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ELECTROMAGNETICSSTEVEN W. ELLINGSON

VOLUME 1 (BETA)

ELECTROMAGNETICS

VOLUME 1 (BETA)

Publication of this book was made possible in part by the Virginia Tech University Libraries’ Open Education Faculty Initiative Grant program: http://guides.lib.vt.edu/oer/grants

VT PublishingBlacksburg, Virginia

ELECTROMAGNETICSSTEVEN W. ELLINGSON

VOLUME 1 (BETA)

Copyright © 2018, Steven W. Ellingson

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Publication Cataloging InformationEllingson, Steven W., author Electromagnetics [Volume 1 Beta] / Steven W. Ellingson Pages cm ISBN 978-0-9979201-2-3 DOI https://doi.org/10.7294/W4WQ01ZM 1. Electromagnetism. 2. Electromagnetic theory. I. Title QC760.E445 2018 621.3

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Contents

Preface xii

1 Preliminary Concepts 1

1.1 What is Electromagnetics? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Electromagnetic Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Fundamentals of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 Guided and Unguided Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5 Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.6 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.7 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Electric and Magnetic Fields 17

2.1 What is a Field? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Electric Field Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Permittivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.4 Electric Flux Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.5 Magnetic Flux Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.6 Permeability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.7 Magnetic Field Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.8 Electromagnetic Properties of Materials . . . . . . . . . . . . . . . . . . . . . . . . . . 25

vi

CONTENTS vii

3 Transmission Lines 28

3.1 Introduction to Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.2 Types of Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.3 Transmission Lines as Two-Port Devices . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.4 Lumped-Element Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.5 Telegrapher’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.6 Wave Equation for a Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.7 Characteristic Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.8 Wave Propagation on a Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.9 Lossless and Low-Loss Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.10 Coaxial Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.11 Microstrip Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.12 Voltage Reflection Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.13 Standing Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.14 Standing Wave Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.15 Input Impedance of a Terminated Lossless Transmission Line . . . . . . . . . . . . . . 48

3.16 Input Impedance for Open- and Short-Circuit Terminations . . . . . . . . . . . . . . . 49

3.17 Applications of Open- and Short-Circuited Transmission Line Stubs . . . . . . . . . . 50

3.18 Measurement of Transmission Line Characteristics . . . . . . . . . . . . . . . . . . . . 51

3.19 Quarter-Wavelength Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3.20 Power Flow on Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.21 Impedance Matching: General Considerations . . . . . . . . . . . . . . . . . . . . . . . 57

3.22 Single-Reactance Matching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

3.23 Single-Stub Matching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

4 Vector Analysis 65

4.1 Vector Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.2 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

viii CONTENTS

4.3 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

4.4 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

4.5 Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.6 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.7 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.8 Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.9 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

4.10 The Laplacian Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

5 Electrostatics 89

5.1 Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

5.2 Electric Field Due to Point Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5.3 Charge Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

5.4 Electric Field Due to a Continuous Distribution of Charge . . . . . . . . . . . . . . . . 92

5.5 Gauss’ Law: Integral Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5.6 Electric Field Due to an Infinite Line Charge using Gauss’ Law . . . . . . . . . . . . . 96

5.7 Gauss’ Law: Differential Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

5.8 Force, Energy, and Potential Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

5.9 Independence of Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

5.10 Kirchoff’s Voltage Law for Electrostatics: Integral Form . . . . . . . . . . . . . . . . . 102

5.11 Kirchoff’s Voltage Law for Electrostatics: Differential Form . . . . . . . . . . . . . . . 103

5.12 Electric Potential Due to Point Charges . . . . . . . . . . . . . . . . . . . . . . . . . . 104

5.13 Electric Potential due to a Continuous Distribution of Charge . . . . . . . . . . . . . . 105

5.14 Electric Field as the Gradient of Potential . . . . . . . . . . . . . . . . . . . . . . . . . 106

5.15 Poisson’s and Laplace’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

5.16 Potential Field Within a Parallel Plate Capacitor . . . . . . . . . . . . . . . . . . . . . 109

5.17 Boundary Conditions on the Electric Field Intensity . . . . . . . . . . . . . . . . . . . 110

5.18 Boundary Conditions on the Electric Flux Density . . . . . . . . . . . . . . . . . . . . 112

CONTENTS ix

5.19 Charge and Electric Field for a Perfectly Conducting Region . . . . . . . . . . . . . . 114

5.20 Dielectric Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

5.21 Dielectric Breakdown . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

5.22 Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

5.23 The Thin Parallel Plate Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

5.24 Capacitance of a Coaxial Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

5.25 Electrostatic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

6 Steady Current and Conductivity 126

6.1 Convection and Conduction Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

6.2 Current Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

6.3 Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

6.4 Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

6.5 Conductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

6.6 Power Dissipation in Conducting Media . . . . . . . . . . . . . . . . . . . . . . . . . . 134

7 Magnetostatics 138

7.1 Comparison of Electrostatics and Magnetostatics . . . . . . . . . . . . . . . . . . . . . 138

7.2 Gauss’ Law for Magnetic Fields: Integral Form . . . . . . . . . . . . . . . . . . . . . . 138

7.3 Gauss’ Law for Magnetism: Differential Form . . . . . . . . . . . . . . . . . . . . . . . 140

7.4 Ampere’s Circuital Law (Magnetostatics): Integral Form . . . . . . . . . . . . . . . . . 141

7.5 Magnetic Field of an Infinitely-Long Straight Current-Bearing Wire . . . . . . . . . . 141

7.6 Magnetic Field Inside a Straight Coil . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

7.7 Magnetic Field of a Toroidal Coil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

7.8 Magnetic Field of an Infinite Current Sheet . . . . . . . . . . . . . . . . . . . . . . . . 149

7.9 Ampere’s Law (Magnetostatics): Differential Form . . . . . . . . . . . . . . . . . . . . 150

7.10 Boundary Conditions on the Magnetic Flux Density (B) . . . . . . . . . . . . . . . . . 151

7.11 Boundary Conditions on the Magnetic Field Intensity (H) . . . . . . . . . . . . . . . . 152

x CONTENTS

7.12 Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

7.13 Inductance of a Straight Coil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

7.14 Inductance of a Coaxial Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

7.15 Magnetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

7.16 Magnetic Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

8 Time-Varying Fields 165

8.1 Comparison of Static and Time-Varying Electromagnetics . . . . . . . . . . . . . . . . 165

8.2 Electromagnetic Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

8.3 Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

8.4 Induction in a Motionless Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

8.5 Transformers: Principle of Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

8.6 Transformers as Two-Port Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

8.7 The Electric Generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

8.8 The Maxwell-Faraday Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

8.9 Displacement Current and Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . 181

9 Plane Wave Propagation in Lossless Media 185

9.1 Maxwell’s Equations in Differential Phasor Form . . . . . . . . . . . . . . . . . . . . . 185

9.2 Wave Equations for Source-Free and Lossless Regions . . . . . . . . . . . . . . . . . . 186

9.3 Types of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

9.4 Uniform Plane Waves: Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

9.5 Uniform Plane Waves: Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

9.6 Wave Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

9.7 Wave Power in a Lossless Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

A Constitutive Parameters of Some Common Materials 204

A.1 Permittivity of Some Common Materials . . . . . . . . . . . . . . . . . . . . . . . . . . 204

A.2 Permeability of Some Common Materials . . . . . . . . . . . . . . . . . . . . . . . . . 205

CONTENTS xi

A.3 Conductivity of Some Common Materials . . . . . . . . . . . . . . . . . . . . . . . . . 206

B Mathematical Formulas 207

B.1 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

B.2 Vector Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

B.3 Vector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

C Physical Constants 209

Preface

About This Book

[m0146]

What this book is about: This book isintended to serve as a primary textbook for aone-semester first course in undergraduateengineering electromagnetics, including thefollowing topics: electric and magnetic fields;electromagnetic properties of materials;electromagnetic waves; and devices that operateaccording to associated electromagneticprinciples including resistors, capacitors,inductors, transformers, generators, andtransmission lines.

This book employs the “transmission lines first”approach, in which transmission lines areintroduced using a lumped-element equivalentcircuit model for a differential length oftransmission line, leading to one-dimensionalwave equations for voltage and current.1 This issufficient to address transmission line conceptsincluding characteristic impedance, inputimpedance of terminated transmission lines, andimpedance matching techniques. Attention thenturns to electrostatics, magnetostatics,time-varying fields, and waves, in that order.

Target audience: This book is intended forelectrical engineering students in the third yearof a bachelor of science degree program. It isassumed that readers are familiar with thefundamentals of electric circuits and linearsystems, which are normally taught in the secondyear of the degree program. It is also assumed

1Are you an instructor who is not a fan of the “trans-mission lines first” approach? Then see “What are those

little numbers in square brackets?” later in this sec-tion.

that readers have received training in basicengineering mathematics, including complexnumbers, trigonometry, vectors, partialdifferential equations, and multivariate calculus.Review of the relevant principles is provided atvarious points in the text. In a few cases (e.g.,phasors, vectors) this review consists of aseparate stand-alone section.

Notation, examples, and highlights.Section 1.7 summarizes the mathematicalnotation used in this book. Examples are setapart from the main text as follows:

Example 0.1. This is an example.

“Highlight boxes” are used to identify key ideasas follows:

This is a key idea.

What are those little numbers in squarebrackets? This book is a product of theOpen Electromagnetics Project. This projectprovides a large number of sections (“modules”)which are assembled (“remixed”) to create newand different versions of the book. The text“[m0146]” that you see at the beginning of thissection uniquely identifies the module within thelarger set of modules provided by the project.This identification is provided because differentremixes of this book exist, each consisting of adifferent subset and arrangement of thesemodules. Prospective authors can use thisidentification as an aid in creating their ownremixes.

Why do some sections of this book seemto repeat material presented in previoussections? In some remixes of this book, authors

Electromagnetics Vol 1 (Beta). c© 2018 S.W. Ellingson. CC BY SA 4.0 https://doi.org/10.7294/W4WQ01ZM

https://doi.org/10.7294/W4WQ01ZM

xiii

might choose to eliminate or reorder modules.For this reason, the modules are written to“stand alone” as much as possible. As a result,there may be some redundancy between sectionsthat would not be present in a traditional(non-remixable) textbook. While this may seemawkward to some at first, there are clear benefits:In particular, it never hurts to review relevantpast material before tackling a new concept.And, since the electronic version of this book isbeing offered at no cost, there is not much gainedby eliminating this useful redundancy.

Why cite Wikipedia pages as additionalreading? Many modules cite Wikipedia entriesas sources of additional information. Wikipediarepresents both the best and worst that theinternet has to offer. Most authors of traditionaltextbooks would agree that citing Wikipediapages as primary sources is a bad idea, since peerreview is crowd-sourced and content is subject tochange over time. On the other hand, manyWikipedia pages are excellent, and serve asuseful sources of relevant information that is notstrictly within the scope of the curriculum.Furthermore, students benefit from seeing thesame material presented differently, in a broadercontext, and with additional references cited byWikipedia pages. We trust instructors andstudents to realize the potential pitfalls of thistype of resource, and to be alert for problems.

Acknowledgments: Here’s a list of talentedand helpful people who have contributed to thisbook:

The staff of VT Publishing, University Libraries,Virginia Tech:Managing Editor: Anita WalzAdvisors: Peter Potter, Corinne GuimontCover: Robert Browder, Anita Walz

Virginia Tech students:Figure designer: Michaela GoldammerFigure designer: Kruthika KikkeriFigure designer: Youmin Qin

External Reviewers:Samir El-Ghazaly, University of ArkansasStephen Gedney, University of Colorado Denver

Randy Haupt, Colorado School of MinesKarl Warnick, Brigham Young University

About the OpenElectromagnetics Project

[m0148]

The Open Electromagnetics Project wasestablished at Virginia Tech in 2017 with thegoal of creating no-cost openly-licensedtextbooks for courses in undergraduateengineering electromagnetics. While a number ofvery fine traditional textbooks are available onthis topic, we feel that it has becomeunreasonable to insist that students payhundreds of dollars per book when high-qualityalternatives can be provided using modern mediaat little or no cost to the student. This project isequally motivated by the desire for the freedomto adopt, modify, and improve educationalresources. This work is distributed under aCreative Commons BY SA license which allows –and we hope encourages – others to adopt,modify, improve, and expand the scope our work;subsequently “returning the favor.”

About the Author

[m0153]

Steven W. Ellingson ([email protected]) is anAssociate Professor at Virginia Tech inBlacksburg, Virginia in the United States. Hereceived PhD and MS degrees in ElectricalEngineering from the Ohio State University anda BS in Electrical & Computer Engineering fromClarkson University. He was employed by the USArmy, Booz-Allen & Hamilton, Raytheon, andthe Ohio State University ElectroScienceLaboratory before joining the faculty of VirginiaTech, where he teaches courses inelectromagnetics, radio frequency systems,wireless communications, and signal processing.His research includes topics in wireless

mailto:[email protected]

xiv PREFACE

communications, radio science, and radiofrequency instrumentation. Professor Ellingsonserves as a consultant to industry andgovernment and is the author of Radio SystemsEngineering (Cambridge University Press, 2016).

Chapter 1

Preliminary Concepts

1.1 What isElectromagnetics?

[m0037]

The topic of this book is applied engineeringelectromagnetics. This topic is often described as“the theory of electromagnetic fields and waves”,which is both true and misleading. The truth isthat electric fields, magnetic fields, their sources,waves, and the behavior these waves are alltopics covered by this book. The misleading partis that our principal aim shall be to close the gapbetween basic electrical circuit theory and themore general theory that is required to addresscertain topics that are of broad and commoninterest in the field of electrical engineering. (Fora preview of topics where these techniques arerequired, see the list at the end of this section.)

In basic electrical circuit theory, the behavior ofdevices and systems is abstracted in such a waythat the underlying electromagnetic principles donot need to be considered. Every student ofelectrical engineering encounters this, and isgrateful since this greatly simplifies analysis anddesign. For example, a resistor is commonlydefined as a device which exhibits a particularvoltage V = IR in response to a current I, andthe resistor is therefore completely described bythe value R. This is an example of a “lumpedelement” abstraction of an electrical device.Much can be accomplished knowing nothing elseabout resistors; no particular knowledge of thephysical concepts of electrical potential,

conduction current, or resistance is required.However this simplification makes it impossibleto answer some frequently-encountered questions.Here are just a few:

• What determines R? How does one goabout designing a resistor to have aparticular resistance?

• Practical resistors are rated forpower-handling capability; e.g., discreteresistors are frequently identified as“1/8-W”, “1/4-W”, and so on. How doesone determine this and how can this beadjusted in the design?

• Practical resistors exhibit significantreactance as well as resistance. Why? Howis this determined? What can be done tomitigate this?

• Most things which are not resistors alsoexhibit significant resistance and reactance;for example, electrical pins andinterconnects. Why? How is thisdetermined? What can be done to mitigatethis?

The answers to the these questions must involveproperties of materials and the geometry in whichthose materials are arranged. These are preciselythe things that disappear in lumped elementdevice models, so it is not surprising that suchmodels leave us in the dark on these issues. Itshould also be apparent that what is true for theresistor is also going to be true for other devicesof practical interest, including capacitors (and



2 CHAPTER 1. PRELIMINARY CONCEPTS

devices unintentionally exhibiting capacitance),inductors (and devices unintentionally exhibitinginductance), transformers (and devicesunintentionally exhibiting mutual impedance),and so on. From this perspective,electromagnetics may be viewed as ageneralization of electrical circuit theory thataddresses these considerations. Conversely basicelectric circuit theory may be viewed a specialcase of electromagnetic theory which applieswhen these considerations are not important.Many instances of this “electromagnetics asgeneralization” vs. “lumped-element theory asspecial case” dichotomy appear in the study ofelectromagnetics.

There is more to the topic, however. There aremany devices and applications in whichelectromagnetic fields and waves are primaryengineering considerations that must be dealtwith directly. Examples include persistentstorage of data (e.g., hard drives), electricalgenerators and motors, antennas, printed circuitboard stackup and layout, fiber optics, andsystems for radio, radar, remote sensing, andmedical imaging. Considerations such as signalintegrity and electromagnetic compatibility(EMC) similarly require explicit consideration ofelectromagnetic principles.

Although electromagnetic considerations pertainto all frequencies, these considerations becomeincreasingly difficult to avoid with increasingfrequency. This is because the wavelength of anelectromagnetic field decreases with increasingfrequency.1 When wavelength is large comparedto the size of the region of interest (e.g., acircuit), then analysis and design is not muchdifferent from zero-frequency (“DC”) analysisand design.

For example, the free space wavelength at 3 MHzis about 100 m, so a planar circuit havingdimensions 10 cm × 10 cm is just 0.1% of awavelength across at this frequency. Although anelectromagnetic wave may be present, it hasabout the same value over the region of space

1Most readers have encountered the concepts of fre-quency and wavelength previously, but if needed see Sec-tion 1.3 for a quick primer.

occupied by the circuit. In contrast, the freespace wavelength at 3 GHz is about 10 cm, sothe same circuit is one full wavelength across atthis frequency. In this case, different parts of thiscircuit observe the same signal with verydifferent magnitude and phase.

Some of the behaviors associated withnon-negligible dimensions are undesirable,especially if not taken into account in the designprocess. However these behaviors can also beexploited to do some amazing and useful things –for example, to launch an electromagnetic wave(i.e., to create an antenna), or to create filtersand impedance matching devices consisting onlyof metallic shapes, free of discrete capacitors orinductors.

Electromagnetic considerations become not onlyunavoidable but central to analysis and designabove a few hundred MHz, and especially in themillimeter-wave, infrared (IR), optical, andultraviolet (UV) bands.2 The discipline ofelectrical engineering encompasses applications inthese frequency ranges even though – ironically –such applications may not operate according toprinciples that can be considered “electrical”!Fortunately, electromagnetic theory applies.

Another common way to answer the question“What is electromagnetics?” is to identify thetopics that are commonly addressed within thisdiscipline. Here’s a list of topics – some of whichhave already been mentioned – in which explicitconsideration of electromagnetic principles iseither important or essential:3

• Antennas

• Coaxial cable

• Design and characterization of commondiscrete passive components includingresistors, capacitors, and inductors

• Distributed filters

2See Section 1.2 for a quick primer on the electromag-netic spectrum and this terminology.

3Presented in alphabetical order so as to avoid the ap-pearance of any bias on the part of the author!

1.2. ELECTROMAGNETIC SPECTRUM 3

• Electromagnetic compatibility (EMC)

• Fiber optics

• Generators

• Magnetic resonance imaging (MRI)

• Magnetic storage (of data)

• Microstrip transmission lines

• Modeling of non-ideal behaviors of discretecomponents

• Motors

• Non-contact sensors

• Photonics

• Printed circuit board stackup & layout

• Radar

• Radio wave propagation

• Radio frequency electronics

• Signal integrity

• Transformers

• Waveguides

In summary:

Applied engineering electromagnetics is thestudy of those aspects of electrical engineer-ing in situations in which the electromagneticproperties of materials and the geometry inwhich those materials are arranged is impor-tant. This requires an understanding of elec-tromagnetic fields and waves, which are ofprimary interest in some applications.

Finally, here are two broadly-defined learningobjectives that should now be apparent: (1)Learn the techniques of engineering analysis anddesign that apply when electromagneticprinciples are an important consideration, and(2) Better understand the physics underlying theoperation of electrical devices and systems, sothat when issues associated with these physicalprinciples emerge one is prepared to recognizeand grapple with them.

1.2 ElectromagneticSpectrum

[m0075]

Electromagnetic fields exist at frequencies fromDC (0 Hz) to at least 1020 Hz – that’s at least 20orders of magnitude!

At DC, electromagnetics consists of two distinctdisciplines: electrostatics, concerned with electricfields; and magnetostatics, concerned withmagnetic fields.

At higher frequencies, electric and magneticfields interact to form propagating waves. Waveshaving frequencies within certain ranges aregiven names based on how they manifest asphysical phenomena; these names are (in order ofincreasing frequency): radio, infrared (IR),optical (also known as “light”), ultraviolet (UV),X-rays, and gamma-rays (γ-rays); see Table 1.1and Figure 1.1.

The term electromagnetic spectrum refers tothe various forms of electromagnetic phenom-ena that exist over the continuum of frequen-cies.

The speed (properly known as “phase velocity”)at which electromagnetic fields propagate in freespace is given the symbol c, and has the value∼= 3.00× 108 m/s. This value is often referred toas the “speed of light”. While it is certainly thespeed of light in free space, it is also speed of anyelectromagnetic wave in free space. Givenfrequency f , wavelength is given by theexpression

λ =c

fin free space

Table 1.1 shows the free space wavelengthsassociated with each of the regions of theelectromagnetic spectrum.

This book presents a version of electromagnetictheory which is based on classical physics. Thisapproach works well for most practical problems.However at very high frequencies wavelengths


Regime Frequency Range Wavelength Rangeγ-Ray > 3× 1019 Hz < 0.01 nmX-Ray 3× 1016 Hz – 3× 1019 Hz 10–0.01 nmUltraviolet (UV) 2.5× 1015 – 3× 1016 Hz 120–10 nmOptical 4.3× 1014 – 2.5× 1015 Hz 700–120 nmInfrared (IR) 300 GHz – 4.3× 1014 Hz 1 mm – 700 nmRadio 3 kHz – 300 GHz 100 km – 1 mm

Table 1.1: The electromagnetic spectrum. Note that the indicated ranges are arbitrary, but consistentwith common usage.

become small enough that quantum effects maybe important. This is usually the case in theX-ray band and above. In some applicationsthese effects become important at frequencies aslow as the optical, IR, or radio bands. (A primeexample is the photoelectric effect ; see“Additional References”, below.) Thus, cautionis required when applying the classical version ofelectromagnetic theory presented here, especiallyat these higher frequencies.

Theory presented in this book is applicableto DC, radio, IR, and optical waves; and toa lesser extent to UV waves, X-rays, and γ-rays. Certain phenomena in these frequencyranges – in particular quantum mechanicaleffects – are not addressed in this book.

The radio portion of the electromagneticspectrum alone spans 12 orders of magnitude infrequency (and wavelength), and so, notsurprisingly, exhibits a broad range ofphenomena. This is shown in Figure 1.1. For thisreason the radio spectrum is further subdividedinto bands as shown in Table 1.2. Also shown inTable 1.2 are commonly-used band identificationacronyms and some typical applications.

Similarly the optical band is partitioned into thefamiliar “rainbow” of red through violet, asshown in Figure 1.1 and Table 1.3. Otherportions of the spectrum are sometimes similarlysubdivided in certain applications.

Additional Reading:

• “Electromagnetic spectrum” on Wikipedia.

400 nm

500 nm

600 nm

700 nm

1000 m

100 m

10 m

1 m

10 cm

1 cm

1 mm1000 µm

100 µm

10 µm

1 µm1000 nm

100 nm

10 nm

0.1 Å

0.1 nm1Å

1 nm

Wavelength

1017

1016

1015

1014

1013

1012

1011

1010

109

108

106

Frequency (Hz)

1018

1019

UHF

VHF7-13

FM

VHF2-6

1000 MHz

500 MHz

100 MHz

50 MHz

Gamma-rays

X-rays

Ultraviolet

VisibleNear IR

Infra-red

Thermal IR

Far IR

Microwaves

Radar

Radio, TV

AM

Long-waves

107

c© V. Blacus CC BY-SA 3.0

Figure 1.1: Electromagnetic Spectrum.

https://en.wikipedia.org/wiki/Electromagnetic_spectrum

https://creativecommons.org/licenses/by-sa/3.0/

1.3. FUNDAMENTALS OF WAVES 5

Band Frequencies Wavelengths Typical ApplicationsEHF 30-300 GHz 10–1 mm WLAN (60 GHz), Data LinksSHF 3–30 GHz 10–1 cm Terrestrial & Satellite Data Links, RadarUHF 300–3000 MHz 1–0.1 m TV Broadcasting, Cellular, WLANVHF 30–300 MHz 10–1 m FM & TV Broadcasting, LMRHF 3–30 MHz 100–10 m Global terrestrial comm., CB RadioMF 300–3000 kHz 1000–100 m AM BroadcastingLF 30–300 kHz 10–1 km Navigation, RFIDVLF 3–30 kHz 100–10 km Navigation

Table 1.2: The radio portion of the electromagnetic spectrum, according to a common scheme fornaming ranges of radio frequencies. WLAN: Wireless local area network, LMR: Land mobile radio,RFID: Radio frequency identification.

Band Frequencies WavelengthsViolet 668–789 THz 450–380 nmBlue 606–668 THz 495–450 nmGreen 526–606 THz 570–495 nmYellow 508–526 THz 590–570 nmOrange 484–508 THz 620–590 nmRed 400–484 THz 750–620 nm

Table 1.3: The optical portion of the electromag-netic spectrum.

• “Photoelectric effect” on Wikipedia.

1.3 Fundamentals of Waves

[m0074]

In this section we formally introduce the conceptof a wave and explain some basic characteristics.

To begin, let us consider not electromagneticwaves, but rather sound waves. To be clear,sound waves and electromagnetic waves arecompletely distinct phenomena: Sound waves arevariations in pressure, whereas electromagneticwaves are variations in electric and magneticfields. However the mathematics that governsound waves and electromagnetic waves are verysimilar, so the analogy provides useful insight.Furthermore sound waves are intuitive for mostpeople because they are readily observed. So,

here we go:

Imagine standing in an open field and that it iscompletely quiet. In this case the air pressureeverywhere is about 101 kPa (101,000 N/m2) atsea level, and we refer to this as the quiescent airpressure. Sound can be described as thedifferential air pressure p(x, y, z, t), which wedefine as the absolute air pressure at the spatialcoordinates (x, y, z) minus the quiescent airpressure. So, when there is no sound, p = 0. Thefunction p as an example of a scalar field.4

Let’s also say you are standing at x = y = z = 0and you have brought along a friend who isstanding at x = d; i.e., a distance d from youalong the x axis. Also, for simplicity, let usconsider only what is happening along the x axis;i.e., p(x, t).

At t = 0, you clap your hands once. This forcesthe air between your hands to press outward,creating a region of increased pressure (i.e.,p > 0) that travels outward. As the region ofincreased pressure moves outward, it leavesbehind a region of low pressure where p < 0. Airmolecules immediately move toward this regionof lower pressure, and so the air pressure close toyour hands quickly returns to the quiescentvalue, p = 0. This traveling disturbance in p(x, t)is the sound of the clap. The disturbancecontinues to travel outward until it reaches your

4Although it’s not important in this section, you canread about the concept of a “field” in Section 2.1.

https://en.wikipedia.org/wiki/Photoelectric_effect


p(x,t)

x

p(x,t)

x

c© Y. Qin CC BY 4.0

Figure 1.2: The differential pressure p(x, t) (top)a short time after the clap and (bottom) a slightlylonger time after the clap.

friend, who then hears the clap.

At each point in time, you can make a plot ofp(x, t) versus x for the current value of t. Attimes t < 0, we have simply p(x, t) = 0. A shorttime after t = 0, the peak pressure is located atslightly to the right of x = 0. This is shown inFigure 1.2. The pressure is not a simple impulse,because interactions between air moleculesconstrain the pressure to be continuous overspace. So instead we see a rounded pulserepresenting the rapid build-up and similarlyrapid decline in air pressure. A short time laterp(x, t) looks very similar, except the pulse is nowlocated further away.

Now: What precisely is p(x, t)? Completelyskipping over the derivation, the answer is thatp(x, t) is the solution to the acoustic waveequation (see “Additional References” at the endof this section):

∂2p

∂x2− 1

c2s

∂2p

∂t2= 0 (1.1)

where cs is the speed of sound, which is about340 m/s at sea level. Just to emphasize thequality of the analogy between sound waves andelectromagnetic waves, know that the acoustic

wave equation is mathematically identical to theequation that governs electromagnetic waves.

Although “transient” phenomena – analogous toa clap – are of interest in electromagnetics, aneven more common case of interest is the waveresulting from a sinusoidally-varying source. Wecan demonstrate this kind of wave in the contextof sound as well. Here we go:

In the previous scenario, you pick up a trumpetand blow a perfect A note. The A note is 440 Hz,meaning that the air pressure emerging from yourtrumpet is varying sinusoidally at a frequency of440 Hz. Let’s say you can continue to blow thisnote long enough for the entire field to be filledwith the sound of your trumpet. Now what doesthe pressure-versus-distance curve look like? Twosimple observations will settle that question:

• p(x, t) at any constant position x is going tobe a sinusoid as a function of x. This isbecause the acoustic wave equation is linearand time invariant, so a sinusoidal excitation(i.e., your trumpet) results in a sinusoidalresponse at the same frequency (i.e., thesound heard by your friend).

• p(x, t) at any constant time t is also going tobe a sinusoid as a function of x. This isbecause the sound is propagating away fromthe trumpet and toward your friend, andanyone in between will also hear the A note,but with a phase shift determined by thedifference in distances.

This is enough information to know that thesolution must have the form:

p(x, t) = Am cos (ωt− βx+ ψ) (1.2)

where ω = 2πf , f = 440 Hz, and Am, β, and ψremain to be determined.

You can readily verify that Equation 1.2 satisfiesthe acoustic wave equation when

β =ω

cs(1.3)

In this problem we find β ∼= 8.13 rad/m. Thismeans that at any given time, the difference in

https://creativecommons.org/licenses/by/4.0/

1.3. FUNDAMENTALS OF WAVES 7

phase measured between any two pointsseparated by a distance of 1 m is 8.13 rad. Theparameter β goes by at least three names: phasepropagation constant, wavenumber, and spatialfrequency. The term “spatial frequency” isparticularly apt, since β plays precisely the samerole for distance (x) as ω plays for time (t) –This is apparent from Equation 1.2. However,“wavenumber” is probably the morecommonly-used term.

The wavenumber β (rad/m) is the rate atwhich the phase of a sinusoidal wave pro-gresses with distance.

Note that Am and ψ are not determined by thewave equation, but instead are properties of thesource. That is, Am is determined by how hardwe blow, and ψ is determined by the time atwhich we began to blow and the location of thetrumpet. For simplicity, let us assume that webegin to blow at time t≪ 0; i.e., in the distantpast so that the sound pressure field has achievedsteady state by t = 0, and that ψ = 0. Also let usset Am = 1 in whatever units we choose toexpress p(x, t). We then have:

p(x, t) = cos (ωt− βx) (1.4)

Now we have everything we need to make plotsof p(x) at various times.

Figure 1.3(a) shows p(x, t = 0). As expected,p(x, t = 0) is periodic in x. The associated periodis referred to as the wavelength λ. Since λ is thedistance required for the phase of the wave toprogress by 2π rad, and because phase isprogressing at a rate of β rad/m, it must be that:

λ =2π

β(1.5)

In the present example, we find λ ∼= 77.3 cm.

Wavelength λ = 2π/β is the distance requiredfor the phase of a sinusoidal wave to progressby one complete cycle (i.e., 2π rad) at anygiven time.

Now let us consider the situation at t = +1/4f ,which is t = 568 µs or ωt = π/2. We see in

p(x,t)

x

λ

p(x,t)

x

p(x,t)

x


Figure 1.3: The differential pressure p(x, t) for(a) t = 0, (b) t = 1/4f for “−βx”, as indicatedin Equation 1.4 (wave traveling to right); and (c)t = 1/4f for “+βx” (wave traveling to left).



Figure 1.3(b) that the waveform has shifted adistance λ/4 to the right. It is in this sense thatwe say the wave is propagating in the +xdirection. Furthermore, we can now compute aphase velocity vp: We see that a point of constantphase has shifted a distance λ/4 in time 1/4f , so

vp = λf (1.6)

In the present example, we find vp ∼= 340 m/s;i.e., we have found that the phase velocity isequal to the speed of sound cs. It is in this sensethat we say that the phase velocity is the speedat which the wave propagates.5

Phase velocity vp = λf (m/s) is the speedat which a point of constant phase in a sinu-soidal waveform travels.

Recall that earlier we arbitrarily made thecontribution of βx to the phase negative (seeEquation 1.2). What’s the deal with that? Tosort this out, let’s change the sign of βx and seeif it still satisfies the wave equation – one findsthat it does. Next we repeat the previousexperiment and see what happens. The result isshown in Figure 1.3(c). Note that points ofconstant phase have traveled an equal distance,but now in the −x direction. In other words, thisalternative choice of sign for βx within theargument of the cosine function represents a wavewhich is propagating in the opposite direction.This leads us to the following realization:

If the phase of the wave is decreasing withβx, then the wave is propagating in the +xdirection. If the phase of the wave is increas-ing with βx, then the wave is propagating inthe −x direction.

Since the prospect of sound traveling toward thetrumpet is clearly nonsense in the presentsituation, we may neglect that possibility.However, what happens if there is a wall locatedin the distance, behind your friend? Then we

5It is worth noting here is that “velocity” is technicallya vector; i.e., speed in a given direction. Nevertheless, thisquantity is actually just a speed, and this particular abuseof terminology is generally accepted.

expect an echo from the wall, which would be asecond wave propagating in the reverse direction,and for which the argument of the cosinefunction would contain the term “+βx”.

Finally, let us return to electromagnetics.Electromagnetic waves satisfy precisely the samewave equation (i.e., Equation 1.1) as do soundwaves, except that the phase velocity is muchgreater. Interestingly, though, the frequencies ofelectromagnetic waves are also much greater thanthose of sound waves, so we can end up withwavelengths having similar orders of magnitude.In particular, an electromagnetic wave withλ = 77.3 cm (the wavelength of the “A” note inthe preceding example) lies in the radio portionof the electromagnetic spectrum.

An important difference between sound andelectromagnetic waves is that electromagneticwaves are vectors; that is, they have direction aswell as magnitude. Furthermore, we often needto consider multiple electromagnetic vector waves(in particular, both the electric field and themagnetic field) in order to completelyunderstand the situation. Nevertheless theconcepts of wavenumber, wavelength, phasevelocity, and direction of propagation apply inprecisely the same manner to electromagneticwaves as they do to sound waves.

Additional Reading:

• “Wave Equation” on Wikipedia.

• “Acoustic Wave Equation” on Wikipedia.

1.4 Guided and UnguidedWaves

[m0040]

Broadly speaking, waves may be either guided orunguided.

Unguided waves include those which are radiatedby antennas as well as those which areunintentionally radiated. Once initiated, these

https://en.wikipedia.org/wiki/Wave_equation

https://en.wikipedia.org/wiki/Acoustic_wave_equation

1.5. PHASORS 9

waves propagate in an uncontrolled manner untilthey are redirected by scattering or dissipated bylosses associated with the materials.

Examples of guided waves are those that existwithin structures such as transmission lines,waveguides, and optical fibers. We refer to theseas guided because they are constrained to followthe path defined by the structure.

Antennas and unintentional radiators emitunguided waves. Transmission lines, waveg-uides, and optical fibers carry guided waves.

