long and assuming that the line current is the same small circular concentric coplanar in every cross section, loops free wound uniformly and densely along respectively. core, Large square and small circular concentric coplanar loops. Fig. 6.33 shows two concentric current space; for Problem 6.16. /o wire loops lying in the same plane, in free space, a large square loop of side length a and a small circular one of radius b (b The a). wise, direction. The square loop is R. Determine the induced current circular loop, neglecting its in the own magnetic field. Electromagnetic induction in a nonlinear mag- of i(t) = Iq sin cot, where = 10 6 rad/s. The secondary N2 = 4 wire turns encircling the intensity and has only co H carries a low- = Iq sin cot, and the resistance of the circular loop the entire fed by a low-frequency time-harmonic = 0.1 A coil frequency time-harmonic current of intensity i(t) is primary winding, and is open-circuited. The idealized hysteresis loop of the core material is that in Fig. 6.13(c), with B m = 0.5 T and m — 170 A/m. Sketch roughly the voltage waveform across the secondary coil terminals within one period of time-harmonic variation of the current in the primary circuit (T = 2 n/co), that is, for 0 < cot < 2tz loops are oriented in the same, counterclock- 6.17. in magnetic field due to induced current, find the emf induced in the loop for situations in (a) Fig. 6.32(a) and (b) Fig. 6.32(b), as well as the 6.16. and large square . A the form of a thin toroidal ferromagnetic core Rotating rod in a uniform magnetic field. conducting rod rotates uniformly with angular velocity w about an axis that splits it onto with two windings, like the one in Fig. 5.19. two unequal parts of lengths The length and area of the in a respectively. flux density B, as netic circuit. Consider a magnetic circuit in core are / = The primary 50 cross-sectional cm and coil, 5=1 with N\ = cm 2 , 850 turns of wire 6.18. l\ and I2 ( l\ / h) uniform time-invariant magnetic field of of rotation is shown in Fig. 6.34. The axis perpendicular to the rod, and F 304 Chapter 6 vector total emf B is Slowly Time-Varying Electromagnetic Field parallel to the axis, (a) Find the induced emf if (b) /i = What in the rod. (c) I2 , I2 — and 0, is linearly sliding bar Consider the system with a metallic bar moving in a uniform static magnetic field described in Example 6.11 l\ — magnetic in a the total (d) motor - with a 6.20. Electric 0, respectively? and assume that an ideal voltage (Fig. 6.15), B generator of time-constant emf £ is added in series with the resistor (of resistance R), as i shown h h field. R parameters £, Figure 6.34 Conducting rod itive uniformly rotating about an from Example excentric axis in a uniform static magnetic 6.19. field; for Problem sis 6.1 8. (b) the disk surface. is A B , with B defined by taking leads off field 1 to (a) the total induced emf = as follows, (a) For the 0), find the current in the cir- and mechanical force acting on the bar. For the bar sliding (uniformly), sketch the I Fme ch)- (d) What are the ranges of Fm ech in which the system in Fig. 6.36 function of induced values of operates as a motor (Fmec h < 0 - the main mode of operation of the system) and as a gen- in the disk Coulomb and perform the analy- on the bar (Fme ch)> for —00 < Fme ch < 00, and sketch this dependence (note that v is a with respect to the reference direction from the disk center to the rim, (b) the 6.11), force currents in the disk and thickness of the rod, compute given and pos- of the algebraic intensity of the mechanical and 2, the rim and the due all on the velocity of the bar (v), and negative (movement away from and toward the voltage generator), respectively, (c) Determine the mechanical power of the bar movement (P m ech) in terms center of the disk (via the rod), respectively. Neglecting the magnetic B be for v both positive perpendicular to pair of terminals, , new system of this dependence of w about its axis together with an attached axial copper rod, between the poles of a large permanent magnet producing a uniform magnetic field of flux density a and cuit (7) 6.35 a that rotates at a constant angular velocity , (disregard the concrete numerical values bar at rest (v shows Faraday’s wheel, consisting of a copper disk of radius Faraday’s wheel. Fig. Let the values of the system in Fig. 6.36. \ erator (Pmec h electric vector (due to excess charge) at an arbitrary point of the disk, and (c) the voltage (V = V 12 ) across the open terminals of the field intensity > 0 ), respectively? (e) Compute Fmech for which the mechanical power of the motor m ech (| I ) is maximum, operation of the motor, its (f) If for the safe current has to be smaller in magnitude than Imax (so that the motor does not burn out), what is the corre- wheel. (g) Establish and power balance for the system, in both motor and generator mode of operation. sponding range of velocity v? discuss the £ Figure 6.36 Electric motor - with a linearly magnetic field dc voltage generator for Problem 6.20. sliding metallic bar in a static (as in Fig. 6.1 5), added in the and circuit; a Figure 6.35 Faraday's wheel (a copper between the permanent magnet), with open terminals; for Problem 6.19. disk uniformly rotating poles of a large 6.21. Computation For the problem, for a sliding-bar electric motor, electric let 7 motor from the previous V, R = 2 £2, a = 1 m, =5 305 Problems 'f’mech the = 2.5 power N, and v = 10 m/s. Find B and R that the voltage generator {£, intensity motor - with a rotating bar netic field. A metallic bar of length a one end to a in a is that mag- attached vertical metallic rod, the R rotates with w t = shown in Fig. 6.38(a). At and conductor lie in it, as 0, the loop same plane (plane of drawing). Neglecting the magnetic field due to induced currents in emf the loop, determine (a) the induced in the loop and (b) the instantaneous mechani- A T mec h is an instant axis, it reference direction of vector rectangular wire loop of edge about its axis parallel to the conductor and at a dis- tance c from about can rotate so that its other end slides without friction along a circular horizontal metallic rail (of radius a ), as portrayed in Fig. 6.37. voltage generator of time-invariant emf £ and internal resistance R is connected between the rail and the rod (axis), and the whole system is situated in a uniform static magnetic field, whose field lines are perpendicular to the plane of the rail and flux density is B. The algebraic intensity of an externally applied mechanical torque on the rail is Tme c h, for the at its which, as an A a constant angular velocity delivers to the rest of the structure. 6.22. Electric I. lengths a and b and resistance ) power of loop rotation. [To compute the magnetic flux through the loop, adopt the intecal gration surface consisting of a cylindrical part of radius r\ and a flat part of width indicated in Fig. 6.38(b), where — and r\, as can be found using the cosine rule. See also the flux computation in Fig. 7.9(c).] r\ r2 in Fig- 6.37. The losses in the bar, rod, and rail, and the magnetic field due to the current in the circuit (/) can be neglected, (a) What are I and rmec h for the bar at rest? (b) If the bar rotates uniformly, find its angular velocity (w). (c) Sketch the dependence of the mechanical power of the bar rotation (P me ch) on Tmec h, for -oo < Tmec h < oo, and mark the ranges of system operation as a motor and Compute all as a generator, respectively, (d) relevant powers in the system, and discuss the overall power balance. (b) (a) Figure 6.38 Rectangular loop rotating due in the magnetic field to an infinitely long wire with a steady current: (a) side view at an instant t = 0 and (b) top (cross-sectional) view at an arbitrary instant t, with a suggested integration surface (in two 6.24. Figure 6.37 Electric motor -with a uniformly forming an electric circuit with dc voltage generator; for Problem 6.22. field, 6.23. Rotating loop near an An infinitely uated in infinite dc line current. long straight wire conductor air carries a time-invariant bounded by the loop; for Problem 6.23. System for measurement of fluid velocity with an ideal voltmeter. If in the system for measurement of fluid velocity based on motional electromagnetic induction described in Example 6.14 the voltmeter is ideal, i.e., the intensity of the current flowing through its terminals is so low that it can be assumed to be zero, express (a) the electric field due to excess charge in the region between the capacitor plates and (b) the voltage indicated by the voltmeter in terms of the fluid velocity, v, and other parameters of the system. rotating metallic bar in a static magnetic a parts) sit- current of 6.25. Thevenin generator for fluid flow in a magnetic field. Consider the Thevenin equivalent generator for the system with motional 306 Chapter 6 Slowly Time-Varying Electromagnetic Field electromagnetic induction from Example 6.14 that replaces the rest of the system with respect to the voltmeter in Fig. 6.18, as indicated in Fig. 6.39. nal resistance Show (a) of that the emf and intercomputed generator, this as the open-circuit voltage of the circuit (in Fig. 6.18) it represents and input resistance of the circuit (with the excitations shut down), all = vBd and Rj = equal £j d/(crS), respectively, Obtain the expression for the velocity of the fluid in Eq. (6.108) using the generator from (a). (b) Figure 6.40 System for measurement of conducting Rj 1 + V £t Figure 6.39 Thevenin equivalent M*i- inside a cylindrical capacitor with steady current; for Problem 6.26. generator at N terminals (M and N) nonuniform, given by v(x) = vo[l — (2x/d) 2 ], —d / 2 < x < d/ 2, where vo is a constant, as well of the voltmeter in Fig. 6.1 8; for -/v as that a variable resistor (rheostat) Problem 6.25. nected to the capacitor plates 6.26. Fluid flow dc current. fluid flow velocity using motional electromagnetic induction is con- in place of the voltmeter in Fig. 6.18, as shown in Fig. 6.41. (a) Find the parameters of the Thevenin equiv- through a cylindrical capacitor with A nonmagnetic liquid of conductiva very alent generator - to replace the rest of the long coaxial cable of conductor radii a and structure with respect to the rheostat, as in ity a flows between the conductors of b (a < depicted in Fig. 6.40. Both the b), as Fig. 6.39. (b) If the resistance of the rheostat R inner conductor and the inner surface of the , what is the voltage across is it? outer conductor of the cable are insulated by At one end, a thin layer of perfect dielectric. the cable is fed by an ideal current genera- tor of time-constant current intensity / while g , the other end is short-circuited. cal capacitor of length cable such that its is / A cylindri- placed inside the electrodes (thin cylindrical and plates), with radii a 6, are tightly pressed against the respective (insulated) cable conductors. A voltmeter, whose internal resistance, Figure 6.41 Structure in Fig. 6.18 but with a nonuniform fluid velocity, including the resistance of the interconnecting conductors, tor. The uniform, is /?y, is connected to the capaci- is v. intensity vector, (c) the current density vec- and (d) the Coulomb electric field intensity vector (due to excess charge) - between the electrodes of the itor, as in the region cylindrical capac- well as (e) the voltage indicated by the voltmeter. 6.27. and a variable resistor to the capacitor plates; for connected Problem 6.27. Calculate (a) the magnetic flux density vector, (b) the induced electric field tor, v(jc), velocity of the fluid, considered to be Nonunif'orm fluid flow and motional induction. Consider the structure from Example 6.14, and assume that the velocity of the fluid is i. 6.28. Measurement of ity. fluid velocity For the structure described problem, let S - 0.5 m2 , d = and conductivin the previous 10 cm, and B = When the resistance of the rheostat is set to R = Ro — 40 mfi, its voltage is V = Vq = 30 mV, whereas V = 2Vo if R = 4 Rq. Based 0.1 T. on these data, compute the central velocity (for x — 0), vo, and conductivity, a, of the fluid, using the Thevenin equivalent generator (from the previous problem). m 307 Problems 6.29. Moving contour in area A rectangular contour of side magnetic field. lengths a and b moves along the x-axis with a constant velocity v in a time-harmonic magnetic field of angular frequency co, which can be considered to be low, and flux density B(x, t) = Bq cos kx cos cot, where Bo and k are constants, and vector B is normal to the plane of the contour, as shown At in Fig. 6.42. t = 0, the center of the contour coincides with the coordinate origin (x = 0). Find the emf induced in the contour. What 0 b — ®B(x,f) 5=1 length / = 20 cm, and two windings each having N = 100 wire turns and resistance R = 10 £2. The first winding is connected via a switch K to a dc voltage generator of emf £ = 9 V and internal resistance R\ =20 Q. The second winding is terminated in a ballistic galvanometer, whose resistance = 5 £2. The switch K is first open, and is there is no residual magnetization in the core. The switch is then closed, and the charge flow indicated by the galvanometer is Q = 600 /zC. a nonuniform dynamic is cm 2 mean , the relative permeability of the core? V I X a Figure 6.42 Rectangular contour moving in a nonuniform low-frequency time-harmonic magnetic 6.30. Rotating loop near an Problem 6.29. infinite ac line current. Repeat Problem 6.23 but for a low-frequency time-harmonic of Iq cos cot current intensity i(t) Fig. 6.38(a), where Small loop in the magnetic ing large loop. Assume = co time-constant, of intensity I, w. mary and secondary pair of axis of its sides. symmetry At an 6.34. Problem 6.16 is and that this loop that instant t is = gap of a mag- in the from Example assume that the 5.14, cross section of the core in Fig. 5.30(a) parallel to a 0, it is in Thin conducting disk circuit co circle of radius a (S the completely same plane with the small circular loop (as in The magnetic field due to induced radius a, filled = 7ra 2 ), that the air is a gap is with a thin conducting disk of thickness Iq, meability /zo, like the conductivity a, and per- one in Fig. 6.24, N and that currents in the circular loop can be neglected. the current in the coil (with Find (a) the induced emf in the circular loop and (b) the instantaneous and time-average low-frequency time-harmonic, with intensity torque of magnetic forces acting on meability of the ferromagnetic portion of the Two field of the rotating large the (oq i(t) it. rotating loops. If in the previous prob- same ( coo < circuit are determine / The length and and /z r , is relative per- respectively. Neglecting the magnetic field due to eddy currents in the loop also rotates in disk and core, find Iq such that the time-average power of Joule’s losses dissipated in the disk is in the direction with an angular velocity <w), = Iq cos cot. turns of wire) magnetic lem the small loop situated (a) the maximum, and time-average power of Joule’s (ohmic) losses in the small loop and (b) the time-average mechanical power of its rotation. 6.33. circuits, netic circuit. For the simple linear magnetic Fig. 6.33). 