1.5 Phasors

[m0033]

In many areas of engineering, signals arewell-modeled as sinusoids. Also, devices thatprocess these signals are often well-modeled aslinear time-invariant (LTI) systems. Theresponse of an LTI system to any linearcombination of sinusoids is another linearcombination of sinusoids having the samefrequencies.6 In other words, (1) sinusoidalsignals processed by LTI systems remainsinusoids and are not somehow transformed intosquare waves or some other waveform; and (2) wemay calculate the response of the system for onesinusoid at a time, and then add the results tofind the response of the system when multiplesinusoids are applied simultaneously. Thisproperty of LTI systems is known assuperposition.

The analysis of systems that process sinusoidalwaveforms is greatly simplified when thesinusoids are represented as phasors. Here is thekey idea:

A phasor is a complex-valued number thatrepresents a real-valued sinusoidal waveform.Specifically, a phasor has the magnitude andphase of the sinusoid it represents.

6 A “linear combination” of functions fi(t) where i =1, 2, 3, ... is

∑i aifi(t) where the ai’s are constants.

Amej

mag

nitu

de

ImC

ReCAm

phase ()

-jAm

Am-jBm

Bm

Figure 1.4: Examples of phasors, displayed hereas points in the real-imaginary plane.

Figures 1.4 and 1.5 show some examples ofphasors and the associated sinusoids.

It is important to note that a phasor by itself isnot the signal. A phasor is merely mathematicalshorthand for the actual, real-valued physicalsignal.

Here is a completely general form for a physical(hence, real-valued) quantity varying sinusoidallywith angular frequency ω = 2πf :

A(t;ω) = Am(ω) cos (ωt+ ψ(ω)) (1.7)

where Am(ω) is magnitude at the specifiedfrequency, ψ(ω) is phase at the specifiedfrequency, and t is time. Also, we require∂Am/∂t = 0; that is, that the time variation ofA(t) is completely represented by the cosinefunction alone. Now we can equivalently expressA(t;ω) as a phasor C(ω):

C(ω) = Am(ω)ejψ(ω) (1.8)

To convert this phasor back to the physical signalit represents, we (1) restore the time dependenceby multiplying by ejωt, and then (2) take the realpart of the result. In mathematical notation:


t

t

t

Am

Amejψ

Am-jBm

t

-jAm

note slightly larger amplitude

note now a sine function

Figure 1.5: Sinusoids corresponding to the phasorsshown in Figure 1.4.

A(t;ω) = ReC(ω)ejωt

(1.9)

To see why this works, simply substitute theright hand side of Equation 1.8 intoEquation 1.9. Then:

A(t) = ReAm(ω)ejψ(ω)ejωt

= ReAm(ω)ej(ωt+ψ(ω))

= Re Am(ω) [cos (ωt+ ψ(ω))

+j sin (ωt+ ψ(ω))]= Am(ω) cos (ωt+ ψ(ω))

as expected.

It is common to write Equation 1.8 as follows,dropping the explicit indication of frequencydependence:

C = Amejψ (1.10)

This does not normally cause any confusion sincethe definition of a phasor requires that values ofC and ψ are those that apply at whateverfrequency is represented by the suppressedsinusoidal dependence ejωt.

Table 1.4 shows mathematical representations ofthe same phasors demonstrated in Figure 1.4(and their associated sinusoidal waveforms inFigure 1.5). It is a good exercise is to confirmeach row in the table, transforming from left toright and vice-versa.

It is not necessary to use a phasor to represent asinusoidal signal. We choose to do so becausephasor representation leads to dramaticsimplifications. For example:

• Calculation of the peak value from datarepresenting A(t;ω) requires a time-domainsearch over one period of the sinusoid.However if you know C, the peak value ofA(t) is simply |C|, and no search is required.

• Calculation of ψ from data representingA(t;ω) requires correlation (esentially,integration) over one period of the sinusoid.However if you know C, then ψ is simply thephase of C, and no integration is required.

1.5. PHASORS 11

A(t) CAm cos (ωt) AmAm cos (ωt+ ψ) Ame

jψ

Am sin (ωt) = Am cos(ωt− π

2

)−jAm

Am cos (ωt) +Bm sin (ωt) = Am cos (ωt) +Bm cos(ωt− π

2

)Am − jBm

Table 1.4: Some examples of physical (real-valued) sinusoidal signals and the corresponding phasors.Am and Bm are real-valued and constant with respect to t.

These are just two examples. In fact,mathematical operations applied to A(t;ω) canbe equivalently performed as operations on C,and the latter are typically much easier than theformer. To demonstrate this, we first make twoimportant claims and show that they are true.

Claim 1: Let C1 and C2 be two complex-valuedconstants (independent of t). Also,ReC1e

jωt= Re

C2e

jωtfor all t. Then,

C1 = C2.

Proof: Evaluating at t = 0 we findRe C1 = Re C2. Since C1 and C2 areconstant with respect to time, this must be truefor all t. At t = π/(2ω) we find

ReC1e

jωt= Re C1 · j = −Im C1

and similarly

ReC2e

jωt= Re C2 · j = −Im C2

therefore Im C1 = Im C2. Once again: SinceC1 and C2 are constant with respect to time, thismust be true for all t. Since the real andimaginary parts of C1 and C2 are equal, C1 = C2.

What does this mean? We have just shown thatif two phasors are equal, then the sinusoidalwaveforms that they represent are also equal.

Claim 2: For any real-valued linear operator Tand complex-valued quantity C,T (Re C) = Re T (C).

Proof: Let C = cr + jci where cr and ci arereal-valued quantities, and evaluate the right side

of the equation:

Re T (C) = Re T (cr + jci)= Re T (cr) + jT (ci)= T (cr)

= T (Re C)

What does this mean? The operators that wehave in mind for T include addition,multiplication, differentiation, integration, and soon. Here’s an example with differentiation:

Re

∂

∂ωC

= Re

∂

∂ω(cr + jci)

=

∂

∂ωcr

∂

∂ωRe C =

∂

∂ωRe (cr + jci) =

∂

∂ωcr

In other words, differentiation of a sinusoidalsignal can be accomplished by differentiating theassociated phasor, so there is no need totransform a phasor back into its associatedreal-valued signal in order to perform thisoperation.

Summarizing:

Claims 1 and 2 together entitle us to performoperations on phasors as surrogates for thephysical, real-valued, sinusoidal waveformsthey represent. Once we are done, we cantransform the resulting phasor back into thephysical waveform it represents using Equa-tion 1.9, if desired.

However a final transformation back to the timedomain is usually not desired, since the phasor


tells us everything we can know about thecorresponding waveform.

A skeptical student might question the value ofphasor analysis on the basis that signals ofpractical interest are sometimes notsinusoidally-varying, and therefore phasoranalysis seems not to apply generally. It iscertainly true that many signals of practicalinterest are not sinusoidal, and many are farfrom it. Nevertheless, phasor analysis is broadlyapplicable. There are basically two reasons whythis is so:

• Many signals, although not strictlysinusoidal, are “narrowband” and thereforewell-modeled as sinusoidal. For example, acellular telecommunications signal mighthave a bandwidth on the order of 10 MHzand a center frequency of about 2 GHz. Thismeans the difference in frequency betweenthe band edges of this signal is just 0.5% ofthe center frequency. The frequencyresponse associated with signal propagationor with hardware can often be assumed tobe constant over this range of frequencies.With some caveats, doing phasor analysis atthe center frequency and assuming theresults apply equally well over thebandwidth of interest is often a pretty goodapproximation.

• It turns out that phasor analysis is easilyextensible to any physical signal, regardlessof bandwidth. This is so because anyphysical signal can be decomposed into alinear combination of sinusoids – this isknown as Fourier analysis. The way to findthis linear combination of sinusoids is bycomputing the Fourier series, if the signal isperiodic; or the Fourier Transform,otherwise. Phasor analysis applies to eachfrequency independently, and the results canbe added together to obtain the result forthe complete signal. The process ofcombining results after phasor analysisresults is nothing more than integration over

frequency; i.e.:

∫ +∞

−∞

A(t;ω)dω

Using Equation 1.9, this can be rewritten:

∫ +∞

−∞

ReC(ω)ejωt

dω

In fact we can go one step further usingClaim 2:

Re

∫ +∞

−∞

C(ω)ejωtdω

The quantity in the curly braces is simplythe Fourier transform of C(ω). Thus we seethat we can analyze a signal ofarbitrarily-large bandwidth simply bykeeping ω as an independent variable whilewe are doing phasor analysis, and if we everneed the physical signal, we just take thereal part of the Fourier transform of thephasor. So not only is it possible to analyzeany time-domain signal using phasoranalysis, it is also often far easier than doingthe same analysis on the time-domain signaldirectly.

Summarizing:

Phasor analysis does not limit us to sinu-soidal waveforms. Phasor analysis is applica-ble to signals which are sufficiently narrow-band, and can also be applied to signals ofarbitrary bandwidth via Fourier analysis.

Additional Reading:

• “Phasor” on Wikipedia.

• “Fourier analysis” on Wikipedia.

1.6 Units

[m0072]

https://en.wikipedia.org/wiki/Phasor

https://en.wikipedia.org/wiki/Fourier_analysis

1.6. UNITS 13

Prefix Abbreviation Multiply by:exa E 1018

peta P 1015

tera T 1012

giga G 109

mega M 106

kilo k 103

milli m 10−3

micro µ 10−6

nano n 10−9

pico p 10−12

femto f 10−15

atto a 10−18

Table 1.5: Prefixes used to modify units.

The term “unit” refers to the measure used toexpress a physical quantity. For example, themean radius of the Earth is about6,371,000 meters; in this case the unit is themeter.

A number like “6,371,000” becomes a bitcumbersome to write, so it is common to use aprefix to modify the unit. For example, theradius of the Earth is more commonly said to be6371 kilometers, where one kilometer isunderstood to mean 1000 meters. It is commonpractice to use prefixes, such as “kilo-”, thatyield values in the range 0.001 to 10, 000. A listof standard prefixes appears in Table 1.5.

Writing out the names of units can also becometedious. For this reason it is common to usestandard abbreviations; e.g., “6731 km” asopposed to ‘6371 kilometers”, where “k” is thestandard abbreviation for the prefix “kilo” and“m” is the standard abbreviation for “meter”. Alist of commonly-used base units and theirabbreviations is shown in Table 1.6.

To avoid ambiguity it is important to alwaysindicate the units of a quantity; e.g., writing“6371 km” as opposed to “6371”. Failure to doso is a common source of error andmisunderstandings. An example is theexpression:

l = 3t

Unit Abbreviation Quantifies:ampere A electric currentcoulomb C electric chargefarad F capacitancehenry H inductancehertz Hz frequencyjoule J energymeter m distancenewton N forceohm Ω resistancesecond s timetesla T magnetic flux densityvolt V electric potentialwatt W powerweber Wb magnetic flux

Table 1.6: Some units that are commonly used inelectromagnetics.

where l is length and t is time. It could be that lis in meters and t is in seconds, in which case “3”really means “3 m/s”. However if it is intendedthat l is in kilometers and t is in hours, then “3”really means “3 km/h” and the equation isliterally different. To patch this up, one mightwrite “l = 3t m/s”; however note that this doesdoes not resolve the ambiguity we just identified– i.e., we still don’t know the units of theconstant “3”. Alternatively one might write“l = 3t where l is in meters and t is in seconds”,which is unambiguous but becomes quiteawkward for more complicated expressions. Abetter solution is to write instead:

l = (3 m/s) t

or even better:

l = at where a = 3 m/s

since this separates this issue of units from theperhaps more-important fact that l isproportional to t and the constant ofproportionality (a) is known.

The meter is the fundamental unit of length inthe International System of Units, known by itsFrench acronym “SI” and sometimes informallyreferred to as the “metric system”.


In this work we will use SI units exclusively.

Although SI is probably the most popular forengineering use overall, other systems remain incommon use. For example, the English system,where the radius of the Earth might alternativelybe said to be about 3959 miles, continues to beused in various applications and to a lesser orgreater extent in various regions of the world. Analternative system in common use in physics andmaterial science applications is the CGS(“centimeter-gram-second”) system. The CGSsystem is similar to SI, but with some significantdifferences: For example, the base unit of energyis the CGS system is not the “joule” but ratherthe “erg”, and the values of some physicalconstants become unitless. Therefore – onceagain – it is very important to include unitswhenever values are stated.

SI defines seven fundamental units from which allother units can be derived. These fundamentalunits are distance in meters (m), time in seconds(s), current in amperes (A), mass in kilograms(kg), temperature in kelvin (K), particle count inmoles (mol), and luminosity in candela (cd). SIunits for electromagnetic quantities such ascoulombs (C) for charge and volts (V) for electricpotential are derived from these fundamentalunits.

A frequently-overlooked feature of units is theirability to assist in error-checking mathematicalexpressions. For example, the electric fieldintensity may be specified in volts per meter(V/m), so an expression for the electric fieldintensity that yields units of V/m is said to be“dimensionally correct” (but not necessarilycorrect), whereas an expression that cannot bereduced to units of V/m cannot be correct.

Additional Reading:

• “International System of Units” onWikipedia.

• “Centimetre-gram-second system of units”on Wikipedia.

1.7 Notation

[m0005]

The list below describes notation used in thisbook.

• Vectors : Boldface is used to indicate avector; e.g., the electric field intensity vectorwill typically appear as E. Quantities not inboldface are scalars. When writing by hand,

it is common to write “E” or “−→E ” in lieu of

“E”.

• Unit vectors : A circumflex is used toindicate a unit vector; i.e., a vector havingmagnitude equal to one. For example, theunit vector pointing in the +x direction willbe indicated as x. In discussion, thequantity “x” is typically spoken “x hat”.

• Time: The symbol t is used to indicate time.

• Position: The symbols (x, y, z), (ρ, φ, z), and(r, θ, φ) indicate positions using thecartesian, cylindrical, and polar coordinatesystems, respectively. It is sometimesconvenient to express position in a mannerwhich is independent of a coordinate system;in this case we typically use the symbol r.For example r = xx+ yy + zz in thecartesian coordinate system.

• Phasors: A tilde is used to indicate a phasorquantity; e.g., a voltage phasor might beindicated as V , and the phasorrepresentation of E will be indicated as E.

• Curves, surfaces, and volumes: Thesegeometrical entities will usually be indicatedin script; e.g., an open surface might beindicated as S and the curve bounding thissurface might be indicated as C. Similarlythe volume enclosed by a closed surface Smay be indicated as V.

• Integrations over curves, surfaces, andvolumes will usually be indicated using asingle integral sign with the appropriate

https://en.wikipedia.org/wiki/International_System_of_Units

https://en.wikipedia.org/wiki/Centimetre-gram-second_system_of_units

1.7. NOTATION 15

subscript; for example:∫

C

· · · dl is an integral over the curve C

∫

S

· · · ds is an integral over the surface S∫

V

· · · dv is an integral over the volume V.

• Integrations over closed curves and surfaceswill be indicated using a circle superimposedon the integral sign; for example:∮

C

··· dl is an integral over the closed curve C

∮

S

··· ds is an integral over the closed surface S

A “closed curve” is one which forms anunbroken loop; e.g., a circle. A “closedsurface” is one which encloses a volume withno openings; e.g., a sphere.

• The symbol “∼=” means “approximatelyequal to”. This symbol is used whenequality exists, but is not being expressedwith exact numerical precision. For example,the ratio of the circumference of a circle toits diameter is π, where π ∼= 3.14.

• The symbol “≈” also indicates“approximately equal to”, but in this casethe two quantities are unequal even ifexpressed with exact numerical precision.For example, ex = 1 + x+ x2/2 + ... as ainfinite series, but ex ≈ 1 + x for x≪ 1.Using this approximation e0.1 ≈ 1.1, which isin good agreement with the actual valuee0.1 ∼= 1.1052.

• The symbol “∼” indicates “on the order of”,which is a relatively weak statement ofequality indicating that the indicatedquantity is within a factor of 10 or so theindicated value. For example, µ ∼ 105 for aclass of iron alloys, with exact values beingbeing higher or lower within a factor of 5 orso.

• The symbol “,” means “is defined as” or “isequal as the result of a definition”.

• Complex numbers : j ,√−1.

• See Appendix C for notation used to identifycommonly-used physical constants.

[m0026]


Image Credits

Fig. 1.1 c© V. Blacus, https://commons.wikimedia.org/wiki/File:Electromagnetic-Spectrum.svg,CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).

Fig. 1.2: c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0074 fClap.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/).

Fig. 1.3: c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0074 fTrumpet.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/).

https://commons.wikimedia.org/wiki/File:Electromagnetic-Spectrum.svg


https://commons.wikimedia.org/wiki/File:M0074_fClap.svg


https://commons.wikimedia.org/wiki/File:M0074_fTrumpet.svg


Chapter 2

Electric and Magnetic Fields

2.1 What is a Field?

[m0001]

A field is the continuum of values of a quan-tity as a function of position and time.

The quantity that the field describes may be ascalar or a vector, and the scalar part may beeither real- or complex-valued.

In electromagnetics, the electric field intensity Eis a real-valued vector field that may vary as afunction of position and time, and so might beindicated as “E(x, y, z, t)”, “E(r, t)”, or simply“E”. When expressed as a phasor, this quantityis complex-valued but exhibits no timedependence, so we might say instead “E(r)” or,

again, simply “E”.

An example of a scalar field in electromagneticsis the electric potential, V ; i.e., V (r, t).

A wave is a time-varying field that continues toexist in the absence of the source that created it,and is therefore able to transport energy.

2.2 Electric Field Intensity

[m0002]

Electric field intensity is a vector field which weassign the symbol E and which has units ofelectrical potential per distance; in SI units, volts

F

c© M. Goldammer CC BY SA 4.0

Figure 2.1: A positively-charged particle experi-ences a repulsive force F in the presence of anotherparticle which is also positively-charged.

per meter (V/m). Before offering a formaldefinition, it is useful to consider the broaderconcept of the electric field.

Imagine that the universe is empty except for asingle particle of positive charge. Next imaginethat a second positively-charged particle appears;the situation is now as shown in Figure 2.1.Since like charges repel, the second particle willbe repelled by the first particle, and vice versa.Specifically, the first particle is exerting a forceon the second particle. If the second particle isfree to move, it will do so; this is an expression ofkinetic energy. If the second particle is somehowheld in place, we say the second particle possessesan equal amount of potential energy. Thispotential energy is no less “real”, since we canconvert it to kinetic energy simply by releasingthe particle, thereby allowing it to move.

Now let us revisit the original (one particle)scenario. In that scenario, we could make a mapin which every position in space is assigned avalue which describes the force that a particlehaving a specified charge q would experience if itwere to appear there. The result looks somethinglike Figure 2.2. This map of force vectors is




18 CHAPTER 2. ELECTRIC AND MAGNETIC FIELDS

c© M. Goldammer CC BY SA 4.0

Figure 2.2: A map of the force that would beexperienced by a second particle having a specifiedcharge. Here the magnitude and direction of theforce is indicated by the size and direction of thearrow.

essentially a description of the electric fieldassociated with the first particle.

There are many ways in which the electric fieldmay be quantified. Electric field intensity E issimply one of these ways. We define E(r) to bethe force F(r) experienced by a test particlehaving charge q, divided by q; i.e.,

E(r) , limq→0

F(r)

q(2.1)

Note that it is required for the charge to becomevanishingly small (as indicated by taking thelimit) in order for this definition to work. This isbecause the source of the electric field is charge,and so the test particle contributes to the totalelectric field. Therefore the test charge must besmall enough not to significantly perturb thefield being evaluated, so q must be effectivelyzero. This makes Equation 2.1 unsatisfying froman engineering perspective, and we’ll addressthat later in this section.

According the definition asserted byEquation 2.1, the units of E are those of forcedivided by charge. The SI units for force andcharge are newtons (N) and coulombs (C)respectively, so E has units of N/C. However we

typically express E in units of V/m, not N/C.What’s going on? The short answer is that1 V/m = 1 N/C:

N

C=

N ·mC ·m =

J

C ·m =V

m

where we have used the fact that 1 N·m =1 joule (J) of energy, and 1 J/C = 1 V.

Electric field intensity (E, N/C or V/m) is avector field that quantifies the force experi-enced by a charged particle due to the influ-ence of charge not associated with that par-ticle.

The analysis of units doesn’t do much to answerthe question of why we should prefer to expressE in V/m as opposed to N/C. We now tacklethat question. Figure 2.3 shows a simple thoughtexperiment that demonstrates the concept ofelectric field intensity in terms of an electriccircuit. This circuit consists of a parallel-platecapacitor in series with a 9 V battery.1 The effectof the battery, connected as shown, is to force anaccumulation of positive charge on the upperplate, and an accumulation of negative charge onthe lower plate. If we consider the path from theposition labeled “A”, along the wire and throughthe battery to the position labeled “B”, thechange in electric potential is +9 V. It must alsobe true that the change in electric potential aswe travel from B to A through the capacitor is−9 V, since the sum of voltages over any closedcircuit is zero. Said differently, the change inelectric potential between the plates of thecapacitor, starting from node A and ending atnode B, is +9 V. Now, note that the spacingbetween the plates in the capacitor is 1 mm.Thus the rate of change of the potential betweenthe plates is 9 V divided by 1 mm, which is9000 V/m. This is literally the electric fieldintensity between the plates. That is, if oneplaces a particle with an infinitesimally-smallcharge between the plates (point “C”), and thenmeasures the ratio of force to charge, one finds itis 9000 N/C pointing toward A. We come to thefollowing remarkable conclusion:

1It is not necessary to know anything about capacitorsto get to the point, so no worries!


2.3. PERMITTIVITY 19

A

B+

-

9V

1mm


Figure 2.3: A simple circuit used to describe theconcept of electric field intensity. In this example,E at point C is 9000 V/m directed from B towardA.

E points in the direction in which electric po-tential is most rapidly decreasing, and themagnitude of E is the rate of change in elec-trical potential with distance in this direc-tion.

Additional Reading:

• Electric field on Wikipedia

2.3 Permittivity

[m0008]

Permittivity describes the effect of material indetermining the magnitude of the electric field inresponse to electric charge. For example, one canobserve from laboratory experiments that aparticle having charge q gives rise to the electricfield

E = R q1

4πR2

1

ǫ(2.2)

where R is distance from the charge and Rpoints away from the charge. Note that Eincreases with q, which makes sense since electriccharge is the source of E. Also note that E isinversely proportional to 4πR2, indicating that Edecreases in proportion to the area of a spheresurrounding the charge – a principle commonly

known as the inverse square law. The remainingfactor 1/ǫ is the constant of proportionalitywhich captures the effect of material. Givenunits of V/m for E and C for Q, we find that ǫmust have units of farads per meter (F/m). (Tosee this, note that 1 F = 1 C/V.)

Permittivity (ǫ, F/m) describes the effect ofmaterial in determine the electric field inten-sity in response to charge.

In free space (that is, a perfect vacuum), we findthat ǫ = ǫ0 where:

ǫ0 ∼= 8.854× 10−23 F/m (2.3)

The permittivity of air is only slightly greater,and usually can be assumed to be equal to thatof free space. In most other materials, thepermittivity is significantly greater; that is, thesame charge results in a weaker electric fieldintensity.

It is common practice to describe thepermittivity of materials relative to thepermittivity of free space. This relativepermittivity is given by:

ǫr ,ǫ

ǫ0(2.4)

Values of ǫr for a few representative materials isgiven in Appendix A.1. Note that ǫr ranges from1 (corresponding to a perfect vacuum) to about60 or so in engineering applications, and isgenerally less than 10 for commonly-encounteredmaterials. Also note that relative permittivity issometimes referred to as dielectric constant. Thisterm is a bit misleading however, sincepermittivity is a meaningful concept for manymaterials which are not dielectrics.

Additional Reading:

• “Permittivity” on Wikipedia.

• Appendix A.1 (“Permittivity of SomeCommon Materials”).

• “Inverse square law” on Wikipedia.


https://en.wikipedia.org/wiki/Electric_field

https://en.wikipedia.org/wiki/Permittivity

https://en.wikipedia.org/wiki/Inverse-square_law


2.4 Electric Flux Density

[m0011]

Electric flux density, assigned the symbol D, isan alternative to electric field intensity (E) as away to quantify an electric field. Although thisalternative description is not strictly necessary, itoffers some actionable insight and somenotational convenience, as we shall point out atthe end of this section.

First, what is electric flux density? Recall that aparticle having charge q gives rise to the electricfield intensity

E = R q1

4πR2

1

ǫ(2.5)

where R is distance from the charge and Rpoints away from the charge. Note that E isinversely proportional to 4πR2, indicating that Edecreases in proportion to the area of a spheresurrounding the charge. Now integrate both sidesof Equation 2.5 over a sphere S of radius R:

∮

S

E(r) · ds =∮

S

[R q

1

4πR2

1

ǫ

]· ds (2.6)

Factoring out constants that do not vary withthe variables of integration, the right-hand side is

q1

4πR2

1

ǫ

∮

S

ds

The remaining integral is simply the area of S,which is 4πR2. Thus we find:∮

S

E(r) · ds = q

ǫ(2.7)

The integral of a vector field over a specifiedsurface is known as flux (see “AdditionalReading” at the end of this section). Thus, wehave found that the flux of E through the sphereS is equal to a constant, namely q/ǫ, and thatthis constant is the same regardless of the radiusR of the sphere. Said differently, the flux of E isconstant with distance, and does not vary as Eitself does. Let us capitalize on this observationby making the following small modification toEquation 2.7:

∮

S

[ǫE] · ds = q (2.8)

In other words, the flux of the quantity ǫE isequal to the enclosed charge, regardless of theradius of the sphere over which we are doing thecalculation. This leads to the following definition:

The electric flux density D = ǫE, havingunits of C/m2, is a description of the elec-tric field in terms of flux, as opposed to forceor change in electric potential.

It may appear that D is redundant informationgiven E and ǫ, but this is true only inhomogeneous media. The concept of electric fluxdensity becomes important – and decidedly notredundant – when we encounter boundariesbetween media having different permittivities.As we shall see in Section 5.18, boundaryconditions on D constrain the component of theelectric field which is perpendicular to theboundary separating two material regions. If oneignores the “flux” character of the electric fieldrepresented by D and instead considers only E,then only the tangential component of theelectric field is constrained. In fact, when one ofthe two materials comprising the boundarybetween two material regions is a perfectconductor, then the electric field is completelydetermined by the boundary condition on D.This greatly simplifies the problem of finding theelectric field in a region bounded or partiallybounded by materials that can be modeled asperfect conductors, including many metals. Inparticular, this principle makes it easy to analyzecapacitors.

We conclude this section with a warning: Eventhough the SI units for D are C/m2, D describesan electric field and not a surface charge density.It is certainly true that one may describe theamount of charge distributed over a surface usingunits of C/m2. However D is not necessarily adescription of actual charge, and there is noimplication that the source of the electric field isa distribution of surface charge. On the otherhand it is true that D can be interpreted as anequivalent surface charge density that would giverise to the observed electric field, and in somecases this equivalent charge density turns out tobe the actual charge density.

2.5. MAGNETIC FLUX DENSITY 21

S N


Figure 2.4: Evidence of a vector field from obser-vations of the force perceived by the bar magnetson the right in the presence of the bar magnets onthe left.

Additional Reading:

• “Flux” on Wikipedia.

2.5 Magnetic Flux Density

[m0003]

Magnetic flux density is a vector field which weidentify using the symbol B and which has SIunits of tesla (T). Before offering a formaldefinition, it is useful to consider the broaderconcept of the magnetic field.

Magnetic fields are an intrinsic property of somematerials; most notably permanent magnets.The basic phenomenon is probably familiar, andis shown in Figure 2.4. A bar magnet has “poles”identified as “N” (“north”) and “S” (”south”).The N-end of one magnet attracts the S-end ofanother magnet but repels the N-end of the othermagnet, and so on. The existence of a vectorfield is apparent since the observed force acts ata distance and is asserted in a specific direction.In the case of a permanent magnet, the source ofthe magnetic field is a combination of effectsoccurring at the scale of the atoms and electronscomprising the material. These mechanismsrequire some additional explanation which weshall defer for now.

Magnetic fields also appear in the presence of


Figure 2.5: Evidence that current can also createa magnetic field.

current. For example, a coil of wire bearing acurrent is found to influence permanent magnets(and vice versa) in the same way that permanentmagnets effect each other. This is shown inFigure 2.5. From this we infer that theunderlying mechanism is the same – i.e., thevector field generated by a current-bearing coil isthe same phenomenon as the vector fieldassociated with a permanent magnet. Whateverthe source, we are now interested in quantifyingits behavior.

To begin, let us consider the effect of a magneticfield on a electrically-charged particle. First,imagine a region of free space with no electric ormagnetic fields. Next imagine that a chargedparticle appears. This particle will experience noforce. Next a magnetic field appears; perhapsthis is due to a permanent magnet or a current inthe vicinity. This situation is shown in Figure 2.6(top). Still, no force is applied to the particle. Infact, nothing happens until the particle in set inmotion. Figure 2.6 (bottom) shows an example:Suddenly, the particle perceives a force. We’ll getto the details about direction and magnitude in amoment, but the main idea is now evident: Amagnetic field is something that applies a forceto a charged particle in motion, distinct from (infact, in addition to) the force associated with an


https://en.wikipedia.org/wiki/Flux



B

B

v>

F

v=

F=


Figure 2.6: The force perceived by a charged par-ticle which is (top) motionless and (bottom) mov-ing with velocity v = zv, which is perpendicular tothe plane of this document and toward the reader.

electric field.

Now it is worth noting that a single chargedparticle in motion is the simplest form of current,and since motion is required for the magneticfield to influence the particle, it appears thatcurrent is not only a source of the magnetic field,but also that the magnetic field exerts a force oncurrent. Summarizing:

A magnetic field quantifies the force exertedon permanent magnets and currents in thepresence of other permanent magnets andcurrents.

So, how to quantify a magnetic field? Theanswer from classical physics involves anotherexperimentally-derived equation that predictsforce as a function of charge, velocity, and avector field B representing the magnetic field.Here it is: The force applied to a particle bearingcharge q is

F = qv ×B (2.9)

where v is the velocity of the particle and “×”denotes the cross product. The cross product oftwo vectors is in a direction perpendicular toeach of the two vectors, so the force exerted by

the magnetic field is perpendicular to both thedirection of motion and the direction in whichthe magnetic field points. Why perpendicular toboth? For that matter, why should the forcedepend on how fast the particle is traveling?Those are great questions, but unfortunatelythose are questions for which classical physicsprovides no obvious answers! These are simplytwo of the ways in which the magnetic field isstrange.

Dimensional analysis of Equation 2.9 reveals thatB has units of (N·s)/(C·m). In SI, thiscombination of units is known as the tesla (T).

We have referred to B as magnetic flux density,and therefore tesla is a unit of magnetic fluxdensity. A fair question to ask at this point is:What makes this a flux density? The shortanswer is that this terminology is somewhatarbitrary, and in fact is not even uniformlyaccepted. In engineering electromagnetics, thepreference for referring to B as a “flux density”is because we frequently find ourselvesintegrating B over a mathematical surface. Anyquantity which is obtained by integration over asurface is referred to as “flux”, and so it becomesnatural to think of B as a flux density; i.e., asflux per unit area. The SI unit for magnetic fluxis the weber (Wb). Therefore, B mayalternatively be described as having units ofWb/m2, and 1 Wb/m2 = 1 T.

Magnetic flux density (B, T or Wb/m2) is adescription of the magnetic field that can bedefined as the solution to Equation 2.9.

When describing magnetic fields, we occasionallyrefer to the concept of a field line, defined asfollows:

A magnetic field line is the curve in spacetraced out by following the direction in whichthe magnetic field vector points.

This concept is illustrated in Figure 2.7 for apermanent bar magnet and Figure 2.8 for acurrent-bearing coil.


2.6. PERMEABILITY 23


Figure 2.7: The magnetic field of a bar magnet,illustrating field lines.


Figure 2.8: The magnetic field of a current-bearing coil, illustrating field lines.

Magnetic field lines are remarkable for thefollowing reason:

A magnetic field line always forms a closedloop.

This is true in a sense even for field lines whichseem to form straight lines (for example, thosealong the axis of the bar magnet and the coil inFigures 2.7 and 2.8), since a field line that travelsto infinity in one direction reemerges frominfinity in the opposite direction.

Additional Reading:

• “Magnetic Field” on Wikipedia.

2.6 Permeability

[m0009]

Permeability describes the effect of material indetermining the magnetic flux density. All elsebeing equal, magnetic flux density increases inproportion to permeability.

To illustrate the concept, consider that a particlebearing charge q moving at velocity v gives riseto a magnetic flux density:

B(r) = µqv

4πR2× R (2.10)

where R is the unit vector pointing from thecharged particle to the field point r, R is thisdistance, and “×” is the cross product. Notethat B increases with charge and speed, whichmakes sense since moving charge is the source ofthe magnetic field. Also note that B is inverselyproportional to 4πR2, indicating that |B|decreases in proportion to the area of a spheresurrounding the charge, also known as theinverse square law. The remaining factor, µ, isthe constant of proportionality which capturesthe effect of material. We refer to µ as thepermeability of the material. Since B can beexpressed in units of Wb/m2 and the units of vare m/s, we see that µ must have units of henries



https://en.wikipedia.org/wiki/Magnetic_field


per meter (H/m). (To see this, note that 1 H ,1 Wb/A.)

Permeability (µ, H/m) describes the effectof material in determining the magnetic fluxdensity.

In free space, we find that the permeabilityµ = µ0 where:

µ0 = 4π × 10−7 H/m (2.11)

It is common practice to describe thepermeability of materials in terms of theirrelative permeability:

µr ,µ

µ0(2.12)

which gives the permeability relative to theminimum possible value; i.e., that of free space.Relative permeability for a few representativematerials is given in Appendix A.2.

Note that µr is approximately 1 for all but asmall class of materials. These are known asmagnetic materials, and may exhibit values of µras large as ∼ 106. A commonly-encounteredcategory of magnetic materials is ferromagneticmaterial, of which the best-known example isiron.

Additional Reading:

• “Permeability (electromagnetism)” onWikipedia.

• Section 7.16 (“Magnetic Materials”).

• Appendix A.2 (“Permeability of SomeCommon Materials”).

2.7 Magnetic Field Intensity

[m0012]

Magnetic field intensity H is an alternativedescription of the magnetic field in which theeffect of material is factored out. For example,

the magnetic flux density B (reminder:Section 2.5) due to a point charge q moving atvelocity v can be written:

B = µqv

4πR2× R (2.13)

where R is the unit vector pointing from thecharged particle to the field point r, R is thisdistance, “×” is the cross product, and µ is thepermeability of the material. We can rewriteEquation 2.13 as:

B , µH (2.14)

with:H =

qv

4πR2× R (2.15)

so H in homogeneous media does not depend onµ.

Dimensional analysis of Equation 2.15 revealsthat the units for H are amperes per meter(A/m). However, H does not represent surfacecurrent density,2 as the units might suggest.While it is certainly true that a distribution ofcurrent (A) over some linear cross-section (m)can be described as a current density havingunits of A/m, H is associated with the magneticfield and not a particular current distribution.Said differently, H can be viewed as a descriptionof the magnetic field in terms of an equivalent(but not actual) current.

The magnetic field intensity H (A/m), de-fined using Equation 2.14, is a description ofthe magnetic field independent from its de-pendence on material properties.

It may appear that H is redundant informationgiven B and µ, but this is true only inhomogeneous media. The concept of magneticfield intensity becomes important – anddecidedly not redundant – when we encounterboundaries between media having differentpermeabilities. As we shall see in Section 7.11,boundary conditions on H constrain thecomponent of the magnetic field which is tangent

2The concept of current density is not essential to un-derstand this section; however a primer can be found inSection 6.2.

https://en.wikipedia.org/wiki/Permeability_(electromagnetism)

2.8. ELECTROMAGNETIC PROPERTIES OF MATERIALS 25

to the boundary separating two material regions.If one ignores the characteristics of the magneticfield represented by H and instead considers onlyB, then only the perpendicular component of themagnetic field is constrained.

The concept of magnetic field intensity also turnsout to be useful in a certain problems in which µis not a constant, but rather is a function ofmagnetic field strength. In this case the magneticbehavior of the material is said to be nonlinear.For more on this, see Section 7.16.

Additional Reading:

• “Magnetic field” on Wikipedia.

2.8 ElectromagneticProperties of Materials

[m0007]

In electromagnetic analysis one is principallyconcerned with three properties of matter. Theseproperties are quantified in terms of constitutiveparameters, which describe the effect of materialin determining an electromagnetic quantity inresponse to a source. Here are the three principalconstitutive parameters:

• Permittivity (ǫ, F/m) quantifies the effect ofmatter in determining the electric field inresponse to electric charge. Permittivity isaddressed in Section 2.3.

• Permeability (µ, H/m) quantifies the effectof matter in determining the magnetic fieldin response to current. Permeability isaddressed in Section 2.6.

• Conductivity (σ, S/m) quantifies the effectof matter in determining the flow of currentin response to an electric field. Conductivityis addressed in Section 6.3.

The electromagnetic properties of most com-mon materials in most common applicationscan be quantified in terms of the constitutiveparameters ǫ, µ, and σ.

To keep the theory from becoming too complex,we usually require the constitutive parametersthemselves to exhibit a few basic properties.These properties are as follows:

• Homogeneity. A material which ishomogeneous is uniform over the space itoccupies; that is, the value of its constitutiveparameters is constant for all points in thespace in which the material exists. Acounter-example would be some materialwhich is composed of multiplechemically-distinct compounds which arenot thoroughly mixed; for example, soil.

• Isotropy. A material which is isotropicbehaves in precisely the same way regardlessof how it is oriented with respect to sourcesand fields occupying the same space. Acounter-example is quartz, whose atoms arearranged in a uniformly-spaced crystallinelattice. As a result, the electromagneticproperties of quartz can be changed simplyby rotating the material with respect to theapplied sources and fields.

• Linearity. A material is said to be linear ifits properties are constant and independentof the magnitude of the sources and fieldsapplied to the material. For example,discrete capacitors have capacitance which isdetermined in part by the permittivity ofthe material separating the terminals(Section 5.23). This material isapproximately linear when the appliedvoltage V is below the rated workingvoltage; i.e., ǫ is constant and so capacitancedoes not vary significantly with respect toV . When V is greater than the workingvoltage, the dependence of ǫ on V becomesmore pronounced, and then capacitancebecomes a function of V . In anotherpractical example, it turns out that µ forferromagnetic materials is nonlinear such

https://en.wikipedia.org/wiki/Magnetic_field


that the precise value of µ depends on themagnitude of the magnetic field.

• Time-invariance. An example of a class ofmaterials which is not necessarilytime-invariant is piezoelectric materials,whose electromagnetic properties varysignificantly depending on the mechanicalforces applied to them (a property which canbe exploited to make sensors andtransducers).

Linearity and time-invariance (LTI) areparticularly important properties to considerbecause they are requirements for the property ofsuperposition. For example: In a LTI material wemay calculate the field E1 due to a point chargeq1 at r1, calculate the field E2 due to a pointcharge q2 at r2, and then the field when bothcharges are simultaneously present is simplyE1 +E2. The same is not true for materialswhich are not LTI. Devices which are nonlinearand therefore not LTI – such as thevoltage-stressed capacitor described above – nolonger follow the rules of basic circuit theory,which presumes that superposition applies. Thiscondition makes analysis and design much moredifficult.

No practical material is truly homogeneous,isotropic, linear, and time-invariant. However formost materials in most applications, thedeviation from this ideal condition is not largeenough to significantly affect engineering analysisand design. In other cases, materials may besignificantly non-ideal in one of these respects,but may still be analyzed with appropriatemodifications to the theory.

[m0054]

2.8. ELECTROMAGNETIC PROPERTIES OF MATERIALS 27

Image Credits

Fig. 2.1: c© M. Goldammer,https://commons.wikimedia.org/wiki/File:M0002 fTwoChargedParticles.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 2.2: c© M. Goldammer, https://commons.wikimedia.org/wiki/File:M0002 fForceMap.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 2.3: c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0002 fBatCap.svg,CC BY 3.0 (https://creativecommons.org/licenses/by/3.0/).

Fig. 2.4: c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0003 fBarMagnet.svg,CC BY 3.0 (https://creativecommons.org/licenses/by/3.0/).

Fig. 2.5: c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0003 fCoilBarMagnet.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/).

Fig. 2.6: c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0003 fFqvB.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/).

Fig. 2.7: c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0003 fFieldLinesBarMagnet.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/).

Fig. 2.8: c© Y. Qin, https://commons.wikimedia.org/wiki/File:0003 fFieldLinesCoil.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/).

https://commons.wikimedia.org/wiki/File:M0002_fTwoChargedParticles.svg


https://commons.wikimedia.org/wiki/File:M0002_fForceMap.svg


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https://commons.wikimedia.org/wiki/File:M0003_fBarMagnet.svg


https://commons.wikimedia.org/wiki/File:M0003_fCoilBarMagnet.svg


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https://commons.wikimedia.org/wiki/File:M0003_fFieldLinesBarMagnet.svg


https://commons.wikimedia.org/wiki/File:0003_fFieldLinesCoil.svg


Chapter 3

Transmission Lines

3.1 Introduction toTransmission Lines

[m0028]

A transmission line is a structure intended totransport electromagnetic signals or power.

A rudimentary transmission line is simply a pairof wires with one wire serving as “ground” andthe other wire bearing an electrical potentialwhich is defined relative to ground. Transmissionlines having random geometry, such as the testleads shown in Figure 3.1, are useful only at verylow frequencies and when loss, reactance, andimmunity to electromagnetic interference (EMI)are not a concern.

by Dmitry G

Figure 3.1: These leads used to connect testequipment to circuits in a laboratory are a veryrudimentary form of transmission line, suitableonly for very low frequencies.

However many circuits and systems operate atfrequencies where the length or cross-sectionaldimensions of the transmission line may be asignificant fraction of a wavelength. In this casethe transmission line is no longer “transparent”to the circuits at either end. Furthermore loss,reactance, and EMI are significant problems inmany applications. These concerns motivate theuse of particular types of transmission lines, andmake it necessary to understand how to properlyconnect the transmission line to the rest of thesystem.

In electromagnetics, the term “transmission line”refers to a structure which is intended to supporta guided wave. A guided wave is anelectromagnetic wave that is contained within orbound to the structure, and which does notradiate away from the structure. This conditionis normally met if the length and cross-sectionaldimensions of the transmission line are smallrelative to a wavelength – say λ/100 (i.e., 1% ofthe wavelength). For example, tworandomly-arranged wires serve well enough tocarry a signal at f = 10 MHz over a lengthl = 3 cm, since l is only 0.1% of the wavelengthλ = c/f = 30 m. However if l is increased to 3 m,or if f is increased to 1 GHz, then l is now 10%of the wavelength. In this case one shouldconsider using a transmission line that forms aproper guided wave.

Preventing unintended radiation is not the onlyconcern. Once we have established a guided waveon a transmission line, it is important that powerapplied to the transmission line be delivered tothe circuit or device at the other end, and not



3.2. TYPES OF TRANSMISSION LINES 29

reflected back into the source. For the randomwire f = 10 MHz, l = 3 cm example above, thereis little need for concern, since we expect a phaseshift of roughly 0.001 · 360 = 0.36 over thelength of the transmission line, which is about0.72 for a round trip. So, to a goodapproximation, the entire transmission line is atthe same electrical potential and thus transparentto the source and destination. However if l isincreased to 3 m, or if f is increased to 1 GHz,then the associated round-trip phase shiftbecomes 72. In this case a reflected signaltraveling in the opposite direction will add tocreate a total electrical potential which varies inboth magnitude and phase with position alongthe line. Thus the impedance looking toward thedestination via the transmission line will bedifferent than the impedance looking toward thedestination directly. The modified impedancewill depend on the cross-sectional geometry,materials, and length of the line.

Cross-sectional geometry and materials alsodetermine the loss and EMI immunity of thetransmission line.

Summarizing:

Transmission lines are designed to supportguided waves with controlled impedance, lowloss, and a degree of immunity from EMI.

3.2 Types of TransmissionLines

[m0144]

Two common types of transmission lines arecoaxial line (Figure 3.2) and microstrip line(Figure 3.3). Both are examples of transverseelectromagnetic (TEM) transmission lines. ATEM line employs a single electromagnetic wave“mode” having electric and magnetic fieldvectors in directions perpendicular to the axis ofthe line, as shown in Figures 3.4 and 3.5. TEMtransmission lines appear primarily in radiofrequency applications.

c© Tkgd2007 CC BY 3.0 (modified)

Figure 3.2: Structure of a coaxial transmissionline.

gound plane

dielectric slabmetallic trace

c© SpinningSpark CC BY SA 3.0 (modified)

Figure 3.3: Structure of a microstrip transmissionline.

E

Figure 3.4: Structure of the electric and magneticfields within coaxial line. In this case, the wave ispropagating away from the viewer.

Figure 3.5: Structure of the electric and magneticfields within microstrip line. (The fields outsidethe line are possibly significant, complicated, andnot shown.) In this case, the wave is propagatingaway from the viewer.



30 CHAPTER 3. TRANSMISSION LINES

TEM transmission lines such as coaxial lineand microstrip line are designed to support asingle electromagnetic wave which propagatesalong the length axis of the transmission linewith electric and magnetic field vectors per-pendicular to the direction of propagation.

Not all transmission lines exhibit TEM fieldstructure. In non-TEM transmission lines theelectric and magnetic field vectors that are notnecessarily perpendicular to the axis of the line,and the structure of the fields is complex relativeto the field structure of TEM lines. An exampleof a transmission line that exhibits non-TEMfield structure is the waveguide (see example inFigure 3.6). Waveguides are most prevalent atradio frequencies, and tend to appear inapplications where it is important to achieve verylow loss, or where power levels are very high.Another example is common “multimode”optical fiber (Figure 3.7). Optical fiber exhibitscomplex field structure because the wavelengthof light is very small compared to the physicaldimensions of practical fiber, making theexcitation and propagation of non-TEM wavesdifficult to avoid. (This issue is overcome in adifferent form of optical fiber, known as “singlemode” fiber, which is much more difficult andexpensive to manufacture.)

Higher-order transmission lines, includingradio-frequency waveguides and multimodeoptical fiber, are designed to guide waves thathave relatively complex structure.

Additional Reading:

• Coaxial cable on Wikipedia.

• Microstrip on Wikipedia.

• Waveguide (electromagnetism) onWikipedia.

• Optical fiber on Wikipedia.

c© Averse CC BY SA 2.0 Germany

Figure 3.6: A network of radio frequency waveg-uides in an air traffic control radar.

c© BigRiz CC BY SA 3.0 Unported

Figure 3.7: Strands of optical fiber.

https://en.wikipedia.org/wiki/Coaxial_cable

https://en.wikipedia.org/wiki/Microstrip

https://en.wikipedia.org/wiki/Waveguide_(electromagnetism)

https://en.wikipedia.org/wiki/Optical_fiber

https://creativecommons.org/licenses/by-sa/2.0/de/deed.en

https://creativecommons.org/licenses/by-sa/3.0/deed.en

3.3. TRANSMISSION LINES AS TWO-PORT DEVICES 31

ZL

ZS

VS

VS

VS

ZS

ZS

ZL

ZL

l

l

c© Omegatron CC BY SA 3.0 Unported (modified)

Figure 3.8: Symbols representing transmissionlines: Top: As a generic two-conductor direct con-nection. Middle: As a generic two-port “blackbox”. Bottom: As a coaxial cable.

3.3 Transmission Lines asTwo-Port Devices

[m0077]

Figure 3.8 shows common methods to representtransmission lines in circuit diagrams. In eachcase the source is represented using a Theveninequivalent circuit consisting of a voltage sourceVS in series with an impedance ZS .

1 Intransmission line analysis, the source may also bereferred to as the generator. The termination onthe receiving end of the transmission line isrepresented, without loss of generality, as animpedance ZL. This termination is often referredto as the load, although in practice it can be anycircuit that exhibits an input impedance of ZL.

The two-port representation of a transmissionline may be completely described by its length l

1For a refresher on this concept, see “Additional Read-ing” at the end of this section.

along with some combination of the followingparameters:

• Phase propagation constant β, having unitsof rad/m. This parameter also representsthe wavelength in the line through therelationship λ = 2π/β.

• Attenuation constant α, having units of1/m. This parameter quantifies the effect ofloss in the line.

• Characteristic impedance Z0, having units ofΩ. This is the ratio of potential (“voltage”)to current when the line is perfectlyimpedance-matched at both ends.

These parameters depend on the materials andgeometry of the line, as is explained elsewhere inthis book.

Note that a transmission line is typically nottransparent to the source and load. In particular,the load impedance may be ZL, but theimpedance presented to the source may or maynot be equal to ZL. Similarly, the sourceimpedance may be ZS , but the impedancepresented to the load may or may not be equal toZS . The effect of the transmission line on thesource and load impedances will depend on theparameters identified above.

Additional Reading:

• Thevenin’s theorem on Wikipedia.

3.4 Lumped-Element Model

[m0029]

It is possible to ascertain the relevant behaviorsof a transmission line using elementary circuittheory applied to a differential-lengthlumped-element model of the transmission line.The concept is illustrated in Figure 3.9, whichshows a generic transmission line, aligned withits length axis along the z axis. The transmission


https://en.wikipedia.org/wiki/Thevenin's_theorem


line is divided into segments having small butfinite length ∆z. Each segment is modeled as anidentical two-port having the equivalent circuitrepresentation shown in Figure 3.10. Theequivalent circuit consists of 4 components asfollows:

• The resistance R′∆z represents theseries-combined ohmic resistance of the twoconductors. This should account for bothconductors since the current in the actualtransmission line must flow through bothconductors. The prime notation reminds usthat R′ is resistance per unit length; i.e.,Ω/m, and it is only after multiplying bylength that we get a resistance in Ω.

• The conductance G′∆z represents theleakage of current directly from oneconductor to the other. When G′∆z > 0,the resistance between the conductors is lessthan infinite and therefore current may flowbetween the conductors. This amounts to aloss of power separate from the lossassociated with R′ above. G′ has units ofS/m. Further note that G′ is not equal to1/R′ as defined above: G′ and R′ aredescribing entirely different physicalmechanisms (and in principle either couldbe defined as either a resistance or aconductance).

• The capacitance C ′∆z represents thecapacitance of the transmission linestructure. Capacitance is the tendency tostore energy in electric fields, and dependson the cross-sectional geometry and themedia separating the conductors. C ′ hasunits of F/m.

• The inductance L′∆z represents theinductance of the transmission linestructure. Inductance is the tendency tostore energy in magnetic fields, and (likecapacitance) depends on the cross-sectionalgeometry and the media separating theconductors. L′ has units of H/m.

In order to use the model, one must have valuesfor R′, G′, C ′, and L′. Methods for computing

z

zz z z

Figure 3.9: Interpretation of a transmission line asa cascade of discrete series-connected two-ports.

R'Δz L'Δz

G'Δz C'Δz


Figure 3.10: Lumped-element equivalent circuitmodel for each of the two-ports in Figure 3.9.

these parameters are addressed elsewhere in thisbook.

3.5 Telegrapher’s Equations

[m0079]

In this section we derive the equations thatgovern the potential v(z, t) and current i(z, t)along a transmission line which is oriented alongthe z axis. For this we will employ thelumped-element model developed in Section 3.4.

To begin, we define voltages and currents asshown in Figure 3.11. We assign the variablesv(z, t) and i(z, t) to represent the potential andcurrent on the left side of the segment, withreference polarity and direction as shown in thefigure. Similarly we assign the variablesv(z +∆z, t) and i(z +∆z, t) to represent thepotential and current on the right side of thesegment, again with reference polarity anddirection as shown in the figure. ApplyingKirchoff’s voltage law from the left port, through


3.5. TELEGRAPHER’S EQUATIONS 33

R'Δz L'Δz

G'ΔzC'Δz

v(z,t) i(z+Δz,t)i(z,t)

+

_

+

_

v(z+Δz,t)


Figure 3.11: Lumped-element equivalent circuittransmission line model, annotated with sign con-ventions for potentials and currents.

R′∆z and L′∆z, and returning via the rightport, we obtain:

v(z, t)− (R′∆z) i(z, t)− (L′∆z)∂

∂ti(z, t)

− v(z +∆z, t) = 0 (3.1)

Moving terms containing factors of ∆z to theright and then dividing through by ∆z, we obtain

− v(z +∆z, t)− v(z, t)

∆z=

R′ i(z, t) + L′ ∂

∂ti(z, t) (3.2)

Then taking the limit as ∆z → ∞:

− ∂

∂zv(z, t) = R′ i(z, t) + L′ ∂

∂ti(z, t) (3.3)

Applying Kirchoff’s current law at the right port,we obtain:

i(z, t)−(G′∆z) v(z+∆z, t)−(C ′∆z)∂

∂tv(z+∆z, t)

− i(z +∆z, t) = 0 (3.4)

Arranging factors containing currents to appearon the left and then dividing through by ∆z, weobtain

− i(z +∆z, t)− i(z, t)

∆z=

G′ v(z +∆z, t) + C ′ ∂

∂tv(z +∆z, t) (3.5)

Taking the limit as ∆z → ∞:

− ∂

∂zi(z, t) = G′ v(z, t) + C ′ ∂

∂tv(z, t) (3.6)

Equations 3.3 and 3.6 are the Telegrapher’sEquations. These coupled simultaneous dif-ferential equations can be solved for v(z, t)and i(z, t) given R′, G′, L′, C ′ and suitableboundary conditions.

The time-domain Telegrapher’s Equations areusually more than we need or want. If we areonly interested in the response to a sinusoidalstimulus, then considerable simplification ispossible using phasor representation.2 First wedefine phasors V (z) and I(z) through the usualrelationship:

v(z, t) = ReV (z) ejωt

(3.7)

i(z, t) = ReI(z) ejωt

(3.8)

Now we see:

∂

∂zv(z, t) =

∂

∂zReV (z) ejωt

= Re

[∂

∂zV (z)

]ejωt

In other words, ∂v(z, t)/∂z expressed in phasor

representation is simply ∂V (z)/∂z; and

∂

∂ti(z, t) =

∂

∂tReI(z) ejωt

= Re

∂

∂t

[I(z)ejωt

]

= Re[jωI(z)

]ejωt

In other words, ∂i(z, t)/∂t expressed in phasor

representation is jωI(z). Therefore Equation 3.3expressed in phasor representation is:

− ∂

∂zV (z) = [R′ + jωL′] I(z) (3.9)

Following the same procedure, Equation 3.6expressed in phasor representation is found to be:

− ∂

∂zI(z) = [G′ + jωC ′] V (z) (3.10)

2For a refresher on phasor analysis, see Section 1.5.



Equations 3.9 and 3.10 are the Telegrapher’sEquations in phasor representation.

The principal advantage of these equations overthe time-domain versions is that we no longermust need to contend with derivatives withrespect to time; only derivatives with respect todistance remain. This considerably simplifies theequations.

Additional Reading:

• Telegrapher’s equations on Wikipedia.

• Kirchhoff’s circuit laws on Wikipedia.

3.6 Wave Equation for aTransmission Line

[m0027]

Consider a transmission line aligned along the zaxis. The phasor form of the Telegrapher’sEquations (Section 3.5) relate the potential

phasor V (z) and the current phasor I(z) to eachother and to the lumped-element modelequivalent circuit parameters R′, G′, C ′, and L′.These equations are

− ∂

∂zV (z) = [R′ + jωL′] I(z) (3.11)

− ∂

∂zI(z) = [G′ + jωC ′] V (z) (3.12)

An obstacle to using these equations is that werequire both equations to solve for either thepotential or the current. In this section we reducethese equations to a single equation, a waveequation, that is more convenient to use andwhich provides some additional physical insight.

We begin by differentiating both sides ofEquation 3.11 with respect to z, yielding:

− ∂2

∂z2V (z) = [R′ + jωL′]

∂

∂zI(z) (3.13)

Then using Equation 3.12 to eliminate I(z), weobtain

− ∂2

∂z2V (z) = − [R′ + jωL′] [G′ + jωC ′] V (z)

(3.14)This equation is normally written as follows:

∂2

∂z2V (z)− γ2 V (z) = 0 (3.15)

where we have made the substitution:

γ2 = (R′ + jωL′) (G′ + jωC ′) (3.16)

The positive square root of γ2 is known as thepropagation constant :

γ ,√

(R′ + jωL′) (G′ + jωC ′) (3.17)

The propagation constant γ (units of m−1)captures the effect of materials, geometry,and frequency in determining the variationin potential and current with distance on atransmission line.

Following essentially the same procedure butbeginning with Equation 3.12, we obtain

∂2

∂z2I(z)− γ2 I(z) = 0 (3.18)

Equations 3.15 and 3.18 are the wave equa-tions for V (z) and I(z), respectively.

Note that both V (z) and I(z) satisfy the samelinear homogeneous differential equation. Thisdoes not mean that V (z) and I(z) are equal.

Rather, it means that V (z) and I(z) can differby no more than a multiplicative constant. SinceV (z) is potential and I(z) is current, thatconstant must be an impedance. This impedanceis known as the characteristic impedance and isdetermined in Section 3.7.

The general solutions to Equations 3.15 and 3.18are

V (z) = V +0 e

−γz + V −0 e

+γz (3.19)

https://en.wikipedia.org/wiki/Telegrapher's_equations

https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

3.7. CHARACTERISTIC IMPEDANCE 35

I(z) = I+0 e−γz + I−0 e

+γz (3.20)

where V +0 , V −

0 , I+0 , and I−0 are complex-valuedconstants. It is shown in Section 3.8 thatEquations 3.19 and 3.20 represent sinusoidalwaves propagating in both the +z and −zdirections along the length of the line. Thevalues of the constants are determined byboundary conditions; i.e., constraints on V (z)

and I(z) at some position(s) along the line.These constants may represent sources, loads, orsimply discontinuities in the materials and/orgeometry of the line.

The reader is encouraged to verify that theEquations 3.19 and 3.20 are in fact solutions toEquations 3.15 and 3.18, respectively, for anyvalues of the constants V +

0 , V −0 , I+0 , and I−0 .

3.7 CharacteristicImpedance

[m0052]

Characteristic impedance is the ratio of voltageto current for a wave which is propagating insingle direction on a transmission line. This is animportant parameter in the analysis and designof circuits and systems using transmission lines.In this section we formally define this parameterand derive an expression for this parameter interms of the equivalent circuit model introducedin Section 3.4.

Consider a transmission line aligned along the zaxis. Employing some results from Section 3.6,recall that the phasor form of the wave equationin this case is

∂2

∂z2V (z)− γ2 V (z) = 0 (3.21)

where

γ ,√(R′ + jωL′) (G′ + jωC ′) (3.22)

Equation 3.21 relates the potential phasor V (z)to the equivalent circuit parameters R′, G′, C ′,and L′. The same equation relates the current

phasor I(z) to these equivalent circuitparameters:

∂2

∂z2I(z)− γ2 I(z) = 0 (3.23)

Since both V (z) and I(z) satisfy the same linearhomogeneous differential equation, they maydiffer by no more than a multiplicative constant.Since V (z) is potential and I(z) is current, thatconstant can be expressed in units of impedance.Specifically, this is the characteristic impedance,so-named because it depends only on thematerials and cross-sectional geometry of thetransmission line – i.e., things which determine γ– and not length, excitation, termination, orposition along the line.

To derive the characteristic impedance, firstrecall that the general solutions toEquations 3.21 and 3.23 are

V (z) = V +0 e

−γz + V −0 e

+γz (3.24)

I(z) = I+0 e−γz + I−0 e

+γz (3.25)

where V +0 , V −

0 , I+0 , and I−0 are complex-valuedconstants whose values are determined byboundary conditions; i.e., constraints on V (z)

and I(z) at some position(s) along the line. Alsowe will make use of the telegrapher’s equations(Section 3.5):

− ∂

∂zV (z) = [R′ + jωL′] I(z) (3.26)

− ∂

∂zI(z) = [G′ + jωC ′] V (z) (3.27)

We begin by differentiating Equation 3.24 withrespect to z, which yields

∂

∂zV (z) = −γ

[V +0 e

−γz − V −0 e

+γz]

(3.28)

Now we use this this to eliminate ∂V (z)/∂z inEquation 3.26, yielding

γ[V +0 e

−γz − V −0 e

+γz]= [R′ + jωL′] I(z)

(3.29)

Solving the above equation for I(z) yields:

I(z) =γ

R′ + jωL′

[V +0 e

−γz − V −0 e

+γz]

(3.30)


Comparing this to Equation 3.25, we note

I+0 =γ

R′ + jωL′V +0 (3.31)

I−0 =−γ

R′ + jωL′V −0 (3.32)

We now make the substitution

Z0 =R′ + jωL′

γ(3.33)

and observe

V +0

I+0=

−V −0

I−0, Z0 (3.34)

As anticipated, we have found that coefficients inthe equations for potentials and currents arerelated by an impedance, namely, Z0.Subsequently, Z0 may be defined in terms of theratio of the associated constants.

Characteristic impedance can be written entirelyin terms of the equivalent circuit parameters bysubstituting Equation 3.22 into Equation 3.33,yielding:

Z0 =

√R′ + jωL′

G′ + jωC ′(3.35)

The characteristic impedance Z0 (Ω) is theratio of potential to current in a wave travel-ing in a single direction along the transmis-sion line.

Take care to note that Z0 is not the ratio of V (z)

to I(z) in general; rather, Z0 relates only thepotential and current waves traveling in the samedirection.

Additional Reading:

• Characteristic impedance on Wikipedia.

3.8 Wave Propagation on aTransmission Line

[m0080]

In Section 3.6 it is shown that expressions for thephasor representations of the potential andcurrent along a transmission line are

V (z) = V +0 e

−γz + V −0 e

+γz (3.36)

I(z) = I+0 e−γz + I−0 e

+γz (3.37)

where γ is the propagation constant and itassumed that the transmission line is alignedalong the z axis. In this section we demonstratethat these expressions represent sinusoidal waves,and point out some important features. Beforeattempting this section the reader should befamiliar with the contents of Sections 3.4, 3.6,and 3.7. A refresher on fundamental waveconcepts (Section 1.3) may also be helpful.

We first define real-valued quantities α and β tobe the real and imaginary components of γ; i.e.,

α , Re γ (3.38)

β , Im γ (3.39)

and subsequently

γ = α+ jβ (3.40)

Then we observe

e±γz = e±(α+jβ)z = e±αz e±jβz (3.41)

It may be easier to interpret this expression byreverting to the time domain:

Ree±γzejωt

= e±αz cos (ωt± βz) (3.42)

Thus, e−γz represents a damped sinusoidal wavetraveling in the +z direction, and e+γz representsa damped sinusoidal wave traveling in the −zdirection.

Let’s define V +(z) and I+(z) to be the potentialand current associated with a wave propagatingin the +z direction. Then:

V +(z) , V +0 e

−γz (3.43)

or equivalently in the time domain:

v+(z, t) = ReV +(z) ejωt

= ReV +0 e

−γzejωt

=∣∣V +

0

∣∣ e−αz cos (ωt− βz + ψ) (3.44)

https://en.wikipedia.org/wiki/Characteristic_impedance

3.8. WAVE PROPAGATION ON A TRANSMISSION LINE 37

λ

z

v+(z,t=0)

Figure 3.12: The potential v+(z, t) of the wavetraveling in the +z direction at t = 0 for ψ = 0.

where ψ is the phase of V +0 . Figure 3.12 shows

v+(z, t). From fundamental wave theory werecognize

β , Re γ (rad/m) is the phase propagationconstant, which is the rate at which phasechanges as a function of distance.

Subsequently the wavelength in the line is

λ =2π

β(3.45)

Also from fundamental wave theory we recognize

α , Im γ (1/m) is the attenuation con-stant, which is the rate at which magnitudediminishes as a function of distance.

Sometimes the units of α are indicated as“Np/m” (“nepers” per meter), where the term“neper” is used to indicate the units of theotherwise unitless real-valued exponent of theconstant e.

Note that α = 0 for a wave that does notdiminish in magnitude with increasing distance,in which case the transmission line is said to belossless. If α > 0 then the line is said to be lossy(or possibly “low loss” if the loss can beneglected), and in this case the rate at which the

magnitude decreases with distance increases withα.

Next let us consider the speed of the wave. Toanswer this question, we need to be a bit morespecific about what we mean by “speed”. At themoment we mean phase velocity; that is, thespeed at which a point of constant phase seemsto move through space. In other words, whatdistance ∆z does a point of constant phasetraverse in time ∆t? To answer this question, wefirst note that the phase of v+(z, t) can bewritten generally as

ωt− βz + φ

where φ is some constant. Similarly the phase atsome time ∆t later and some point ∆z furtheralong can be written as

ω (t+∆t)− β (z +∆z) + φ

The phase velocity vp is ∆z/∆t when these twophases are equal; i.e., when

ωt− βz + φ = ω (t+∆t)− β (z +∆z) + φ

Solving for vp = ∆z/∆t, we obtain:

vp =ω

β(3.46)

Having previously noted that β = 2π/λ, theabove expression also yields the expected result

vp = λf (3.47)

The phase velocity vp = ω/β = λf is thespeed at which a point of constant phase trav-els along the line.

Returning now to consider the current associatedwith the wave traveling in the +z direction:

I+(z) = I+0 e−γz (3.48)

We can rewrite this expression in terms of thecharacteristic impedance Z0, as follows:

I+(z) =V +0

Z0e−γz (3.49)


Similarly we find that the current I−(z)

associated with V −(z) for the wave traveling inthe −z direction is

I−(z) =−V −

0

Z0e−γz (3.50)

The negative sign appearing in the aboveexpression emerges as a result of the signconventions used for potential and current in thederivation of the telegrapher’s equations(Section 3.5). The physical significance of thischange of sign is that wherever the potential ofthe wave traveling in the −z direction is positive,then the current at the same point is flowing inthe −z direction.

It is frequently necessary to consider thepossibility that waves travel in both directionssimultaneously. A very important case where thisarises is when there is reflection from adiscontinuity of some kind; e.g., from atermination which is not perfectlyimpedance-matched. In this case the totalpotential V (z) and total current I(z) can beexpressed as the general solution to the waveequation; i.e., as the sum of the “incident”(+z-traveling) wave and the reflected(−z-traveling) waves:

V (z) = V +(z) + V −(z) (3.51)

I(z) = I+(z) + I−(z) (3.52)

The existance of waves propagatingsimultaneously in both directions gives rise to aphenomenon known as a standing wave.Standing waves and the calculation of thecoefficients V −

0 and I−0 due to reflection areaddressed in Sections 3.13 and 3.12 respectively.

3.9 Lossless and Low-LossTransmission Lines

[m0083]

Quite often the loss in a transmission line issmall enough that it may be neglected. In thiscase several aspects of transmission line theory

may be simplified. In this section we presentthese simplifications.

First, recall that “loss” refers to the reduction ofmagnitude as a wave propagates through space.In the lumped-element equivalent circuit model(Section 3.4), the parameters R′ and G′ of therepresent physical mechanisms associated withloss. Specifically, R′ represents the resistance ofconductors, whereas G′ represents theundesirable current induced between conductorsthrough the spacing material. Also recall that thepropagation constant γ is, in general, given by

γ ,√(R′ + jωL′) (G′ + jωC ′) (3.53)

With this in mind, we now define “low loss” asmeeting the conditions:

R′ ≪ ωL′ (3.54)

G′ ≪ ωC ′ (3.55)

When these conditions are met, the propagationconstant simplifies as follows:

γ ≈√(jωL′) (jωC ′)

=√−ω2L′C ′

= jω√L′C ′ (3.56)

and subsequently

α , Re γ ≈ 0 (low-loss approx.) (3.57)

β , Im γ ≈ ω√L′C ′ (low-loss approx.) (3.58)

vp = ω/β ≈ 1√L′C ′

(low-loss approx.) (3.59)

Similarly:

Z0 =

√R′ + jωL′

G′ + jωC ′≈√L′

C ′(low-loss approx.)

(3.60)

Of course if the line is strictly lossless (i.e.,R′ = G′ = 0) then these are not approximations,but rather the exact expressions.

In practice these approximations are quitecommonly used, since practical transmission linestypically meet the conditions expressed in

3.10. COAXIAL LINE 39

Equations 3.54 and 3.55, and the resultingexpressions are much simpler. We furtherobserve that Z0 is approximately independent offrequency when these conditions hold.

However also note that “low loss” does not mean“no loss”, and it is common to apply theseexpressions even when R′ and/or G′ is largeenough to yield significant loss. For example, acoaxial cable used to connect an antenna on atower to a radio near the ground typically hasloss that is important to consider in the analysisand design process, but nevertheless satisfiesEquations 3.54 and 3.55. In this case the low-lossexpressions for β and Z0 are used, but α mightnot be approximated as zero.

3.10 Coaxial Line

[m0143]

Coaxial transmission line consists of metallicinner and outer conductors separated by a spacermaterial is shown in Figure 3.13. The spacermaterial is typically a low-loss dielectric materialhaving permeability µ ≈ µ0 and permittivity ǫswhich may range from very near ǫ0 (air-filledline) to a factor of 2–3 times ǫ0. The outerconductor is alternatively referred to as the“shield”, since it typically provides a high degreeof isolation from nearby objects andelectromagnetic fields. Coaxial line issingle-ended3 in the sense that the conductorgeometry is asymmetric and the shield isnormally attached to ground at both ends. Thesecharacteristics make coaxial line attractive forconnecting circuits in separate enclosures, andfor connecting antennas to receivers andtransmitters.

Coaxial line exhibits TEM field structure asshown in Figure 3.14.

Expressions for the equivalent circuit parametersC ′ and L′ for coaxial line can be obtained from

3The references in “Additional Reading” at the end ofthis section may be helpful if you are not familiar with thisconcept.

rσs

b

a

Figure 3.13: Cross-section of a coaxial transmis-sion line, indicating design parameters.