6.32. coils, and a galvanometer (BG) in the prirespectively; for Problem 6.33. ballistic of a rotat- field uniformly rotates with an angular velocity its Figure 6.43 Linear ferromagnetic core with two voltage generator and that the current in the large square loop from about = flowing through the infinitely long wire conductor in 6.31. field; for 6.35. find that maximum Eddy currents in Faraday’s wheel, the time-average power of power. (a) Compute Joule’s losses due in Faraday’s wheel described in Problem 6.19. (b) What is the mag- to induced (eddy) currents Charge flow through the secondary coil on a magnetic core. Fig. 6.43 shows a thin lin- netic field that these currents ear ferromagnetic core with cross-sectional center of the wheel? produce at the . 308 6.36. Chapter 6 Eddy der. Slowly Time-Varying Electromagnetic Field currents in an infinite conducting cylin- An infinitely long solenoid with N' wire per unit of length its is with a rectangular cross section, placed coax- around the solenoid and centrally with its length. The inner and outer radii of the toroid are b and c (a < b < c), and its turns of wound about a ially con- respect to ducting ferromagnetic (infinitely long) cylinder of radius permeability a, a The winding and conductivity /x, height harmonic current of intensity i(t) = Iq sinaV, and the medium outside the solenoid is air. Determine the time-average per-unit(a) length power of ohmic losses due to eddy currents induced in the cylinder, neglecting the magnetic field they produce, and then (b) find this magnetic field at the axis of the cylinder (use the procedure from Example 4.14). 6.37. Hollow disk field. A a, intensity < and permeability /xo is i(t) mately melts = Iq it) power of Joule’s b), conductivity field, TV turns of wire cos cot. The channel is como metal, namely, induction heating, that and ulti- (b) the total time-average losses dissipated to heat in the metal. situated in a uniform slowly time-varying magnetic The solenoid has and permeability /xo- Neglecting the end effects (i.e., computing the induced electric field of a solenoid as if it were infinitely long) and the magnetic field due to eddy currents, find (a) the distribution of eddy currents in the channel (these currents produce local heating in the thin hollow conducting disk of radii a and b, thickness 8 (8 <£ a h. pletely filled with a metal of conductivity a triangular-pulse magnetic in is with a low-frequency time-harmonic current of carries a low-frequency time- such that the field lines are perpendicular to the disk, as shown H{t), is in Fig. 6.44. The intensity of this field, a periodic alternating triangular-pulse time function of amplitude sketched in Fig. 6.13(d). netic field Hm and period T, Neglecting the mag- produced by eddy currents, find (a) the distribution of these currents throughout the disk and (b) the total instantaneous and time-average powers of Joule’s losses asso- and then (c) compute the due to eddy currents at the disk ciated with them, magnetic field center (point O). Figure 6.45 Induction furnace: eddy currents accompanying the induced electric field of a very long solenoid with a low-frequency time-harmonic current in the winding heat and melt metal in a toroidal channel of rectangular cross section placed around the solenoid; for Problem 6.38. 6.39. Eddy currents in a thin conducting spherical shell. Consider the solenoid described in 6.22, and assume that a thin conduct- Example Figure 6.44 Thin hollow conducting ing spherical shell of radius b (b disk in a uniform slowly time-varying magnetic is field, sketched whose <$(<$«; b ), conductivity intensity, H(t), in Fig. 6.1 3(d); for is Problem 6.37. placed inside shows an induc- solenoid, with circular cross section of radius that they produce. nel (/ » a), and a toroidal chan- (carrying a metal piece that is heated), thickness fiQ such that the sphere center on the solenoid to / a), axis. average power of Joule’s losses tion furnace consisting of a very long air-filled a and length < and permeability Determine the timein the shell, due eddy currents, neglecting the magnetic field lies 6.38. Induction furnace. Fig. 6.45 it, or, 6.40. Loss power in a laminated ferromagnetic core. laminated conducting ferromagnetic core in A 309 Problems the form of a packet of N insulated stacked thin d (d a), and conductivity a is placed in a uniform slowly time-varying magnetic field, such that the field plates of length b, width a, thickness lines are perpendicular to the packet cross shown section, as sity Figure 6.46 Continuously of this in Fig. 6.25(c). field, B(t), is The inhomogeneous flux den- conducting situated a periodic alternating triangular-pulse time function of amplitude Bm and period T, like the function Fig. 6.13(d). End effects and magnetic field due in strip two mutually orthogonal time-harmonic sketched in magnetic fields of to equal amplitudes and 90° out of total phase; for eddy currents can be neglected, (a) Find the time-average power of Joule’s losses in the core, (b) How large has to be N for a given Problem 6.42. total thickness of the packet, c, so that the total loss power does not exceed a given value field, PI of flux density one of the on a rotating strip in a magnetic For the conducting strip rotating in a uniform time-constant magnetic field from Example 6.24, find the instantaneous torque of magnetic forces on the strip - (a) by integrating torques of magnetic forces on elementary strips (with eddy currents) of width dx in Fig. 6.26(b) (see Example 4.22) and (b) from energy con- 6 . 41 . Torque What is Example (Fig. 6.47). At t = 0, Neglecting the end effects and the magnetic field. servation (see B strips is parallel to the field lines. field due to eddy currents in the conductor, induced electric field vector at an arbitrary point of each of the two strips, (b) the instantaneous power of Joule’s losses in each of the strips, (c) the total instantaneous torque of find (a) the magnetic forces on the conductor, and (d) the time-average torque on the conductor. 6.12), respectively, (c) the time-average torque on the strip? 6 . 42 . Inhomogeneous strip in two orthogonal mag- shows a very long, thin nonmagnetic conducting strip of length /, width a, and thickness 8 (8 <$; a /) lying in the section of a xy-plane of a Cartesian coordinate system. conductor The conductivity of the strip varies with the composed x coordinate, and is netic fields. Fig. 6.46 -a/2 stant. <x< a/2, The given by a(x) — Aoqx2 /a 1 Figure 6.47 Cross strip two rotating in a uniform where ao is a positive conis exposed to two uniform low-frequency time-harmonic magnetic of crossed strips , time-invariant magnetic field; for Problem 6.43. fields = Bocoscvtx and 82(0 = Bosincatz, respectively. Neglecting the end effects and the magnetic field due to eddy currents, compute (a) the time-average power of Joule’s losses dissipated in the strip and (b) the time-average torque of magnetic forces acting on the strip. of flux density vectors Bi(t) 6 43 . . Eddy currents in two crossed rotating strips. Fig. 6.47 shows a cross section of conductor, of length meability /a 0 , /, a very long conductivity a, and per- consisting of two crossed right angle) thin strips, each of width a thickness 8 (8 <$C a «; formly rotates in angular velocity co, air in (at a and /). The conductor uniabout its axis, with an a uniform magnetostatic 6.44. Continuously inhomogeneous rotating cylinder. Repeat Example 6.25 but for a continuously inhomogeneous cylinder whose conductivity is given by the following function of the radial coordinate r [Fig. 6.27(b)]: a(r) = a^r/a, 0 6.45. < r < a, where ao is a positive constant. Eddy currents in a rotating cylindrical shell. An infinitely long thin conducting cylindrical shell of radius a, thickness 8 (8 <£ a), conductivity a, and permeability juo is formly about by an externally applied its axis rotated in air uni- mechanical torque, T[nech per unit length of the shell and against the torque of forces of a uniform time-invariant magnetic field of flux , 310 Chapter 6 Slowly Time-Varying Electromagnetic Field density 5, in which the shell resides, as in Fig. 6.48. to Neglecting the magnetic eddy currents in the shell, find (a) the lar velocity of rotation of the shell (a>) the per-unit-length mechanical Figure 6.48 Rotating cylindrical conducting in a shell uniform time-invariant magnetic field; for Problem 6.45. rotate the shell (P' mech ). shown field due angu- and (b) power used to 7 W Inductance and Magnetic Energy \ Introduction: n this chapter, we An introduce and study the con- equally important concept of energy in I cepts of self- and mutual inductance. In general, magnetic inductance can be interpreted as a measure of trans- just as configurations of former electromagnetic induction in a system of conducting contours (circuits) with slowly timevarying currents in a linear magnetic medium. Briefly, self-inductance is a measure of the magnetic flux and induced emf in a single isolated contour (or in one of the contours in a system) due to its own current. Similarly, a current in one contour causes magnetic flux through another contour and induced emf in it, and mutual inductance is used to characterize this coupling between the contours. Some conductor configurations, called inductors, are specially designed to have a desired (large) inductance (they can have many turns of wire and can be loaded with magnetic cores). Along with the resistor and capacitor, the inductor represents another fundamental element in circuit theory and a basic building block for ac electric circuits. The inductance of an tric inductor is dual to the capacitance of a capacitor. fields is also discussed. We shall see that, charged bodies store elec- energy, configurations of current-carrying con- ductors store magnetic energy. Using circuit-theory terminology, inductors (magnetically coupled or uncoupled) in a circuit contain magnetic energy, in a manner analogous to capacitors as electric energy “containers.” We shall introduce magnetic energy density as well. In the case of conductors in the presence of linear magnetic materials, magnetic energy will be related to self- and mutual inductances of the conductors. For systems that contain ferromagnetic materials with pronounced nonlinearity and hysteresis behavior, on the other side, special attention be paid to establishing a clear physical understanding and precise mathematical characterization of the energy balance in the system, including will so-called hysteresis losses in the material. The material of this chapter represents a culmination of our investigations of steady and slowly 311 312 Chapter 7 Inductance and Magnetic Energy time-varying magnetic fields and electromagnetic On the other hand, it par- conductors and magnetic energy density in the field will bediscussed next. Finally, the concept of to a large extent, the electrostatic analysis of self-inductance will be revisited from the energy induction (Chapters 4-6). allels, previous work will be referenced and used here, in Examples will include inductance and energy computation for a large variety of theoretically and practically important electromagnetic structures with slowly time-varying and time- both the theoretical narrative and examples. invariant currents and fields, ranging from various capacitors and other systems of charged conduct- standpoint. ing bodies, including electric energy considerations (Chapter We 2). Therefore, a substantial portion of the and mutual inductance in two separate sections, and then analyze magnetically coupled circuits based on both concepts. Magnetic energy of current-carrying contours and coils with cores of different shapes and material compositions to several types of trans- shall first study self- mission lines and circuits with magnetically coupled inductors. SELF-INDUCTANCE 7.1 Consider a stationary conducting wire contour (loop), C, in a linear, homogeneous or inhomogeneous, magnetic medium, and assume that a slowly time-varying current of intensity is i shown established in the contour, as in Fig. 7.1. This current produces a magnetic field whose flux density vector, B, at any point of space and any instant of time is linearly proportional to i. The magnetic field, as well as an induced electric field (see Section 6.1), of intensity Ej nci, exist both around the contour and inside the wire itself. The magnetic flux through a surface S bounded by C is given by Eq. (4.95) and is also linearly proportional to i. The proportionality 1 Figure 7.1 Current contour (loop) in a linear magnetic medium. constant, self-inductance (unit: H) (7.1) is termed the self-inductance or inductance of the contour. just More precisely, the inductance defined by Eq. (7.1) is the so-called external inductance, since it takes into account only the flux O of the magnetic field that exists outside the conductor of the loop. There inside the conductor. is We also an internal inductance of the loop, due to the flux shall introduce the concept of internal inductance in terms of the magnetic energy stored inside the conductor in a later section. Because the surrounding medium magnetically linear, is L depends only on the medium perme- ability and on the shape and dimensions of the contour, and not on the current intensity tions of It i. C . always positive, provided, of course, that the reference orienta- is and S (i.e., the reference directions of i and O) are interconnected by the right-hand rule, as in Fig. 7.1. The induced emf along portional to di/dt. It is the loop electromagnetic induction, and emf due to self-induction C is given by Eq. (6.33) and is linearly proO by means of Faraday’s law of we can ^ind — write | dO (7.2) dt This emf is the circuit 1 referred to as the (it Since the current along the contour. is is i interrelated with the flux emf due to self-induction (or self-induced caused by the magnetic slowly time-varying, its intensity field (i) is due emf) in to the current in the circuit only a function of time and does not change Section 7.1 313 Self-Inductance and not due to currents in other circuits). The unit for the self-inductance henry (H). From Eqs. (7.1) and (7.2), which represent two equivalent definitions of L, H = Wb/A = V- s/A. One henry is a very large unit. Typical values of self-inductances in practice are on the order of mH, ,u.H, and nH. itself, is HISTORICAL ASIDE Joseph 1878), an sicist Henry (1797American phy- and one of greatest scientists the and in- ventors in the area of electricity magne- and tism ever, was a professor of natural philosophy at Princeton and the first secretary of the Smith- sonian Institution. From 1819 to 1822, he attended the Albany Academy, where he was appointed professor of mathematics and natural philosophy in 1826. Although his teaching duties were extremely heavy, Henry soon became the foremost American scientist of his time. He discovered electromagnetic induc- tion independently of Faraday fact, Henry had performed (1791-1867). In the key experiments ahead of that led to the discovery of induction Faraday, in 1830, but Faraday published his dis- covery first, in 1831. On the other side, Henry is fully credited for the discovery of self-induction. It was his idea to wind many layers of insulated wire on an iron core and thus obtain electromagnets of unmatched power. In a demonstration at Yale University in 1831, his electromagnet lifted more than a ton of iron (previous electromagnets were capable of lifting only a couple of kilograms). Experimenting with electromagnets, he observed a large spark that was generated whenever the circuit was broken, and thus discovered selfinduction (in 1831). He realized that, in general, a time-varying current in a circuit not only induces electromotive force in another circuit (mutual induction), but also in itself (self-induction). He also defined the associated property of a circuit, its self-inductance, as a measure of to “self-induce” electromotive force. its ability He found that “coiling” of the wire greatly enhances the self-inductance of the circuit. The same year, he demonstrated his electric motor based on a continuous oscillating motion of a straight electromagnet with two coils at its ends. The electromagnet rocked back and forth on a horizontal axis with its ends being alternately attracted and repelled by two vertical permanent magnets and its polarity being reversed automatically in the same rhythm by alternately connecting the coils to electrochemical cells (sources). Although still an experimental laboratory device, Henry’s motor was much closer to a mechanically useful practical machine than Faraday’s 1821 motor. It was just one year later, in 1832, that William Sturgeon (1783-1850) invented the commutator and the first motor with continuous rotary motion, which was a rotary analogue of Henry’s oscillating motor and essentially a “prototype” of our modern dc motors. In another stunning demonstration to his students at the Albany Academy, in 1831, Henry strung a mile of wire all around the inside of the lecture hall to connect an electromagnet to a battery. The magnet was placed close to one end of a pivot mounted steel bar, whose other end, in turn, was next to a bell. After the circuit was closed and a current “sent” from the battery to the coil of the electromagnet, a steel bar swung toward the magnet, striking the bell on the other end. Breaking the made connection, next, the electromagnet lose force and release the bar, which to strike again. This was then was the world’s its free first electri- cal relay (electromechanical switch). In addition, by connecting and disconnecting the battery to the circuit in a particular pattern, the steel bar, made to ring the same and this is nothing else but telegraphy. Although, obviously, Henry invented the telegraph in 1831, it is Samuel Morse a mile away, could be series of signals (1791-1872) on the who is bell, credited for this invention [Henry did not patent any of his devices, and actually helped Morse to put his telegraph model to practical use and transmit the first telegraph message using the Morse code (invented in 1838) from 314 Chapter 7 Inductance and Magnetic Energy Baltimore to Washington, D.C. on In 1832, May 24, 1844], Henry became professor of natural phi- losophy (physics) at Princeton University (then College of New Institution was established Jersey). After the Smithsonian 1846, in Henry was second president (elected in 1868) of the National Academy of Sciences, and he held both positions until his death. In his honor, we use henry (H) as the unit for inductance. Brady-Handy Photograph (Portrait: Library of Congress, Collection) named its first secretary (director). He was also the 6 • Figure 7.2 Circuit-theory representation of an inductor and equivalent controlled voltage generator. Note that, while the emf definition of self-inductance in Eq. (7.2) does not make any sense for steady currents, the flux definition in Eq. (7.1) can be used in practically the same way under both dynamic and static conditions. Namely, a dc current I that is assumed to flow in the contour produces the flux O through the contour in the same way a (slowly) time-varying current does, and the inductance obtained as L = <t>/7 is equal to that obtained, for the same contour, using either Eq. (7.1) or (7.2) and time-varying current. On the other hand, the flux O due to the current I is time-invariant, and thus no emf is generated in the contour. Every conducting loop or circuit has some inductance (self-inductance), usually as an undesirable side effect, which can often be neglected. In practical applications, however, we frequently design and use conductors that are specially arranged and shaped (such as a conducting wire wound as a coil), and sometimes loaded with magnetic cores, to supply a certain (large) amount of inductance. Such a device, with its inductance L as its basic property, is called an inductor. Just as a capacitor can store electric energy, an inductor can store magnetic energy, as we shall see in a later section. Resistors, capacitors, and inductors are basic circuit elements that are combined, together with voltage and current generators, to form arbitrary RLC circuits. Fig. 7.2 shows the circuit-theory representation of an inductor. When a timeis varying current of intensity i flows through the inductor terminals, the emf induced in the inductor, given by Eq. (7.2), where L is the inductance of the inductor. With respect to its terminals, the inductor can now be replaced by an equivalent ideal voltage generator whose emf equals e m <\ (this is a voltage by the time derivative of a current), as indicated emf of this is the same in Fig. 7.2. generator controlled The reference as the reference direction of the current voltage v across the inductor terminals in Fig. 7.2 V element law for an inductor — — Cj n d is i direction (see Fig. 7.1). The hence di (7.3) L, dr' This is the element law (current-voltage characteristic) for an inductor. It tells us change of i in time, with L as the proportionality constant. Its form is just opposite to the element law for a capacitor, Eq. (3.45), and we say that an inductor and a capacitor are dual elements. These two laws, the element law for a resistor (Ohm’s law), Eq. (3.72), Kirchhoff’s current and voltage laws, Eqs. (3.42) and (1.92), and element laws for an ideal voltage generator, Eq. (3.1 16), and an ideal current generator, Eq. (3.127), represent a full set of basic that v is linearly proportional to the rate of equations of circuit theory for analysis of linear ac circuits. Note that the model in Fig. 7.2 actually represents an ideal inductor, which does not include any parasitic effects, such as parasitic capacitances between the adjacent turns in a coil and Joule’s losses (in wires and magnetic cores). These effects are present to a greater or lesser extent in many practical situations. The only all effect real inductors, but can be neglected in modeled by an ideal inductor is the emf Section 7.1 due On to self-induction, e j nc j. the other side, it is assumed Self-Inductance in circuit theory that the magnetic flux, induced emf, and magnetic energy are concentrated only in the inductors in a circuit, while the magnetic field and the associated flux, emf, and energy due to the connecting conductors are assumed to be negligible. In practice, there is always an induced emf distributed along conductors in the circuit (in the ac regime), and it depends on the shape and dimensions of the conductors. However, this emf, i.e., the inductance of the conductors, can, again, be neglected in practical applications. We zero resistance of the conductors have been many assumptions of zero capacitance and recall that similar made while introducing circuit-theory The representations of a capacitor and a resistor in Figs. 2.15 and 3.9, respectively. between the elements in circuit layouts) as if they were all ideal short-circuits, where the shape, dimensions, and material properties of the interconnects are assumed to be completely irrelevant for the operation (and analysis and design) of the circuit. circuit-theory model, therefore, deals with connecting conductors (lines Finally, inductors filled Fig. 6.13) are with magnetically nonlinear materials (permeability) of the material depend H whereas is always proportional to depends on the current intensity, is the coil of on the applied magnetic field intensity, H, Consequently, the inductance of the inductor i. L= which (e.g., nonlinear circuit elements. In such cases, the magnetic properties L(i), (7.4) nonlinear inductor analogous to the relations in Eqs. (2.114) and (3.74) for a nonlinear capacitor and a nonlinear resistor, respectively. Example 7.1 Inductance of a Very Long Air-Filled Solenoid An air-filled solenoidal coil has N = 200 turns of wire. The length of the solenoid is / = 20 cm and the surface area of its cross section is S = 4 cm 2 The solenoid can be considered as very long, so that the end effects can be neglected. Under these circumstances, find the inductance . of the coil. Solution Let us assume a slowly time-varying current of intensity i in the coil, find the magnetic flux through all of its turns, and use Eq. (7.1) for the inductance. By neglecting the end effects, we also assume that the magnetic field produced by the current i is uniform inside the entire solenoid (as for an infinitely long solenoid). This field is given by Eq. (6.48), which leads to the following expressions for the magnetic flux through a surface spanned over a single turn of the coil and the total flux through the coil: — ^single turn I^qHS where the reference direction for <t> respect to the reference direction for > is i. < t > = adopted ^VOgjugie turn in = 7~ '» (7-5) accordance to the right-hand rule with The inductance of the coil is hence (7.6) l i that is, L= 100 /u,H for the given numerical data. cross section (provided that the solenoid Example 7.2 Coil It is the same for any shape of the solenoid very long). on an Inhomogeneous Thick Toroidal Core A coil with N turns of wire made is is wound over a thick toroidal core of rectangular cross section of two ferromagnetic layers of permeabilities g\ and /z 2 , as portrayed in Fig. 7.3. The L - very long solenoid 315 ) Chapter 7 Inductance and Magnetic Energy c b dr a : m2 N Ml Figure 7.3 Evaluation of the inductance of a ds with two ferromagnetic Example b : wound on coil a thick toroidal core linear h : layers; for r 6 7.2. inner and outer radii of the toroid are a and surface between the layers Solution If b (a is < b < c, its height is h, and the radius of the boundary Calculate the inductance of the c). a steady current of intensity / assumed is to flow in the coil (Fig. 7.3), the netic field intensity H(r) inside the core (in both ferromagnetic layers) The magnetic <1> r =N flux through the c B(r)dS =N Jr=a r coil is is mag- given by Eq. (5.83). then [see Fig. 7.3 and Eqs. (5.84) and (5.92)] b r n\H(r) dS / coil. + M 2 #(r) dS = Jb Ja N ih — — 2 c / 2n c\ b / ( mi In \ a + M2 M 2 In — b ). ) (7.7) and the inductance of the coil L= N2 h — = -r— I Example 7.3 mi 2jt External Inductance b ( <t> \ p.u.l. In a c + M2 In of a Thin - (7.8) b Two-Wire Line Find the external inductance per unit length of a thin symmetrical two-wire transmission in air. The conductor Solution upon radii are The two-wire itself at a and the distance between their axes can be considered as an line both ends of the line at infinity. Fig. 7.4 infinitely is d (d line » a). long wire loop that closes shows the cross section of the line. The procedure of evaluation of the line inductance (external self-inductance) per unit length is analogous (dual) to the procedure of evaluation of the capacitance per unit length of a thin two-wire line in Example 2.15. We assume that a slowly time-varying current of intensity i (or a steady current of intensity /) is established in the line (Fig. 7.4). At a point plane containing the axes of conductors, the magnetic flux density vectors B) and currents (of the same magnitude and opposite where x is B\ + fi 2 Figure 7.4 Evaluation of the external inductance per unit length of a thin two-wire transmission line in air (cross section of the structure); for Example 7.3. to is (7.9) -- the distance from the axis of the in the directions) in individual conductors of the line are collinear, so that the resultant magnetic flux density B= M B 2 due first conductor. Section 7.1 As the structure is infinitely long, the magnetic flux through the we consider only a part of surface of length flat that l is Seif-Inductance 317 that is l long and compute spanned between the line it is the right choice because B is perpendicular to the surand oriented as required by the right-hand rule with respect to the adopted reference direction of the line current. This flux is given by conductors (such integration surface face) <t> = B Idx = 1 Mo il dx d—a unit = Mo U In n 2 7Z , In n a d - (7.10) a dS [see the integration in Eq. (2.140)]. meter) of the line comes out Hence, the external inductance per unit length (per each to be <D u — LP (7.11) where O' is the flux per unit length of the As a numerical example, for d/a = external inductance amounts to Example 7.4 Consider an L= coaxial cable. L' - thin two-wire H/m) line. 30, L' = 1.36 \x H/m. If the line is 100 m long, its total 136 ^H. External Inductance air-filled a jt il p.u.l. The the inner radius of the outer conductor of a Coaxial Cable is a and Obtain the expression for the external radius of the inner conductor of the cable is b (b > a). inductance per unit length of the cable. Let us assume a dc current of intensity I flowing through the line conductors, as The associated magnetic flux density vector, B, is circular with respect to Solution shown in Fig. 7.5. the cable axis and its magnitude, B(r), in the dielectric is given by the second expression in Eq. (4.61), where r is the distance from the axis and a < r < b. Computing the magnetic flux through a flat surface of length / spanned between the conductors (Fig. 7.5) then yields the following expression for the external inductance per unit length of the cable: 1 B{r) ll dS = 1 b f W>/ TljM-r where the dielectric can be any nonmagnetic material (m voltage and capacitance computation in Example 2.10. As Mr= , = . Mo)- Mo ^ = 0.6 mm and b = 3.62 mm (dielectric is (7.12) a- The procedure a numerical example, the inductance per unit length of an having a b , ln polyethylene), is is dual to the RG-11 coaxial cable, computed to be L' = 359 nH/m. Comparing the expression for L! in Eq. (7.12) and the expression for the capacitance per — 2tzsq/ In (b/a) [see Eq. (2.123)], we note that the following same cable, unit length of the C Figure 7.5 Evaluation of the external inductance per unit length of a coaxial cable; for Example 7.4. L' - coaxial cable line (unit: . 318 Chapter 7 Inductance and Magnetic Energy relationship exists between the two parameters: duality of L' and C' L' C' for air — (7.13) eoMo- dielectric Comparing then the expressions in Eqs. (7.11) and (2.141), for a thin two-wire line in same relationship tivity exists matter of a we fact, as we also note that the shall prove same is true in a later chapter, the between L' and C' for any two-conductor transmission For a transmission dielectric. As air. and permeability, L'C' line — line with air having homogeneous linear dielectric of arbitrary permit- ep.. Using these relations, we can now very easily obtain the inductance per unit length of transmission lines (with homogeneous linear dielectrics) for which we already have the capacitance per unit length (found in Chapter 2). For example, from the expression in Eq. (2.146) for the capacitance per unit length of the transmission line consisting of a thin wire conductor and a grounded conducting plane in Fig. 2.24(a), we can directly write the expression for the external inductance per unit length of this line: , B Example _ MO — — 2n 2/? £qMo , a Superconducting Contour 7.5 (7.14) In C in a Magnetic A superconducting planar contour of area S and inductance netostatic field of flux density B. The contour perpendicular to the plane of the contour, as intensity flows along the contour. I\ The Solution magnetic total flux contour inductance L magnetostatic Example 7.5. in a field; for is situated in a uniform then turned such that new in steady the mag- B is steady current of its plane becomes state. first state is = LIi+BS, 4>i uniform is in the through the contour Superconducting contour of L positioned such that the vector in Fig. 7.6. In this state, a The contour parallel to B. Find the current I2 in the Figure 7.6 is first Field (7.15) where the first term is the flux due to the current (/] ) in the contour (self-flux) and the second term is the flux of the external magnetic field (B), i.e., the flux due to other currents (or permanent magnets) in the system (mutual flux). second In the state, 0 2 = LI2 since the mutual flux The total is (7.16) , S = 0). through a superconducting contour cannot be changed [see zero (B magnetic flux Eq. (6.135)], which means that 0i Combining Eqs. (7. 