Figure 3.14: Structure of the electric and magneticfields within coaxial line. In this case, the wave ispropagating away from the viewer.


basic electromagnetic theory. It is shown inSection 5.24 that the capacitance per unit lengthis

C ′ =2πǫs

ln (b/a)(3.61)

where a and b are the radii of the inner and outerconductors, respectively. Using analysis shown inSection 7.14, the inductance per unit length is

L′ ≈ µ0

2πln

(b

a

)(3.62)

The loss conductance G′ depends on theconductance σs of the spacer material, and isgiven by

G′ =2πσs

ln (b/a)(3.63)

This expression is derived in Section 6.5.

The loss resistance per unit length, R′, isrelatively difficult to describe in a general way.One obstacle is that the inner and outerconductors typically consist of different materialsor compositions of materials. The innerconductor is not necessarily a singlehomogeneous material; instead, the innerconductor may consist of a variety of materialsselected by trading-off between conductivity,strength, weight, and cost. Similarly the outerconductor is not necessarily homogeneous; for avariety of reasons the outer conductor mayinstead be a metal mesh, braid, or a composite ofmaterials. Another complicating factor is thefact that the resistance of the conductor variessignificantly with frequency, whereas C ′, L′, andG′ exhibit relatively little variation from theirelectro- and magnetostatic values. These factorsmake it difficult to devise a single expression forR′ that is both as simple as those shown abovefor the other parameters, and generallyapplicable. Fortunately, it turns out that thelow-loss conditions R′ ≪ ωL′ and G′ ≪ ωC ′ arenormally applicable,4 so that R′ and G′ areimportant only if it is necessary to compute loss.

Since the low-loss conditions are normally met, aconvenient expression for the characteristicimpedance is obtained from Equations 3.61 and

4See Section 3.9 for a reminder about this concept.

3.62 for L′ and C ′ respectively:

Z0 ≈√L′

C ′(low-loss)

≈ 1

2π

√µ0

ǫslnb

a(3.64)

The spacer permittivity can be expressed asǫs = ǫrǫ0 where ǫr is the relative permittivity ofthe spacer material. Since

√µ0/ǫ0 is a constant,

the above expression is commonly written

Z0 ≈ 60 Ω√ǫr

lnb

a(low-loss) (3.65)

Thus, it is possible to express Z0 directly interms of parameters describing the geometry (aand b) and material (ǫr) used in the line, withoutthe need to first compute the values ofcomponents in the lumped-element equivalentcircuit model.

Example 3.1. RG-59 Coaxial Cable.

RG-59 is a very common type of coaxialline. Figure 3.15 shows a section of RG-59cut away so as to reveal its structure. Theradii are a ∼= 0.292 mm and b ≈ 1.855 mm(mean), yielding L′ ≈ 370 nH/m. Thespacer material is polyethylene havingǫr ∼= 2.25, yielding C ′ ≈ 67.7 pF/m. Theconductivity of polyethylene isσs ∼= 5.9× 10−5 S/m, yieldingG′ ≈ 200 µS/m. Typical resistance per unitlength R′ is on the order of 0.1 Ω/m nearDC, increasing approximately in proportionto the square root of frequency.

From the above values, we find that RG-59satisfies the low-loss criteria R′ ≪ ωL′ forf ≫ 43 kHz and G′ ≪ ωC ′ for f ≫ 470 kHz.Under these conditions, we findZ0 ≈

√L′/C ′ ∼= 74 Ω. Thus, the ratio of the

potential to the current in a wave travelingin a single direction on RG-59 is about 74 Ω.

The phase velocity of RG-59 is found to bevp ≈ 1/

√L′C ′ ∼= 2× 108 m/s, which is about

67% of c, the speed of electromagneticradiation in free space. In other words, a

3.11. MICROSTRIP LINE 41

c© Arj CC BY SA 3.0

Figure 3.15: RG-59 coaxial line. A: Insulatingjacket. B: Braided outer conductor. C: Dielectricspacer. D: Inner conductor.

signal that takes 1 ns to traverse a distance lin free space requires about 1.5 ns totraverse a length-l section of RG-59. Sincevp = λf , a wavelength in RG-59 is 67% of awavelength in free space.

Using the expression

γ ,√

(R′ + jωL′) (G′ + jωC ′) (3.66)

with R′ = 0.1 Ω/m, and then taking the realpart to obtain α, we find α ∼ 0.01 m−1. So,for example, the magnitude of the potentialor current is decreased by about 50% bytraveling a distance of about 70 m. In otherwords, e−αl = 0.5 for l ∼ 70 m at relativelylow frequencies, and increases withincreasing frequency.

Additional Reading:

• “Coaxial cable” on Wikipedia. Includesdescriptions and design parameters for avariety of commonly-encountered coaxialcables.

• “Single-ended signaling” on Wikipedia.

• Sec. 8.7 (“Differential Circuits”) inS.W. Ellingson, Radio Systems Engineering,Cambridge Univ. Press, 2016.

hW

l

r

t

0

c© 7head7metal7 CC BY SA 3.0 (modified)

Figure 3.16: Microstrip transmission line struc-ture and design parameters.

3.11 Microstrip Line

[m0082]

Microstrip transmission line consists of a narrowmetallic trace separated from a metallic groundplane by a slab of dielectric material, as shown inFigure 3.16. This is a natural way to implementa transmission line on a printed circuit board,and so accounts for an important and expansiverange of applications. The reader should beaware that microstrip is distinct from stripline,which is a very different type of transmissionline; see “Additional Reading” at the end of thissection for disambiguation of these terms.

Microstrip line is single-ended5 in the sense thatthe conductor geometry is asymmetric and theone conductor – namely, the ground plane – alsonormally serves as ground for the source andload.

The spacer material is typically a low-lossdielectric material having permeability µ ≈ µ0

and relative permittivity ǫr in the range 2 toabout 10 or so.

Microstrip line nominally exhibits TEM fieldstructure. This structure is shown in Figure 3.17.Note that electric and magnetic fields exist bothin the dielectric and in the space above thedielectric, which is typically (but not always!)air. This complex field structure makes itdifficult to describe microstrip line concisely in

5The reference in “Additional Reading” at the end ofthis section may be helpful if you are not familiar with thisconcept.


https://en.wikipedia.org/wiki/Coaxial_cable

https://en.wikipedia.org/wiki/Single-ended_signaling



h

W

Figure 3.17: Structure of the electric and magneticfields within microstrip line. (The fields outsidethe line are possibly significant, complicated, andnot shown.) In this case, the wave is propagatingaway from the viewer.

terms of the equivalent circuit parameters of thelumped-element model. Instead, expressions forZ0 directly in terms of h/W and ǫr are typicallyused instead. A variety of these expressions arein common use, representing differentapproximations and simplifications. Awidely-accepted and broadly-applicableexpression is:6

Z0 ≈ 42.4 Ω√ǫr + 1

×

ln

[1 +

4h

W ′

(Φ+

√Φ2 +

1 + 1/ǫr2

π2

)]

(3.67)

where

Φ ,14 + 8/ǫr

11

(4h

W ′

)(3.68)

and W ′ is W adjusted to account for thethickness t of the microstrip line. Typicallyt≪W and t≪ h, for which W ′ ≈W . Simplerapproximations for Z0 are also commonlyemployed in the design and analysis of microstriplines. These expressions are limited in the rangeof h/W for which they are valid, and can usuallybe shown to be special cases or approximationsof Equation 3.67. Nevertheless they aresometimes useful for quick “back of theenvelope” calculations.

Accurate expressions for wavelength λ, phasepropagation constant β, and phase velocity vp

6This is from Wheeler 1977, cited in “Additional Read-ing” at the end of this section.

are similarly difficult to obtain for waves inmicrostrip line. An approximate techniqueemploys a result from the theory of uniformplane waves in unbounded media (Equation 9.38from Section 9.2):

β = ω√µǫ (3.69)

It turns out that the electromagnetic fieldstructure in the space between the conductors iswell-approximated as that of a uniform planewave in unbounded media having the samepermeability µ0 but a different relativepermittivity, which we shall assign the symbolǫr,eff (for “effective relative permittivity”). Then

β ≈ ω√µ0 ǫr,eff ǫ0 (low-loss microstrip)

= β0√ǫr,eff (3.70)

In other words, the phase propagation constantin microstrip can be approximated as thefree-space phase propagation β0 , ω

√µ0ǫ0 times

a correction factor√ǫr,eff . Then ǫr,eff may be

crudely approximated as follows:

ǫr,eff ≈ ǫr + 1

2(3.71)

i.e., ǫr,eff is roughly the average of the relativepermittivity of the dielectric slab and the relativepermittivity of free space. The assumptionemployed here is that ǫr,eff is approximately theaverage of these values because some fraction ofthe power in the guided wave is in the dielectric,and the rest is above the dielectric. Variousapproximations are available to improve on thisapproximation; however in practice variations inthe value of ǫr for the dielectric due tomanufacturing processes typically make a moreprecise estimate irrelevant.

Using this concept, we obtain

λ =2π

β=

2π

β0√ǫr,eff

=λ0√ǫr,eff

(3.72)

where λ0 is the free-space wavelength c/f .Similarly the phase velocity vp, can be estimatedusing the relationship

vp =ω

β=

c√ǫr,eff

(3.73)

3.12. VOLTAGE REFLECTION COEFFICIENT 43

i.e., the phase velocity in microstrip is slowerthan c by a factor of

√ǫr,eff .

Example 3.2. 50 Ω Microstrip in FR4Printed Circuit Boards.

FR4 is a low-loss fiberglass epoxy dielectricwhich is commonly used to make printedcircuit boards (see “Additional Reading” atthe end of this section). FR4 circuit boardmaterial is commonly sold in a slab havingthickness h ∼= 1.575 mm with ǫr ∼= 4.5. Letus consider how we might implement amicrostrip line having Z0 = 50 Ω using thismaterial. Since h and ǫr are fixed, the onlyparameter remaining to set Z0 is W . A bitof experimentation with Equation 3.67reveals that h/W ≈ 1/2 yields Z0 ≈ 50 Ω forǫr = 4.5. Thus W should be about 3.15 mm.The effective relative permittivity is

ǫr,eff ≈ (4.5 + 1)/2 = 2.75

so the phase velocity for the wave guided bythis line is about c/

√2.75; i.e., 60% of c.

Similarly the wavelength of this wave isabout 60% of the free space wavelength.

Additional Reading:

• “Microstrip” on Wikipedia.

• “Printed circuit board” on Wikipedia.

• “Stripline” on Wikipedia.

• “Single-ended signaling” on Wikipedia.

• Sec. 8.7 (“Differential Circuits”) inS.W. Ellingson, Radio Systems Engineering,Cambridge Univ. Press, 2016.

• H.A. Wheeler, “Transmission LineProperties of a Strip on a Dielectric Sheet ona Plane,” IEEE Trans. Microwave Theory &Techniques, Vol. 25, No. 8, Aug 1977,pp. 631–47.

• “FR-4” on Wikipedia.

ZL

Z0

I

=0

V

+

L

L

z

z

V+

V

Figure 3.18: A wave arriving from the left incidenton a termination located at z = 0.

3.12 Voltage ReflectionCoefficient

[m0084]

We now consider the scenario shown inFigure 3.18. Here a wave arriving from the leftalong a lossless transmission line havingcharacteristic impedance Z0 arrives at atermination located at z = 0. The impedancelooking into the termination is ZL, which may bereal-, imaginary-, or complex-valued. Thequestions are: Under what circumstances is areflection – i.e., a leftward traveling wave –expected, and what is precisely is that wave?

The potential and current of the incident waveare related by the constant value of Z0. Similarly,the potential and current of the incident waveare related by Z0. Therefore it suffices toconsider either potential or current. Choosingpotential, we may express the incident wave as

V +(z) = V +0 e

−jβz (3.74)

where V +0 is determined by the source of the

wave, and so is effectively a “given”. Anyreflected wave must have the form

V −(z) = V −0 e

+jβz (3.75)

Therefore the problem is solved by determiningthe value of V −

0 given V +0 , Z0, and ZL.

Considering the situation at z = 0, note that by

https://en.wikipedia.org/wiki/Microstrip

https://en.wikipedia.org/wiki/Printed_circuit_board

https://en.wikipedia.org/wiki/Stripline

https://en.wikipedia.org/wiki/Single-ended_signaling

https://en.wikipedia.org/wiki/FR-4


definition we have

ZL ,VL

IL(3.76)

where VL and IL are the potential across andcurrent through the termination, respectively.Also the potential and current on either side ofthe z = 0 interface must be equal; thus:

V +(0) + V −(0) = VL (3.77)

I+(0) + I−(0) = IL (3.78)

where I+(z) and I−(z) are the currents

associated with V +(z) and V −(z), respectively.Since the voltage and current are related by Z0,Equation 3.78 may be rewritten as follows:

V +(0)

Z0− V −(0)

Z0= IL (3.79)

Evaluating the left sides of Equations 3.77 and3.79 at z = 0, we find:

V +0 + V −

0 = VL (3.80)

V +0

Z0− V −

0

Z0= IL (3.81)

Substituting these expressions into Equation 3.76we obtain:

ZL =V +0 + V −

0

V +0 /Z0 − V −

0 /Z0

(3.82)

Solving for V −0 we obtain

V −0 =

ZL − Z0

ZL + Z0V +0 (3.83)

Thus the answer to the question posed earlier isthat

V −0 = ΓV +

0 , where (3.84)

Γ ,ZL − Z0

ZL + Z0(3.85)

The quantity Γ is known as the voltage reflectioncoefficient. Note that when ZL = Z0, Γ = 0 andtherefore V −

0 = 0. In other words,

If the terminating impedance is equal to thecharacteristic impedance of the transmissionline, then there is no reflection.

If on the other hand ZL 6= Z0, then |Γ| > 0,V −0 = ΓV +

0 , and a leftward-traveling reflectedwave exists.

Since ZL may be real-, imaginary-, orcomplex-valued, Γ too may be real-, imaginary-,or complex-valued. Therefore V −

0 may bedifferent from V +

0 in magnitude, sign, or phase.

Note also that Γ is not the ratio of I−0 to I+0 .The ratio of the current coefficients is actually−Γ (This is relatively straightforward to prove,and is left as an exercise for the student). It isquite simple to show this with a simplemodification to the above procedure, and is leftas an exercise for the student.

Summarizing:

The voltage reflection coefficient Γ, given byEquation 3.85, determines the magnitude andphase of the reflected wave given the inci-dent wave, the characteristic impedance ofthe transmission line, and the terminatingimpedance.

[m0085]

We now consider values of Γ that arise forcommonly-encountered terminations.

Matched Load (ZL = Z0). In this case thetermination may be a device with impedance Z0,or the termination may be another transmissionline having the same characteristic impedance.When ZL = Z0, Γ = 0 and there is no reflection.

Open Circuit. An “open circuit” is the absenceof a termination. This condition impliesZL → ∞, and subsequently Γ → +1. Since thecurrent reflection coefficient is −Γ, the reflectedcurrent wave is 180 out of phase with theincident current wave, making the total currentat the open circuit equal to zero, as expected.

Short Circuit. “Short circuit” means ZL = 0,

3.13. STANDING WAVES 45

and subsequently Γ = −1. In this case the phaseof Γ is 180, and therefore the potential of thereflected wave cancels the potential of theincident wave at the open circuit, making thetotal potential equal to zero, as it must be. Sincethe current reflection coefficient is −Γ, thereflected current wave is in phase with theincident current wave, making the magnitude ofthe total current at the short circuit greater thanzero, as expected.

Purely Reactive Load. A purely reactive load,including that presented by a capacitor orinductor, has ZL = jX where X is reactance. Inparticular, an inductor is represented by X > 0and a capacitor is represented by X < 0. We find

Γ =−Z0 + jX

+Z0 + jX(3.86)

The numerator and denominator have the samemagnitude, so |Γ| = 1. Let φ be the phase of thedenominator (+Z0 + jX). Then the phase of thenumerator is π − φ. Subsequently the phase of Γis (π − φ)− φ = π − 2φ. Thus we see that thephase of Γ is no longer limited to be 0 or 180,but can be any value in between. The phase ofreflected wave is subsequently shifted by thisamount.

Other Terminations. Any other termination,including series and parallel combinations of anynumber of devices, can be expressed as a value ofZL which is, in general, complex-valued.Nevertheless, the value of |Γ| is limited to therange 0 to 1. To see this, note:

Γ =ZL − Z0

ZL + Z0=ZL/Z0 − 1

ZL/Z0 + 1(3.87)

Note that the smallest possible value of |Γ|occurs when the numerator is zero; i.e., whenZL = Z0. Therefore the smallest value of |Γ| iszero. The largest possible value of |Γ| occurswhen ZL/Z0 → ∞ (i.e., an open circuit) or whenZL/Z0 = 0 (a short circuit); the result in eithercase is |Γ| = 1. Thus

0 ≤ |Γ| ≤ 1 (3.88)

3.13 Standing Waves

[m0086]

A standing wave consists of waves moving inopposite directions. These waves add to make adistinct magnitude variation as a function ofdistance that does not vary in time.

To see how this can happen, first consider thatan incident wave V +

0 e−jβz, which is traveling in

the +z axis along a lossless transmission line.Associated with this wave is a reflected waveV −0 e

+jβz = ΓV +0 e

+jβz, where Γ is the voltagereflection coefficient. These waves add to makethe total potential

V (z) = V +0 e

−jβz + ΓV +0 e

+jβz

= V +0

(e−jβz + Γe+jβz

) (3.89)

The magnitude of V (z) is most easily found by

first finding |V (z)|2, which is:

V (z)V ∗(z)

=|V +0 |2


) (e−jβz + Γe+jβz

)∗

=|V +0 |2


) (e+jβz + Γ∗e−jβz

)

=|V +0 |2

(1 + |Γ|2 + Γe+j2βz + Γ∗e−j2βz

)(3.90)

Let φ be the phase of Γ; i.e.,

Γ = |Γ| ejφ (3.91)

Then, continuing from the previous expression:

|V +0 |2

(1 + |Γ|2 + |Γ| e+j(2βz+φ) + |Γ| e−(2βz+φ)

)

=|V +0 |2

(1 + |Γ|2 + |Γ|

[e+j(2βz+φ) + e−(2βz+φ)

])

(3.92)

The quantity in square brackets can be reducedto a cosine function using the identity

cos θ =1

2

[ejθ + e−jθ

]

yielding:

|V +0 |2

[1 + |Γ|2 + 2 |Γ| cos (2βz + φ)

](3.93)


Recall that this is |V (z)|2. |V (z)| is therefore thesquare root of the above expression:

∣∣∣V (z)∣∣∣ = |V +

0 |√1 + |Γ|2 + 2 |Γ| cos (2βz + φ)

(3.94)Thus we have found that the magnitude of theresulting total potential varies sinusoidally alongthe line. This is referred to as a standing wavebecause the variation of the magnitude of thephasor resulting from the interference betweenthe incident and reflected waves does not varywith time.

We may perform a similar analysis of thecurrent, leading to:

∣∣∣I(z)∣∣∣ =

|V +0 |Z0

√1 + |Γ|2 − 2 |Γ| cos (2βz + φ)

(3.95)Again we find the result is a standing wave.

Now let us consider the outcome for a few specialcases.

Matched load. When the impedance of thetermination of the transmission line, ZL, is equalto the characteristic impedance of thetransmission line, Z0, Γ = 0 and there is noreflection. In this case the above expressionsreduce to |V (z)| = |V +

0 | and |I(z)| = |V +0 |/Z0, as

expected.

Open or Short-Circuit. In this case Γ = ±1and we find:

∣∣∣V (z)∣∣∣ = |V +

0 |√2 + 2 cos (2βz + φ) (3.96)

∣∣∣I(z)∣∣∣ =

|V +0 |Z0

√2− 2 cos (2βz + φ) (3.97)

where φ = 0 for an open circuit and φ = π for ashort circuit. The result for an open circuittermination is shown in Figure 3.19(a)(potential) and 3.19(b) (current). The result fora short circuit termination is identical except theroles of potential and current are reversed. Ineither case, note that voltage maxima correspondto current minima, and vice versa.

Also, note that the period of the standing wave isλ/2; i.e., one-half of a wavelength. This can be

z

V+

02

V+

0

V(z)

z=0

(a) Potential.

I(z)

z=0

z

I+

02

I+

0

(b) Current.

Figure 3.19: Standing wave associated with anopen-circuit termination.

3.14. STANDING WAVE RATIO 47

1.25

1.00

0.75

0.50

0.00

0 2 3z

1.50

2.00

2

32

52

1.75

0.25

|V|

0/ V

[rad]

ΓΓ

Γ=1

Γ

Γ

by Inductiveload (modified)

Figure 3.20: Standing waves associated withloads exhibiting various reflection coefficients.

confirmed as follows: First note that thefrequency argument of the cosine function of thestanding wave is 2βz. This can be rewritten as2π (β/π) z, so the frequency of variation is β/πand the period of the variation is π/β. Sinceβ = 2π/λ, we see that the period of the variationis λ/2. Furthermore, this is true regardless of Γ.

Mismatched loads. A common situation isthat the termination is neither perfectly-matched(Γ = 0) nor an open/short circuit (|Γ| = 1).Examples of the resulting standing waves areshown in Figure 3.20.

Additional Reading:

• “Standing Wave” on Wikipedia.

3.14 Standing Wave Ratio

[m0081]

Precise matching of transmission lines toterminations is often not practical or possible.Whenever a significant mismatch exists, astanding wave (Section 3.13) is apparent. Thequality of the match is commonly expressed interms of the standing wave ratio (SWR) of thisstanding wave. SWR is defined as the ratio of

the maximum magnitude of the standing wave tominimum magnitude of the standing wave. Interms of the potential:

SWR ,maximum |V |maximum |V |

(3.98)

SWR can be calculated using a simpleexpression, which we shall now derive. Recallfrom Section 3.13 that

∣∣∣V (z)∣∣∣ = |V +

0 |√1 + |Γ|2 + 2 |Γ| cos (2βz + φ)

(3.99)The maximum value occurs when the cosinefactor is equal to +1, yielding:

max∣∣∣V∣∣∣ = |V +

0 |√

1 + |Γ|2 + 2 |Γ| (3.100)

Note that the argument of the square rootoperator is equal to (1 + |Γ|)2; therefore:

max∣∣∣V∣∣∣ = |V +

0 | (1 + |Γ|) (3.101)

Similarly the minimum value is achieved whenthe cosine factor is equal to −1, yielding:

min∣∣∣V∣∣∣ = |V +

0 |√

1 + |Γ|2 − 2 |Γ| (3.102)

So:

min∣∣∣V∣∣∣ = |V +

0 | (1− |Γ|) (3.103)

Therefore:

SWR =1 + |Γ|1− |Γ| (3.104)

This relationship is shown graphically inFigure 3.21. Note that SWR ranges from 1 forperfectly-matched terminations (Γ = 0) toinfinity for open- and short-circuit terminations(|Γ| = 1).

It is sometimes of interest to find the magnitudeof the reflection coefficent given SWR. SolvingEquation 3.104 for |Γ| we find:

|Γ| = SWR− 1

SWR+ 1(3.105)

https://en.wikipedia.org/wiki/Standing_wave


0

2

4

6

8

10

0 0.2 0.4 0.6 0.8 1

Γ

!"

Figure 3.21: Relationship between SWR and |Γ|.

SWR is often referred to as the voltage standingwave ratio (VSWR), although repeating theanalysis above for the current reveals that thecurrent SWR is equal to potential SWR, so theterm “SWR” suffices.

SWR < 2 or so is usually considered a “goodmatch”, although some applications require SWR< 1.1 or better, and other applications aretolerant to SWR of 3 or greater.

Example 3.3. Reflection coefficient forvarious values of SWR.

What is the reflection coefficient for theabove-cited values of SWR? UsingEquation 3.105, we find:

• SWR = 1.1 corresponds to |Γ| = 0.0476.

• SWR = 2.0 corresponds to |Γ| = 1/3.

• SWR = 3.0 corresponds to |Γ| = 1/2.

3.15 Input Impedance of aTerminated LosslessTransmission Line

[m0087]

ZL

Z0

=0

z

z#= lz

l

Zin

Figure 3.22: A transmission line driven by a sourceon the left and terminated by an impedance ZL onthe right.

Consider Figure 3.22, which shows a losslesstransmission line being driven from the left andwhich is terminated by an impedance ZL on theright. If ZL is equal to the characteristicimpedance Z0 of the transmission line, then theinput impedance Zin will be equal to ZL.Otherwise Zin depends on both ZL and thecharacteristics of the transmission line. In thissection we determine a general expression for Zinin terms of ZL, Z0, the phase propagationconstant β, and the length l of the line.

Impedance is defined at the ratio of potential tocurrent, so:

Zin(l) ,V (z = −l)I(z = −l)

(3.106)

Using the coordinate system indicated inFigure 3.22, the interface between source andtransmission line is located at z = −l. Nowemploying expressions for V (z) and I(z) fromSection 3.13 with z = −l, we find:

Zin(l) =V +0

(e+jβl + Γe−jβl

)

V +0 (e+jβl − Γe−jβl) /Z0

= Z0e+jβl + Γe−jβl

e+jβl − Γe−jβl(3.107)

Multiplying both numerator and denominator bye−jβl:

Zin(l) = Z01 + Γe−j2βl

1− Γe−j2βl(3.108)

Recall that Γ in the above expression is:

Γ =ZL − Z0

ZL + Z0(3.109)

3.16. INPUT IMPEDANCE FOR OPEN- AND SHORT-CIRCUIT TERMINATIONS 49

Summarizing:

Equation 3.108 is the input impedance of alossless transmission line having characteris-tic impedance Z0 and which is terminatedinto a load ZL. The result also depends onthe length and phase propagation constant ofthe line.

It’s worth noting that Zin(l) is periodic in l.Since the argument of the complex exponentialfactors is 2βl, the frequency at which Zin(l) isβ/π; and since β = 2π/λ, the associated period isλ/2. Not surprisingly, this is the same periodexhibited by the standing wave (Section 3.13),and underscores that the variation in impedancewith length is due to the interference of incidentand reflected waves.

Also worth noting is that Equation 3.108 can bewritten entirely in terms of ZL and Z0, since Γdepends only on these two parameters. Here’sthat version of the expression:

Zin(l) =ZL + jZ0 tanβl

Z0 + jZL tanβl(3.110)

This expression can be derived by substitutingEquation 3.109 into Equation 3.108, and is left asan exercise for the student.

3.16 Input Impedance forOpen- andShort-CircuitTerminations

[m0088]

Let us now consider the input impedance of atransmission line which is terminated in an open-or short-circuit. Such a transmission line issometimes referred to as a stub. First: Whyconsider such a thing? From a “lumped element”circuit theory perspective, this would not seem tohave any particular application. However, thefact that this structure exhibits an inputimpedance that depends on length (Section 3.15)

enables some very useful applications. First,however, let us consider the answer to thequestion at hand: What is the input impedancewhen the transmission line is open- orshort-circuited.

For a short circuit, ZL = 0, Γ = −1, and we find


1− Γe−j2βl

= Z01− e−j2βl

1 + e−j2βl(3.111)

Multiplying numerator and denominator by e+jβl

we obtain

Zin(l) = Z0e+jβl − e−jβl

e+jβl + e−jβl(3.112)

Now we invoke the following trigonometricidentities:

cos θ =1

2

[e+jθ + e−jθ

](3.113)

sin θ =1

j2

[e+jθ − e−jθ

](3.114)

Employing these identities, we obtain:

Zin(l) = Z0j2 (sinβl)

2 (cosβl)(3.115)

and finally:

Zin(l) = +jZ0 tan (βl) (3.116)

Figure 3.23(a) shows what’s going on. Asexpected, Zin = 0 when l = 0, since this amountsto a short circuit with no transmission line. AlsoZin varies periodically with increasing length,with period λ/2. This is precisely as expectedfrom standing wave theory (Section 3.13). Whatis of particular interest now is that as l → λ/4 wesee Zin → ∞: Remarkably, the transmission linehas essentially transformed the short circuittermination into an open circuit!

For an open circuit termination, ZL → ∞,Γ = +1, and we find


1− Γe−j2βl

= Z01 + e−j2βl

1− e−j2βl(3.117)


0

0 0.25 0.5 0.75 1

imag

inar

y pa

rt o

f inp

ut im

peda

nce

length [wavelengths]

(a) Short-circuit termination (ZL = 0).

0

0 0.25 0.5 0.75 1

imag

inar

y pa

rt o

f inp

ut im

peda

nce

length [wavelengths]

(b) Open-circuit termination (ZL → ∞).

Figure 3.23: Input reactance (ImZin) of a stub.ReZin is always zero.

Following the same procedure detailed above forthe short-circuit case, we find

Zin(l) = −jZ0 cot (βl) (3.118)

Figure 3.23(b) shows the result for open-circuittermination. As expected, Zin → ∞ for l = 0,and the same λ/2 periodicity is observed. Whatis of particular interest now is that at l = λ/4 wesee Zin = 0: In this case the transmission linehas transformed the open circuit termination intoa short circuit.

Now taking stock of what we have determined:

The input impedance of a short- or open-circuited lossless transmission line is com-pletely imaginary-valued and is given byEquations 3.116 and 3.118, respectively.

The input impedance of a short- or open-circuited lossless transmission line periodicin length with period λ/2 and alternates be-tween open- and short-circuit conditions witheach λ/4-increase in length.

Additional Reading:

• “Stub (electronics)” on Wikipedia.

3.17 Applications of Open-and Short-CircuitedTransmission LineStubs

[m0145]

The theory of open- and short-circuitedtransmission lines – often referred to as stubs –was addressed in Section 3.16. These structureshave important and wide-ranging applications.

In particular, these structures can be used toreplace discrete inductors and capacitors in

https://en.wikipedia.org/wiki/Stub_(electronics)

3.18. MEASUREMENT OF TRANSMISSION LINE CHARACTERISTICS 51

certain applications. To see this, consider theshort-circuited line (Figure 3.23(b) ofSection 3.16): Note that each value of l which isless than λ/4 corresponds to a particular positivereactance; i.e., the transmission line “looks” likean inductor. Also note that lengths between λ/4and λ/2 result in reactances which are negative;i.e., the transmission line “looks” like acapacitor. Thus, it is possible to replace aninductor or capacitor with a short-circuitedtransmission line of the appropriate length. Theinput impedance of such a transmission line isidentical to that of the inductor or capacitor atthe design frequency. The variation of reactancewith respect to frequency will not be identical,which may or may not be a concern dependingon the bandwidth and frequency responserequirements of the application. Open-circuitedlines may be used in a similar way.

This property of open- and short-circuitedtransmission lines makes it possible to implementimpedance matching circuits (see Section 3.23),filters, and other devices entirely fromtransmission lines, with fewer or no discreteinductors or capacitors required. Transmissionlines do not suffer the performance limitations ofdiscrete devices at high frequencies and are lessexpensive. A drawback of transmission line stubsin this application is that the lines are typicallymuch larger than the discrete devices they areintended to replace.

Example 3.4. Emitter Induction UsingShort-Circuited Line.

In the design of low-noise amplifiers usingbipolar transistors in common-emitterconfiguration, it is often useful to introducea little inductance between the emitter andground. This is known as “inductivedegeneration”, “emitter induction”, orsometimes by other names. It can bedifficult to find suitable inductors, especiallyfor operation in the UHF band and higher.However a microstrip line can be used toachieve the desired inductive impedance.Determine the length of a stub that

implements a 2.2 nH inductance at 6 GHzusing microstrip line with characteristicimpedance 50 Ω and phase velocity 0.6c.

Solution. At the design frequency, theimpedance looking into this section of linefrom the emitter should be equal to that ofa 2.2 nH inductor, which is+jωL = +j2πfL = +j82.9 Ω. The inputimpedance of a short-circuited stub oflength l which is grounded (thus,short-circuited) at the opposite end is+jZ0 tanβl (Section 3.16). Setting thisequal to +j82.9 Ω and noting thatZ0 = 50 Ω, we find that βl ∼= 1.028 rad. Thephase propagation constant is (Section 3.8):

β =ω

vp=

2πf

0.6c∼= 209.4 rad/m (3.119)

Therefore the length of the microstrip line isl = (βl) /β ∼= 4.9 mm.

Additional Reading:


3.18 Measurement ofTransmission LineCharacteristics

[m0089]

This section presents a simple technique formeasuring the characteristic impedance Z0,electrical length βl, and phase velocity vp of alossless transmission line. This technique requirestwo measurements: The input impedance Zinwhen the transmission line is short-circuted, andZin when the transmission line is open-circuited.

In Section 3.16 it is shown that the inputimpedance Zin of a short-circuited transmissionline is

Z(SC)in = +jZ0 tanβl

and when a transmission line is terminated in an



open circuit, the input impedance is

Z(OC)in = −jZ0 cotβl

Observe what happens when we multiply theseresults together:

Z(SC)in · Z(OC)

in = Z20

that is, the product of the measurements Z(OC)in

and Z(SC)in is simply the square of the

characteristic impedance. Therefore

Z0 =

√Z

(SC)in · Z(OC)

in (3.120)

If we instead divide these measurements, we find

Z(SC)in

Z(OC)in

= − tan2 βl

Therefore

tanβl =

[−Z

(SC)in

Z(OC)in

]2(3.121)

If l is known in advance to be less than λ/2, thenβl can be determined by taking the inversetangent. If l is of unknown length and longerthan λ/2, one must take care to account for theperiodicity of tangent function; in this case itmay not be possible to unambiguously determineβl. Although we shall not present the methodhere, it is possible to resolve this ambiguity bymaking multiple measurements over a range offrequencies.

Once βl is determined, it is simple to determine lgiven β, β given l, and then vp. For example, thephase velocity of a transmission line may bedetermined by first finding βl for a known lengthof transmission line using the above procedure,calculating β = (βl) /l, and then vp = ω/β.

3.19 Quarter-WavelengthTransmission Line

[m0091]

Quarter-wavelength sections of transmission lineplay an important role in many systems at radioand optical frequencies. The remarkableproperties of open- and short-circuitedquarter-wave line are presented in Section 3.16,and should be reviewed before reading further.In this section we perform a more generalanalysis, considering not just open- andshort-circuit terminations but any terminatingimpedance, and then address some applications.

The general expression for the input impedanceof a lossless transmission line is (Section 3.15):


1− Γe−j2βl(3.122)

Note that when l = λ/4:

2βl = 2 · 2πλ

· λ4= π

Subsequently:

Zin(λ/4) = Z01 + Γe−jπ

1− Γe−jπ

= Z01− Γ

1 + Γ

(3.123)

Recall that (Section 3.15):

Γ =ZL − Z0

ZL + Z0(3.124)

Substituting this expression and then multiplyingnumerator and denominator by ZL + Z0, oneobtains

Zin(λ/4) = Z0(ZL + Z0)− (ZL − Z0)

(ZL + Z0) + (ZL − Z0)

= Z02Z0

2ZL

(3.125)

Thus

Zin(λ/4) =Z20

ZL(3.126)

Note that the input impedance is inverselyproportional to the load impedance. For thisreason, a transmission line of length λ/4 issometimes referred to as a quarter-wave inverteror simply as a impedance inverter.

3.19. QUARTER-WAVELENGTH TRANSMISSION LINE 53

ZL

l=λ/4

Figure 3.24: Impedance-matching using a quarter-wavelength transmission line.