1 5)— (7. 17), Note field, (7.17) . the current in the contour in the second state that the increment in current amount is hence = /t + ^. /2 the contour exactly at the = 02 between the two (7-18) states, A / = BS/L, is induced in that completely compensates, via the associated magnetic the change of the mutual flux (from BS to zero), maintaining thus a constant total flux through the contour. Problems : MATLAB 7.2 7. 1-7.9; Conceptual Questions (on Companion Website): 7.1-7.10; Exercises (on Companion Website). MUTUAL INDUCTANCE Consider now two stationary conducting wire contours, C) and C2 in a linear (homogeneous or inhomogeneous) magnetic medium, as shown in Fig. 7.7. Let the first contour carry a slowly time-varying current of intensity /]. As a result, a , Section 7.2 319 Mutual Inductance magnetic field, of flux density Bi, is produced everywhere, and Bi, which is a function of both the spatial coordinates and time, is linearly proportional to i\. Some of the lines of Bi pass through the second contour, i.e., through a surface S2 bounded by C2 These lines constitute the magnetic flux through the second contour due to the current q, which can be expressed as . 02 = ( Bi -dS 2 (7.19) . Js2 The flux 02 is linearly proportional to q 02 where the proportionality constant L 21 two contours and is obtained as as well: = ^21*1, is called the (7-20) mutual inductance between the (7.21) mutual inductance (unit: H) M is also used to denote mutual inductance. It is a measure between the contours. Its magnitude depends on the shape, size, and mutual position of the contours, and on the magnetic properties (permeability) of the medium. The mutual inductance can be both positive and negative, depending on the adopted reference orientation of each of the contours for their given mutual position. Namely, if a positive current q in the contour C 1 gives rise to a positive magnetic flux 02 for the orientation of the surface S 2 that is in accor- Note that the symbol of magnetic coupling dance to the right-hand rule with respect to the orientation of the contour C2 the mutual inductance is positive. Otherwise, it is negative. 2 The mutual inductance is , also expressed in henrys. By applying Faraday’s law of electromagnetic induction, Eq. (6.34), to the flux we obtain the induced emf (due to transformer induction) in the Eq. (7.20), contour C2 in : ^ind2 d0 2 — (7.22) dt Eqs. (7.21) and (7.22) should be regarded as equivalent definitions of mutual They indicate that the mutual inductance between two magnetically coupled contours (circuits) can be evaluated by assuming a current (q ) to flow in one contour (primary circuit) and computing or measuring the magnetic flux ( 02 ) or the induced emf (ej n d 2 ) in the other contour (secondary circuit). Again, note inductance. that, while the flux definition of L 21 in Eq. (7.21) can be used for both time-varying Figure 7.7 Two magnetically coupled conducting contours. 2 In some texts, of contours is the information about the sign of the mutual inductance for specific reference orientations not included in its definition, i.e., mutual inductance is defined as always being nonnegative. emf due to mutual induction 320 Chapter 7 Inductance and Magnetic Energy and steady currents, the emf definition Eq. (7.22) makes sense for time-varying in currents only. Conversely, if we assume that a current '2 z flows along the contour mine the associated magnetic flux Oi through the contour between the contours is given by C 1 , C2 and deter- the mutual inductance (7.23) emf e m ^\ along with an analogous expression using the induced the contour C\. Because of reciprocity (in a linear system, transfer functions remain the same if the source location and the location at which the response to the source is observed are interchanged), reciprocity of mutual ^12 = (7.24) ^21- induction If the medium around the contours is air (or any nonmagnetic medium), the induced electric field intensity vector due to the current in the first contour is given by [see Eq. (6.5) and Fig. 7.7] so that the induced ®ind2 i = emf in Ejndl • MO — Ejndl dli dz'i (7.25) An the second contour can be written as [see Eq. (6.33)] d>2 MO = 47T dli ll • dq dl2 R dq = —Li\ dt (7.26) ”d7 This yields the following expression for the mutual inductance between the contours Ci and Neumann C2 : formula for mutual (7.27) inductance which is known as the Neumann 3 formula for mutual inductance. It implies the evaluation of a double line integral along the contours and underscores the fact that the mutual inductance only a property of the geometrical shape and the is physical arrangement of coupled contours, as well as of the permeability of the medium [if the contours are situated in a homogeneous linear magnetic medium of permeability p, the constant p 0 needs to be replaced by p in Eq. (7.27)]. It is obvious from Eq. (7.27) that the change of the reference orientation of one of the contours means the change of the reference direction changes the polarity (sign) of the mutual inductance L 21 It is also obvious that interchanging the subscripts 1 and 2 in Eq. (7.27) does not change the algebraic value (which includes both the magnitude and the in Fig. 7.7 (but not both of them), which of one of the vectors dlj and dl 2 , . sign) of the double integral (since the dot product is commutative and the two line integrations can be performed in an arbitrary order). This proves the identity in Eq. (7.24). The Neumann formula represents a good basis for numerical evaluation of the mutual inductance of contours of arbitrary shapes, where the involved line integrals along the contours are computed using numerical 3 German Franz Ernst Neumann (1798-1895), a mineralogist, physicist, and mathematician, a profes- sor of mineralogy and physics at the University of Konigsberg. electromagnetic induction, and derived, in Neumann 1845, the formula for the equal parallel coaxial polygons of wire, the generalization of which inductance of arbitrary wire contours (loops), as integration methods. we use it today. is contributed to the theory of mutual inductance between two the Neumann formula for mutual a Section 7.2 Mutual Inductance between a Loop and an Example 7.6 Infinite Find the mutual inductance between the rectangular loop and the 321 Mutual Inductance Wire infinitely long straight wire in Fig. 6.12. Solution The infinitely long wire can be considered as a loop that closes upon itself at and the rectangular loop as C2 the current i\ and magnetic flux <f>2 in Eq. (7.21) are actually the current i carried by the wire in Fig. 6.12 and the flux through the rectangular contour given by Eq. (6.64), respectively. The mutual inductance between the contours for their reference orientations given in Fig. 6.12 (upward orientation for the infinitely long wire and clockwise direction for the rectangular loop) is hence infinity. Designating it as C\ , L2 i = $ Mo — =— b 2 that the same result is + (7.28) . c 2 Tt l\ Note c In obtained by applying the emf definition of mutual inductance Eq. (7.22) to the expression for the induced emf in the rectangular contour in Eq. (6.65). Note also that computing the mutual inductance as L\ 2 in this case would be prohibitively in It would require either finding the induced electric field intensity vector Ei n d 2 an arbitrary point of the infinite wire due to an assumed current i2 in the rectangular contour, and then integrating it along the wire (which, in fact, is the Neumann formula for complicated. at L 12 ) or finding and integrating the magnetic flux density vector plane bounded by the loop Example 7.7 Two coils are C 1 B 2 due to i2 across a half- . Mutual Inductance of wound uniformly and a thin toroidal core (such as the Two on Coils a Thin Toroidal Core densely in two layers, one on top of the other, about one shown in Fig. 5.11). The core is made from a linear ferromagnetic material of relative permeability Mr = 500. Its length is / = 50 cm and 2 cross-sectional area 5 = 1 cm The number of wire turns is N\ = 400 for the first coil and . N2 — 600 for the second one. Compute the mutual inductance of the coils. Let us adopt the same reference orientations for the two coils (this will give us mutual inductance of the coils) and assume that the first coil carries a current of intensity i\. This current produces a magnetic field of the same intensity {H\) everywhere inside the core. From Eq. (5.53), H\ = N\i\/l so that the total magnetic flux through the second coil ($ 2 ) is given by [see also Eq. (5.96)] Solution a positive , ^single turn — pH\S (m — MrMo) and the mutual inductance between the 4*2 coils — ^2 ^single turn — comes out p-n x . (7.29) to be n2 s (7.30) l Finally, substituting the coil numerical data gives L 21 = 30 mH. coil (<t>i). Magnetic Coupling between a Toroidal Coil and a Loop N = 500 turns of wire and densely wound about a ferromagnetic core. The relative permeability of the material is Mr = 1000. The inner and outer radii of the toroid are a = 2 cm and b = 4 cm (thick toroid), and its height is h = 1 cm. A wire loop is placed around the toroid, as shown in Fig. 7.8. There is a low-frequency time-harmonic current of intensity i = Iq cos cot An open-circuited toroidal coil with a rectangular cross section has that are uniformly two coupled thin toroidal core Note that the same result is obtained by assuming a current i2 to flow in the second and computing the mutual inductance L 12 from the total magnetic flux through the first Example 7.8 L 21 - coils on a t 322 Chapter 7 Inductance and Magnetic Energy Figure 7.8 Magnetic coupling between a coil wound over a and a wire thick toroidal core loop encircling the toroid; for Example 7.8. flowing along the loop, where /o terminals of the coil. = 2A and a> = 7 Let us denote the wire loop as Solution no current in the coil Eqs. (6.62) and (7.22)] is (it is x 10 3 rad/s. Find the voltage between the circuit 1 and the toroidal coil as circuit 2. open-circuited) and the voltage across the coil terminals v — ^ind 2 — dz'i E21 = (r'l There is [see (7.31) i). dr However, the mutual inductance L21 is extremely difficult to find using Eq. (7.21). Namely, the field Bi due to q, which would have to be integrated through each of the turns of the toroidal coil in order to compute the flux $2, is not only highly nonuniform but also impossible to find analytically for a loop of arbitrary, irregular shape. On instead. we can use the identity in Eq. (7.24) and find the inductance L \2 we assume a current of intensity q in the toroidal coil, and no cur- the other side, To this end, rent in the loop ( i\ = the flux 0), find <Jq through a surface bounded by the loop, and use Eq. (7.23). Under these circumstances, there is no field outside the coil [see Example 4.12], so that the flux <Jq equals the negative of the flux <J> through a cross section of the toroid given by Eq. (5.84), where the minus comes from sign (negative) different reference directions of the fluxes in Figs. 7.8 and 5.26. Hence, L12 12 Finally, = ~~ combining Eqs. coil terminals in Fig. 7.8 v — = -— = h (7.31), (7.24), comes out = L \2 —7“ = MrMO Nh r 2n 1 to and b ln- -693 a (7.32), the n H. (7.32) time-harmonic voltage between the be —cdL\ 2 Iq sincvf = 9.7 sin(7 x 10 3 f) V ( (7.33) in s). at Note Z21 - transfer impedance between two circuits (unit: Q) Eo = that the amplitude of the voltage v can IZ21I/0, where Z21 be written as = 0 L 21 = cuL 12 = — — In C 2n — = —4.85 £2. (7.34) a 4 can be measured by a voltmeter connected ) by directly measuring Vo, we can indirectly measure the amplitude (or rms value) of the current (Iq) in an arbitrary conductor, without inserting an ammeter in its circuit. In such cases, the coil in Fig. 7.8 is used as a test transformer, where the current to be determined is transformed to the voltage that is indicated by the voltmeter This amplitude (or the corresponding rms value to the terminals of the coil. Thus, by means of magnetic coupling and transformer electromagnetic induction. The constant Z21 in Eq. (7.34), expressed in ohms, is said to be the mutual impedance or transfer impedance 4 Note that amplitudes most instruments show rms (root-mean-square) values of measured quantities instead of their (maximum values). u ' Section 7.2 between the two the circuits is, directions of Example i circuits = Z21 (Z 12 negative here, as the mutual inductance between ). It is because of the particular reference orientations of the contours and reference and v in Fig. 7.8. Mutual Inductance 7.9 Two Two-Wire of p.u.l. shows a cross section of a system composed of two Fig. 7.9(a) 323 Mutual Inductance The Lines infinitely long thin two-wire has conductors marked as 1 and T, and the second one as 2 and 2'. The distances between axes of the first and second conductor of the first line and each of the conductors of the second line are r\ 2 r\ 2 H'2> and H'2'> respectively. Find the mutual inductance per unit length between the two lines. lines running parallel to each other in air. first line r , , We assume that the first line (line 1-1') carries a current of intensity q Solution , Second as indicated Our goal is to find the magnetic flux $2 due to this current through a surface S bounded by the conductors of the second line (line 2-2'), and to apply Eq. (7.21) the mutual inductance between the lines. Let us first find the flux Oa through S due to the conductor 1 alone. The corresponding in Fig. 7.9(b). of length for / magnetic flux density vector, Bi magnitude is [Eq. (4.22)] circular with respect to the axis of the conductor is , B = r is the distance from the simply as the flat which ri 2 and a flat part this part of integration (note the integration over the rest of integrate Bi, however, r r 12 not convenient to adopt S more complicated shown in Fig. 7.9(c), is adopted. It consists of a cylindrical (strip) whose width is equal to r\ 2 — The flux through is zero, because Instead, a Bi is tangential to the surface (Bi analogy with the electric potential computation in S (flat part) ' J/r=r\2 it is 2'. 2 ' (b) is perpendicular to the integration surface (Bi reduces to the one in Eq. (6.64), and we have = its ' the cylindrical part of the surface The To surface spanned across the conductors 2 and surface, the cross section of part of radius axis. and (7.35) t where line (a) B t dS = is || r r 12 quite simple to perform, because Bi dS). Hence, the ' dr l J r\2 _L = , In 2n is now computation practically flux Mo ill 2nr dS) in Fig. 1.23). — n r\ (7.36) 2 (c) Similarly, the flux <t>b due conductor V alone can be found by intethrough a surface conveniently placed with respect to the current grating the corresponding magnetic field i\ in the However, instead of repeating the above computation for a new position of the conductor (now conductor 1'), we can use the final result in Eq. (7.36) with just changing to this conductor. mutual inductance per unit length of two parallel long thin two-wire infinitely the notation: <*>b = — Moil I, r V2' ln 2n r V2 lines in air: (a) cross section (7.37) • being opposite to the direction of the current in the conductor in the conductor V 1. three-dimensional view showing superposition, the total magnetic flux through the line 2-2' at the length / of the system amounts to a part of the system of length l that only considered, and ^ ^ +A = $2 = 4>a 4>b The corresponding of the system with given geometrical data, (b) where the minus sign comes from the reference direction of the current q By Figure 7.9 Evaluation of the flux per unit length (for Moill — — In H2'TL'2 ~, , 2n o\ (7.38) . r\ 2 r\' 2 > integration of the magnetic field one meter) of the system is due to the conductor = (* 2 )^. 1. = ^, obtained as (7-39) bounded by the conductors of the for Example line 2-2!) 7.9. so that the mutual inductance per unit length of the two lines for the orientations of the lines given in Fig. 7.9 is L21 — (L 2 l)p. u .i. — ^21 / _ ^2 _ q MO 2n H2'H'2 j ' r\ 2 r\' 2 1 over a conveniently chosen surface *2 is (c) (7.40) L'2] - two two-wire parallel thin lines (unit: H/m) . 324 Chapter 7 ’ Inductance and Magnetic Energy Depending on the actual mutual position of the lines, L'2l can be both positive (when > r \2 r Y2') and negative (when r^ryj < ri2 r l'2')- Th e unit for ^21 s H/m. f\2' r \'2 * This formula is of great importance in assessing (usually undesired) magnetic coupling and mutual electromagnetic induction between transmission lines power in practical applications different combinations of parallel two-wire coupling between a phone line and nearby (e.g., or between pairs of wire conductors line an electronic device). in for other types of transmission lines that can be approximated by wire It can be used also lines in some con- rough computation of mutual inductances within multiconductor microstrip and strip transmission lines that give rise to the so-called inductive “cross talk” between conductors in printed circuit boards in computers). siderations (e.g., Problems 7.10-7.15; Conceptual Questions (on Companion Website): : MATLAB 73 ANALYSIS OF MAGNETICALLY COUPLED Having now in 7.1 1-7.21; Companion Website). Exercises (on hand the concepts of both self- CIRCUITS and mutual inductance, let us assume that slowly time-varying currents exist in both contours (circuits) of Fig. 7.7 at the same time, as shown in Fig. 7.10. now caused by both principle, we have is and i\ z' 2 . The total magnetic flux Oj through the first circuit Using Eqs. (7.1) and (7.23) and the superposition Ol with L 1 = Liz't + Li2i2> being the self-inductance of the between the circuits. Similarly, — L 12 and L 2 token, the induced emf is in first circuit = Lnh + ^2*2 the self-inductance of the second circuit. each of the circuits is =- d<t>] ~ dT dz‘i = ~ L] composed of both By the same self-induction as ~ d /2 Ll2 ~dt~ d<J>2 (7.43) ~d7 d/ di\ = -T 21-77 - l 2~T72 T— d dt at ^ind2 is (7.42) > and mutual-induction terms, which can be written eindl and L 12 the mutual inductance the total magnetic flux through the second circuit $2 where L 21 (7.41) (7.44) t Finally, from Fig. 7.10 [see also Figs. 6.6 and 6.1 1(b)], voltages across the terminals of the circuits are given by current-voltage characteristic V\ = -<Mndl —+L =T —+T — =L d/l / circuits V2 = d/ ~eind2 J with slowly time-varying currents. (7.45) dr d/ 2 l , / 2 21 dr Figure 7.10 Analysis of two 12-77, dr of two magnetically coupled magnetically coupled circuits d/ 2 , , 1 dr (7.46) Analysis of Magnetically Section 7.3 In words, the voltage across the terminals of each of the circuits nation of time derivatives of currents in both with L\, circuits, is Coupled Circuits 325 a linear combi- L 2 and L \2 = L 21 , as linearity (proportionality) constants. Shown coupled in Fig. 7.11 is the circuit-theory representation of the two magnetically circuits of Fig. 7.10. This is a two-port network whose current-voltage char- given by Eqs. (7.45) and (7.46). It consists of two coupled ideal inductors, where, in addition to modeling the emf due to self-induction in each inductor, the acteristic is effect of the emf due to mutual induction between the inductors is The mutual inductance between the inductors is customarily written also modeled. as Figure 7.11 Circuit-theory representation of L 12 = ±ky/L\L2, where k is (7.47) two coupled inductors. a positive dimensionless constant called the coefficient of (magnetic) coupling of the inductors (circuits) and defined as k — 0<k<l. (7.48) VLiLi’ coefficient of magnetic coupling between two circuits (dimensionless) We k shall see in the next section that is always less than or eventually equal to unity. The coefficient k in Fig. 7.11 thus provides the information about the magnitude of the mutual inductance between the circuits of Fig. 7.10. The sign of L 12 however, depends on the adopted reference directions of currents and and therefore cannot be given as a single piece of information (positive or 2 negative) along with k, independently from the current directions. That is why we use a so-called two-dot notation to include the information about the sign of L 12 in the representation in Fig. 7.11, by placing two big dots near the particular ends of the two inductors. According to this notation (convention), if both currents (z'i and 2 ) enter the inductors at ends marked by a big dot (as in Fig. 7.11), the mutual inductance, for that particular combination of reference directions of currents, is positive (note that Z42 > 0 in Fig. 7.10). The same is true if both currents leave the inductors at marked (dotted) ends. Otherwise, if one current enters and the other leaves the inductor at marked ends, the mutual inductance is negative (note that a change of the reference direction of one of the currents in Figs. 7.10 and 7.11 would result in Z42 becoming negative). Finally, if k = 0, the two inductors in Fig. 7.11 become decoupled and independently described by the current-voltage characteristic in Eq. (7.3) for a single inductor [Eqs. (7.45) and z'i , z' , z' (7.46) with L \2 = 0]. Note that Eqs. (7.41) and (7.42) can [<*>] where [O] and [z] are be represented = in matrix form: (7.49) [L][i], column matrices whose elements are the fluxes and curand [L] is a symmetrical square matrix rents of the coupled circuits, respectively, of inductances, [L] (Ln = L\ and = L 22 = L 2 ). Note L\\ L 12 _L 2 1 L 22_ N x N matrix. number ( N ) L\ L\2 L\2 _ L2 \ (7.50) also that these equations, as well as Eqs. (7.43)- (7.46) for the electromotive forces with an arbitrary = and voltages, can be generalized to the system is an of coupled contours (circuits), in which case [L] inductance matrix ) 326 Chapter 7 Inductance and Magnetic Energy Example 7.10 Two Coupled Inductors Connected in Series and Parallel L 2 and Find the equivalent inductance of two coupled inductors of inductances L\ and coupling coefficient k if they are connected in (a) series, as in Fig. 7.12(a), and (b) parallel, as in Fig. 7.12(b). Solution (a) Voltages vj and their connection v = vj by Eqs. (7.45) and V 2 of individual inductors are given through the inductors in Fig. 7.12(a) is the same, i\ = = 12 i, (7.46). The current so that the voltage across is + V2 = — + L — + T — + Z/2— = (^1 + ^2 + 2L — dz dz L\ dz inductor dz 12 ) dr dr dz 12 \2 dr dr (7.51 . dr inductor 2 1 By comparison with the current-voltage characteristic for a single inductor, Eq. (7.3), conclude that the total (equivalent) inductance of the connection in series equals L— equivalent inductance of two coupled inductors L\ + L 2 + 2L\2, we (7.52) in series where the mutual inductance between the inductors, Fig. 7.12(a), is positive (the current enters both inductors for the situation given in ends marked by a big at their dot) and, using Eq. (7.47), amounts to L12 (b) = kyjL\L 2 (7.53) . The voltage across the inductors in Fig. 7.12(b) is the same, v\ =V 2 — v. Assuming it to be known, Eqs. (7.45) and (7.46) provide the following system of equations with d/j /dr and dz' 2 /dr as unknowns: dil L, ~di Its solution , is ~ L\2 L,2 — T|2 dr 1 Ln d/2 and dr d *l I + L1 , ~ii d, 2 ’ = dzj d ~t~ ~di expressing v in terms of dz/ + dz'2 ~di ^1 ~ ^12 ~ L L -L\ 2 2 _ and (b) parallel; for Example 7.10. in = i\ + z'2, (7.56) x T 12 dz' (7.57) dr i (a) series is i dr, x coupled inductors connected (7.55) x L + L 2 — 2L X2 L L 2 - L 2n L + L 2 — 2L\2 Figure 7.12 Evaluation of the (7.54) ' x ~ L,\ Lj 2 equivalent inductance of two V ~ii the terminals of the parallel connection of inductors dz By A and z^2 dz'i As the current through we have d h = v +Lu ^i 1 Section 7.3 we conclude that the two coupled inductors of Analysis of Magnetically Coupled 327 Circuits can be replaced by a single Fig. 7.12(b) equivalent inductor of inductance L1L2 ^12 (7.58) L\+ L 2 — 2L\ 2 The coupled inductors particular (given) placement of big dots in Fig. 7.12(b) tells us that the inductance L \2 is that for k in parallel mutual negative, and hence Eq. (7.47) yields Li 2 Note equivalent inductance of two = 0 and L\ 2 = = -kJUL 2 (7.59) . Eqs. (7.52) and (7.58) reduce to expressions for L 2 (that are 0, equivalent inductances of two ordinary inductors of inductances L\ and not magnetically coupled together) connected in series and parallel, respectively. Note also that these expressions (with L 12 = 0) have the same form as the correspond- ing expressions for two connected resistors of resistances R\ and and R2 [see Eqs. (3.86) (3.94)]. Example Compute the Example 7.7. Solution Coupling Coefficient of 7.1 coefficient of coupling Using the expression of the core due to the current i\ in Two Coils between the on coils a Toroidal on the Core thin toroidal core from Eq. (7.29) for the magnetic flux through the cross section in the first coil, the self-inductance of the first coil ‘I* single turn is pN^S (7.60) h Similarly, the self-inductance of the between the coils is given by Eq. second coil is L 2 = pN^S/l. The mutual inductance Hence, their coupling coefficient turns out to be \Ln\ i maximum coupling is a _ NjN2 (7.61 Voltage Transformation by ings shown respectively. circuit in the in Fig. 7.13(a). maximum coupling - coils a toroidal core consequence of the magnetic so that the entire flux passes through every turn of both Consider a magnetic ) y/NjNj flux due to the current of each of the coils being concentrated entirely inside the core (flux leakage from the core Example 7.12 a thin toroidal core (7.30). ^L L2 This L - coil on is negligible), coils. Two Coupled Coils form of a thin toroidal ferromagnetic core with two wind= 1000 and N2 = 500 turns of wire, The windings have N\ The ferromagnetic material can be considered to be linear. Losses in the Figure 7.13 circuit (a) Magnetic with two windings and (b) equivalent schematic diagram with two coupled inductors; for Example 7.12. on . 328 Chapter 7 2 1 2 Inductance and Magnetic Energy windings and the core can be neglected. The primary winding = voltage generator of emf e g is connected to an ideal ac V (r in s). The secondary winding is open-circuited. 100 cos 377f Find the voltage across the secondary winding. Solution Fig. 7.13(b) shows the equivalent schematic diagram with two coupled inducwhere L \ 2 is positive for the reference orientations of windings in Fig. 7.13(a). As the secondary winding is open-circuited, Eqs. (7.45) and (7.46) give tors, = v\ L\-j- and v2 = L2 Using Eqs. (7.60) and (7.30), vi coupled write L _ that is, numbers of wire turns. = vj v"2 can say that the circuit port to N\ N\N2 _ its Assume The voltage across = = eg in Fig. 7.13 50cos377r Current Transformation by equal to the hence (rins). (7.64) Two Coupled its primary Coils that the primary winding in the magnetic circuit in Fig. 7.13(a) secondary winding From Solution is the equivalent schematic diagram , g = 15 sin 377/ shown — +L — =0 d/i mA (/ is connected to in s), while the in Fig. 7.14 and Eq. (7.46), d/7 r 2 (v 2 =0). (7.65) at at means / and determine the current of the secondary winding. short-circuited, L 12 that h_ by coils N _ N\N2 L2 _ _ Z^i 2 *2 i.e., V is is operates as a transformer of voltage from an ideal ac current generator of current intensity coupled the secondary winding N2 /N\ Example 7.13 current transformation (7.63) N2 secondary port, with the transformation (multiplication) factor being equal to the wire turns ratio This N\ the ratio of the voltages across the primary and secondary windings ratio of the We ^ X L\2 V2 coils (7.62) dr we can voltage transformation by = 0). (*2 \ dr N (7.66) the ratio of the currents in the primary and secondary windings inverse ratio of the numbers of wire is equal to the negative turns. Solving for the current in the secondary winding, we have /v = — —1 *2 We /'1 N2 =— /V N2 / g = — 30 sin 377/ mA (/ins). (7.67) see that the circuit in Fig. 7.14 performs a transformation of current between ports, with the magnitude transformation factor being the reciprocal of transformation in Eq. (7.64). —— lJL + p Figure 7.14 Current 1 transformation by two magnetically coupled for Example coils; c — <N s ^ 2 7.1 3. — — II l2 ' its that for the voltage 0 . Analysis of Magnetically Section 7.3 Coupling Coefficient of Example 7.14 Fig. 7.15 Two shows a cross section of two very long solenoidal N2 of a linear ferromagnetic material of relative permeability coils is air-filled. two Neglecting the end 329 coil is /u, r , is the same, wound on /, a core with and made while the space between the effects, calculate the coefficient of coupling between the coils. Solution (Fig. 7.15). Let us adopt counter-clockwise reference directions of currents in both coils first that a current, of intensity /1 (steady current), exists only in the first Assume (inner) coil, while I2 the coils positioned coaxially their length , The inner respectively. , Circuits Coaxial Solenoids respect to each other. Radii of solenoids are a\ and a 2 numbers of wire turns are Ni and Coupled first coil and is = 0. Under this assumption, the magnetic field is nonzero only inside given by [see Eq. (6.48)] Hi = Nih (7.68) ' l Therefore, the magnetic fluxes through a single turn of both coils are the same. single turn Using 2 — MrAoNl/lTT^ — (7.69) | Figure 7.15 Cross section found to be this expression, the self-inductance of the first coil is of two coupled solenoids; for Ll <f>l = = Ni<t> single turn I1 t /J.qN^7Z = (7.70) h ~h 7 [note that this can also be obtained by multiplying the inductance in Eq. (7.6) by the same expression (for Single turn), the i21 4>2 = = mutual inductance between the ^2 'h single turn 1 7T = coils is /r r ]. Using obtained as (7.71) 7 On the other hand, the assumption of a current, of intensity I2 = , existing only in the second (outer) coil (while 7i 0) in Fig. 7.15 gives a nonzero magnetic field everywhere inside the second coil, of intensity 2 i2 n H2 = (7.72) l The magnetic flux density in the core is B 2 = ix t h,qH2 while B 2q = ij-qH2 in the air-filled space between the coils, which leads to the following for the total magnetic flux through the , second coil: 0. 