Quarter-wave lines play a very important role inRF engineering. As impedance inverters, theyhave the useful attribute of transforming smallimpedances into large impedances, and vice-versa– we’ll come back to this idea later in thissection. First, let’s consider how quarter-wavellines are used for impedance matching. Lookwhat happens when we solve Equation 3.126 forZ0:

Z0 =√Zin(λ/4) · ZL (3.127)

This equation indicates that we may match theload ZL to a source impedance (represented byZin(λ/4)) simply by making the characteristicimpedance equal to the value given by the aboveexpression and setting the length to λ/4. Thescheme is shown in Figure 3.24.

Example 3.5. 300-to-50 Ω match using anquarter-wave section of line.

Design a transmission line segment thatmatches 300 Ω to 50 Ω at 10 GHz using aquarter-wave match. Assume microstrip linefor which propagation occurs withwavelength 60% that of free space.

Solution. The line is completely specifiedgiven its characteristic impedance Z0 andlength l. The length should be one-quarterwavelength with respect to the signalpropagating in the line. The free-spacewavelength λ0 = c/f at 10 GHz is ∼= 3 cm.Therefore the wavelength of the signal in theline is λ = 0.6λ0 ∼= 1.8 cm, and the length ofthe line should be l = λ/4 ∼= 4.5 mm.

ZL

λ/4

Z1(completely

real-valued)

Figure 3.25: Impedance-matching a complex-valued load impedance using quarter-wavelengthtransmission line.

The characteristic impedance is given byEquation 3.127:

Z0 =√300 Ω · 300 Ω ∼= 122.5 Ω (3.128)

This value would be used to determine thewidth of the microstrip line, as discussed inSection 3.11.

It should be noted that for this scheme to yield areal-valued characteristic impedance, the productof the input (source) and load impedances mustbe a real-valued number. For example, thismethod is not suitable if ZL has a significantimaginary-valued component and match to areal-valued impedance is desired. In that casethe two-stage strategy shown in Figure 3.25 maybe used. In this scheme, the load impedance isfirst transformed to a real-valued impedanceusing a length l1 of transmission line. This isaccomplished using Equation 3.122 (quite simpleusing a numerical search) or using the Smithchart (see “Additional Reading” at the end ofthis section). The characteristic impedance Z01

of this transmission line is not critical and can beselected for convenience. Normally the smallestvalue of l1 is desired. This value will always beless than λ/4 since Zin(l1) is periodic in l1 withperiod λ/2; i.e., there are two changes in the signof the imaginary component of Zin(l1) as l1 isincreased from zero to λ/2. After eliminating theimaginary component of ZL in this manner, thereal component of the resulting impedance maythen be transformed using the quarter-wavematching technique described above.


Example 3.6. Matching a patch antennato 50 Ω.

A particular patch antenna exhibits a sourceimpedance of ZA = 35 + j35 Ω. Interfacethis antenna to 50 Ω using the techniquedescribed above. (See “Microstrip antenna”in “Additional Reading” at the end of thissection for some optional reading on patchantennas.) For the section of transmissionline adjacent to the patch antenna, usecharacteristic impedance Z01 = 50 Ω.Determine the lengths l1 and l2 of the twosegments of transmission line, and thecharacteristic impedance Z02 of the second(quarter-wave) segment.

Solution. The length of the first section ofthe transmission line (adjacent to theantenna) is determined usingEquation 3.122:

Z1(l1) = Z011 + Γe−j2β1l1

1− Γe−j2β1l1(3.129)

where β1 is the phase propagation constantfor this section of transmission line and

Γ ,ZA − Z01

ZA + Z01

∼= −0.0059− j0.4142 (3.130)

We seek the value of smallest positive valueof β1l1 for which the imaginary part ofZ1(l1) is zero. This can determined using aSmith chart (see “Additional Reading” atthe end of this section) or simply by a fewiterations of trial-and-error. Either way wefindZ1(β1l1 = 0.779 rad) ∼= 20.709 + j0.031 Ω,which we deem to be close enough to beacceptable. Note that β1 = 2π/λ, where λ isthe wavelength of the signal in thetransmission line. Therefore

l1 =β1l1β1

=β1l12π

λ ∼= 0.124λ (3.131)

The length of the second section of thetransmission line, being a

quarter-wavelength transformer, should bel2 = 0.25λ. Using Equation 3.127, thecharacteristic impedance Z02 of this sectionof line should be

Z02∼=√(20.709 Ω) (50 Ω) ∼= 32.2 Ω

(3.132)

By the way, patch antennas often appear insystems implemented on printed circuitboards, because they themselves are easilyimplemented on printed circuit boards. Agreat follow-up to this example would be tocomplete the solution in microstrip(Section 3.11) by determining the width oflines given a particular substrate anddielectric thickness.

Discussion. A bit of follow-up isappropriate here. The total length of thematching structure is l1 + l2 ∼= 0.374λ. Apatch antenna would typically have sides oflength about λ/2 = 0.5λ, so we see that thematching structure is nearly as big as theantenna itself. At frequencies where patchantennas are commonly used, and especiallyat frequencies in the UHF (300–3000 MHz)band, patch antennas often dominate thesize of the system, so it is not attractive tohave the matching structure also require asimilar amount of space. Thus we would bemotivated to find a smaller matchingstructure.

Although quarter-wave matching techniques aregenerally effective and commonly used, they haveone important contraindication, noted above:They often result in structures which are largerthan necessary. That is, any structure whichemploys a quarter-wave match will be at leastλ/4 long. Other transmission line matchingtechniques – and in particular, single stubmatching (Section 3.23) – typically result instructures which are significantly smaller.

The impedance inversion property ofquarter-wavelength lines has applications beyondimpedance matching. The following exampledemonstrates one such application:

3.19. QUARTER-WAVELENGTH TRANSMISSION LINE 55

Example 3.7. RF/DC decoupling intransistor amplifiers.

Transistor amplifiers for RF applicationsoften receive DC current at the sameterminal which delivers the amplified RFsignal, as shown in Figure 3.26 The powersupply is essentially a current source and sohas a low output impedance. If the powersupply is directly connected to thetransistor, then the RF will flowpredominantly in the direction of the powersupply as opposed to following the desiredpath, which exhibits a higher impedance.This can be addressed using an inductor inseries, which presents low impedance at DCand high impedance at RF. Unfortunatelydiscrete inductors are often not practical athigh RF frequencies. This is becausepractical inductors also exhibit parallelcapacitance, which tends to decreaseimpedance.

A solution is to replace the inductor with atransmission line having length λ/4 asshown in Figure 3.27. A wavelength at DCis infinite, so the transmission line isessentially transparent to the power supply.At radio frequencies the line transforms thelow impedance of the power supply to animpedance which is very large relative to theimpedance of the desired RF path, asdesired. Furthermore, transmission lines onprinted circuit boards are much cheaperthan discrete inductors (and are always instock!).

Additional Reading:

• “Quarter-wavelength impedance transformer”on Wikipedia.

• “Smith chart” on Wikipedia.

• “Microstrip antenna” on Wikipedia.

RF out(moderate Z)

DC power in

(low Z)

Figure 3.26: Decoupling of DC input power andRF output signal at the output of a common-emitter RF amplifier, using a discrete inductor.

RF out

(moderate $)

DC power in

(low $)

λ/4

Figure 3.27: Decoupling of DC input power andRF output signal at the output of a common-emitter RF amplifier, using a quarter-wavelengthtransmission line.

https://en.wikipedia.org/wiki/Quarter-wave_impedance_transformer

https://en.wikipedia.org/wiki/Smith_chart

https://en.wikipedia.org/wiki/Microstrip_antenna


3.20 Power Flow onTransmission Lines

[m0090]

It is often important to know the powerassociated with transmission line waves. Thepower of the waves incident upon, reflected by,and absorbed by a load are each of interest. Inthis section we shall work out expression forthese powers and consider some implications interms of reflection coefficient (Γ) and standingwave ratio (SWR).

Let’s begin by considering a lossless transmissionline which is oriented along the z axis. Thetime-average power associated with a sinusoidalwave having potential v(z, t) and current i(z, t) is

Pav(z) ,1

T

∫ t0+T

t0

v(z, t) i(z, t) dt (3.133)

where T , 2π/f is one period of the wave and t0is the start time for the integration. Since thetime-average power of a sinusoidal signal istime-invariant, t0 may be set equal to zerowithout loss of generality.

Let us now calculate the power of a waveincident from z < 0 on a load impedance ZL atz = 0. We may express the associated potentialand current as follows:

v+(z, t) =∣∣V +

0

∣∣ cos (ωt− βz + φ) (3.134)

i+(z, t) =

∣∣V +0

∣∣Z0

cos (ωt− βz + φ) (3.135)

And so the associated time-average power is

P+av(z) =

1

T

∫ T

0

v+(z, t) i+(z, t) dt

=

∣∣V +0

∣∣2

Z0· 1T

∫ T

0

cos2 (ωt− βz + φ) dt

(3.136)

Employing a well-known trigonometric identity:

cos2 θ =1

2+

1

2cos 2θ (3.137)

we may rewrite the integrand as follows

cos2 (ωt− βz + φ) =1

2+

1

2cos (2 [ωt− βz + φ])

(3.138)Then integrating over both sides of this quantity

∫ T

0

cos2 (ωt− βz + φ) dt =T

2+ 0 (3.139)

The second term of the integral is zero because itis the integral of cosine over two completeperiods. Subsequently we see that the positiondependence (here, the dependence on z) iseliminated. In other words, the power associatedwith the incident wave is the same for all z < 0,as expected. Substituting into Equation 3.136 weobtain:

P+av =

∣∣V +0

∣∣2

2Z0(3.140)

This is the time-average power associated withthe incident wave, measured at any point z < 0along the line.

Equation 3.140 gives the time-average powerassociated with a wave traveling in a singledirection along a lossless transmission line.

Using precisely the same procedure, we find thatthe power associated with the reflected wave is

P−av =

∣∣ΓV +0

∣∣2

2Z0= |Γ|2

∣∣V +0

∣∣2

2Z0(3.141)

or simply

P−av = |Γ|2 P+

av (3.142)

Equation 3.142 gives the time-average powerassociated with the wave reflected from animpedance mismatch.

Now: What is the power PL delivered to the loadimpedance ZL? The simplest way to calculatethis power is to use the principle of conservationof power. Applied to the present problem, thisprinciple asserts that the power incident on theload must equal the power reflected plus thepower absorbed; i.e.,

P+av = P−

av + PL (3.143)

3.21. IMPEDANCE MATCHING: GENERAL CONSIDERATIONS 57

Applying the previous equations, this can berewritten

PL =(1− |Γ|2

)P+av (3.144)

Equations 3.144 gives the time-average powertransferred to a load impedance, and is equalto the difference between the powers of theincident and reflected waves.

Example 3.8. How important is it tomatch 50 Ω to 75 Ω?

Two impedances which commonly appear inradio engineering are 50 Ω and 75 Ω. It isnot uncommon to find that it is necessary toconnect a transmission line having a 75 Ωcharacteristic impedance to a device, circuit,or system having a 50 Ω input impedance,or vice-versa. If no attempt is made tomatch these impedances, what fraction ofthe power will be delivered to thetermination, and what fraction of power willbe reflected? What is the SWR?

Solution. The voltage reflection coefficientgoing from 50 Ω transmission line to a 75 Ωload is

Γ =75− 50

75 + 50= 0.2

The fraction of power reflected is|Γ|2 = 0.04, which is 4%. The fraction of

power transmitted is 1− |Γ|2, which is 96%.Going from a 50 Ω transmission line to a75 Ω termination changes only the sign of Γ,and therefore the fractions of reflected andtransmitted power remain 4% and 96%,respectively. In either case (fromSection 3.14):

SWR =1 + |Γ|1− |Γ| = 1.5

This is often acceptable, but may not begood enough in some particularapplications. Suffice it to say that it is not a

“given” that an impedance-matching deviceis required to connect a 50 Ω to 75 Ω.

3.21 Impedance Matching:General Considerations

[m0092]

“Impedance matching” refers to the problem oftransforming a particular impedance ZL into amodified impedance Zin. The problem ofimpedance matching arises because it is notconvenient, practical, or desirable to have alldevices in a system operate at the same inputand output impedances. Here are just a few ofthe issues:

• It is not convenient or practical to marketcoaxial cables having characteristicimpedance equal to every terminatingimpedance that might be encountered.

• Different types of antennas operate atdifferent impedances, and the impedance ofmost antennas vary significantly withfrequency.

• Different types of amplifiers operate mosteffectively at different output impedances.For example, amplifiers operating as currentsources operate most effectively with lowoutput impedance, whereas amplifiersoperating as voltage sources operate mosteffectively with high output impedances.

• Independently of the above issue, techniquesfor the design of transistor amplifiers rely onintentionally mismatching impedances; i.e.,matching to an impedance different thanthat which maximizes power transfer orminimizes reflection. In other words, variousdesign goals are met by applying particularimpedances to the input and output ports ofthe transistor.7

7For a concise introduction to this concept, see Chap-ter 10 of S.W. Ellingson, Radio Systems Engineering, Cam-bridge Univ. Press, 2016.


For all of these reasons, electrical engineersfrequently find themselves with the task oftransforming a particular impedance ZL into amodified impedance Zin.

The reader is probably already familiar with abroad class of approaches to the impedancematching problem that employ discretecomponents and do not require knowledge ofelectromagnetics.8 To list just a few of these:transformers, resistive (lossy) matching,single-reactance matching, and two-reactance(“L” network) matching. However all of thesehave limitations. Perhaps the biggest set oflimitations pertains to the performance ofdiscrete components at high frequencies. Hereare just a few of the most common problems:

• Practical resistors actually behave as idealresistors in series with ideal inductors,

• Practical capacitors actually behave as idealcapacitors in series with ideal resistors, and

• Practical inductors behave as ideal inductorsin parallel with ideal capacitors, and inseries with ideal resistors.

All of this makes the use of discrete componentsincreasingly difficult with increasing frequency.

One possible solution to these types of problemsis to more precisely model each component, andthen to account for the non-ideal behavior byincorporating the appropriate models in theanalysis and design process. Alternatively, onemay consider ways to replace particulartroublesome components – or, in some cases, alldiscrete components – with transmission linedevices. The later approach is particularlyconvenient in circuits implemented on printedcircuit boards at frequencies in the UHF bandand higher, since the necessary transmission linestructures are easy to implement as microstripline, and are relatively compact since thewavelength is relatively small. However,applications employing transmission lines as

8For an overview, see Chapter 9 of S.W. Ellingson, Ra-dio Systems Engineering, Cambridge Univ. Press, 2016.

components in impedance matching devices canbe found at lower frequencies as well.

3.22 Single-ReactanceMatching

[m0093]

An impedance matching structure can bedesigned using a section of transmission linecombined with a discrete reactance, such as acapacitor or an inductor. In the strategypresented here, the transmission line is used totransform the real part of the load impedance oradmittance to the desired value, and then thereactance is used to modify the imaginary part tothe desired value. (Note the difference betweenthis approach and the quarter-wave techniquedescribed in Section 3.19: In that approach, thefirst transmission line is used to zero theimaginary part.) There are two versions of thisstrategy, which we will now consider separately.

The first version is shown in Figure 3.28. Thepurpose of the transmission line is to transformthe load impedance ZL into a new impedance Z1

for which ReZ1 = ReZin. This can be doneby solving the equation (from Section 3.15)

Re Z1 = Re

Z0

1 + Γe−j2βl

1− Γe−j2βl

(3.145)

for l, using a numerical search, or using theSmith chart.9 The characteristic impedance Z0

and phase propagation constant β of thetransmission line are independent variables andcan be selected for convenience. Normally thesmallest value of l that satisfies Equation 3.145 isdesired. This value will be ≤ λ/4 because thereal part of Z1 is periodic in l with period λ/4.

After matching the real component of theimpedance in this manner, the imaginarycomponent of Z1 may then be transformed to thedesired value (ImZin) by attaching a reactance

9For more about the Smith chart, see “Additional Read-ing” at the end of this section.

3.22. SINGLE-REACTANCE MATCHING 59

ZL

Z1

Z0

%

jXS

Zin

Figure 3.28: Single-reactance matching with a se-ries reactance.

Xs in series with the transmission line input,yielding Zin = Z1 + jXS . Therefore we choose

Xs = Im Zin − Z1 (3.146)

The sign of Xs determines whether thisreactance is a capacitor (Xs < 0) or inductor(Xs > 0), and the value of this component isdetermined from Xs and the design frequency.

Example 3.9. Single reactance in series.

Design a match consisting of a transmissionline in series with a single capacitor orinductor that matches a source impedance of50Ω to a load impedance of 33.9 + j17.6 Ωat 1.5 GHz. The characteristic impedanceand phase velocity of the transmission lineare 50Ω and 0.6c respectively.

Solution. From the problem statement:Zin , ZS = 50 Ω and ZL = 33.9 + j17.6 Ωare the source and load impedancesrespectively at f = 1.5 GHz. Thecharacteristic impedance and phase velocityof the transmission line are Z0 = 50 Ω andvp = 0.6c respectively.

The reflection coefficient Γ (i.e., ZL withrespect to the characteristic impedance ofthe transmission line) is

Γ ,ZL − Z0

ZL + Z0

∼= −0.142 + j0.239 (3.147)

The length l of the primary line (that is, theone that connects the two ports of thematching structure) is determined using theequation:

Re Z1 = Re

Z0

1 + Γe−j2βl

1− Γe−j2βl

(3.148)

where here Re Z1 = Re ZS = 50 Ω. So amore-specific form of the equation that canbe solved for βl (as a step toward finding l)is:

1 = Re

1 + Γe−j2βl

1− Γe−j2βl

(3.149)

By trial and error (or, using the Smith chartif you prefer) we find βl ∼= 0.408 rad for theprimary line, yielding Z1

∼= 50.0 + j29.0 Ωfor the input impedance after attaching theprimary line.

We may now solve for l as follows: Sincevp = ω/β (Section 3.8), we find

β =ω

vp=

2πf

0.6c∼= 52.4 rad (3.150)

Therefore l = (βl) /β ∼= 7.8 mm.

The impedance of the series reactanceshould be jXs

∼= −j29.0 Ω to cancel theimaginary part of Z1. Since the sign of thisimpedance is negative, it must be acapacitor. The reactance of a capacitor is−1/ωC, so it must be true that

− 1

2πfC∼= −29.0 Ω (3.151)

Thus we find the series reactance is acapacitor of value C ∼= 3.7 pF.

The second version of the single-reactancestrategy is shown in Figure 3.29. The differencein this scheme is that the reactance is attached inparallel. In this case it is easier to work theproblem using admittance (i.e., reciprocalimpedance) as opposed to impedance; this isbecause the admittance of parallel reactances issimply the sum of the associated admittances;


YL

Y1

Y0

&jBp

Yin

Figure 3.29: Single-reactance matching with aparallel reactance.

i.e.,

Yin = Y1 + jBp (3.152)

where Yin = 1/Zin, Y1 = 1/Z1, and Bp is thediscrete parallel susceptance; i.e., the imaginarypart of the discrete parallel admittance.

So, the procedure is as follows: The transmissionline is used to transform YL into a newadmittance Y1 for which ReY1 = ReYin.First we note that

Y1 ,1

Z1= Y0

1− Γe−j2βl

1 + Γe−j2βl(3.153)

where Y0 , 1/Z0 is characteristic admittance.Again the characteristic impedance Z0 and phasepropagation constant β of the transmission lineare independent variables and can be selected forconvenience. In the present problem we aim tosolve the equation

Re Y1 = Re

Y0

1− Γe−j2βl

1 + Γe−j2βl

(3.154)

for the smallest value of l, using a numericalsearch or using the Smith chart. After matchingthe real component of the admittances in thismanner, the imaginary component of theresulting admittance may then be transformed tothe desired value by attaching the susceptanceBp in parallel with the transmission line input.Since we desire jBp in parallel with Y1 to be Yin,the desired value is

Bp = Im Yin − Y1 (3.155)

The sign of Bp determines whether this is acapacitor (Bp < 0) or inductor (Bp > 0), and thevalue of this component is determined from Bpand the design frequency.

Example 3.10. Single reactance in parallel.

Design a match consisting of a transmissionline in parallel with a single capacitor orinductor that matches a source impedance of50Ω to a load impedance of 33.9 + j17.6 Ωat 1.5 GHz. The characteristic impedanceand phase velocity of the transmission lineare 50Ω and 0.6c respectively.

Solution. From the problem statement:Zin , ZS = 50 Ω and ZL = 33.9 + j17.6 Ωare the source and load impedancesrespectively at f = 1.5 GHz. Thecharacteristic impedance and phase velocityof the transmission line are Z0 = 50 Ω andvp = 0.6c respectively.

The reflection coefficient Γ (i.e., ZL withrespect to the characteristic impedance ofthe transmission line) is

Γ ,ZL − Z0

ZL + Z0

∼= −0.142 + j0.239 (3.156)

The length l of the primary line (that is, theone that connects the two ports of thematching structure) is the solution to:

Re Y1 = Re

Y0

1− Γe−j2βl

1 + Γe−j2βl

(3.157)

where here Re Y1 = Re 1/ZS = 0.02 mhoand Y0 = 1/Z0 = 0.02 mho. So the equationto be solved for βl (as a step toward findingl) is:

1 = Re

1− Γe−j2βl

1 + Γe−j2βl

(3.158)

By trial and error (or the Smith chart) wefind βl ∼= 0.126 rad for the primary line,yielding Y1 ∼= 0.0200− j0.0116 mho for the

3.23. SINGLE-STUB MATCHING 61

input admittance after attaching theprimary line.

We may now solve for l as follows: Sincevp = ω/β (Section 3.8), we find

β =ω

vp=

2πf

0.6c∼= 52.4 rad (3.159)

Therefore l = (βl) /β ∼= 2.4 mm.

The admittance of the parallel reactanceshould be jBp ∼= +j0.0116 mho to cancelthe imaginary part of Y1. The associatedimpedance is 1/jBp ∼= −j86.3 Ω. Since thesign of this impedance is negative, it mustbe a capacitor. The reactance of a capacitoris −1/ωC, so it must be true that

− 1

2πfC∼= −86.3 Ω (3.160)

Thus we find the parallel reactance is acapacitor of value C ∼= 1.2 pF.

Additional Reading:


3.23 Single-Stub Matching

[m0094]

In Section 3.22 we considered impedancematching schemes consisting of a transmissionline combined with a reactance which is placedeither in series or in parallel with thetransmission line. In many problems the requireddiscrete reactance is not practical because it isnot a standard value, or because of non-idealbehavior at the desired frequency (seeSection 3.21 for more about this), or because onemight simply wish to avoid the cost and logisticalissues associated with an additional component.Whatever the reason, a solution is to replace thediscrete reactance with a transmission line“stub”; that is, a transmission line which hasbeen open- or short-circuited. Section 3.16

Connector

M'() *(),

Stub

Spinningspark CC BY SA 3.0

Figure 3.30: A practical implementation ofa single-stub impedance match using microstriptransmission line. Here, the stub is open-circuited.

explains how it is that a stub can replace adiscrete reactance. Figure 3.30 shows a practicalimplementation of this idea, implemented inmicrostrip. This section explains the theory, andwe’ll return to this implementation at the end ofthe section.

Figure 3.31 shows the scheme. This scheme isususally implemented using the parallel reactanceapproach, as depicted in the figure. Although aseries reactance scheme is also possible inprinciple, it is usually not as convenient. This isbecause most transmission lines use one of theirtwo conductors as a local datum; e.g., the groundplane of a printed circuit board for microstripline is tied to ground, and the outer conductor(“shield”) of a coaxial cable is usually tied toground. This is contrast to a discrete reactance(such as a capacitor or inductor), which does notrequire that either of its terminals be tied toground. This issue is avoided in theparallel-attached stub because theparallel-attached stub and the transmission lineto which it is attached both have one terminal atground.

The single-stub matching procedure is essentiallythe same as the single parallel reactance method,except the parallel reactance is implementedusing a short- or open-circuited stub as opposeda discrete inductor or capacitor. Since parallelreactance matching is most easily done usingadmittances, it is useful to expressEquations 3.116 and 3.118 (input impedance ofan open- and short-circuited stub, respectively,




YL

Y1

Y01

-1

jBp

Yin

stub characteristics: Y 02-

2 2

stub may be open- or short-circuited

stub

Figure 3.31: Single-stub matching.

from Section 3.16) in terms of susceptance:

Bp = −Y02 cot (β2l2) short-circuited stub(3.161)

Bp = +Y02 tan (β2l2) open-circuited stub(3.162)

As in the main line, the characteristic impedanceZ02 = 1/Y02 is an independent variable and ischosen for convenience.

A final question is: When to choose theshort-circuited stub, and when to choose theopen-circuited stub? Given no other basis forselection, the termination that yields the shorteststub is chosen. An example of an “other basis forselection” that frequently comes up is whetherDC might be present on the line. If DC is presentwith the signal of interest, then a short circuittermination without some kind of remediation forthe DC component would certainly be a bad idea.

Example 3.11. Single stub matching.

Design a single-stub match that matches asource impedance of 50Ω to a loadimpedance of 33.9 + j17.6 Ω. Usetransmission lines having characteristicimpedances of 50Ω throughout, and leaveyour answer in terms of wavelengths.

Solution. From the problem statement:Zin , ZS = 50 Ω and ZL = 33.9 + j17.6 Ωare the source and load impedancesrespectively. Z0 = 50 Ω is the characteristicimpedance of the transmission lines to beused. The reflection coefficient Γ (i.e., ZLwith respect to the characteristic impedanceof the transmission line) is

Γ ,ZL − Z0

ZL + Z0

∼= −0.142 + j0.239 (3.163)

The length l1 of the primary line (that is,the one that connects the two ports of thematching structure) is the solution to theequation (from Section 3.22):

Re Y1 = Re

Y01

1− Γe−j2β1l1

1 + Γe−j2β1l1

(3.164)

where here Re Y1 = Re 1/ZS = 0.02 mhoand Y01 = 1/Z0 = 0.02 mho. Also note

2β1l1 = 2

(2π

λ

)l1 = 4π

l1λ

(3.165)

where λ is the wavelength in thetransmission line. So the equation to besolved for l1 is:

1 = Re

1− Γe−j4πl1/λ

1 + Γe−j4πl1/λ

(3.166)

By trial and error (or using the Smith chart;see “Additional Reading” at the end of thissection) we find l1 ∼= 0.020λ for the primaryline, yielding Y1 ∼= 0.0200− j0.0116 mho forthe input admittance after attaching theprimary line.

We now seek the shortest stub having aninput admittance of ∼= +j0.0116 mho tocancel the imaginary part of Y1. For anopen-circuited stub, we need

Bp = +Y0 tan 2πl2/λ ∼= +j0.0116 mho(3.167)

The smallest value of l2 for which this istrue is ∼= 0.084λ. For a short-circuited stub,

3.23. SINGLE-STUB MATCHING 63

we need

Bp = −Y0 cot 2πl2/λ ∼= +j0.0116 mho(3.168)

The smallest positive value of l2 for whichthis is true is ∼= 0.334λ; i.e., much longer.Therefore, we choose the open-circuited stubwith l2 ∼= 0.084λ. Note the stub is attachedin parallel at the source end of the primaryline.

Finally, returning to Figure 3.30: Single-stubmatching is a very common method forimpedance matching using microstriptransmission lines at frequences in the UHF band(300-3000 MHz) and above. In Figure 3.30 thetop (visible) traces comprise one conductor,whereas the ground plane (underneath, so notvisible) comprises the other conductor. The endof the stub is not connected to the ground plane,so the termination is an open circuit. A shortcircuit termination is accomplished by connectingthe end of the stub to the ground plane using avia; that is, a plated-through that electricallyconnects the top and bottom layers.

Additional Reading:



• “Via (electronics)” on Wikipedia.

[m0151]



https://en.wikipedia.org/wiki/Via_(electronics)


Image Credits

Fig. 3.1: Dmitry G, https://en.wikipedia.org/wiki/File:Mastech test leads.JPG,public domain.

Fig. 3.2: c© Tkgd2007, https://commons.wikimedia.org/wiki/File:Coaxial cable cutaway.svg,CC BY 3.0 (https://creativecommons.org/licenses/by/3.0/).Minor modifications from the original.

Fig. 3.3: c© SpinningSpark, https://en.wikipedia.org/wiki/File:Microstrip structure.svg,CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).Minor modifications from the original.

Fig. 3.6: c© Averse, https://en.wikipedia.org/wiki/File:Diplexer1.jpg,CC BY SA 2.0 Germany (https://creativecommons.org/licenses/by-sa/2.0/de/deed.en).

Fig. 3.7: c© BigRiz, https://en.wikipedia.org/wiki/File:Fibreoptic.jpg,CC BY SA 3.0 Unported (https://creativecommons.org/licenses/by-sa/3.0/deed.en).

Fig. 3.8: c© Omegatron, https://commons.wikimedia.org/wiki/File:Transmission line symbols.svg,CC BY SA 3.0 Unported (https://creativecommons.org/licenses/by-sa/3.0/deed.en).Modified.

Fig. 3.10: c© Omegatron, https://commons.wikimedia.org/wiki/File:Transmission line element.svg,CC BY SA 3.0 Unported (https://creativecommons.org/licenses/by-sa/3.0/deed.en).Modified.

Fig. 3.11: c© Omegatron, https://commons.wikimedia.org/wiki/File:Transmission line element.svg,CC BY SA 3.0 Unported (https://creativecommons.org/licenses/by-sa/3.0/deed.en).Modified.

Fig. 3.15: c© Arj, https://en.wikipedia.org/wiki/File:RG-59.jpg,CC BY SA 3.0 Unported (https://creativecommons.org/licenses/by-sa/3.0/deed.en).

Fig. 3.16: c© 7head7metal7, https://commons.wikimedia.org/wiki/File:Microstrip scheme.svg,CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).Minor modifications from the original.

Fig. 3.20: by Inductiveload, https://commons.wikimedia.org/wiki/File:Standing Wave Ratio.svg,public domain.Modifications from the original.

Fig. 3.30: c© Spinningspark, https://commons.wikimedia.org/wiki/File:Stripline stub matching.svg,CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).Minor modifications from the original: curly braces deleted.

https://en.wikipedia.org/wiki/File:Mastech_test_leads.JPG

https://commons.wikimedia.org/wiki/File:Coaxial_cable_cutaway.svg


https://en.wikipedia.org/wiki/File:Microstrip_structure.svg


https://en.wikipedia.org/wiki/File:Diplexer1.jpg

https://creativecommons.org/licenses/by-sa/2.0/de/deed.en

https://en.wikipedia.org/wiki/File:Fibreoptic.jpg


https://commons.wikimedia.org/wiki/File:Transmission_line_symbols.svg


https://commons.wikimedia.org/wiki/File:Transmission_line_element.svg


https://commons.wikimedia.org/wiki/File:Transmission_line_element.svg


https://en.wikipedia.org/wiki/File:RG-59.jpg


https://commons.wikimedia.org/wiki/File:Microstrip_scheme.svg


https://commons.wikimedia.org/wiki/File:Standing_Wave_Ratio.svg

https://commons.wikimedia.org/wiki/File:Stripline_stub_matching.svg


Chapter 4

Vector Analysis

4.1 Vector Arithmetic

[m0006]

A vector is a mathematical object that has botha scalar part (i.e., a magnitude and possibly aphase) as well as a direction. Many physicalquantities are best described as vectors. Forexample, the rate of movement through spacecan be described as speed ; i.e., as a scalar havingunits of m/s. However this quantity is morecompletely described as velocity ; i.e., as a vectorwhose scalar part is speed and direction indicatesthe direction of movement. Similarly, force is avector whose scalar part indicates magnitude(units of N) and direction indicates the directionin which the force is applied. Electric andmagnetic fields are also best described as vectors.

In mathematical notation, a real-valued vector Ais said to have a magnitude A = |A| anddirection a such that

A = Aa (4.1)

where a is a unit vector (i.e., a real-valued vectorhaving magnitude equal to one) having the samedirection as A. If a vector is complex-valued,then A is similarly complex-valued.

Cartesian Coordinate System. Fundamentalsof vector arithmetic are most easily gasped usingthe Cartesian coordinate system. This system isshown in Figure 4.1. Note carefully the relativeorientation of the x, y, and z axes. Thisorientation is important. For example: there aretwo directions that are perpendicular to the

y

x

.

origin

c© K. Kikkeri CC BY SA 4.0

Figure 4.1: Cartesian coordinate system.

z = 0 plane (in which the x- and y-axes lie), butthe +z axis is specified to be one of these inparticular.

Position-Fixed vs. Position-Free Vectors.It is often convenient to describe a position inspace as a vector whose magnitude is thedistance from the origin of the coordinate systemand whose direction is measured from the origintoward the position of interest. This is shown inFigure 4.2. These position vectors are“position-fixed” in the sense that they aredefined with respect to a single point in space; inthis case, the origin. Position vectors can also be




66 CHAPTER 4. VECTOR ANALYSIS

y

/

0

r1

r2


Figure 4.2: Position vectors. The vectors r1 andr1 are position-fixed and refer to particular loca-tions.

defined as vectors that are defined with respectto some other point in space; in which case theyare considered position-fixed to that position.

Position-free vectors, on the other hand, are notdefined with respect to a particular point inspace. An example is shown in Figure 4.3.Particles 1 m apart may both be traveling at2 m/s in the same direction. In this case thevelocity of each particle can be described usingthe same vector, even though the particles arelocated at different points in space.

Position-free vectors are said to be equal if theyhave the same magnitudes and directions.Position-fixed vectors, on the other hand, mustbe also be referenced to the same position (e.g.,the origin) to be considered equal.

Basis Vectors. Each coordinate system isdefined in terms of three basis vectors whichconcisely describe all possible ways to traversethree-dimensional space. A basis vector is a

v

particle 1

1

v

particle 2

2


Figure 4.3: Two particles exhibiting the samevelocity. In this case, the velocity vectors v1 andv2 are position-free and equal.

position-free unit vector which is perpendicularto all other basis vectors for that coordinatesystem. The basis vectors x, y, and z of theCartesian coordinate system are shown inFigure 4.4. In this notation, x indicates thedirection in which x increases most rapidly, yindicates the direction in which y increases mostrapidly, and z indicates the direction in which zincreases most rapidly. Alternatively, you mightinterpret x, y, and z as unit vectors that areparallel to the x-, y-, and z-axes and point in thedirection in which values along each axisincrease.