2 = NAT 2 vn [B 2 na\2 + a\- a\) 2 m = MN\-Kh{^a\ 1 — + Bd 2on(a^2 - af)] j f The self-inductance of the second h2= coil is 4>2 k = (7.73) hence — - ixqNItt[(ii x 1 )a\ + a\] (7.74) 7 whereas computing <t>i through the first coil and L 12 = <4>i/72 would, of course, give the same result for the mutual inductance as in Eq. (7.71). Finally, the coefficient of coupling between the coils, Eq. (7.48), comes out to be 1^121 a/L\L 2 _ Mr I V Mr Note that for very large relative permeabilities (coupling is second [Mrfli coil very strong), which is - 1 («2/«l) /r r (e.g., /u, r = 2 (7.75) ’ 1000), k is very close to unity a consequence of the magnetic flux 4>2 due to 12 in the being practically entirely concentrated » a\ - a\ in Eq. (7.73)]. + in the core of the first coil for /j, T 3> 1 coaxial Example 7.14. h 330 Chapter 7 Inductance and Magnetic Energy Magnetically Coupled Circuits Containing Two-Wire Lines Example 7.15 and 2-2') in air, where b = 20 cm The first line is connected at one end to an ideal voltage generator of time-harmonic emf e (r) = £ ocos&>r, where £ o = 3 V g g g and co = 10 6 rad/s, and the other end of the line is short-circuited. The second line is two Fig. 7.16(a) depicts and c = mm. The 5 parallel thin two-wire lines (1-1' = 0.3 mm. radii of all wires are a short-circuited at both ends. Neglecting internal inductances, losses in the wires, capacitive coupling between the currents in the Solution Using Eq. end and propagation effects, /zH/m 1.565 In %=— n and a respectively, h M where we neglect (r\ 2 = - 1.125 /zH/m. (7.76) a ~ , = In 2 In Cz 2c and r \2 = = ryy CY it nH/m. n-7-7 -i / 277 - -7 (7.77) 71 We neglect the end effects and the propagation effects, so that the total self-inductances and mutual inductance of the magnetically coupled circuits formed by the lines are obtained by multiplying the per-unit-length inductances by the corresponding lengths of the lines, L\ = L\3b = 939 nH, L 2 = L'2 b = 225 nH, and L 12 = L'v b = 55.4 nH (the length of the shorter of the two lines, b is relevant for the mutual inductance between the lines). Fig. 7.16(b) shows the equivalent schematic diagram of the two coupled circuits (we neglect the losses in wires and capacitive coupling between the circuits). According to this , Figure 7.16 Analysis of magnetically coupled containing two thin lines: (a) = ryi ) , 2ll two-wire = ' MO MO — —r— = — r circuits In internal inductances of the lines. Eq. (7.40) gives the mutual inductance per unit length of the lines Lj ]7 = (b) amplitudes of effects, find the (7.11), self-inductances per unit length of the lines in Fig. 7.16(a) are —n — = = L\ lines, lines. diagram. structure geometry and (b) equivalent schematic diagram with two dh L + Ln '~it The di2 La £p ~it dii ~ir d +i2 h = ir 0 (7.78) . solution of these two equations for the time derivatives of currents in the circuits coupled inductors; for Example 7.1 5. L2 di\ dt which L\ L-2 ~~ % 7 dM £go cos cot and dr L\ L 12 — L,2 L\2 £go cos cot is (7.79) , then integrated with respect to time to obtain the solution for the currents. is Integration in time of time-harmonic quantities results in an additional factor 1 expressions for amplitudes, so that the amplitudes (peak-values) of the currents the circuits in /co in the (and the lines) are, respectively, = A) 1 ^ 12 co{L\L 2 Example 7.16 p.u.l. symmetrical thin two-wire line as shown is h ( The wire Solution Assume is in Fig. of a = Two-Wire Line above positioned in air 7.17(b) is a PMC Plane » (in the is cl (cl for the » a). Compute the presence of the plane). magnetic i is established in the line we can remove the ferromagcurrent conductors. The equivalent field, Fig. 5.14, images of the original thus obtained, which consists of two (magnetically coupled) parallel thin two-wire lines with the the upper (original) line (7.80) over a ferromagnetic (n /-to) plane, axes of both wires with respect that a slowly time-varying current of intensity By image theory 0.8 A. ^12) » a), and the distance between the wire axes netic plane by introducing positive system £ 12 %) = radii are a, the height of the inductance per unit length of the line [Fig. 7.17(a)]. /02 ^ A to the plane and co(L\L 2 Inductance in Fig. 7.17(a). A 3.24 is same current (/) in air. The total magnetic flux per unit length of therefore [see Eq. (7.41)] <t>' — L\i + L\ 2 i , (7.81) d 331 Magnetic Energy of Current-Carrying Conductors Section 7.4 where L) is the self-inductance p.u.l. of the upper line when isolated in free space and L'u is the mutual inductance p.u.l. of the upper and lower lines in the equivalent system in air. These inductances are computed using Eqs. (7.11) and (7.40), respectively, so that the inductance p.u.l. of the original line in Fig. 7.17(a) comes out to be L' = — =L I 1 + L'u = l ^ , In 7T d — , h In ——h 2 2 —y/d + 4 h 2 + 4h 2 (7.82) i n 2 a dy/ ii o h Mo 2 ah Mr (a) Obviously, L'12 represents the influence of the ferromagnetic material in the lower half-space on the ^°° line inductance. AO' Problems 7.16-7.24; Conceptual Questions (on Companion Website): 7.22-7.28; : MATLAB Exercises (on Companion Website). MAGNETIC ENERGY OF CURRENT-CARRYING CONDUCTORS 7.4 Every system of conducting loops with currents contains a certain amount of energy, called magnetic energy, in a manner analogous to a system of charged conducting bodies storing electric energy. In other words, current-carrying (coupled or uncoupled) inductors in an ac or dc circuit store magnetic energy, much like charged capacitors store electric energy. By the principle of conservation of energy, the magnetic energy of a system of loops (or inductors) with currents equals the work done to the system in the process of establishing these currents from zero to their final values. Thus, by simulating this process of “loading” the loops by currents, we can find general expressions for computing the magnetic energy in terms of the current intensities and associated magnetic fluxes (final values) of the loops. Let us consider first a single loop with a slowly time -varying current of intensity i. The current flow is maintained in the circuit by an ideal voltage generator of emf e g as shown in Fig. 7.18. As the magnetic flux through the loop, O, changes in time, an emf ej nd is induced in the loop, given by Faraday’s law of electromagnetic induction, Eq. (6.34). Using Kirchhoff’s voltage law, Eq. (3.119), we can write (b) Figure 7.17 Evaluation of the inductance per unit length of a two-wire line above a PMC) plane: (a) original ferromagnetic (or system and (b) equivalent free-space system with two magnetically coupled two-wire lines; for Example 7.1 6. , eg = Ri - e ind (7.83) , R is the resistance of the loop. The emf e m & (emf due to self-induction) opposes the current change in the loop (Lenz’s law), that is, it acts against the emf e g Therefore, an amount of work must be done in establishing the current in the loop to overcome this induced emf. This work is done by en external agent, with the energy transfer to the circuit being modeled by the generator (of emf e g ) in Fig. 7.18. To investigate the energy balance in the circuit during a differentially short time interval dr, we multiply both sides of Eq. (7.83) by i d t, where . eg i d t = 2 Ri d t + (-<?j n d/ dr). (7.84) By means of Eq. (3.121), the term on the left-hand side of this equation is the work done by the emf of the generator during the time interval dr. It equals the energy of external sources delivered to the circuit during that fraction of time. From Joule’s term on the right-hand side of the equation represents term in the equation (including the minus sign) is the work done against the emf ej n d in dr. Eq. (7.84) thus expresses the principle of conservation of energy for the circuit in Fig. 7.18, law, Eq. (3.77), the first Joule’s (ohmic) losses in the loop during dr. Finally, the last Figure 7.18 Wire loop of inductance R L and resistance with a slowly time-varying current of intensity i, maintained by an ideal voltage generator of emf eg . 2 . 332 Chapter 7 Inductance and Magnetic Energy work done by the generator during dr is partly converted into and partly used to overcome the induced emf in the loop. While the lost to heat, the second part is given to the emf Cj n d, i.e., to the flux O of telling us that the Joule’s losses first part is the loop. Ultimately, it represents the energy delivered to the magnetic field of the dr. Using Eq. (6.34), — e; nc dr = dO, so that the elementary work done by external sources to the magnetic field of the loop can be current i in the loop during the time i written as dWm = — ejncfidr = rdO. If the medium surrounding the loop dWm = where L is i = d<t> the inductance of the loop. of the loop, expressed in joules magnetically linear, the use of Eq. (7.1) yields is id(Li) = Lidi, (7.86) The magnetic field, on and ble of storing the received energy, (7.85) this stored energy the other hand, is Consequently, the work (J). capa- is the magnetic energy dWm in Eq. (7.86) represents an increment of the magnetic energy of the circuit in Fig. 7.18. When the current in the loop is zero, the loop has stored in the magnetic field of the loop work done value /, when its current no energy. The is is obtained by adding up all from elementary works dVTm n = o total energy equals therefore the net to the magnetic field in changing the loop current and W = f dWm = L fidi=L^z Ji= Jo From i - ur i — 0 to its final : (7.87) 2 is also the energy of a linear inductor of an arbitrary electric circuit. Employing Eq. (7.1), the equivalent expressions for the energy of an inductor are the circuit-theory standpoint, this inductance L in 11 1 energy of inductor II (unit: J ) . = d> 2 — 2L (7.88) note the complete analogy (duality) with the corresponding expressions for the electric energy (Vke ) of a capacitor Generally, for a system with energy of - 0/ 2 2 We 1 11 in Eq. (2.192). > 1) magnetically coupled loops (inductors), N (N N magnetically (7.89) coupled loops where the magnetic fluxes through the loops are given by linear relationships in Eqs. (7.41)-(7.42) or (7.49). Using these relationships, the energy of the system can be expressed only in terms of the currents and self- and mutual inductances of the loops. For example, for N = 2, the magnetic energy of two coupled loops can be written in the form magnetic energy of two coupled inductors (7.90) sum of energies of the two from each other (when L \2 = 0), because L can be both positive and negative. Note that the expression in Eq. (7.89) is entirely analogous to the expression for the electric energy of a linear multibody system in This energy can be both larger and smaller than the loops when magnetically Eq. (2.195). isolated \ Magnetic Energy of Current-Carrying Conductors Section 7.4 333 The magnetic energy can also be expressed in terms of distributions of the curJ, and the magnetic vector potential, A, throughout the volume rent density vector, of current-carrying loops. Namely, the magnetic flux <t>£ through the loop Ck of the along Ck [see Eq. (4.121)], so that system with loops equals the circulation of Eq. (7.89) becomes A N A ik dlfc (7.91) i i k can be brought inside the integral sign in the line integral does not change along the wire (currents are slowly time-varying). As J dvk, Eq. (4.10), we then have where the current because ik dU = it Wm = A • ik dv k {7.92) , with v k standing for the volume of the kth loop. Finally, the sum of N integrals over volumes of individual loops can be joined together into a single volume integral: U W n J Adv. (7.93) Since any slowly time-varying volume current distribution can be considered as consisting of an infinite number of filamentary current loops, this is a general expression for evaluation of the magnetic energy of a system of current conductors of arbitrary shapes in a linear medium. In general, the volume integration is performed over all parts of the system populated by current (v curre nt)- We again note the duality with the corresponding expression for the electric energy, namely, with the volume W e in terms of the charge density (p) and the (V) for an electrostatic system with a volume charge. integral expression in Eq. (2.196) for electric scalar potential Magnetic Energy of Two Coupled Two-Wire Lines Example 7.17 Find the time-average magnetic energy of the system of two coupled two-wire lines shown in Fig. 7.16 and described Solution From Example 7.15. Eqs. (7.79), instantaneous intensities of currents in the lines have the fol- = Iqi sin cot and 2 (0 = ~h)2 smart, where the current amplitudes /01 and are given in Eqs. (7.80). Using Eq. (7.90), the instantaneous magnetic energy contained in lowing forms: /02 in i\(t) z' the magnetic field of the two lines Wm (t) 1 , = - = - Li/qj L\i{{t) is + 1 sin 1 - , L 2 q{t) + Li2*i (0*2(0 2 art + 1 - L 2 Iq2 sin2 <*>* ~ L,\2h\hi sin 2 (7.94) where the self-inductances and mutual inductance of the coupled circuits in Fig. 7.16(b) are also found in Example 7.15. Eq. (6.95) tells us that the time-average value of the function sin 2 art is 1/2, so that the time-average magnetic energy of the system (Wm ) ave = 2 1 Q ui 0l + ~ L 2 I22 - L 12 /01 /02 = is 2.43 Hi. (7.95) ) This energy can also be obtained using the equivalent inductance, L, seen by the generator in Fig. 7.16. To find L, we rewrite the first magnetic energy volume currents y current equation of Eqs. (7.79) in the following in terms of ' 334 Chapter 7 ' Inductance and Magnetic Energy form: d/i !Al dh , \ (7.96) /dr L2 dr J Hence, L= In other words, this is = e2 925.4 nH. (7.97) the inductance of an inductor that, as far as the generator can replace the two coupled inductors coupled two-wire l\ 2 Li lines is the same as the energy of the equivalent inductor, can be used. Since the current of the equivalent inductor Wm (r) is is and Eq. (7.88) q, Ei^t), — concerned, Consequently, the energy of the two in Fig. 7.16(b). (7.98) ^ which, averaged in time, gives the same result as in Eq. (7.95). Proof that the Example 7.18 Maximum Coupling Coefficient is Unity Prove that the largest possible value for the magnitude of the mutual inductance of two magnetically coupled circuits equals the geometric mean of the self-inductances of circuits. Solution The we have inequality that to prove, IL 12 simple fact that the stored magnetic energy this, we start QE L 2 1 , it consequence of the a is field of the two circuits L\i\f(x), (7.99) with the expression for the energy stored in the magnetic Eq. (7.90) and write in < I always positive (or eventually zero). To show is in the following form: 2 12 Li f + 2^ E \ii/ L\ Wm = -Ui\ 1 2 1 i\ 2 * where = fix) ax 2 + bx + x 1, h_ and E\2 — b 2 (7.100) h From for all the fact that values of Wm > 0, we conclude that the quadratic function fix) x. This, in turn, is satisfied only the if E 2 L 2n 0 L\ L\ is minimum ^ 12 ^ 12 E\ + must be nonnegative of/, given by 1 - - L ci +I El - (7.101) nonnegative. Hence, /min —0 > which concludes our proof. or '12 We IL 12 I < v/LiL 2 — » k < 1, (7.102) realize that these inequalities can also be written in terms of the coefficient of coupling of the two circuits, defined by Eq. (7.48), as k coefficient cannot be larger than unity we noted earlier, in Section 7.3, < 1 (that the coupling but without a proof). Problems: 7.25 and 7.26; Conceptual Questions (on Companion Website): 7.29 and MATLAB Exercises (on Companion Website). 7.30; 7.5 In MAGNETIC ENERGY DENSITY analogy with the concept of electric energy density (Section 2.16), we shall now define and use the magnetic energy density to describe the actual localization and distribution of the magnetic energy of a system of current-carrying conductors, the total consider amount first of which is given by Eqs. (7.89) and (7.93). To this end, let us magnetic field of uniform intensity in a thin toroidal a simple case of a 0 Section 7.5 Magnetic Energy Density 335 ferromagnetic core with a winding that carries a slowly time-varying current of intensity i. Let the number of wire turns of the winding be N, the length of the core /, and its cross section S in area. From Eq. (5.53), the current in the winding can H in the core as i = Hl/N. On be expressed in terms of the magnetic field intensity the other hand, the magnetic flux through the winding (all of its turns) can be writ= NBS, where B is the magnetic flux density in the core. ten as 0 = s n gie turn Substituting these two expressions in Eq. (7.85) leads to the following expression for the work of external sources needed for a change d<f> in the magnetic flux of the i in the winding): winding (not counting Joule’s losses dWrM= idO = where v — SI or, the volume of the core, The net work i.e., the volume of the (7.103) domain where the changing the flux from zero to its final value equivalently, the flux density from zero to the final value B is hence magnetic O is — d(NBS) =HdBSl = HdBv, N field exists. in B f dWm = v JB = Dividing it by v, [ HdB. (7.104) Jo we get the magnetic energy density (energy per unit volume) of the core (in J/m 3 ), wm = More precisely, for —= v B \ HdB. (7.105) Jo arbitrary an arbitrary magnetic material of the core, this is the energy per volume of the material spent by external sources in establishing the field, and not the energy stored in the field per unit volume of the material. Namely, as we unit shall see later in this section, in the case of materials that exhibit hysteresis effects (see Fig. 5.21), this energy can only partially be returned by the field to the sources in H to zero, because of losses occurring during the magnetization (while establishing H and demagnetization (while reducing H of the material. These losses, appearing as heat, are a consequence of the reverse process of reducing the field intensity ) ) microscopic frictions encountered as elementary magnetic domains (see Fig. 5.2) change their size and rotate in the magnetization-demagnetization process of the known as hysteresis losses. For materials with no hysteresis losses, on the other side, Eq. (7.105) gives the energy that is contained in the field per unit volume of the core and can be obtained from it at any time in its entirety by reducing to zero the current in the coil. While using the term magnetic energy density for the integral expression in Eq. (7.105), we shall always keep in mind this distinction in its actual meaning between materials with pronounced hysteresis behavior and those for which hysteresis effects are not present or can be neglected. Although derived for a special case of a coil on a thin toroidal core, the result in Eq. (7.105) can be generalized to an arbitrary magnetic field, which can be visualized as a collection of elementary flux tubes (toroids) formed by the lines of vector B [see Fig. 4.26 and the proof of the law of conservation of magnetic flux, Eq. (4.99)]. In general, the magnetic energy of a differentially small cell of volume dv in an material and are ! i j i j arbitrary (nonuniform) magnetic field in an arbitrary (nonlinear) material is ; dWm = w m dv, (7.106) where the energy density is given by Eq. (7.105). By summing up the energies of all of the cells, that is, by integrating the energy dWm over the entire domain with the i magnetic energy medium density, (unit: J/rrr’J 336 Chapter 7 Inductance and Magnetic Energy magnetic field (volume v), we Wm = obtain the total magnetic energy of the system: dv J = id: //dfljdv. (7.107) In the case of a linear magnetic material of permeability the integral with respect to 5 in the B— /jl, /xH, so that expression for the magnetic energy density in Eq. (7.105) can easily be solved, B magnetic energy linear B — = - BH — -nH 2 5d5 = Wr density, M Jo medium 2/z Jo 1 1 2 2 , 2 (7.108) . Note that these expressions are entirely analogous to the corresponding expressions Eq. (2.199) for the electric energy density w e in a linear dielectric of permittivity 5 e. The expression for the total magnetic energy becomes in 1 2 B Hdv. l (7.109) For nonlinear magnetic materials, the integral in Eq. (7.105), in general, cannot be solved analytically (in a closed form). Having in mind a typical initial magnetization curve of a nonlinear ferromagnetic material that in H dB is (e.g., that in Fig. 5.20), proportional to the area of a thin strip of “length” A/m) and “width” d B (width measured the 5-axis at the “height” B in we note H (length measured T) positioned between the curve and with respect to the H- axis, as indicated in Fig. 7.19(a). This means that the integral in Eq. (7.105) represents the sum of areas of all such strips as the point P' with abscissa and ordinate B moves in the integration pro- H cess from the coordinate origin to its final magnetic energy density in the material magnetization curve and the 5-axis, that OPQ in Fig. 7.19(a). position (P). We conclude, thus, that the proportional to the area between the is is, to the area of the curvilinear triangle The proportionality constant (w m / area) can be expressed terms of the ratio of the magnetic H field intensity in H in A/m that corresponds to a cer- and the length (e.g., a 1-cm division along the and the similar ratio (5/length) for the 5-axis. This conclusion, of course, applies also to linear materials, which can always be considered as a special case of nonlinear ones. In a linear case, the hypotenuse of the triangle OPQ becomes straight, and the area of the triangle is computed as 5/7/2, Fig. 7.19(b), which is the same result as in Eq. (7.108). tain physical length along the axis may represent - axis a 100- A/m field intensity) In ferromagnetic materials that exhibit hysteresis effects, the function value at the point P in Fig. 7.19(a) to zero, O 5 (b) Figure 7.19 Correspondence between the magnetic energy density given by the integral in Eq. (7.105) and the area between the magnetization curve and the B-ax is for (a) nonlinear and (b) linear magnetic materials. As in 5 H B{H) is reduced from its does not go to zero, but to the remanent not only nonlinear, but also has multiple branches. Thus, if is the electrostatic case, the magnetic energy of a system of current-carrying conductors might alter- natively be viewed to reside in the system current, and not the magnetic field [see the corresponding discussion on the two energy localization viewpoints (field-based and charge-based) for an electrostatic system, in Section 2.16]. With this approach, the magnetic energy density would be evaluated as being equal to J A/2 at points where the current exists in the system [from Eq. (7.93)] and would be zero we choose netic energy localization of a system in terms of the magnetic field distribution (energy exists magwherever and whenever the As noted elsewhere. Although this viewpoint in field exists) is also “correct” and has its merit, and use the associated energy density expressions the discussion of the electric energy localization, the field-based approach in is to describe the Eq. (7.108). much better suited to modeling electromagnetic wave propagation and the associated energy flow, where both electric and magnetic fields of a wave exist (and carry energy) independent of the charge and current that produced them at previous instants of time. Section 7.5 Magnetic Energy Density given Figure 7.20 Evaluation of hysteresis losses in a ferromagnetic material as the difference between the energy given to the to the sources in field and the energy returned the process of magnetization and demagnetization of the material. magnetic flux density B = B r (see Fig. 5.21). As B also decreases, d B is negative. means that the energy d B per unit volume of the material given to the magnetic field in this process is negative, that is, the field is returning its energy to the external sources (e.g., to the generator in Fig. 7.18). However, this returned energy, H This which is is proportional to the area of the curvilinear triangle the material (curvilinear triangle the material in the process of losses), and the difference OPR. Consequently, is RPQ shown in Fig. 7.20, smaller than the energy spent by the sources in increasing the magnetic field in OPQ). The its difference in energy is lost to heat in magnetization and demagnetization (hysteresis in area in Fig. 7.20 is the area of the curvilinear triangle the energy of hysteresis losses per unit proportional to the area of the triangle volume of the material OPR. Finally, let us consider a full hysteresis cycle in the material. The density of energy spent on changing the magnetic field in this cycle is given by the integral in Eq. (7.105) with the integration being carried out all around the hysteresis loop. Dividing the loop (contour Ch) into four characteristic segments, as indicated in H dB 0 along the first line segment (H > 0 and d B > 0) and the third segment (H < 0 and dB < 0), while dB < 0 along the second line segment (H > 0 and dB < 0) and the fourth one (H < 0 and dB > 0). The areas between line segments 1 and 3 and the 5-axis are therefore a measure of the energy density given to the field at a point in the material, while the areas between line segments 2 and 4 and the 5-axis correspond to the energy density returned to the sources at the same point. The difference, the area enclosed by the hysteresis loop, Sh, represents hysteresis losses in the material. In other words, the energy density Fig. 7.21, we note that > H Figure 7.21 Evaluation of hysteresis losses in a complete magnetization-demagnetization hysteresis cycle of the material. 337 338 Chapter 7 Inductance and Magnetic Energy Figure 7.22 Cases with no hysteresis material with retracing of the in nonlinear losses: magnetization curve initial the periodic magnetization-demagnetization of the material, which results in zero material (or material approximated by loop area, and linear whose magnetization curve can be a linear function). of hysteresis losses, Wh, in one complete magnetization-demagnetization hysteresis cycle is proportional to the area Sh, wh = energy density of hysteresis = H'm H dB a Sh- <p (7.110) JCh losses (see Fig. 7.21) If the field is time-harmonic (ac field), with a frequency/, the time of one cycle of the periodic magnetization-demagnetization of the material, that the field variation, obtained by dividing the energy Thus, in the case of a uniform lost an ac within a cycle by T, i.e., field intensity in the material, time-average power of hysteresis losses for is, the time dur- r P circumscribes once the hysteresis loop, equals the period of T — 1/f. The time-average power of hysteresis losses, (Ph)ave, is ing which the point (Ph)a\e CX by multiplying we can it by/. write (7.111) / field where v is the volume of the material with losses. We now recall that hard ferromagnetic materials, having large Sh used primarily for dc applications, so that / = 0 in Eq. (7.111) (see Fig. 5.23), and hysteresis losses do not represent any problem. Soft ferromagnetic materials have small Sh, and that is why they are very suitable for ac applications. In the limiting case, if no hysteresis is present and the initial magnetization curve is retraced, as depicted = 0 in Eq. (7.111) and the periodic magnetization-demagnetization in Fig. 7.22, process is accomplished with no hysteresis losses. Finally, 5h = 0 for linear magnetic are materials as well (Fig. 7.22). In general, total losses in a ferromagnetic material in an ac field are the sum of hysteresis losses and Joule’s losses due to eddy currents, given by Eq. (6.137). We recall that the time-average power of eddy-current losses is proportional to the frequency squared [Eq. (6.152)]. Energy Distribution Example 7.19 Consider the toroidal of intensity i is coil from Example established in the magnetic energy in coil. in a 5.1 1, Under Thick Linear Toroidal Core and assume that a slowly time-varying current these circumstances, find (a) the distribution of the core and (b) the total energy of the coil. Magnetic Energy Density Section 7.5 339 Solution (a) The As distribution of energy in the core the material of the core the core is linear, (b) use Eq. (7.108). The magnetic wm . field intensity in given by Eq. (5.83), and hence is w m (r) = where described by the magnetic energy density, is we r is the - ixH = (r) < (a 8n 2 r2 r < (7. b ), 112 ) distance from the toroid axis. The energy of the coil, that the total magnetic energy stored in the core, can be is, obtained by integrating the energy density wm /xN2 i2 w m over the volume v of the core we adopt dv [Eq. (7.109)]. form of a differentially thin toroid of radius r and thickness d r, the cross section of which is the thin strip of length h and width dr shown in Fig. 5.26. The volume of this elementary toroid is thus dv = / d5, where l = 2nr (length of the toroid) and d S = h dr, so that the magnetic energy comes out to be Because a function of the coordinate r only, is -f Wr ,(r) 2nrhdr = J r=a The above /xN 2 i2 h /j.N f 4n dv result can also — dr be obtained from Eq. (7.88), as 2 2 i in the h b In-. ( 47r a Wm = 4> total */2, where 7 . 113 ) <l>total is N times the flux through the cross section of the core, found in Eq. (5.84). Energy of a Simple Nonlinear Magnetic Circuit Example 7.20 Calculate the energy spent for establishing the field in the magnetic circuit and described Solution in Example shown in Fig. 5.30 5.13. Final (established) values of the magnetic flux density and field intensity in the ferromagnetic core and the material of the core is air gap, ( B H) , and (Bo, Hq), are found nonlinear and the operating point P in Example 5.13. As the of the circuit, in Fig. 5.30(b), does segment of the idealized initial magnetization curve (which could be we must use Eq. (7.105) for the density of energy spent to change the field in the core from zero to (B, H). Having in mind Fig. 7.19(a), this density is proportional to the area of a polygon formed by the curve and the B-axis, from the coordinate origin to the point P, as shown in Fig. 7.23. This polygon, in turn, is a sum of a triangle and a trapezoid, which gives the following solution for the integral in Eq. (7.105): not belong to the described by an first initial wm = permeability), HdB = jj 1 Hk B k + \i.Hk + H)(B - B k = 272 J/m point Bk = 0.4 T and Hk = 1000 (7.1 14) , trapezoid triangle where 3 ) A/m are the magnetic flux density and field intensity at the K (“knee” point between the two segments of the curve) in Figs. 7.23 and 5.30(b). Air is a linear medium, so that the magnetic energy density in the gap can be obtained Figure 7.23 Evaluation of using Eq. (7.108), the density of energy spent w m0 = BqHq = \ Finally, the total magnetic energy of the 77 kJ/m3 (7.1 . 1 5) amounts This energy can be completely returned by the field only to if the Energy Lost in (7.116) initial magnetization curve is Magnetization and Demagnetization In a thin toroidal core of cross-sectional area field is established of intensity Hm = 1 5 = 1 cm 2 and in Fig. Example 7.20. retraced in the process of reducing the field intensities in the circuit, as in Fig. 7.22. Example 7.21 for establishing the field in the core circuit Wm = w m Sl + WmoSo/o = 290 mJ. ! B length kA/m, and then reduced to / = 20 cm, a magnetic H = 0. In this process, 5.30; for . 