Vectors in the Cartesian CoordinateSystem. In Cartesian coordinates, we maydescribe any vector A as follows:

A = xAx + yAy + zAz (4.2)

where Ax, Ay, and Az are scalar quantitiesdescribing the components of A in each of theassociated directions, as shown in Figure 4.5.This description makes it clear that themagnitude of A is:

|A| =√A2x +A2

y +A2z (4.3)

and therefore we can calculate the associated



4.1. VECTOR ARITHMETIC 67

y

1

3

y

x

z


Figure 4.4: Basis vectors in the Cartesian coor-dinate system.

unit vector as

a =A

|A| =A√

A2x +A2

y +A2z

= xAx(A2x +A2

y +A2z

)−1/2

+ yAy(A2x +A2

y +A2z

)−1/2

+ zAz(A2x +A2

y +A2z

)−1/2

(4.4)

Vector Addition and Subtraction. It iscommon to add and subtract vectors. Forexample, vectors describing two different forcesA and B applied to the same point can bedescribed as a single force vector C which is thesum of A and B; i.e., C = A+B. This additionis quite simple in the Cartesian coordinatesystem:

C = A+B

= (xAx + yAy + zAz) + (xBx + yBy + zBz)

= x (Ax +Bx) + y (Ay +By) + z (Az +Bz)

(4.5)

In other words, the x component of C is the sumof the x components of A and B, and similarly

y

4

5

A6

A7

Ay

A


Figure 4.5: Components of a vector A in theCartesian coordinate system.

for y and z. From the above example, it is clearthat vector addition is commutative; i.e.,

A+B = B+A (4.6)

In other words, vectors may be added in anyorder. Vector subtraction is defined similarly:

D = A−B

= (xAx + yAy + zAz)− (xBx + yBy + zBz)

= x (Ax −Bx) + y (Ay −By) + z (Az −Bz)

(4.7)

In other words, the x component of D is thedifference of the x components of A and B, andsimilarly for y and z. Like scalar subtraction,vector subtraction is not commutative.

Relative Positions and Distances. Acommon task in vector analysis is to describe theposition of one point in space relative to adifferent point in space. Let us identify those twopoints using the position vectors r1 and r2, asindicated in Figure 4.6. We may identify a thirdvector r12 as the position of r2 relative to r1:

r12 = r2 − r1 (4.8)




origin

r8

r2

r89 r8= -r2


Figure 4.6: Relative position (distance and direc-tion) between locations identified by their positionvectors.

Now |r12| is the distance between these points,and r12/ |r12| is a unit vector indicating thedirection to r2 from r1.

Example 4.1. Direction and distancebetween positions.

Consider two positions, identified using theposition vectors r1 = 2x+ 3y + 1z andr2 = 1x− 2y + 3z, both expressed in unitsof meters. Find the direction vector thatpoints from r1 to r2, the distance betweenthese points, and the associated unit vector.

Solution. The vector that points from r1 tor2 is

R = r2 − r1

= (1− 2)x+ (−2− 3)y + (3− 1)z

= −x− 5y + 2z (4.9)

The distance between r1 and r2 is simplythe magnitude of this vector:

|R| =√(−1)

2+ (−5)

2+ (2)

2 ∼= 5.48 m

(4.10)

The unit vector R is simply R normalized

to have unit magnitude:

R = R/ |R|∼= (−x− 5y + 2z) /5.48∼= −0.182x− 0.913y + 0.365z (4.11)

Multiplication of a Vector by a Scalar.Let’s say a particular force is specified by avector F. What is the new vector if this force isdoubled? The answer is simply 2F; that is, twicethe magnitude applied in the same direction.This is an example of scalar multiplication of avector. Generalizing, the product of the scalar αand the vector A is simply αA.

Scalar (“Dot”) Product of Vectors. Anothercommon task in vector analysis is to determinethe similarity in the direction in which twovectors point. In particular, it is useful to have ametric which, when applied to the vectorsA = aA and B = bB, has the followingproperties (see Figure 4.7):

• If A is perpendicular to B, the result is zero.

• If A and B point in the same direction, theresult is AB.

• If A and B point in opposite directions, theresult is −AB.

• Results intermediate to these conditionsdepend on the angle ψ between A and B,measured as if A and B were arranged“tail-to-tail” as shown in Figure 4.8.

In vector analysis, this operator is known as thescalar product (not to be confused with scalarmultiplication) or the dot product. The dotproduct is written A ·B and is given in generalby the expression:

A ·B = AB cosψ (4.12)

Note that this expression yields the special casespreviously identified, which are ψ = π/2, ψ = 0,and ψ = π, respectively. The dot product iscommutative; i.e.,

A ·B = B ·A (4.13)


4.1. VECTOR ARITHMETIC 69


Figure 4.7: Special cases of the dot product.

The dot product is also distributive; i.e.,

A · (B+C) = A ·B+A ·C (4.14)

The dot product has some other useful

A

Bψ


Figure 4.8: Calculation of the dot product.

properties. For example, note:

A ·A = (xAx + yAy + zAz) · (xAx + yAy + zAz)

= x · xA2x + x · yAxAy + x · zAxAz

+ y · xAxAy + y · yA2y + y · zAyAz

+ z · xAxAz + z · yAyAz + z · zA2z

(4.15)

which looks pretty bad until you realize that

x · x = y · y = z · z = 1 (4.16)

and any other dot product of basis vectors iszero. Thus, the whole mess simplifies to:

A ·A = A2x +A2

y +A2z (4.17)

This is the square of the magnitude of A, so wehave discovered that

A ·A = |A|2 = A2 (4.18)

Applying the same principles to the computationof the dot product of potentially different vectorsA and B, we find:

A ·B = (xAx + yAy + zAz) · (xBx + yBy + zBz)

= AxBx +AyBy +AzBz(4.19)

This is a particularly easy way to calculate thedot product, since it eliminates the problem ofdetermining the angle ψ. In fact, an easy way tocalculate this angle is to first calculate the dotproduct using Equation 4.19 and then use theresult to solve Equation 4.12 for ψ.

Example 4.2. Angle between two vectors.

Consider the position vectorsC = 2x+ 3y + 1z and D = 3x− 2y + 2z,both expressed in units of meters. Find theangle between these vectors.

Solution. From Equation 4.12

C ·D = CD cosψ (4.20)

where C = |C|, D = |D|, and ψ is the angle




we seek. From Equation 4.19:

C ·D = CxDx + CyDy + CzDz (4.21)

= 2 · 3 + 3 · (−2) + 1 · 2 m2 (4.22)

= 2 m2 (4.23)

also

C =√C2x + C2

y + C2z∼= 3.742 m (4.24)

D =√D2x +D2

y +D2z∼= 4.123 m (4.25)

so

cosψ =C ·DCD

∼= 0.130 (4.26)

Taking the inverse cosine, we find ψ = 82.6.

Cross Product. The cross product is a form ofvector multiplication that results in a vectorwhich is perpendicular to both of the operands.The definition is as follows:

A×B = nAB sinψAB (4.27)

As shown in Figure 4.9, the unit vector n isdetermined by the “right hand rule”: Using yourright hand, curl your fingers to most directlytraverse the angle ψAB beginning at A andending at B; then n points in the direction ofyour fully-extended thumb.

It should be apparent that the cross product isnot commutative but rather is anticommutative;that is,

A×B = −B×A (4.28)

You can confirm this for yourself using eitherEquation 4.27 or simply by applying theright-hand rule.

The cross product is distributive:

A× (B+C) = A×B+A×C (4.29)

There are two useful special cases of the crossproduct that are worth memorizing. The first isthe cross product of a vector with itself, which iszero:

A×A = 0 (4.30)

n

B

A

ABψ


Figure 4.9: The cross product A×B.

The second is the cross product of vectors whichare perpendicular; i.e., for which ψAB = π/2. Inthis case:

A×B = nAB (4.31)

In particular note:

x× x = y × y = z× z = 0 (4.32)

whereas

x× y = z (4.33)

y × z = x (4.34)

z× x = y (4.35)

A useful diagram that summarizes theserelationships is shown in Figure 4.10.

It is typically awkward to “manually” determinen in Equation 4.27. However in Cartesiancoordinates the cross product may be calculatedas:

A×B = x (AyBz −AzBy)

+ y (AzBx −AxBz)

+ z (AxBy −AyBx)

(4.36)

This may be easier to remember as a matrixdeterminant:

A×B =

∣∣∣∣∣∣

x y zAx Ay AzBx By Bz

∣∣∣∣∣∣(4.37)


4.2. CARTESIAN COORDINATES 71

x y

z

Figure 4.10: Cross products among basis vectorsin the Cartesian system. The cross product of anytwo basis vectors is the third basis vector when theorder of operands is counter-clockwise as shown inthe diagram, and is −1 times the third basis vec-tor when the order of operands is clockwise withrespect to the arrangement in the diagram.

Similar expressions are available for othercoordinate systems.

Vector analysis routinely requires expressionsinvolving both dot products and cross productsin different combinations. Often theseexpressions may be simplified, or otherwise mademore convenient, using the vector identities listedin Appendix B.3.

Additional Reading:

• “Vector field” on Wikipedia.

• “Vector algebra” on Wikipedia.

4.2 Cartesian Coordinates

[m0004]

The Cartesian coordinate system is introduced inSection 4.1. Concepts described in that section –i.e., the dot product and cross product – aredescribed in terms of the Cartesian system. Inthis section we identify some additional featuresof this system that are useful in subsequent work,and also set the stage for alternative systems;namely the cylindrical and spherical coordinatesystems.

Integration Over Length. Consider a vector

field A = xA(r), where r is a position vector.What is the integral of A over some curve Cthrough space? To answer this question, we firstidentify a differential-length segment of the curveand note that this section of the curve can bedescribed as

dl = xdx+ ydy + zdz (4.38)

The contribution to the integral for that sectionof the curve is simply A · dl, and then weintegrate to obtain the result; i.e.,

∫

C

A · dl (4.39)

For example, if A = xA0 (i.e., A(r) is a constant)and if C is a straight line from x = x1 and x = x2along some constant y and z, then dl = xdx,A · dl = A0dx, and subsequently the aboveintegral is

∫ x2

x1

A0dx = A0 (x2 − x1) (4.40)

In particular notice that if A0 = 1 then thisintegral gives the length of C. Although theformalism seems unnecessary in this simpleexample, it becomes very useful when integratingover paths that vary in more than one directionand with more complicated operands.

Note that the Cartesian system was anappropriate choice for preceding example becausethis allowed two of the three basis directions (i.e.,y and z) to be immediately eliminated from thecalculation. Said differently, the precedingexample is expressed with the minimum numberof varying coordinates in the Cartesian system.Here’s a counter-example: If C had been a circlein the z = 0 plane, then the problem would haverequired two basis directions to be considered;namely, both x and y. In this case anothersystem – namely cylindrical coordinates(Section 4.3) – minimizes the number of varyingcoordinates (to just one; namely φ).

Integration Over Area. Now we ask thequestion: What is the integral of some vectorfield A over some surface S? The answer is

∫

S

A · ds (4.41)

https://en.wikipedia.org/wiki/Vector_field

https://en.wikipedia.org/wiki/Vector_algebra


We refer to ds as the differential surface element,which is a vector having magnitude equal to thedifferential area ds; and is normal(perpendicular) to each point on the surface.There are actually two such directions: We’llchoose +z for now and return to clear up theambiguity a bit later. Now, as an example, ifA = z and S is the surface bounded byx1 ≤ x ≤ x2, y1 ≤ y ≤ y2, then

ds = z dx dy (4.42)

since dxdy is differential surface area in the z = 0plane and z is normal to the z = 0 plane. SoA · ds = dxdy, and subsequently the aboveintegral becomes

∫ x2

x1

∫ y2

y1

dx dy = (x2 − x1) (y2 − y1) (4.43)

i.e., this is a calculation of area.

Once again, we see the Cartesian system was anappropriate choice for this example because thischoice minimizes the number of varyingcoordinates; in the above example, the surface ofintegration is described by a constant value of zwith variable values of x and y. If the surfacehad instead been a cylinder or a sphere, not onlywould all three basis directions be variable, butalso the surface normal n would be variable,making the problem dramatically morecomplicated.

Now let’s return to the issue of the direction ofds. What’s the problem with choosing −z as thisdirection? Simply that the resulting area wouldbe negative. “Negative area” is simply theexpected (“positive”) area, except with respectto the opposite-facing normal vector. In thisproblem the sign of the area is simply atechnicality, but in some problems inelectromagnetics this sign becomes important.This usually happens when the quantity ofinterest is a flux. If A were a flux density, thenthe integration over area that we just performedindicates the magnitude and direction of flux,and so the direction chosen for ds defines thedirection of positive flux. Here’s just oneexample: Section 2.4 describes the electric fieldin terms of a flux (i.e., as electric flux density D)

in which case positive flux flows away from apositively-charged source.

Integration Over Volume. Another commontask in vector analysis is integration of somequantity over a volume. Since the procedure isthe same for scalar or vector quantities, we shallconsider integration of a scalar quantity A(r) forsimplicity. First, we note that the contributionfrom a differential volume element

dv = dx dy dz (4.44)

is A(r) dv, so the integral over the volume V is

∫

V

A(r) dv (4.45)

For example, if A(r) = 1 and V is a cubebounded by x1 ≤ x ≤ x2, y1 ≤ y ≤ y2, andz1 ≤ z ≤ z2, then the above integral becomes

∫ x2

x1

∫ y2

y1

∫ z2

z1

dx dy dz = (x2 − x1) (y2 − y1) (z2 − z1)

(4.46)i.e., this is a calculation of volume.

The Cartesian system was an appropriate choicefor this example because V is a cube, which iseasy to describe in Cartesian coordinates andrelatively difficult to describe in any othercoordinate system.

Additional Reading:

• “Cartesian coordinate system” onWikipedia.

4.3 Cylindrical Coordinates

[m0096]

Cartesian coordinates (Section 4.2) are notconvenient in certain cases. One of these is whenthe problem has cylindrical symmetry. Forexample: In the Cartesian coordinate system, thecross-section of a cylinder concentric with thez-axis requires two coordinates to describe: x

https://en.wikipedia.org/wiki/Cartesian_coordinate_system

4.3. CYLINDRICAL COORDINATES 73

y

x

:

z

;

:

z

<

z


Figure 4.11: Cylindrical coordinate system andassociated basis vectors.

and y. However this cross section can bedescribed using a single parameter – the radius ρ– in the cylindrical coordinate system. Thisresults in a dramatic simplification of themathematics in some applications.

The cylindrical system is defined with respect tothe Cartesian system in Figure 4.11. In lieu of xand y, the cylindrical system uses ρ, the distancemeasured from the closest point on the z axis;1

and φ, the angle measured in a plane of constantz, beginning at the +x axis (φ = 0) with φincreasing toward the +y direction.

The basis vectors in the cylindrical system are ρ,φ, and z. As in the Cartesian system, the dotproduct of like basis vectors is equal to one, andthe dot product of unlike basis vectors is equal tozero. The cross products of basis vectors are asfollows:

ρ× φ = z (4.47)

φ× z = ρ (4.48)

z× ρ = φ (4.49)


1Note that some textbooks use “r” in lieu of ρ for thiscoordinate.

? @

z

Figure 4.12: Cross products among basis vectorsin the cylindrical system. (See Figure 4.10 for in-structions on the use of this diagram.)

The cylindrical system is usually less useful thanthe Cartesian system for identifying absolute andrelative positions. This is because the basisdirections depend on position. For example, ρ isdirected radially outward from the z axis, soρ = x for points along the x-axis but ρ = y forpoints along the y axis. Similarly, the direction φvaries as a function of position. To overcome thisawkwardness, it is common to set up a problemin cylindrical coordinates in order to exploitcylindrical symmetry, but at some point toconvert to cartesian coordinates. Here are theconversions:

x = ρ cosφ (4.50)

y = ρ sinφ (4.51)

and z has the same meaning in both systems.The conversion from Cartesian to cylindrical is asfollows:

ρ =√x2 + y2 (4.52)

φ = arctan (y, x) (4.53)

where arctan is the four-quadrant inverse tangentfunction; i.e., arctan(y/x) in the first quadrant(x > 0, y > 0), but possibly requiring anadjustment for the other quadrants because thesigns of both x and y are individuallysignificant.2

Similarly it is often necessary to represent basisvectors of the cylindrical system in terms ofCartesian basis vectors, and vice-versa.Conversion of basis vectors is straightforwardusing dot products to determine the components

2Note that this function is available in MATLAB andOctave as atan2(y,x).



· ρ φ zx cosφ − sinφ 0y sinφ cosφ 0z 0 0 1

Table 4.1: Dot products between basis vectors inthe cylindrical and Cartesian coordinate systems.

of the basis vectors in the new system. Forexample, x in terms of the basis vectors of thecylindrical system is

x = ρ (ρ · x) + φ(φ · x

)+ z (z · x) (4.54)

This requires dot products between basis vectorsin the two systems, which are summarized inTable 4.1. Using this table, we find

x = ρ cosφ− φ sinφ (4.55)

y = ρ sinφ+ φ cosφ (4.56)

and of course z requires no conversion. Goingfrom Cartesian to cylindrical, we find

ρ = x cosφ+ y sinφ (4.57)

φ = −x sinφ+ y cosφ (4.58)

Integration Over Length. Adifferential-length segment of a curve in thecylindrical system is described in general as

dl = ρdρ+ φρdφ+ z dz (4.59)

Note that the contribution of the φ coordinate todifferential length is ρdφ, not simply dφ. This isbecause φ is an angle, not a distance. To see whythe associated distance is ρdφ, consider thefollowing: The circumference of a circle of radiusρ is 2πρ. If only a fraction of the circumference istraversed, the associated arclength is thecircumference scaled by φ/2π, where φ is theangle formed by the traversed circumference.Therefore the distance is 2πρ · φ/2π = ρφ, andthe differential distance is ρdφ.

As always the integral of a vector field A(r) overa curve C is ∫

C

A · dl (4.60)

dl

y

xC DG


Figure 4.13: Example in cylindrical coordinates:The circumference of a circle.

To demonstrate the cylindrical system, let uscalculate the integral of A(r) = φ when C is acircle of radius ρ0 in the z = 0 plane, as shown inFigure 4.13. In this example, dl = φ ρ0 dφ sinceρ = ρ0 and z = 0 are both constant along C.Subsequently A · dl = ρ0dφ and the aboveintegral is ∫ 2π

0

ρ0 dφ = 2πρ0 (4.61)

i.e., this is a calculation of circumference.

Note that the cylindrical system is anappropriate choice for the preceding examplebecause the problem can be expressed with theminimum number of varying coordinates in thecylindrical system. If we had attempted thisproblem in the Cartesian system, we would findthat both x and y vary over C, and in a relativelycomplex way.3

Integration Over Area. Now we ask thequestion: What is the integral of some vectorfield A over a circular surface S in the z = 0plane, having radius ρ0? This is shown inFigure 4.14. The differential surface vector inthis case is

ds = z (dρ) (ρdφ) = z ρ dρ dφ (4.62)

3Nothing will drive this point home more firmly thantrying it. It can be done, but it’s a lot more work...


4.3. CYLINDRICAL COORDINATES 75

y

xdρ

ρdϕ

ds


Figure 4.14: Example in cylindrical coordinates:The area of a circle.

The quantities in parentheses are the radial andangular dimensions, respectively. The directionof ds indicates the direction of positive flux – seethe discussion in Section 4.2 for an explanation.In general, the integral over a surface is

∫

S

A · ds (4.63)

To demonstrate, let’s consider A = z; in this caseA · ds = ρ dρ dφ and the integral becomes∫ ρ0

0

∫ 2π

0

ρ dρ dφ =

(∫ ρ0

0

ρ dρ

)(∫ 2π

0

dφ

)

=

(1

2ρ20

)(2π)

= πρ20(4.64)

which we recognize as the area of the circle, asexpected. The corresponding calculation in theCartesian system is quite difficult in comparison.

Whereas the previous example considered aplanar surface, we might consider instead acurved surface. Here we go: What is the integralof a vector field A = ρ over a cylindrical surfaceS concentric with the z axis, having radius ρ0and extending from z = z1 to z = z2? This isshown in Figure 4.15. The differential surfacevector in this case is

ds = ρ (ρ0dφ) (dz) = ρρ0 dφ dz (4.65)

y

x

z

ρz2

z1

ρdϕ

dz

0


Figure 4.15: Example in cylindrical coordinates:The area of the curved surface of a cylinder.

The integral is∫

S

A · ds =∫ 2π

0

∫ z2

z1

ρ0 dφdz

= ρ0

(∫ 2π

0

dφ

)(∫ z2

z1

dz

)

= 2πρ0 (z2 − z1)

(4.66)

which is the area of S, as expected. Once again,the corresponding calculation in the Cartesiansystem is quite difficult in comparison.

Integration Over Volume. The differentialvolume element in the cylindrical system is

dv = dρ (ρdφ) dz = ρ dρ dφ dz (4.67)

For example, if A(r) = 1 and the volume V is acylinder bounded by ρ ≤ ρ0 and z1 ≤ z ≤ z2,then∫

V

A(r) dv =

∫ ρ0

0

∫ 2π

0

∫ z2

z1

ρ dρ dφ dz

=

(∫ ρ0

0

ρ dρ

)(∫ 2π

0

dφ

)(∫ z2

z1

dz

)

= πρ20 (z2 − z1)

(4.68)




i.e., area time length, which is volume.

Once again: The procedure above is clearly morecomplicated than is necessary if we are interestedonly in computing volume; however then theoperand is not constant-valued we are no longercomputing volume; then this particularformalism is appropriate and possibly necessary.

Additional Reading:

• “Cylindrical coordinate system” onWikipedia.

4.4 Spherical Coordinates

[m0097]

The spherical coordinate system is defined withrespect to the Cartesian system in Figure 4.16.The spherical system uses r, the distancemeasured from the origin;4 θ, the angle measuredfrom the +z axis toward the z = 0 plane; and φ,the angle measured in a plane of constant z,identical to φ in the cylindrical system.

Spherical coordinates are preferred overCartesian and cylindrical coordinates when thegeometry of the problem exhibits sphericalsymmetry. For example: In the Cartesiancoordinate system, the surface of a sphereconcentric with the origin requires all threecoordinates (x, y, and z) to describe. Howeverthis surface can be described using a singleconstant parameter – the radius r – in thespherical coordinate system. This leads to adramatic simplification in the mathematics incertain applications.

The basis vectors in the spherical system are r,θ, and φ. As always, the dot product of like basisvectors is equal to one, and the dot product ofunlike basis vectors is equal to zero. For thecross-products, we find:

r× θ = φ (4.69)

4Note that some textbooks use “R” in lieu of r for thiscoordinate.

y

x

ϕz

r

θθ

ϕ

r


Figure 4.16: Spherical coordinate system and as-sociated basis vectors.

θ × φ = r (4.70)

φ× r = θ (4.71)


Like the cylindrical system, the spherical systemis usually less useful than the Cartesian systemfor identifying absolute and relative positions.The reason is the same: Basis directions in thespherical system depend on position. Forexample, r is directed radially outward from the

θ

ϕ

r

Figure 4.17: Cross products among basis vectorsin the spherical system. (See Figure 4.10 for in-structions on the use of this diagram.)

https://en.wikipedia.org/wiki/Cylindrical_coordinate_system


4.4. SPHERICAL COORDINATES 77

· r θ φx sin θ cosφ cos θ cosφ − sinφy sin θ sinφ cos θ sinφ cosφz cos θ − sin θ 0

Table 4.2: Dot products between basis vectors inthe spherical and Cartesian coordinate systems.

origin, so r = x for points along the x-axis butr = y for points along the y axis and r = z forpoints along the z axis. Similarly, the directionsof θ and φ vary as a function of position. Toovercome this awkwardness, it is common to setup a problem in spherical coordinates in order toexploit spherical symmetry, but at some point toconvert to Cartesian coordinates. Here are theconversions:

x = r cosφ sin θ (4.72)

y = r sinφ sin θ (4.73)

z = r cos θ (4.74)

The conversion from Cartesian to sphericalcoordinates is as follows:

r =√x2 + y2 + z2 (4.75)

θ = arccos (z/r) (4.76)

φ = arctan (y, x) (4.77)

(4.78)

where arctan is the four-quadrant inverse tangentfunction.5

Dot products between basis vectors in thespherical and Cartesian systems are summarizedin Table 4.2. This information can be used toconvert between basis vectors in the sphericaland Cartesian systems, in the same mannerdescribed in Section 4.3; e.g.

x = r (r · x) + θ(θ · x

)+ φ

(φ · x

)(4.79)

r = x (x · r) + y (y · r) + z (z · r) (4.80)

and so on.

5Note that this function is available in MATLAB andOctave as atan2(y,x).

Example 4.3. Cartesian to sphericalconversion.

A vector field G = x xz/y. Develop anexpression for G in spherical coordinates.

Solution: Simply substitute expressions interms of spherical coordinates forexpressions in terms of Cartesiancoordinates. Use Table 4.2 andEquations 4.72–4.74. Making thesesubstitutions and applying a bit ofmathematical clean-up afterward, oneobtains

G =(r sin θ cotφ+ φ cos θ cotφ− φ

)

· r cos θ cosφ (4.81)

Integration Over Length. Adifferential-length segment of a curve in thespherical system is

dl = r dr + θ r dθ + φ r sin θ dφ (4.82)

Note that θ is an angle, as opposed to a distance.The associated distance is r dθ in the θ direction.Note also that in the φ direction, distance is r dφin the z = 0 plane, and less by the factor sin θ forz <> 0.

As always the integral of a vector field A(r) overa curve C is ∫

C

A · dl (4.83)

To demonstrate line integration in the sphericalsystem, imagine a sphere of radius a centered atthe origin with “poles” at z = +a and z = −a.let us calculate the integral of A(r) = θ where Cis the arc drawn directly from pole to pole alongthe surface of the sphere, as shown inFigure 4.18. In this example, dl = θ a dθ sincer = a and φ (which could be any value) are bothconstant along C. Subsequently A · dl = a dθ andthe above integral is

∫ π

0

a dθ = πa (4.84)

i.e., half the circumference of the sphere, asexpected.


dl

ρ

z

θ

a

pHIJ

pHIJ


Figure 4.18: Example in spherical coordinates:Pole-to-pole distance on a sphere.

The point to be noted here is that the sphericalsystem is an appropriate choice for this examplebecause the problem can be expressed with themaximum number of constant coordinates in thespherical system. If we had attempted thisproblem in the Cartesian system, we would findthat both z and either x or y (or all three) varyover C, and in a relatively complex way.

Integration Over Area. Now we ask thequestion: What is the integral of some vectorfield A over the surface S of a sphere of radius acentered on the origin? This is shown inFigure 4.19. The differential surface vector inthis case is

ds = r (r dθ) (r sin θ dφ) = r r2 sin θ dθ dφ(4.85)

As always, the direction is normal to the surfaceand in the direction associated with positive flux.The quantities in parentheses are the distancesassociated with varying θ and φ, respectively. Ingeneral, the integral over a surface is

∫

S

A · ds (4.86)

In this case, let’s consider A = r; in this case

a

r sinθ dϕ

y

x

z

r dθ


Figure 4.19: Example in spherical coordinates:The area of a sphere.

A · ds = a2 sin θ dθ dφ and the integral becomes

∫ π

0

∫ 2π

0

a2 sin θ dθ dφ = a2(∫ π

0

sin θdθ

)(∫ 2π

0

dφ

)

= a2 (2) (2π)

= 4πa2

(4.87)

which we recognize as the area of a sphere, asexpected. The corresponding calculation in theCartesian or cylindrical systems is quite difficultin comparison.

Integration Over Volume. The differentialvolume element in the spherical system is

dv = dr (rdθ) (r sin θdφ) = r2dr sin θ dθ dφ(4.88)

For example, if A(r) = 1 and the volume V is asphere of radius a centered on the origin, then

∫

V

A(r) dv =

∫ a

0

∫ π

0

∫ 2π

0

r2dr sin θ dθ dφ

=

(∫ a

0

r2dr

)(∫ π

0

sin θ dθ

)(∫ 2π

0

dφ

)

=

(1

3a3)(2) (2π)

=4

3πa3

(4.89)



4.5. GRADIENT 79

which is the volume of a sphere.

Additional Reading:

• “Spherical coordinate system” on Wikipedia.

4.5 Gradient

[m0098]

The gradient operator is a important and usefultool in electromagnetic theory. Here’s the mainidea:

The gradient of a scalar field is a vector whichpoints in the direction in which the field ismost rapidly increasing, with the scalar partequal to the rate of change.

A particularly important application of thegradient is that it relates the electric fieldintensity E(r) to the electric potential field V (r).This is apparent from a review of Section 2.2(“Electric Field Intensity”); see in particular thebattery-charged capacitor example. In thatexample it is demonstrated that:

• The direction of E(r) is the direction inwhich V (r) decreases most quickly, and

• The scalar part of E(r) is the rate of changeof V (r) in that direction. Note that this isalso implied by the units, since V (r) hasunits of V whereas E(r) has units of V/m.

The gradient is the mathematical operation thatrelates the vector field E(r) to the scalar fieldV (r) and is indicated by the symbol “∇” asfollows:

E(r) = −∇V (r) (4.90)

or, with the understanding that we are interestedin the gradient as a function of position r, simply

E = −∇V (4.91)

At this point we should note that the gradient isa very general concept, and that we have merely

identified one application of the gradient above.In electromagnetics there are many situations inwhich we seek the gradient ∇f of some scalarfield f(r). Furthermore we find that otherdifferential operators that are important inelectromagnetics can be interpreted in terms ofthe gradient operator ∇. These includedivergence (Section 4.6), curl (Section 4.8), andthe Laplacian (Section 4.10).

In the Cartesian system:

∇f = x∂f

∂x+ y

∂f

∂y+ z

∂f

∂z(4.92)

Example 4.4. Gradient of a ramp function.

Find the gradient of f = ax (a “ramp”having slope a along the x direction).

Solution. Here, ∂f/∂x = a and∂f/∂y = ∂f/∂z = 0. Therefore ∇f = xa.Note that ∇f points in the direction inwhich f most rapidly increases, and hasmagnitude equal to the slope of f in thatdirection.

The gradient operator in the cylindrical andspherical systems is given in Appendix B.2.

Additional Reading:

• “Gradient” on Wikipedia.

4.6 Divergence

[m0044]

In this section we present the divergenceoperator, which provides a way to calculate theflux associated with a point in space. First, let usreview the concept of flux.

The integral of a vector field over a surface is ascalar quantity known as flux. Specifically, theflux F of a vector field A(r) over a surface S is

∫

S

A · ds = F (4.93)

https://en.wikipedia.org/wiki/Spherical_coordinate_system

https://en.wikipedia.org/wiki/Gradient


Note that A could be fairly described as a fluxdensity ; i.e., a quantity having units equal to theunits of F , but divided by area (i.e., m2). Alsoworth noting is that the flux of a vector fieldwhich has unit magnitude and is normal to allpoints on S is simply the area of S.

It is quite useful to identify some electromagneticquantities as either fluxes or flux densities. Hereare two important examples:

• The electric flux density D, having units ofC/m2, is a description of the electric field asa flux density. (See Section 2.4 for moreabout electric flux density.) The integral ofD over a closed surface yields the enclosedcharge Qencl, having units of C. Thisrelationship is known as Gauss’ Law:

∮

S

D · ds = Qencl (4.94)

(See Section 5.5 for more about Gauss’ Law.)

• The magnetic flux density B, having units ofWb/m2, is a description of the magnetic fieldas a flux density. (See Section 2.5 for moreabout magnetic flux density.) The integralof B over a surface (open or closed) yieldsthe magnetic flux Φ, having units of Wb:

∫

S

B · ds = Φ (4.95)

This is important because, for example, thetime rate of change of Φ is proportional toelectric potential. (See Section 8.3 for moreabout this principle, called Faraday’s Law.)

Summarizing:

Flux is the scalar quantity obtained by inte-grating a vector field, expressed in this caseas a flux density, over a specified surface.

The concept of flux applies to a surface of finitesize. However what is frequently of interest isbehavior at a single point, as opposed to the sumor average over a region of space. For example:Returning to the idea of electric flux density (D),

perhaps we are not concerned about the totalcharge (units of C) enclosed by a surface, butrather the charge density (C/m3) at a point. Inthis case we could begin with Equation 4.94 anddivide both sides of the equation by the volumeV enclosed by S:

∮SD · dsV

=QenclV

(4.96)

Now we let V shrink to zero, giving us anexpression that must be true at whatever pointwe decide to converge upon. Taking the limit asV → 0:

limV→0

∮SD · dsV

= limV→0

QenclV

(4.97)

The quantity on the right hand side is the volumecharge density ρv (units of C/m3) at the point atwhich we converge after letting the volume go tozero. The left hand side is, by definition, thedivergence of D, sometimes abbreviated “div D”.Thus the above equation can be written

div D = ρv (4.98)

Divergence is the flux per unit volumethrough an infinitesimally-small closed sur-face surrounding a point.

We will typically not actually want to integrateand take a limit in order to calculate thedivergence. Fortunately, we do not have to. Itturns out that this operation can be expressed asthe dot product ∇ ·D; where, for example,

∇ , x∂

∂x+ y

∂

∂y+ z

∂

∂z(4.99)

in the Cartesian coordinate system. This is thesame “∇” that appears in definitions of thegradient operator (Section 4.5) and is sameoperator that often arises when considering otherdifferential operators. If we expand D in terms ofits Cartesian components:

D = xDx + yDy + zDz (4.100)

Then

div D = ∇ ·D =∂Dx

∂x+∂Dy

∂y+∂Dz

∂z(4.101)

4.6. DIVERGENCE 81

This seems to make sense for two reasons. First,it is dimensional correct: Taking the derivative ofa quantity having units of C/m2 with respect todistance yields a quantity having units of C/m3.Second, it makes sense that flux from a pointshould be related to the sum of the rates ofchange of the flux density in each basis direction.Summarizing:

The divergence of a vector field A is ∇ ·A.

Example 4.5. Divergence of a uniformfield.

A field A which is constant with respect toposition is said to be uniform. A completelygeneral description of such a field isA = xAx + yAy + zAz where Ax, Ay, andAz are all constants. We see immediatelythat the divergence of such a field must bezero: That is, ∇ ·A = 0, because eachcomponent of A is constant with respect toposition. This also makes sense from theperspective of the “flux through aninfinitesmally-small closed surface”interpretation of divergence: If the flux isuniform, the flux into the surface equals theflux out of the surface, for a net flux of zero.

Example 4.6. Divergence of alinearly-increasing field.

Consider a field A = xA0x where A0 is aconstant. The divergence of A is∇ ·A = A0. If we interpret A as a fluxdensity, then we have found that the net fluxper unit volume is simply the rate at whichthe flux density is increasing with distance.

To compute divergence in other coordinatesystems, we merely need to know either ∇ forthose systems. In the cylindrical system:

∇ = ρ1

ρ

∂

∂ρρ+ φ

1

ρ

∂

∂φρ+ z

∂

∂z(4.102)

and in the spherical system:

∇ = r1

r2∂

∂rr2 + θ

1

r sin θ

∂

∂θsin θ + φ

1

r sin θ

∂

∂φ(4.103)

Alternatively, one may use the explicitexpressions for divergence given in Appendix B.2.

Example 4.7. Divergence of aradially-decreasing field.