340 Chapter 7 Inductance and Magnetic Energy B shown the operating point described the path // a = 0.001 H/m. where the in Fig. 7.24, Find the net magnetic energy spent in the permeability initial is magnetization-demagnetization of the core. The magnetic energy spent Solution H ative in the process of reducing H = Hm is positive, while it is negThe net magnetic energy, VTm spent in the to establish to zero. , entire magnetization-demagnetization process represents hysteresis losses in the core (see Fig. 7.20). Its density is is in wm = [ JO the magnetization- HdB+ where H dB = f B m = ji a Hm = 1 2 2 Jp and Wm T, so that Hm = B m Hm \ 4 = 250 J/m 3 (7.117) , AOPR demagnetization magnetization demagnetization of a ferromagnetic core; for Fig. 7.24 R P Figure 7.24 Evaluation of the net magnetic energy spent OPR in proportional to the area of the shaded triangle given by — w m S/ = 5 mJ. Example 7.21 Time-Average Power of Hysteresis Losses Core in a Consider the core depicted in Fig. 6.13(a) and (c), and assume that a low-frequency timeharmonic current of intensity i = Iosin(2nft ) is established in the coil, where /o = 0.1 A and f —l kHz. Compute the time-average power of hysteresis losses in the core. The energy Solution density of hysteresis losses in one magnetization-demagnetization hysteresis cycle of the material parallelogram in Fig. 6.13(c). is given by Eq. (7.110), with Sh The current amplitude the peak-values of H(t) and B{t) are also the same, that B(t) is same the is Hm = now being the area of the as in Example A/m 100 and 6.10, so that Bm = 0.1 T [note not a time-harmonic function, due to the nonlinearity and hysteresis behavior of the core material]. Thus, computing the parallelogram area, wh = <t we obtain HdB = 2B m Hm = 20 J/m 3 (7.118) . Jch Based on Eq. core (7.111), the time-average power of hysteresis losses in the volume v — SI of the is CPh)ave=MS/ = 0.8W. (7.119) Evaluation of Force from Energy for an Electromagnet Example 7.23 An electromagnet consisting of an iron core in the shape of a horseshoe and a coil is sketched The in Fig. 7.25. cross-sectional area of the core the coil, the electromagnet that there are two is capable of tiny air gaps density in the core and the gaps The lifting a is 5. With a steady current established weight W (made also of iron). between the core and the weight and is fi, in Assuming that the magnetic flux find the lifting force of the electromagnet. consequence of the magnetic field in F m on the lower part of the circuit that is lifting its weight. This force can be determined from the magnetic energy of the system using the principle of conservation of energy. Let us suppose that F m moves the weight by an elementary distance dr upward. We recall the energy considerations in connection with Fig. 7.18. Here, however, the elementary work of external sources i d<t> from Eq. (7.85) is split to the change in the magnetic energy of the system dWm and the work dW/r Solution lifting force of the electromagnet the magnetic circuit in Fig. 7.25, namely, of the force F m along the displacement dr. id<t>= If the magnetic flux the current magnetic force from energy i in is it is maintained the coil is is a the magnetic force, As d\VF = Fm dr, we can dWm + dWF = dWm + Fm dx. at a constant value (4>) varied appropriately, then _ ~ d<t> in this dr <J>=const write (7.120) experiment, which means that = 0 and dwm , the above equation yields Section 7.5 341 Magnetic Energy Density Figure 7.25 Evaluation of the lifting force of an electromagnet from the magnetic energy contained the When system is the weight moved upward by is air in gaps; for Example 7.23. dx, the only change in magnetic energy of the the reduction in energy contained in the two air gaps due to their decreased length. dWm The energy change in Eq. (7.121) thus negative and corresponds to the energy con- is tained in the parts of the air gaps that are dx long and that vanish in our experiment. In other words, it equals the negative of the magnetic energy density by Eq. (7.108) with m = mo> times the change in the (7.122) MO ' v in the air gaps, given -^ dWm = -^25dx —= 2mo — dv — ' wm volume of the gaps: ' (25 is the total cross-sectional area of the gaps). electromagnet is From Eq. (7.121), the lifting force of the dWm _ B^5 dx As this a numerical example, for electromagnet can eration of free fall). lift T and 5 == 0.125 m 2 Fm of m = Fm /g « 10 tons (g = B= a weight (7.123) mo 1 , Such powerful electromagnets are used 100 kN, which means that 9.81 m/s 2 - standard accel- in cranes for lifting large pieces of iron. Magnetic Pressure Example 7.24 (a) For the electromagnet of of the lifting force result in (a), Fm on , Fig. 7.25, determine the magnetic pressure, that the surface of the iron piece that is being compare the maximal values of the magnetic and is, lifted, (b) the pressure Based on the electric pressures attained in practical situations. Solution (a) The force on the parts of the surface of the iron piece that form the air gaps with the is, on the surface of area 25 in total. The corresponding pressure is therefore given by acts core in Fig. 7.25, that Pm Fm /-» rt Zo B2 r\ (7.124) ’ 2/Xq and is called the magnetic pressure. This expression is between a ferromagnetic material (with m Mo) and any boundary surface any other nonmagnetic valid for air (or magnetic pressure (unit: Pa) 342 Chapter 7 Inductance and Magnetic Energy medium), with B that the pressure being the local magnetic flux density in pm netic side of the surface. p (b) - p0 Note near the surface. air actually equals the local magnetic energy density It acts on the We see nonmag- from the ferromagnetic material toward the medium with . that the expression in Eq. (7.124) entirely analogous to the expression in is Eq. (2.133) for the electric pressure on a metallic surface in air. These two expressions provide us now with an opportunity to compare the electric and magnetic pressures and forces. Combining them, we To estimate the get maximal magnetic and maximal electric pressure attained maximal permissible electric field intensity, £, is determined by the dielectric strength of air, Eq. (2.53), and we take therefore £ max = 3 MV/m in this estimation. On the other side, there is no such limit for B. However, maximal magnetic flux densities that are normally attained in typical magnetic circuits are on the order of one tesla and it is reasonable to assume that B ma x = 1 T for the comparison. Hence, ratio of the in practical situations, ratio of and we recall that the 2 Emax (Pm) max 1 (Pe)max e 0M0 10 000 maximal magnetic , electric pressures attained in practice We (7.126) . ( Emax conclude that practically attainable magnetic pressures are several orders of magni- tude stronger than electric ones. This is why magnetic forces, the actual workhorses of our industrial world. Practically all and not electric ones, are devices for electromechani- energy conversion, such as different types of electric motors and generators, are based on magnetic forces and their work and power. cal Problems: 7.27-7.35; Conceptual Questions (on Companion Website): 7.31-7.33; MATLAB 7.6 Exercises (on Companion Website). INTERNAL AND EXTERNAL INDUCTANCE IN TERMS OF MAGNETIC ENERGY In this section, it, now from we revisit the concept of self-inductance and techniques to compute We consider, thus, an arbitrary conductor in the energy standpoint. a linear magnetic medium. From Eq. (or steady) current of intensity inductance) of the conductor, in the magnetic due field to i, i (7.88), if we assume L can be expressed , Wm , a slowly time-varying to flow in the conductor, the inductance (selfin terms of the energy contained as L= inductance from magnetic 2 W n (7.127) energy This expression can be viewed as the third equivalent definition of self-inductance, the other two being the flux definition in Eq. (7.1) and emf definition in Eq. (7.2). It can hence be used, as an alternative general means for computing the inductance of different structures, where the magnetic energy of the structure is computed using the integral expression in Eq. (7.109). Since the energy Wm can be written as the sum of energies localized inside and outside the conductor, VFm = Wmi + Wme , (7.128) Section 7.6 L the inductance and External Inductance Internal in Terms of Magnetic Energy can be decomposed accordingly into the internal inductance, and external inductance, Le , 343 Lu of the conductor: L= + Le Li (7.129) . In other words, — j 2Wmi ^ , Lq and ,rs — 2Wme jz (7.130) , internal and external inductances where the internal and external magnetic energies of the conductor are obtained by integrating the energy density over the conductor interior (volume Vj) and exterior (volume v e ), respectively, that is, W n More precisely, v e is ml all - H dv W and B Hdv. n (7.131) only that part of the conductor exterior which the magnetic field (there In B is no need occupied by is to integrate zero energy density). of the examples of Section 7.1, we only evaluated the external inductance of conductors, by employing the flux (or emf) definition of self-inductance with only the external magnetic flux taken into account. Evaluation of Lj from the internal magnetic flux is also possible, but is physically less clear and mathematically more complicated than from the internal stored magnetic energy. At high frequencies, due to skin effect (see Fig. 6.23), the current and magnetic field in a conductor are confined to a very thin region on the surface of the conductor, which considerably reduces the high-frequency value of the internal inductance of the conductor from its low-frequency (or dc) value. Therefore, in most highfrequency applications, Lj is negligible as compared to L e and assumption that L « L e yields very accurate results. , Example 7.25 Conductor Internal Inductance p.u.l. of a Cylindrical Find the internal inductance per unit length of an infinitely long cylindrical conductor of radius a. The permeability of the conductor Assume Solution tributed over conductor i, that is its is pr. r is not pronounced. that the conductor carries a current of intensity cross section (skin effect not pronounced). < given by the expression for r that i The magnetic a in Eq. (4.56) with /iq is uniformly dis- flux density in the substituted by p and 1 by is, B{r) with Skin effect being the distance from the conductor = pir 2na 2 axis, as (7.132) ’ Figure 7.26 Evaluation of shown in Fig. 7.26. From the first sion in Eqs. (7.131), the internal magnetic energy per unit length of the conductor is expres- obtained by integrating the magnetic energy density over the cross section S of the conductor [see Eqs. (2.206) and (2.207) for the similar integration of the electric energy density], = where dS The is I ^^ the low-frequency internal inductance per unit length of an infinitely long conductor of cylindrical permeability P p ; for Example 7.25. d' = l£' f7' 1 33 > the surface area of an elementary ring of radius r and width dr (Fig. 7.26). equation of Eqs. (7.130) then gives the following expression for the lowfrequency internal inductance per unit length of the conductor: first L = , low-frequency internal 2W'. — (7.134) 8tt inductance cylindrical p.u.l. of a conductor 344 Chapter 7 Inductance and Magnetic Energy We note that this inductance = conductors, L[ Determine the Two-Wire Line low-frequency inductance per unit length of the assuming that the conductors are nonmagnetic. total (internal plus external) line in Fig. 7.4, Solution independent of the conductor radius. For nonmagnetic is 50 nH/m. Internal Inductance p.u.l. of a Thin Example 7.26 two-wire = po/(8n) is much larger than the magnetic energies of individual conductors (for an assumed lowthe line) can be evaluated independently from each other. Therefore, Since the distance between the conductors of the line conductor radii, the internal frequency current in the low-frequency internal inductance per unit length of the line can be found as twice that of a single isolated conductor, Eq. (7.134) with p = po (conductors are nonmagnetic), which yields = L[ + L[ x L[ 2 = = 2L' ^= Using the expression for the external inductance low-frequency per-unit-length inductance 100 nH/m. p.u.l. of the (7.1 35) line, Eq. (7.11), p.u.l. total (7.136) low-frequency total inductance its is of a thin two- wire line Inductance from Energy for a Coaxial Cable Example 7.27 Find the total low-frequency inductance per unit length of the coaxial cable from Example 4.11. Solution If we assume that a dc current of intensity I is Eqs. (4.61)-(4.63). established in the cable conductors, from the cable as in Fig. 4.17, the magnetic flux density at a distance r We use the energy definition of inductance, Eq. axis, B(r), is (7.127). The given by total (internal plus external) magnetic energy per unit length of the cable can be obtained by integrating the magnetic energy density over the cross section of the cable (the magnetic field outside the cable, for r > c, is zero) in the same manner as in Eq. (7.133), ^2nrdr. Jr = 0 As < the function B{r) for 0 (4.62), we break and from b to c. ” m r < c is (7.137) -^M0 given by three different expressions in Eqs. (4.61) and the integration with respect to r up from 0 to a from a to into three parts: . b, This yields Mo l 2 MO I 2 An 16tt W' . b p0 I 4n(c2 a 2 -b c ( 2 ) \c2 4 - 3c c 2 - b2 (7.138) b2 b 4 W’ tv: where W7, W'me and WU 2 are the magnetic energies residing in the inner conductor, dielectric, and outer conductor of the cable per unit of its length, respectively. The total dc (or low-frequency) inductance per unit length of the cable is hence , ,, low-frequency total inductance p.u.l. 2 , Wn 7 of a coaxial 2 Mo[l 2;r |_4 1 c* / 4 c - b 2 \c2 - c b* b 3c 2 -b 2 4 \ ) Mo, b + =-lna 2n , p (7.139) cable where cable. L'- and L'e are the internal and external inductance, respectively, per unit length of the 1 Problems The above result for L'e is, of course, the same using the flux definition of inductance. Note that the magnetic energy 345 which was obtained component of Lj, that corresponding as that in Eq. (7.12), first inductance per unit length Eq. (7.134) with \l = hq. As a numerical example, the internal and external inductances per unit length of a coaxial cable with a = 1 mm, b = 4 mm, and c = 5 amount to L[ = 66.6 nH/m and to the in the inner conductor, equals the internal of an isolated cylindrical conductor in air, mm L; = 277.3 nH/m. Problems’. 7.36-7.38; Conceptual Questions (on MATLAB Exercises (on Companion Website): 7.34; Companion Website). Problems 7.1. Induced emf and voltage of an inductor. A current of intensity i(t) = Iq e~ f/ r where Iq and r > 0 are constants, flows through an c • a — , \ i inductor of inductance L. Calculate (a) the magnetic flux, (b) the induced emf, and M2 N (c) the i j Ml voltage of the inductor. Specify the reference directions/orientations for 7.2. 1 0 Inductance of a solenoid with a two-layer core. If the solenoidal coil described in Example 7.1 is wound over a ferromagnetic core with two coaxial layers of relative perme- = abilities /x r i = 500 in Fig. 7.27, where the cross-sectional area of the inner layer tance of the is and Si /x r 2 = 1 cm 2 , 1000, as J/z/2 i quantities. all <! l i Figure 7.28 The same toroidal Fig. 7.3 coil as in but with ferromagnetic layers stacked on top of one another; for Problem shown 7.3. layers of permeability find the induc- /x and thickness a, analogously to the line with dielectri