It is fairly common to encounter a vectorfield which is directed radially outward froma point and which decreases linearly withdistance; i.e., A = rA0/r where A0 is aconstant. For example, the electric fieldintensity due to a point charge does this; seeSection 2.2. In this case, the divergence ismost easily computed in the sphericalcoordinate system since partial derivativesin all but one direction (r) equal zero.Neglecting terms that include thesezero-valued partial derivatives, we find:

∇ ·A = r1

r2∂

∂r

(r2[A0

r

])=A0

r2(4.104)

In other words: If we interpret A as a fluxdensity, then the flux per unit volume isdecreasing with as the square of distancefrom the origin. This seems reasonable sincethe area of a sphere centered on the originincreases with the square of distance fromthe origin, and the flux density in thepresent problem must be inverselyproportional to this area.

It is useful to know is that divergence, like ∇itself, is a linear operator; that is, for anyconstant scalars a and b and vector fields A andB:

∇ · (aA+ bB) = a∇ ·A+ b∇ ·B (4.105)

Additional Reading:

• “Divergence” on Wikipedia.

https://en.wikipedia.org/wiki/Divergence


4.7 Divergence Theorem

[m0046]

The Divergence Theorem relates an integral overa volume to an integral over the surface boundingthat volume. This is useful in a number ofsituations that arise in electromagnetic analysis.In this section we derive this theorem.

Consider a vector field A representing a fluxdensity, such as the electric flux density D ormagnetic flux density B. The divergence of A is

∇ ·A = f (4.106)

where is f(r) is the flux per unit volume throughan infinitesimally-small closed surfacesurrounding the point at r. Since f is flux perunit volume, we can obtain flux for any largercontiguous volume V by integrating over V; i.e.,

flux through V =

∫

V

f dv (4.107)

In the Cartesian system, V can be interpreted asa three-dimensional grid of infinitesimally-smallcubes having sidelengths dx, dy, and dzrespectively. Note that the flux out of any faceof one of these cubes is equal to the flux into thecube that is adjacent through that face. That is,the portion of the total flux that flows betweencubes cancels when added together. In fact theonly fluxes which do not cancel in the integrationover V are those corresponding to faces which lieon the bounding surface S, since the integrationstops there. Stating this mathematically:

∫

V

f dv =

∮

S

A · ds (4.108)

Thus, we have converted a volume integral into asurface integral.

To obtain the Divergence Theorem, we return toEquation 4.106. Integrating both sides of thatequation over V, we obtain

∫

V

(∇ ·A) dv =

∫

V

f dv (4.109)

Now applying Equation 4.108 to the right handside:

∫

V

(∇ ·A) dv =

∮

S

A · ds (4.110)

The Divergence Theorem (Equation 4.110)states that the integral of the divergence ofa vector field over a volume is equal to theflux of that field through the surface bound-ing that volume.

The principal utility of the Divergence Theoremis to convert problems which are defined in termsof quantities known throughout a volume, intoproblems which are defined in terms of quantitiesknown over the bounding surface; and vice-versa.This theorem is employed several times in thisbook.

Additional Reading:

• “Divergence theorem” on Wikipedia.

4.8 Curl

[m0048]

Curl is an operation that, when applied to avector field, quantifies the circulation of thatfield. The concept of circulation has severalapplications in electromagnetics. Two of theseapplications correspond to directly to Maxwell’sEquations:

• The circulation of an electric field isproportional to the rate of change of themagnetic field. This is a statement of theMaxwell-Faraday Equation (Section 8.8),which includes as a special case Kirchoff’sVoltage Law (Section 5.11).

• The circulation of a magnetic field isproportional to the source current and therate of change of the electric field. This is astatement of Ampere’s Law (Sections 7.9and 8.9)

https://en.wikipedia.org/wiki/Divergence_theorem

4.8. CURL 83

Thus we are motivated to formally definecirculation, and then to figure out how to mostconveniently apply the concept in mathematicalanalysis. The result is the curl operator.

So, we begin with the concept of circulation:

“Circulation” is the integral of a vector fieldover a closed path.

Specifically, the circulation of the vector fieldA(r) over the closed path C is

∮

C

A · dl (4.111)

The circulation of a uniform vector field is zerofor any valid path. For example, the circulationof A = xA0 is zero because non-zerocontributions at each point on C cancel out whensummed over the closed path. On the otherhand, the circulation of A = φA0 over a circularpath of constant ρ and z is a non-zero constant,since the non-zero contributions to the integralat each point on the curve are equal andaccumulate when summed over the path.

Example 4.8. Circulation of the magneticfield intensity surrounding a line current.

Consider a current I (units of A) flowingalong the z axis in the +z direction, asshown in Figure 4.20. It is known that thiscurrent gives rise to a magnetic fieldintensity H = φH0/ρ, where H0 is aconstant having units of A since the units ofH are A/m. (Feel free to consult Section 7.5for the details; however no additionalinformation is needed to follow the examplebeing presented here.) The circulation of Halong any circular path of radius a in aplane of constant z is therefore

∮

C

H · dl =∫ 2π

φ=0

(φH0

a

)·(φ a dφ

)= 2πH0

Note that the circulation of H in this casehas two remarkable features: (1) It isindependent of the radius of the path of

z

I

H


Figure 4.20: Magnetic field intensity due to acurrent flowing along the z axis.

integration; and (2) it has units of A, whichsuggests a current. In fact, it turns out thatthe circulation of H in this case is equal tothe enclosed source current I. Furthermore,it turns out that the circulation of H alongany path enclosing the source current isequal to the source current! These findingsare consequences of Ampere’s Law, as notedabove.

Curl is, in part, an answer to the question: Whatis the circulation at a point in space? In otherwords, what is the circulation as C shrinks to it’ssmallest possible size. The answer in one sense iszero, since the arclength of C is zero in this limit– there is nothing to integrate over. However ifwe ask instead what is the circulation per unitarea in the limit, then the result should be thenon-trivial value of interest. To express thismathematically, we constrain C to lie in a plane,and define S to be the open surface bounded byC in this case. Then we define the scalar part ofthe curl of A to be:

lim∆s→0

∮CA · dl∆s

(4.112)

where ∆s is the area of S, and (important!) werequire C and S to lie in the plane thatmaximizes the above result.

Because S and it’s boundary C lie in a plane, it is



possible to assign a direction to the result. Thechosen direction is the normal n to the plane inwhich C and S lie. Because there are twonormals at each point on a plane, we specify theone which satisfies the right hand rule. This rule,applied to the curl, states that the correctnormal is the one which points through the planein the same direction as the fingers of the righthand when the thumb of your right hand isaligned along C in the direction of integration.Why is this the correct orientation of n? SeeSection 4.9 for the answer to that question. Forthe purposes of this section, it suffices to considerthis to be simply an arbitrary sign convention.

Now with the normal vector n unambiguouslydefined, we can now formally define the curloperation as follows:

curl A , lim∆s→0

n∮CA · dl∆s

(4.113)

where, once again, ∆s is the area of S, and weselect S to lie in the plane that maximizes themagnitude of the above result. Summarizing:

The curl operator quantifies the circulation ofa vector field at a point.

The magnitude of the curl of a vector fieldis the circulation, per unit area, at a pointand such that the closed path of integrationshrinks to enclose zero area while being con-strained to lie in the plane that maximizesthe magnitude of the result.

The direction of the curl is determined by theright-hand rule applied to the associated pathof integration.

Curl is a very important operator inelectromagnetic analysis. However, the definition(Equation 4.113) is usually quite difficult toapply. Remarkably, however, it turns out thatthe curl operation can be defined in terms of the∇ operator; that is, the same ∇ operatorassociated with the gradient, divergence, andLaplacian operators. Here is that definition:

curl A , ∇×A (4.114)

For example: In Cartesian coordinates,

∇ , x∂

∂x+ y

∂

∂y+ z

∂

∂z(4.115)

andA = xAx + yAy + zAz (4.116)

so curl can be calculated as follows:

∇×A =

∣∣∣∣∣∣

x y z∂∂x

∂∂y

∂∂z

Ax Ay Az

∣∣∣∣∣∣(4.117)

or, evaluating the determinant:

∇×A = x

(∂Az∂y

− ∂Ay∂z

)

+ y

(∂Ax∂z

− ∂Az∂x

)

+ z

(∂Ay∂x

− ∂Ax∂y

)(4.118)

Expressions for curl in each of the three majorcoordinate systems is provided in Appendix B.2.

It is useful to know is that curl, like ∇ itself, is alinear operator; that is, for any constant scalars aand b and vector fields A and B:

∇× (aA+ bB) = a∇×A+ b∇×B (4.119)

Additional Reading:

• “Curl (mathematics)” on Wikipedia.

4.9 Stokes’ Theorem

[m0051]

Stokes’ Theorem relates an integral over an opensurface to an integral over the curve boundingthat surface. This relationship has a number ofapplications in electromagnetic theory. Here isthe theorem:

∫

S

(∇×A) · ds =∮

C

A · dl (4.120)

https://en.wikipedia.org/wiki/Curl_(mathematics)

4.10. THE LAPLACIAN OPERATOR 85

c© Cronholm144 (modified) CC BY SA 3.0

Figure 4.21: The relative orientations of the di-rection of integration C and surface normal n inStokes’ Theorem.

where S is the open surface bounded by theclosed path C. The direction of the surfacenormal ds is related to the direction ofintegration along C by the right hand rule,illustrated in Figure 4.21. In this case, the righthand rule states that the correct normal is theone which points through the plane in the samedirection as the fingers of the right hand whenthe thumb of your right hand is aligned along Cin the direction of integration.

Stokes’ Theorem is a purely mathematic result,and not a principle of electromagnetics per se.The relevance of the theorem to electromagnetictheory is primarily as a tool in the associatedmathematical analysis. Usually the theorem isemployed to transform a problem expressed interms of an integration over a surface into anintegration over a closed path, or vice-versa. Formore information on the theorem and itsderivation, see “Additional Reading” at the endof this section.

Additional Reading:

• “Stokes’ Theorem” on Wikipedia.

4.10 The Laplacian Operator

[m0099]

The Laplacian ∇2f of a field f(r) is the

divergence of the gradient of that field:

∇2f , ∇ · (∇f) (4.121)

Note that the Laplacian is essentially a definitionof the second derivative with respect to the threespatial dimensions. For example, in Cartesiancoordinates,

∇2f =∂2f

∂x2+∂2f

∂y2+∂2f

∂z2(4.122)

as can be readily verified by applying thedefinitions of gradient and divergence inCartesian coordinates to Equation 4.121.

The Laplacian relates the electric potential (i.e.,V , units of V) to electric charge density (i.e., ρv,units of C/m3). This relationship is known asPoisson’s Equation (Section 5.15):

∇2V = −ρvǫ

(4.123)

where ǫ is the permittivity of the medium. Thefact that V is related to ρv in this way shouldnot be surprising, since electric field intensity (E,units of V/m) is proportional to the derivative ofV with respect to distance (via the gradient) andρv is proportional to the derivative of E withrespect to distance (via the divergence).

The Laplacian operator can also be applied tovector fields; for example, Equation 4.122 is valideven if the scalar field “f” is replaced with avector field. In the Cartesian coordinate system,the Laplacian of the vector fieldA = xAx + yAy + zAz is

∇2A = x∇2Ax + y∇2Ay + z∇2Az (4.124)

An important application of the Laplacianoperator of vector fields is the wave equation;e.g., the wave equation for E in a lossless andsource-free region is (Section 9.2)

∇2E+ β2E = 0 (4.125)

where β is the phase propagation constant.

It is sometimes useful to know that the Laplacianof a vector field can be expressed in terms of thegradient, divergence, and curl as follows:

∇2A = ∇ (∇ ·A)−∇× (∇×A) (4.126)


https://en.wikipedia.org/wiki/Stokes'_theorem


The Laplacian operator in the cylindrical andspherical coordinate systems is given inAppendix B.2.

Additional Reading:

• “Laplace operator” on Wikipedia.

[m0043]

https://en.wikipedia.org/wiki/Laplace_operator

4.10. THE LAPLACIAN OPERATOR 87

Image Credits

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Fig. 4.2: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fPositionFixed.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

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Fig. 4.9: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fCrossProduct.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

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Fig. 4.14: c© K. Kikkeri,https://commons.wikimedia.org/wiki/File:Cylindrical Coordinates-Area of a Circle.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

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Fig. 4.16: c© K. Kikkeri,https://commons.wikimedia.org/wiki/File:Spherical Coordinate System.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 4.18: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0097 fArc.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

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Fig. 4.20: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:Biot Savart Example.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

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https://commons.wikimedia.org/wiki/File:Biot_Savart_Example.svg


https://en.wikipedia.org/wiki/File:Stokes'_Theorem.svg


Chapter 5

Electrostatics

[m0116]

Electrostatics is the theory of the electric field inconditions in which it’s behavior is independentof magnetic fields, including

• The electric field associated with fixeddistributions of electric charge,

• Capacitance (the ability of a structure tostore energy in an electric field), and

• The energy associated with the electrostaticfield.

• Steady current induced in a conductingmaterial in the presence of an electrostaticfield (essentially, Ohm’s Law)

The term “static” refers to the fact that theseaspects of electromagnetic theory can bedeveloped by assuming sources aretime-invariant; we might say that electrostatics isthe study of the electric field at DC. Howevermany aspects of electrostatics are relevent to AC,radio frequency, and higher-frequencyapplications as well.

Additional Reading:

• Electrostatics on Wikipedia.

5.1 Coulomb’s Law

[m0102]

Consider two charge-bearing particles in freespace, identified as “particle 1” and “particle 2”in Figure 5.1. Let the charges borne by theseparticles be q1 and q2, and let R be the distancebetween them. If the particles bear charges ofthe same sign (i.e., if q1q2 is positive), then theparticles repel; otherwise, they attract. Thisrepulsion or attraction can be quantified as aforce experienced by each particle. Physicalobservations reveal that the magnitude of theforce is proportional to q1q2, and inverselyproportional to R2. For particle 2 we find:

F = RF0q1q2R2

(5.1)

where R is the unit vector pointing from theparticle 1 to the particle 2, and F0 is a constant.The value of F0 must have units of inversepermittivity; i.e., (F/m)−1. This is most easilyseen by dimensional analysis of the aboverelationship, including the suspected factor:

C · CF/m ·m2

=C · CF ·m =

C · CC/V ·m =

C ·Vm

=J

m= N

where we have used the facts that 1 F = 1 C/V,1 V = 1 J/C, and 1 N = 1 J/m. This findingsuggests that F0 ∝ ǫ−1, where ǫ is thepermittivity of the medium in which the particlesexist. Observations confirm that the force is infact inversely proportional to the permittivity,with an additional factor of 1/4π (unitless).Putting this all together we obtain what iscommonly known as Coulomb’s Law :

F = Rq1q24πǫR2

(5.2)


https://en.wikipedia.org/wiki/Electrostatics


90 CHAPTER 5. ELECTROSTATICS

Fq2

q1

particle

2

particle

1

R

-F

R


Figure 5.1: Coulomb’s Law describes the forcepercieved by pairs of charged particles.

Subsequently the force perceived by particle 2 isequal and opposite; i.e., equal to −F.

Separately it is known that F can be described interms of the electric field intensity E1 associatedwith particle 1:

F = q2E1 (5.3)

This is essentially the definition of E1, asexplained in Section 2.2. Combining this resultwith Coulomb’s Law, we obtain a means todirectly calculate the field associated with thefirst particle in the absence of the second particle:

E1 = Rq1

4πǫR2(5.4)

where now RR is the vector beginning at theparticle 1 and ending at the point to beevaluated.

The electric field intensity associated witha point charge (Equation 5.4) is (1) di-rected away from positive charge, (2) pro-portional to the magnitude of the charge, (3)inversely proportional to the permittivity ofthe medium, and (3) inversely proportionalto distance squared.

We have described this result as originating fromCoulomb’s Law, which is based on physicalobservations. However the same result may beobtained directly from Maxwell’s Equationsusing Gauss’ Law (Section 5.5).

Example 5.1. Electric field of a pointcharge at the origin.

A common starting point in electrostaticanalysis is the field associated with aparticle bearing charge q at the origin of thecoordinate system. Because the electric fieldis directed radially away from apositively-charged source particle in alldirections, this field is most convenientlydescribed in the spherical coordinatesystem. Thus, R becomes r, R becomes r,and we have

E(r) = rq

4πǫr2(5.5)

Here’s a numerical example: What is theelectric field intensity at a distance 1 µmfrom a single electron located at the origin,in free space? In this caseq ∼= −1.60× 10−19 C (don’t forget thatminus sign!), ǫ = ǫ0, r = 1 µm, and we find:

E(r) = −r (1.44 kV/m) (5.6)

This is large relative to electric fieldstrengths commonly encountered inengineering applications. The strong electricfield of the electron is not readily apparentbecause electrons in common materials tendto be accompanied by roughly equalamounts of positive charge, such as theprotons of atoms. Sometimes, however, theeffect of individual electrons does becomesignificant in practical electronics through aphenomenon known as shot noise.

Additional Reading:

• “Coulomb’s Law” on Wikipedia.

• “Shot Noise” on Wikipedia.

5.2 Electric Field Due toPoint Charges


https://en.wikipedia.org/wiki/Coulomb's_law

https://en.wikipedia.org/wiki/Shot_noise

5.3. CHARGE DISTRIBUTIONS 91

[m0103]

The electric field intensity associated with asingle particle bearing charge q1, located at theorigin, is (Section 5.1)

E(r) = rq1

4πǫr2(5.7)

If this particle is instead located at some positionr1, then the above expression may be written asfollows:

E(r; r1) =r− r1|r− r1|

q1

4πǫ |r− r1|2(5.8)

or, combining the like terms in the denominator:

E(r; r1) =r− r1

|r− r1|3q14πǫ

(5.9)

Now let us consider the field due to multiple suchparticles. Under the usual assumptions about thepermittivity of the medium (reminder:Section 2.8), the property of superpositionapplies. Using this principle, we conclude:

The electric field resulting from a set ofcharged particles is equal to the sum of thefields associated with the individual particles.

Stated mathematically:

E(r) =

N∑

n=1

E(r; rn) (5.10)

where N is the number of particles. Thus wehave

E(r) =1

4πǫ

N∑

n=1

r− rn

|r− rn|3qn (5.11)

5.3 Charge Distributions

[m0100]

In principle the smallest unit of electric chargethat can be isolated is the charge of a singleelectron, which is ∼= 1.60× 10−19 C. This is very

small, and we rarely deal with electrons one at atime, so it is usually more convenient to describecharge as a quantity which is continuous oversome region of space. In particular, it isconvenient to describe charge as beingdistributed in one of three ways: along a curve,over a surface, or within a volume.

Line Charge Distribution. Imagine thatcharge is distributed along a curve C throughspace. Let ∆q be the total charge along a shortsegment of the curve, and let ∆l be the length ofthis segment. The line charge density ρl at anypoint along the curve is defined as

ρl , lim∆l→0

∆q

∆l=dq

dl(C/m) (5.12)

We may then define ρl(l) to be a function ofposition along the curve, parameterized by l.Then the total charge Q along the curve is

Q =

∫

C

ρl(l) dl (C) (5.13)

In other words, line charge density integratedover length yields total charge.

Surface Charge Distribution. Imagine thatcharge is distributed over a surface. Let ∆q bethe total charge on a small patch on this surface,and let ∆s be the area of this patch. The surfacecharge density ρs at any point on the surface isdefined as

ρs , lim∆s→0

∆q

∆s=dq

ds(C/m2) (5.14)

Let us define ρs to be a function of position onthis surface. Then the total charge over a surfaceS is

Q =

∫

S

ρs ds (C) (5.15)

In other words, surface charge density integratedover a surface yields total charge.

Volume Charge Distribution. Imagine thatcharge is distributed over a volume. Let ∆q bethe total charge in a small cell within thisvolume, and let ∆v be the volume of this cell.The volume charge density ρv at any point in thevolume is defined as

ρv , lim∆v→0

∆q

∆v=dq

dv(C/m3) (5.16)


Since ρv is a function of position within thisvolume, the total charge within a volume V is

Q =

∫

V

ρv dv (C) (5.17)

In other words, volume charge density integratedover a volume yields total charge.

5.4 Electric Field Due to aContinuous Distributionof Charge

[m0104]

The electric field intensity associated with Ncharged particles is (Section 5.2):

E(r) =1

4πǫ

N∑

n=1

r− rn

|r− rn|3qn (5.18)

where qn and rn are the charge and position of

the nth particle. However it is common to have acontinuous distribution of charge as opposed to acountable number of charged particles. In thissection we extend Equation 5.18 using theconcept of continuous distribution of charge(Section 5.3) so that we may address this moregeneral class of problems.

Distribution of Charge Along a Curve.Consider a continuous distribution of chargealong a curve C. The curve can be divided intoshort segments of length ∆l. Then the charge

associated with the nth segment, located at rn, is

qn = ρl(rn) ∆l (5.19)

where ρl is charge density (units of C/m) at rn.Substituting this expression into Equation 5.18,we obtain

E(r) =1

4πǫ

N∑

n=1

r− rn

|r− rn|3ρl(rn) ∆l (5.20)

Taking the limit as ∆l → 0 yields:

E(r) =1

4πǫ

∫

C

r− r′

|r− r′|3ρl(r

′) dl (5.21)

where r′ represents the varying position along Cwith integration.

The simplest example of a curve is a straightline. It is straightforward to use Equation 5.21 todetermine the electric field due to a continuousdistribution of charge along a straight line.However it is much easier to analyze thatparticular distribution using Gauss’ Law, asshown in Section 5.6. The following exampleaddresses a charge distribution for whichEquation 5.21 is more appropriate.

Example 5.2. Electric field along the axisof a ring of uniformly-distributed charge.

Consider a ring of radius a in the z = 0plane, centered on the origin, as shown inFigure 5.2 Let the charge density along thisring be uniform and equal to ρl (C/m).Find the electric field along the z axis.

Solution. The source charge position isgiven in cylindrical coordinates as

r′ = ρa (5.22)

The position of a field point along the z axisis simply

r = zz (5.23)

Thusr− r′ = −ρa+ zz (5.24)

and|r− r′| =

√a2 + z2 (5.25)

Equation 5.21 becomes:

E(z) =1

4πǫ

∫ 2π

0

−ρa+ zz

[a2 + z2]3/2

ρl (a dφ)

(5.26)Pulling factors which do not vary with φ outof the integral, and factoring into separateintegrals for the φ and z components, we

5.4. ELECTRIC FIELD DUE TO A CONTINUOUS DISTRIBUTION OF CHARGE 93

y

x

z

ρa

KeLO

r - rP

QRTUV


Figure 5.2: Calculating the electric field alongthe axis of a ring of charge.

obtain:

ρl a

4πǫ [a2 + z2]3/2

[−a∫ 2π

0

ρ dφ+ zz

∫ 2π

0

dφ

]

(5.27)The second integral is equal to 2π. The firstintegral is equal to zero: To see this, notethat the integral is simply summing valuesof ρ for all possible values of φ. Sinceρ(φ+ π) = −ρ(φ), the integrand for anygiven value of φ is equal and opposite theintegrand π radians later. (This is oneexample of a symmetry argument.) Thus weobtain

E(z) = zρl a

2ǫ

z

[a2 + z2]3/2

(5.28)

It is a good exercise to confirm that thisresult is dimensionally correct. Alsorecommended: Confirm that when z ≫ a,the result is approximately the same as thatexpected from a particle having the sametotal charge as the ring.

Distribution of Charge Over a Surface.Consider a continuous distribution of charge overa surface S. The surface can be divided intosmall patches having area ∆s. Then the charge

associated with the nth patch, located at rn, is

qn = ρs(rn) ∆s (5.29)

where ρs is the surface charge density (units ofC/m2) at rn. Substituting this expression intoEquation 5.18, we obtain

E(r) =1

4πǫ

N∑

n=1

r− rn

|r− rn|3ρs(rn) ∆s (5.30)

Taking the limit as ∆s→ 0 yields:

E(r) =1

4πǫ

∫

S

r− r′

|r− r′|3ρs(r

′) ds (5.31)

where r′ represents the varying position over Swith integration.

Example 5.3. Electric field along the axisof a disk of uniformly-distributed charge.

Consider a circular disk of radius a in thez = 0 plane, centered on the origin, as shownin Figure 5.3. Let the charge density overthis disk be uniform and equal to ρs (C/m

2).Find the electric field along the z axis.

Solution. The source charge position isgiven in cylindrical coordinates as

r′ = ρρ (5.32)

The position of a field point along the z axisis simply

r = zz (5.33)

Thusr− r′ = −ρρ+ zz (5.34)

and|r− r′| =

√ρ2 + z2 (5.35)

Equation 5.31 becomes:

E(z) =1

4πǫ

∫ a

ρ=0

∫ 2π

φ=0

−ρρ+ zz

[ρ2 + z2]3/2

ρs (ρ dρ dφ)

(5.36)To solve this integral, first rearrange thedouble integral into a single integral over φ



followed by integration over ρ:

ρs4πǫ

∫ a

ρ=0

ρ

[ρ2 + z2]3/2

[∫ 2π

φ=0

(−ρρ+ zz) dφ

]dρ

(5.37)Now we address the integration over φshown in the square brackets in the aboveexpression:

∫ 2π

φ=0

(−ρρ+ zz) dφ = −ρ∫ 2π

φ=0

ρdφ+zz

∫ 2π

φ=0

dφ

(5.38)The first integral on the right is zero, for thefollowing reason: As the integral progressesin φ, the vector ρ rotates. Because theintegration is over a complete revolution(i.e., φ from 0 to 2π), the contribution fromeach pointing of ρ is canceled out by anotherpointing of ρ that is in the oppositedirection. Since there is an equal number ofthese canceling pairs of pointings, the resultis zero. Thus:

∫ 2π

φ=0

(−ρρ+ zz) dφ = 0 + zz

∫ 2π

φ=0

dφ

= z2πz (5.39)

Substituting this into Expression 5.37 weobtain:

ρs4πǫ

∫ a

ρ=0

ρ

[ρ2 + z2]3/2

[z2πz] dρ

=zρsz

2ǫ

∫ a

ρ=0

ρ dρ

[ρ2 + z2]3/2

(5.40)

This integral can be solved using integrationby parts and trigonometric substitution.Since the solution is tedious and there is noparticular principle of electromagneticsdemonstrated by this solution, we shallsimply state the result:

∫ a

ρ=0

ρ dρ

[ρ2 + z2]3/2

=−1√ρ2 + z2

∣∣∣∣∣

a

ρ=0

=−1√a2 + z2

+1

|z| (5.41)

y

x

z

ρρ

r - rW

a

X[\]^_`ab


Figure 5.3: Calculating the electric field alongthe axis of a disk of charge.

Substituting this result:

E(z) = zρsz

2ǫ

( −1√a2 + z2

+1

|z|

)

= zρs2ǫ

( −z√a2 + z2

+z

|z|

)

= zρs2ǫ

( −z√a2 + z2

+ sgn z

)(5.42)

where “sgn” is the “signum” function; i.e.,sgn z = +1 for z > 0 and sgn z = −1 forz < 0. Summarizing:

E(z) = zρs2ǫ

(sgn z − z√

a2 + z2

)(5.43)

It is a good exercise to confirm that thisresult is dimensionally correct and yields anelectric field vector which points in theexpected direction and with the expecteddependence on a and z.

A special case of the “disk of charge” scenarioconsidered in the preceding example is an infinitesheet of charge. The electric field from an infinitesheet of charge is a useful theoretical result. Weget the field in this case simply by letting a→ ∞


5.5. GAUSS’ LAW: INTEGRAL FORM 95

in Equation 5.43, yielding:

E(r) = zρs2ǫ

sgn z (5.44)

Again it is useful to confirm that this isdimensionally correct: C/m2 divided by F/myields V/m. Also, note that Equation 5.44 is theelectric field at any point above or below thecharge sheet – not just on z axis. This followsfrom symmetry: From the perspective of anypoint in space, the edges of the sheet are thesame distance (i.e., infinitely far) away.

Distribution of Charge In a Volume.Consider a continuous distribution of chargewithin a volume V. The volume can be dividedinto small cells (volume elements) having volume

∆v. Then the charge associated with the nth

cell, located at rn, is

qn = ρv(rn) ∆v (5.45)

where ρv is volume charge density (units ofC/m3) at rn. Substituting this expression intoEquation 5.18, we obtain

E(r) =1

4πǫ

N∑

n=1

r− rn

|r− rn|3ρv(rn) ∆v (5.46)

Taking the limit as ∆v → 0 yields:

E(r) =1

4πǫ

∫

V

r− r′

|r− r′|3ρv(r

′) dv (5.47)

where r′ represents the varying position over Vwith integration.

5.5 Gauss’ Law: IntegralForm

[m0014]

Gauss’ Law is one of the four fundamental lawsof classical electromagnetics, collectively knownas Maxwell’s Equations. Before diving in, thereader is strongly encouraged to reviewSection 2.4. In that section Gauss’ Law emergesfrom the interpretation of the electric field as aflux density. Section 2.4 does not actuallyidentify Gauss’ Law, but here it is:

Gauss’ Law (Equation 5.48) states that theflux of the electric field through a closed sur-face is equal to the enclosed charge.

Gauss’ Law is expressed mathematically asfollows: ∮

S

D · ds = Qencl (5.48)

where D is the electric flux density ǫE, S is aclosed surface with differential surface normal ds,and Qencl is the enclosed charge. We can see thelaw is dimensionally correct: D has units ofC/m2, thus integrating D over a surface gives aquantity with units of C/m2 · m2 = C, which arethe units of charge.

Gauss’ Law has a number of applications inelectromagnetic theory. One of them, as exploredbelow, is as a method to compute the electricfield in response to a distribution of electriccharge. Note that a method to do this, based onCoulomb’s Law, is described in Sections 5.1, 5.2,and 5.4. Gauss’ Law provides an alternativemethod that is easier or more useful in certainapplications.

Example 5.4. Electric field associated witha charged particle, using Gauss’ Law.

In this example we demonstrate the abilityof Gauss’ Law to predict the field associatedwith a charge distribution. Let us do this forthe simplest possible charge distribution: Aparticle of charge q located at the origin, forwhich we already have the answer(Section 5.1).

Solution. Gauss’ Law applies to anysurface that encloses the charge, so forsimplicity we chose a sphere of radius rcentered at the origin. Note that Qencl onthe right hand side is just q for any surfacehaving r > 0. Gauss’ Law in this casebecomes

∫ π

θ=0

∫ 2π

φ=0

D ·(rr2 sin θ dθ dφ

)= q (5.49)


If we can solve for D, we can get E usingD = ǫE. The simplest way to solve for D isto use a symmetry argument, whichproceeds as follows: In this problem, themagnitude of D can depend only on r, andnot θ or φ. This because the charge has noparticular orientation, and the sphere iscentered on the charge. Similarly, it is clearthat D must point either directly toward ordirectly away from the charge. In otherwords, D = rD(r). Substituting this in theabove equation, we encounter the dotproduct r · r, which is simply 1. Since D(r)and r2 are constants with respect to theintegration, we obtain:

r2D(r)

∫ π

θ=0

∫ 2π

φ=0

sin θ dθdφ = q (5.50)

The remaining integral is simply 4π, thus weobtain:

D(r) =q

4πr2(5.51)

Bringing the known vector orientation of Dback into the equation, we obtain

D = rq

4πr2(5.52)

and finally using D = ǫE we obtain theexpected result

E = rq

4πǫr2(5.53)

Here’s the point you should take away from theabove example:

Gauss’ Law combined with a symmetry ar-gument may be sufficient to determine theelectric field due to a charge distribution.Thus, Gauss’ Law may be an alternative toCoulomb’s Law in some applications.

Additional Reading:

• “Gauss’ Law” on Wikipedia.

5.6 Electric Field Due to anInfinite Line Chargeusing Gauss’ Law

[m0149]

Section 5.5 explains one application of Gauss’Law: That is, to find the electric field due to acharged particle. In this section we presentanother application: The electric field due to aninfinite line of charge. The result serves as auseful “building block” in a number of otherproblems, including determination of thecapacitance of coaxial cable (Section 5.24).Although this problem can be solved using the“direct” approach described in Section 5.4 (andit is an excellent exercise to do so), the Gauss’Law approach demonstrated here turns out to berelatively simple.

Example 5.5. Electric field associated withan infinite line charge, using Gauss’ Law.

Use Gauss’ Law to determine the electricfield intensity due to an infinite line ofcharge along the z axis, having chargedensity ρl (units of C/m), as shown inFigure 5.4.

Solution. Gauss’ Law requires integrationover a surface that encloses the charge. So,our first problem is to determine a suitablesurface. A cylinder of radius a which isconcentric with the z axis, as shown inFigure 5.4, is maximally symmetric with thecharge distribution and so is likely to yieldthe simplest possible analysis. At firstglance it seems that we may have a problemsince the charge extends to infinity in the+z and −z directions, so it’s not clear howto enclose all of the charge. Let’s suppressthat concern for a moment and simplychoose a cylinder of finite length l. Inprinciple we can solve the problem first forthis cylinder of finite size, which containsonly a fraction of the charge, and then laterlet l → ∞ to capture the rest of the charge.

https://en.wikipedia.org/wiki/Gauss'_law

5.6. ELECTRIC FIELD DUE TO AN INFINITE LINE CHARGE USING GAUSS’ LAW 97

(In fact we’ll find when the time comes itwill not be necessary to do that, but weshall prepare for it anyway.)

Here’s Gauss’ Law:∮

S

D · ds = Qencl (5.54)

where D is the electric flux density ǫE, S isa closed surface with outward-facingdifferential surface normal ds, and Qencl isthe enclosed charge.

The first order of business is to constrainthe form of D using a symmetry argument,as follows. Consider the field of a pointcharge q at the origin (Section 5.5):

D = rq

4πr2(5.55)

We can “assemble” an infinite line of chargeby adding particles in pairs. One pair isadded at a time, with one particle on the +zaxis and the other on the −z axis, with eachlocated an equal distance from the origin.We continue to add particle pairs in thismanner until the resulting charge extendscontinuously to infinity in both directions.The principle of superposition indicates thatthe resulting field will be the sum of thefields of the particles (Section 5.2). Thus wesee that D cannot have any component inthe φ direction, because none of the fields ofthe constituent particles have a componentin that direction. Similarly we see that themagnitude of D cannot depend on φ,because none of the fields of the constituentparticles depends on φ and because thecharge distribution is identical (“invariant”)with rotation in φ. Also note that for anychoice of z the distribution of charge aboveand below that plane of constant z isidentical; therefore D cannot be a functionof z and D cannot have any component inthe z direction. Therefore the direction of Dmust be radially outward; i.e., in the ρ

direction, as follows:

D = ρDρ(ρ) (5.56)

Next we observe that Qencl on the righthand side of Equation 5.54 is equal to ρll.Thus we obtain

∮

S

[ρDρ(ρ)] · ds = ρll (5.57)

The cylinder S consists of a flat top, curvedside, and flat bottom. Expanding the aboveequation to reflect this, we obtain

ρll =

∫

top

[ρDρ(ρ)] · (+zds)

+

∫

side

[ρDρ(ρ)] · (+ρds)

+

∫

bottom

[ρDρ(ρ)] · (−zds) (5.58)

Examination of the dot products indicatesthat the integrals associated with the topand bottom surfaces must be zero. In otherwords, the flux through the top and bottomis zero because D is perpendicular to thesesurfaces. We are left with

ρll =

∫

side

[Dρ(ρ)] ds (5.59)

The side surface is an open cylinder ofradius ρ = a, so Dρ(ρ) = Dρ(a), a constantover this surface. Thus:

ρll =

∫

side

[Dρ(a)] ds = [Dρ(a)]

∫

side

ds

(5.60)The remaining integral is simply the area ofthe side surface, which is 2πa · l. Solving forDρ(a) we obtain

Dρ(a) =ρll

2πal=

ρl2πa

(5.61)

Remarkably, we see Dρ(a) is independent ofl, So the concern raised in the beginning ofthis solution – that we wouldn’t be able toenclose all of the charge – doesn’t matter.


z

a

ρ

ρ

lccc


Figure 5.4: Finding the electric field of an infiniteline of charge using Gauss’ Law.

Completing the solution, we note the resultmust be the same for any value of ρ (notjust ρ = a), so

D = ρDρ(ρ) = ρρl2πρ

(5.62)

and since D = ǫE:

E = ρρl

2πǫρ(5.63)

This completes the solution. We have foundthat the electric field is directed radiallyaway from the line charge, and decreases inmagnitude in inverse proportion to distancefrom the line charge. Suggestion: Check toensure that this solution is dimensionallycorrect.

5.7 Gauss’ Law: DifferentialForm

[m0045]

The integral form of Gauss’ Law (Section 5.5) isa calculation of enclosed charge Qencl using the

surrounding density of electric flux:∮

S

D · ds = Qencl (5.64)

where D is electric flux density and S is theenclosing surface. It is also useful to do theinverse calculation: i.e., determine electric fieldassociated with a charge distribution. This issometimes possible using Equation 5.64 if thesymmetry of the problem permits; see examplesin Section 5.5 and 5.6. If the problem does notexhibit the necessary symmetry, then it seemsthat one must fall back to the family oftechniques presented in Section 5.4 requiringdirect integration over the charge, which isderived from Coulomb’s Law.

However even Coulomb’s Law / directintegration approach has a limitation that is veryimportant to recognize: It does not account forthe presence of structures that may influence theelectric field. For example, the electric field dueto a charge in free space is different from theelectric field due to the same charge located neara perfectly-conducting surface. In fact, theseapproaches do not account for the possibility ofany spatial variation in material composition,which rules out their use in many engineeringapplications.

To address this broader scope of problems, werequire an alternative form of Gauss’ Law thatapplies at individual points in space. That is, werequire Gauss’ Law expressed in the form of adifferential equation, as opposed to an integralequation. This facilitates the use of Gauss’ Laweven in problems which do not exhibit sufficientsymmetry and which involve material boundariesand spatial variations in material constitutiveparameters. Given this differential equation andthe boundary conditions imposed by structureand materials, we may then solve for the electricfield in these more complicated scenarios. In thissection we derive the desired differential form ofGauss’ Law. Elsewhere (in particular, inSection 5.15) we use this equation as a tool tofind electric fields in problems involving materialboundaries.

There are in fact two methods to develop the


5.7. GAUSS’ LAW: DIFFERENTIAL FORM 99

desired differential equation. One method is viathe definition of divergence, whereas the other isvia the divergence theorem. Both methods arepresented below because each provides a differentbit of insight. Here we go with the first method:

Derivation via the Definition ofDivergence. Let the geometrical volumeenclosed by S be V, which has volume V (unitsof m3). Dividing both sides of Equation 5.64 byV and taking the limit as V → 0:

limV→0

∮SD · dsV

= limV→0

QenclV

(5.65)

The quantity on the right hand side is the volumecharge density ρv (units of C/m3) at the point atwhich we converge after letting the volume go tozero. The left hand side is, by definition, thedivergence of D, indicated in mathematicalnotation as “∇ ·D” (see Section 4.6). Thus wehave Gauss’ Law in Differential Form:

∇ ·D = ρv (5.66)

To interpret this equation, recall that divergenceis simply the flux (in this case, electric flux) perunit volume.

Gauss’ Law in Differential Form (Equa-tion 5.66) says that the electric flux per unitvolume originating from a point in space isequal to the volume charge density at thatpoint.

Yet another way to understand this equation: Itis simply Gauss’ Law in integral form(Equation 5.64) applied to a vanishingly smallvolume.

Derivation via the Divergence Theorem.Equation 5.66 may also be obtained fromEquation 5.64 using the Divergence Theorem(Section 4.7), which in the present case may bewritten:

∫

V

(∇ ·D) dv =

∮

S

D · ds (5.67)

From Equation 5.64 we see that the right handside of the equation may be replaced with the

enclosed charge:∫

V

(∇ ·D) dv = Qencl (5.68)

Furthermore the enclosed charge can beexpressed as an integration of the volume chargedensity ρv over V:

∫

V

(∇ ·D) dv =

∫

V

ρvdv (5.69)

The above relationship must hold regardless ofthe specific location or shape of V. The only waythis is possible is if the integrands are equal.Thus, ∇ ·D = ρv, and we have obtainedEquation 5.66.

Example 5.6. Determining the chargedensity at a point, given the associatedelectric field.

The electric field intensity in free space is

E(r) = xAx2 + yBz + zCx2z

where A = 3 V/m3, B = 2 V/m2, andC = 1 V/m4. What is the charge density atr = x2− y2 m?

Solution. First, we use D = ǫE to get D.Since the problem is in free space, ǫ = ǫ0.Thus we have that the volume chargedensity is

ρv = ∇ ·D= ∇ · (ǫ0E) = ǫ0∇ ·E

= ǫ0

[∂

∂x

(Ax2

)+

∂

∂y(Bz) +

∂

∂z

(Cx2z

)]

= ǫ0[2Ax+ 0 + Cx2

]

Now calculating the charge density at thespecified location r:

ǫ0[2(3 V/m3)(2 m) + 0 + (1 V/m4)(−2 m)2

]

= ǫ0 (16 V/m)

= 142 pC/m3

To obtain the electric field from the chargedistribution in the presence of boundary


conditions imposed by materials and structure,we must enforce the relevant boundaryconditions. These boundary conditions arepresented in Sections 5.17 and 5.18. Frequently asimpler approach requiring only the boundaryconditions on the electric potential (V (r)) ispossible; this is presented in Section 5.15.

Furthermore, the reader should note thefollowing: Gauss’ Law does not alwaysnecessarily fully constrain possible solutions forthe electric field. For that we might also needKirchoff’s Voltage Law; see Section 5.11.

Before moving on, it is worth noting thatEquation 5.66 can be solved in the special case inwhich there are no boundary conditions tosatisfy; i.e., for charge only in a uniform andunbounded medium. In fact, no additionalelectromagnetic theory is required to do this.Here’s the solution:

D(r) =1

4π

∫

V

r− r′

|r− r′|3ρv(r

′) dv (5.70)

which we recognize as one of the results obtainedin Section 5.4 (after dividing both sides by ǫ toget E). It is reasonable to conclude that Gauss’Law (in either integral or differential form) isfundamental, whereas Coulomb’s Law is merely aconsequence of Gauss’ Law.

Additional Reading:

• “Gauss’ Law” on Wikipedia.

• “Partial differential equation” on Wikipedia.

• “Boundary value problem” on Wikipedia.

5.8 Force, Energy, andPotential Difference

[m0061]

The force Fe experienced by a particle at locationr bearing charge q in an electric field intensity Eis (see Sections 2.2 and/or Section 5.1)

Fe = qE(r) (5.71)

If left alone in free space, this particle wouldimmediately begin to move. The resultingdisplacement represents a loss of potential energythat can quantified using the concept of work,W . The incremental work ∆W done by movingthe particle a short distance ∆l, over which weassume the change in Fe is negligible, is

∆W = −Fe · l∆l (5.72)

where in this case l is the unit vector in thedirection of the motion; i.e., the direction of Fe.The minus sign indicates that potential energy ofthe system consisting of the electric field and theparticle is being reduced. Like a spring that waspreviously compressed and is now released, thesystem is “relaxing”.

To confirm that work defined in this way is anexpression of energy, consider the units: Theproduct of force (units of N) and distance (unitsof m) has units of N·m, and 1 N·m is 1 J ofenergy.

Now: What if the motion of the particle is due tofactors other than the force associated with theelectric field? For example, we might consider“resetting” the system to it’s original conditionby applying an external force to overcome Fe.Equation 5.72 still represents the change inpotential energy of the system, but now lchanges sign. The same magnitude of work isdone, but now this work is positive. In otherwords, positive work requires the application ofan external force that opposes and overcomes theforce associated with the electric field, therebyincreasing the potential energy of the system.With respect to the analogy of a mechanicalspring used above, positive work is achieved bycompressing the spring.

It is also worth noting that the purpose of thedot product in Equation 5.72 is to ensure thatonly the component of motion parallel to thedirection of the electric field is included in theenergy tally. This is simply because motion inany other direction cannot be associated with E,and therefore does not increase or decrease theassociated potential energy.

We can make the relationship between work and

https://en.wikipedia.org/wiki/Gauss'_law

https://en.wikipedia.org/wiki/Partial_differential_equation

https://en.wikipedia.org/wiki/Boundary_value_problem

5.8. FORCE, ENERGY, AND POTENTIAL DIFFERENCE 101

the electric field explicit by substitutingEquation 5.71 into Equation 5.72, yielding

∆W = −qE(r) · l∆l (5.73)

Now let us try to generalize this result. Forexample, Equation 5.73 gives the work only for ashort distance around r. If we wish to know thework done over a larger distance, then we mustaccount for the possibility that E varies alongthe path taken. To do this we may sumcontributions from each point along the pathtraced out by the particle, i.e.,

W =N∑

n=1

∆W (rn) (5.74)

where rn are positions defining the path.Substituting Equation 5.73, we have

W = −qN∑

n=1

E(rn) · l(rn)∆l (5.75)

Taking the limit as ∆l → 0 we obtain

W = −q∫

C

E(r) · l(r)dl (5.76)

where C is the path (previously, the sequence ofrn’s) followed. Now omitting the explicitdependence on r in the integrand for clarity:

W = −q∫

C

E · dl (5.77)

where dl = ldl as usual. Now, we are able todetermine the change in potential energy for acharged particle moving between any two pointsin space, given the electric field.

At this point it is convenient to formally definethe electric potential difference V21 between thestart point (1) and end point (2) of C. V21 isdefined as the work done by traversing C, perunit of charge:

V21 ,W

q(5.78)

This has units of J/C, which is volts (V).Substituting Equation 5.77, we obtain:

V21 = −∫

C

E · dl (5.79)

An advantage of analysis in terms of electricalpotential as opposed to energy is that we will nolonger have to explicitly state the value of thecharge involved.

The potential difference V12 between twopoints in space, given by Equation 5.79, isthe change in potential energy of a chargedparticle divided by the charge of the particle.Potential energy is also commonly known as“voltage” and has units of V.

Example 5.7. Potential difference in auniform electric field.

Consider an electric field E(r) = zE0 whichis constant in both magnitude and directionthroughout the domain of the problem. Thepath of interest is a line beginning at zz1and ending at zz2. What is V21? (It’s worthnoting that the answer to this problem is abuilding block for a vast number ofproblems in electromagnetic analysis.)

Solution. From Equation 5.79 we have

V21 = −∫ z2

z1

(zE0) · zdz = −E0(z2 − z1)

(5.80)In other words, V12 is simply the electricfield intensity times the distance betweenthe points. This may seem familiar: Forexample, compare this to the findings of thebattery-charged capacitor experimentdescribed in Section 2.2. There too we findthat potential difference equals electric fieldintensity times distance, and the signs agree.

The solution to the preceding example is simplebecause the direct path between the two points isparallel to the electric field. If the path betweenthe points had been perpendicular to E, then thesolution is even easier: V12 is simply zero. In allother cases, V12 is proportional to the componentof the direct path between the start and endpoints that is parallel to E, as determined by thedot product.


5.9 Independence of Path

[m0062]

In Section 5.8 we find that the potentialdifference (“voltage”) associated with a path C inan electric field intensity E is

V21 = −∫

C

E · dl (5.81)

where the curve begins at point 1 and ends atpoint 2. Let these points be identified using theposition vectors r1 and r2, respectively.

1 Then:

V21 = −∫

r2

r1, along C

E · dl (5.82)

The associated work done by a particle bearingcharge q is

W21 = qV21 (5.83)

This work represents the change in potentialenergy of the system consisting of the electricfield and the charged particle. So, it must also betrue that

W21 =W2 −W1 (5.84)

where W2 and W1 are the potential energieswhen the particle is at r2 and r1, respectively. Itis clear from the above equation that W21 doesnot depend on C; it depends only on thepositions of the start and end points, and not onany of the intermediate points along C. That is,

V21 = −∫

r2

r1

E · dl , independent of C (5.85)

Since the result of the integration inEquation 5.85 is independent of the path ofintegration, any path which begins at r1 andends at r2 yields the same value of W21 and V21.We refer to this concept as independence of path.

The integral of the electric field over a pathbetween two points depends only on the lo-cations of the start and end points, and isindependent of the path taken between thosepoints.

1See Section 4.1 for a refresher on this concept.

A practical application of this concept is thatsome paths may be easier to use than others, sothere may be an advantage in computing theintegral in Equation 5.85 using some path otherthan the path actually traversed.

5.10 Kirchoff’s Voltage Lawfor Electrostatics:Integral Form

[m0016]

As explained in Section 5.9, the electricalpotential at point r2 relative to r1 in an electricfield E (V/m) is

V21 = −∫

r2

r1

E · dl (5.86)

where the path of integration may be any paththat begins and ends at the specified points.Consider what happens if the selected paththrough space begins and ends at the samepoint; i.e., r2 = r1. In this case the path ofintegration is a closed loop. Since V21 dependsonly on the positions of the start and end points,and because the potential energy at those pointsis the same, we conclude:

∮E · dl = 0 (5.87)

This principle is known as Kirchoff’s VoltageLaw for Electrostatics.

Kirchoff’s Voltage Law for Electrostatics(Equation 5.87) states that the integral of theelectric field over a closed path is zero.

It is worth noting that this law is ageneralization of a principle of which the readeris likely already aware: In electric circuit theory,the sum of voltages over any closed loop in acircuit is zero. This is also known as Kirchoff’sVoltage Law, because it is precisely the sameprinciple. To obtain Equation 5.87 for an electriccircuit, simply partition the closed path into

5.11. KIRCHOFF’S VOLTAGE LAW FOR ELECTROSTATICS: DIFFERENTIAL FORM 103

branches, with each branch representing onecomponent. Then the integral of E over eachbranch is the branch voltage; i.e., units of V/mtimes units of m yields units of V. Then the sumof these branch voltages over any closed loop iszero, as dictated by Equation 5.87.

Finally, be advised that Equation 5.87 is specificto electrostatics. In electrostatics, it is assumedthat the electric field is independent of themagnetic field. This is true if the magnetic fieldis either zero or not time-varying. If themagnetic field is time-varying, thenEquation 5.87 must be modified to account forthe effect of the magnetic field, which is to makethe right hand size potentially different fromzero. The generalized version of this expressionthat correctly accounts for that effect is knownas the Maxwell-Faraday Equation (Section 8.8).

Additional Reading:

• “Maxwell’s Equations” on Wikipedia.

• “Kirchoff’s Circuit Laws” on Wikipedia.

5.11 Kirchoff’s Voltage Lawfor Electrostatics:Differential Form

[m0152]

The integral form of Kirchoff’s Voltage Law forelectrostatics (KVL; Section 5.10) states that anintegral of the electric field along a closed path isequal to zero: ∮

C

E · dl = 0 (5.88)

where E is electric field intensity and C is theclosed curve. In this section we derive thedifferential form of this equation. In someapplications this differential equation, combinedwith boundary conditions imposed by structureand materials (Sections 5.17 and 5.18), can beused to solve for the electric field in arbitrarilycomplicated scenarios. A more immediate reason

for considering this differential equation is thatwe gain a little more insight into the behavior ofthe electric field, disclosed at the end of thissection.

The equation we seek may be obtained usingStokes’ Theorem (Section 4.9), which in thepresent case may be written:

∫

S

(∇×E) · ds =∮

C

E · dl (5.89)

where S is any surface bounded by C, and ds isthe normal to that surface with directiondetermined by right-hand rule. The integral formof KVL tells us that the right hand side of theabove equation is zero, so:

∫

S

(∇×E) · ds = 0 (5.90)

The above relationship must hold regardless ofthe specific location or shape of S. The only waythis is possible for all possible surfaces is if theintegrand is zero at every point in space. Thuswe obtain the desired expression:

∇×E = 0 (5.91)

Summarizing:

The differential form of Kirchoff’s VoltageLaw for electrostatics (Equation 5.91) statesthat the curl of the electrostatic field is zero.

Equation 5.91 is a partial differential equation; asnoted above this equation, combined with theappropriate boundary conditions, can be solvedfor the electric field in arbitrarily complicatedscenarios. Interestingly, it is not the only suchequation available for this purpose: Gauss’ Law(Section 5.7) also does this. Thus we see asystem of partial differential equations emerging,and one may correctly infer that that the electricfield is not necessarily fully constrained by eitherequation alone. The complete system ofequations is commonly known as Maxwell’sEquations.

Additional Reading:



https://en.wikipedia.org/wiki/Maxwell's_equations

https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws




5.12 Electric Potential Dueto Point Charges

[m0064]

The electric field intensity due to a point chargeq at the origin is (see Section 5.1 or 5.5)

E = rq

4πǫr2(5.92)

In Sections 5.8 and 5.9 it was determined thatthe potential difference measured from positionr1 to position r2 is

V21 = −∫

r2

r1

E · dl (5.93)

This method for calculating potential differenceis often a bit awkward. To see why, consider anexample from circuit theory, shown in Figure 5.5.In this example, consisting of a single resistor anda ground node, we’ve identified four quantities:

• The resistance R,

• The current I through the resistor,

• The node voltage V1, which is the potentialdifference measured from ground to the leftside of the resistor, and

• The node voltage V2, which is the potentialdifference measured from ground to the rightside of the resistor.

Let’s say we wish to calculate the potentialdifference V21 across the resistor. There are twoways this can be done:

• V21 = −IR

• V21 = V2 − V1

The advantage of the second method is that it isnot necessary to know I, R, or indeed anythingabout what is happening between the nodes; it isonly necessary to know the node voltages. Thepoint is that it is often convenient to have a

R

IV1 Vd

fhijok

lmflnmj

fhijok

lmflnmj


Figure 5.5: A resistor from a larger circuit, usedas an example to demonstrate the concept of nodevoltages.

common datum – in this example, ground – withrespect to which the potential differences at allother locations of interest can be defined. Whenwe have this, calculating potential differencesreduced to simply subtracting predeterminednode potentials.

So, can we establish a datum in generalelectrostatic problems that works the same way?The answer is yes. The datum is arbitrarilychosen to be a sphere that encompasses theuniverse; i.e., a sphere with radius → ∞.Employing this choice of datum, we can useEquation 5.93 to define V (r), the potential atpoint r, as follows:

V (r) = −∫

r

∞

E · dl (5.94)

The electrical potential at a point, given byEquation 5.94, is defined as the potential dif-ference measured beginning at a sphere of in-finite radius and ending at the point r. Thepotential obtained in this manner is with re-spect to the potential infinitely far away.

In the particular case where E is due to the pointcharge at the origin:

V (r) = −∫

r

∞

[r

q

4πǫr2

]· dl (5.95)

The principle of independence of path(Section 5.9) asserts that the path of integration


5.13. ELECTRIC POTENTIAL DUE TO A CONTINUOUS DISTRIBUTION OF CHARGE 105

doesn’t matter as long as the path begins at thedatum at infinity and ends at r. So, we shouldchoose the easiest such path. The radialsymmetry of the problem indicates that theeasiest path will be a line of constant θ and φ, sowe choose dl = rdr. Continuing:

V (r) = −∫ r

∞

[r

q

4πǫr2

]· [rdr] (5.96)

= − q

4πǫ

∫ r

∞

1

r2dr (5.97)

= +q

4πǫ

1

r

∣∣∣∣r

∞

(5.98)

so

V (r) = +q

4πǫr(5.99)

(Suggestion: Confirm that Equation 5.99 isdimensionally correct.) In the context of thecircuit theory example above, this is the “nodevoltage” at r when the datum is defined to bethe surface of a sphere at infinity. Subsequently,we may calculate the potential difference fromany point r1 to any other point r2 as

V21 = V (r2)− V (r1) (5.100)

and that will typically be a lot easier than usingEquation 5.93.

It is not often that one deals with systemsconsisting of a single charged particle. So, for theabove technique to be truly useful, we need astraightforward way to determine the potentialfield V (r) for arbitrary distributions of charge.The first step in developing a more generalexpression is to determine the result for aparticle located at a point r′ somewhere otherthan the origin. Since Equation 5.99 dependsonly on charge and the distance between the fieldpoint r and r′, we have

V (r; r′) , +q′

4πǫ |r− r′| (5.101)

where, for notational consistency, we use thesymbol q′ to indicate the charge. Now applyingsuperposition, the potential field due to Ncharges is

V (r) =

N∑

n=1

V (r; rn) (5.102)

Substituting Equation 5.101 we obtain:

V (r) =1

4πǫ

N∑

n=1

qn|r− rn|

(5.103)

Equation 5.103 gives the electric potential ata specified location due to a finite number ofcharged particles.

The potential field due to continuousdistributions of charge is addressed inSection 5.13.

5.13 Electric Potential dueto a ContinuousDistribution of Charge

[m0065]

The electrostatic potential field at r associatedwith N charged particles is

V (r) =1

4πǫ

N∑

n=1

qn|r− rn|

(5.104)

where qn and rn are the charge and position of

the nth particle. However it is more common tohave a continuous distribution of charge asopposed to a countable number of chargedparticles. We now consider how to compute V (r)three types of these commonly-encountereddistributions. Before beginning, it’s worth notingthat the methods will be essentially the same,from a mathematical viewpoint, as thosedeveloped in Section 5.4; therefore a review ofthat section may be helpful before attemptingthis section.

Continuous Distribution of Charge Alonga Curve. Consider a continuous distribution ofcharge along a curve C. The curve can be dividedinto short segments of length ∆l. Then the

charge associated with the nth segment, locatedat rn, is

qn = ρl(rn) ∆l (5.105)


where ρl is the line charge density (units of C/m)at rn. Substituting this expression intoEquation 5.104, we obtain

V(r) =1

4πǫ

N∑

n=1

ρl(rn)

|r− rn|∆l (5.106)

Taking the limit as ∆l → 0 yields:

V (r) =1

4πǫ

∫

C

ρl(l)

|r− r′|dl (5.107)

where r′ represents the varying position along Cwith integration along the length l.

Continuous Distribution of Charge Over aSurface. Consider a continuous distribution ofcharge over a surface S. The surface can bedivided into small patches having area ∆s. Then

the charge associated with the nth patch, locatedat rn, is

qn = ρs(rn) ∆s (5.108)

where ρs is surface charge density (units ofC/m2) at rn. Substituting this expression intoEquation 5.104, we obtain

V (r) =1

4πǫ

N∑

n=1

ρs(rn)

|r− rn|∆s (5.109)

Taking the limit as ∆s→ 0 yields:

V (r) =1

4πǫ

∫

S

ρs(r′)

|r− r′| ds (5.110)

where r′ represents the varying position over Swith integration.

Continuous Distribution of Charge in aVolume. Consider a continuous distribution ofcharge within a volume V. The volume can bedivided into small cells (volume elements) havingarea ∆v. Then the charge associated with the

nth cell, located at rn, is

qn = ρv(rn) ∆v (5.111)

where ρv is the volume charge density (units ofC/m3) at rn. Substituting this expression intoEquation 5.104, we obtain

V (r) =1

4πǫ

N∑

n=1

ρv(rn)

|r− rn|∆v (5.112)

Taking the limit as ∆v → 0 yields:

V (r) =1

4πǫ

∫

V

ρv(r′)

|r− r′| dv (5.113)

where r′ represents the varying position over Vwith integration.

5.14 Electric Field as theGradient of Potential

[m0063]

In Section 5.8, it was determined that theelectrical potential difference V21 measured overa path C is given by

V21 = −∫

C

E(r) · dl (5.114)

where E(r) is the electric field intensity at eachpoint r along C. In Section 5.12, we defined thescalar electric potential field V (r) as the electricpotential difference at r relative to a datum atinfinity. In this section, we address the “inverseproblem”: namely, how to calculate E(r) givenV (r). Specifically, we are interested in a direct“pointwise” mathematical transform from one tothe other. Since Equation 5.114 is in the form ofan integral, it should not come as a surprise thatthe desired expression will be in the form of adifferential equation.

We begin by identifying the contribution of aninfinitessmal length of the integral to the totalintegral in Equation 5.114. At point r, this is

dV = −E(r) · dl (5.115)

Although we can proceed using any coordinatesystem, the following derivation is particularlysimple in Cartesian coordinates. In Cartesiancoordinates,

dl = xdx+ ydy + zdz (5.116)

We also note that for any scalar function ofposition, including V (r), it is true that

dV =∂V

∂xdx+

∂V

∂ydy +

∂V

∂zdz (5.117)

5.14. ELECTRIC FIELD AS THE GRADIENT OF POTENTIAL 107

Note the above relationship is not specific toelectromagnetics; it is simply mathematics. Alsonote that dx = dl · x, and so on for dy and dz.Making these substitutions into the aboveequation, we obtain:

dV =∂V

∂x(dl · x) + ∂V

∂y(dl · y) + ∂V

∂z(dl · z)(5.118)

This equation may be rearranged as follows:

dV =

([x∂

∂x+ y

∂

∂y+ z

∂

∂z

]V

)· dl (5.119)

Comparing the above equation toEquation 5.115, we find:

E(r) = −[x∂

∂x+ y

∂

∂y+ z

∂

∂z

]V (5.120)

Note that the quantity in square brackets is thegradient operator “∇” (Section 4.5). Thus wemay write

E = −∇V (5.121)

which is the relationship we seek.

The electric field intensity at a point is thegradient of the electric potential at that pointafter a change of sign (Equation 5.121).

Using Equation 5.121 we can immediately findthe electric field at any point r if we can describeV as a function of r. Furthermore, thisrelationship between V and E has a usefulphysical interpretation. Recall that the gradientof a scalar field is a vector which points in thedirection in which that field increases mostquickly. Therefore:

The electric field points in the direction inwhich the electric potential most rapidly de-creases.

This result should not come as a completesurprise; for example the reader should alreadybe aware that the electric field points away fromregions of net positive charge and toward regionsof net negative charge (Sections 2.2 and/or 5.1).What is new here is that both the magnitude

and direction of the electric field may bedetermined given only the potential field,without having to consider the charge that is thephysical source of the electrostatic field.

Example 5.8. Electric field of a chargedparticle, beginning with the potential field.

In this example we determine the electricfield of a particle bearing charge q located atthe origin. This may be done in a “direct”fashion using Coulomb’s Law (Section 5.1).However here we have the opportunity tofind the electric field using a differentmethod. In Section 5.12 we found the scalarpotential for this source was:

V (r) =q

4πǫr(5.122)

So we may obtain the electric field usingEquation 5.121:

E = −∇V = −∇( q

4πǫr

)(5.123)

Here V (r) is expressed in sphericalcoordinates, so we have (Section B.2):

E = −[r∂

∂r+ θ

1

r

∂

∂θ+ φ

1

r sin θ

∂

∂φ

]( q

4πǫr

)

(5.124)In this case V (r) does not vary with φ or θ,so the second and third terms of thegradient are zero. This leaves

E = −r∂

∂r

( q

4πǫr

)

= −rq

4πǫ

∂

∂r

1

r

= −rq

4πǫ

(− 1

r2

)(5.125)

So we find

E = +rq

4πǫr2(5.126)

as was determined in Section 5.1.


5.15 Poisson’s and Laplace’sEquations

[m0067]

The electric scalar potential field V (r), defined inSection 5.12, is useful for a number of reasonsincluding the ability to conveniently computepotential differences (i.e., V12 = V (r2)− V (r1))and the ability to conveniently determine theelectric field by taking the gradient (i.e.,E = −∇V ). One way to obtain V (r) is byintegration over the source charge distribution,as described in Section 5.13. This method isawkward in the presence of material interfaces,which impose boundary conditions on thesolutions that must be satisfied simultaneously.For example, the electric potential on a perfectlyconducting surface is constant:2 A constraintwhich is not taken into account in any of theexpressions in Section 5.13.

In this section we develop an alternativeapproach to calculating V (r) that accommodatesthese boundary conditions, and therebyfacilitates the analysis of the scalar potential fieldin the vicinity of structures and spatially-varyingmaterial properties. This alternative approach isbased on Poisson’s Equation, which we nowderive.

We begin with the differential form of Gauss’Law (Section 5.7):

∇ ·D = ρv (5.127)

Using the relationship D = ǫE (and keeping inmind our standard assumptions about materialproperties, summarized in Section 2.8) we obtain

∇ ·E =ρvǫ

(5.128)

Next we apply the relationship (Section 5.14):

E = −∇V (5.129)

yielding

∇ · ∇V = −ρvǫ

(5.130)

2This fact is probably already known to the reader frompast study of elementary circuit theory; however this is es-tablished in the context of electromagnetics in Section 5.19.

This is Poisson’s Equation, but it is not in theform in which it is commonly employed. Toobtain the alternative form, consider theoperator ∇ · ∇ in Cartesian coordinates:

∇ · ∇ =

[∂

∂xx+

∂

∂yy +

∂

∂zz

]·[∂

∂xx+

∂

∂yy +

∂

∂zz

]

=∂2

∂x2+

∂2

∂y2+

∂2

∂z2

= ∇2 (5.131)

i.e., the operator ∇ · ∇ is identically theLaplacian operator ∇2 (Section 4.10).Furthermore, this is true regardless of thecoordinate system employed. Thus we obtain ourpreferred form of Poisson’s Equation:

∇2V = −ρvǫ

(5.132)

Poisson’s Equation (Equation 5.132) statesthat the Laplacian of the electric potentialfield is equal to the volume charge density di-vided by the permittivity, with a change ofsign.

Note that Poisson’s Equation is a partialdifferential equation, and therefore can be solvedusing well-known techniques already establishedfor such equations. In fact, Poisson’s Equation isan inhomogeneous differential equation, with theinhomogeneous part −ρv/ǫ representing, asusual, the source of the field. In the presence ofmaterial structure, we identify the relevantboundary conditions at the interfaces betweenmaterials, and the task of finding V (r) is reducedto the purely mathematical task of solving theassociated boundary value problem (see“Additional Reading” at the end of this section).This approach is particularly effective when oneof the materials is a perfect conductor, or can bemodeled as such a material. This is because – asnoted at the beginning of this section – theelectric potential at all points on the surface of aperfect conductor must be equal, resulting in aparticularly simple boundary condition.

In many other applications, the chargeresponsible for the electric field lies outside the

5.16. POTENTIAL FIELD WITHIN A PARALLEL PLATE CAPACITOR 109

domain of the problem; i.e., we have non-zeroelectric field (hence, potentially non-zero electricpotential) in a region which is free of charge. Inthis case, Poisson’s Equation simplifies toLaplace’s Equation:

∇2V = 0 (source-free region) (5.133)

Laplace’s Equation (Equation 5.133) statesthat the Laplacian of the electric potentialfield is zero in a source-free region.

Like Poisson’s Equation, Laplace’s Equation,combined with the relevant boundary conditions,can be used to solve for V (r), but only in regionswhich contain no charge.

An interesting and occasionally-useful insightfrom Laplace’s Equation is as follows: At anypoint where there is no charge, the electric scalarpotential is independent of permittivity of themedium.

Additional Reading:

• “Poisson’s equation” on Wikipedia.


• “Laplace’s equation” on Wikipedia.

5.16 Potential Field Withina Parallel PlateCapacitor

[m0068]

This section presents a simple example thatdemonstrates the use of Laplace’s Equation(Section 5.15) to determine the potential field ina source free region. The example, shown inFigure 5.6, pertains to an important structure inelectromagnetic theory: The parallel platecapacitor. Here we are concerned only with thepotential field V (r) between the plates of the

d

z

ρ

a

V-+VC

V-

Figure 5.6: A parallel plate capacitor, as ademonstration of the use of Laplace’s Equation.

capacitor; you do not need to be familiar withcapacitance or capacitors to follow this section(although you’re welcome to look ahead toSection 5.22 for a preview, if desired). What isrecommended before beginning is a review of thebattery-charged capacitor experiment discussedin Section 2.2 – In this section you’ll see arigorous derivation of what we figured out in aninformal way in that section.

The parallel-plate capacitor in Figure 5.6 consistsof two perfectly-conducting circular disksseparated by a distance d by a spacer materialhaving permittivity ǫ. There is no charge presentin the spacer material, so Laplace’s Equationapplies. That equation is (Section 5.15):

∇2V = 0 (source-free region) (5.134)

Let VC be the potential difference between theplates, which would also be the potentialdifference across the terminals of the capacitor.The radius a of the plates is larger than d byenough that we may neglect what is going on atat the edges of the plates – more on this as wework the problem. Under this assumption, whatis the electric potential field V (r) between theplates?

This problem has cylindrical symmetry, so itmakes sense to continue to use cylindricalcoordinates with the z axis being perpendicularto the plates. Equation 5.134 in cylindricalcoordinates is:

[1

ρ

∂

∂ρ

(ρ∂

∂ρ

)+

1

ρ2∂2

∂φ2+

∂2

∂z2

]V = 0 (5.135)

https://en.wikipedia.org/wiki/Poisson's_equation


https://en.wikipedia.org/wiki/Laplace's_equation


or perhaps a little more clearly written as follows:

1

ρ

∂

∂ρ

(ρ∂V

∂ρ

)+

1

ρ2∂2V

∂φ2+∂2V

∂z2= 0 (5.136)

Since the problem has radial symmetry,∂V/∂φ = 0. Since d≪ a, we expect the fields tobe approximately constant with ρ until we getclose to the edge of the plates. Therefore weassume ∂V/∂ρ is negligible and can be taken tobe zero. Thus we are left with

∂2V

∂z2≈ 0 for ρ≪ a (5.137)

The general solution to Equation 5.137 isobtained simply by integrating both sides twice,yielding

V (z) = c1z + c2 (5.138)

where c1 and c2 are constants that must beconsistent with the boundary conditions. Thus,we must develop appropriate boundaryconditions. Let the node voltage at the negative(z = 0) terminal be V−. Then the voltage at thepositive (z = +d) terminal is V− +VC . Therefore:

V (z = 0) = V− (5.139)

V (z = +d) = V− + VC (5.140)

These are the relevant boundary conditions.Substituting V (z = 0) = V− into Equation 5.138yields c2 = V−. SubstitutingV (z = +d) = V− + VC into Equation 5.138 yieldsc1 = VC/d. Thus, the answer to the problem is

V (z) ≈ VCdz + V− for ρ≪ a (5.141)

Note that the above result is dimensionallycorrect and confirms that the potential deepinside in a “thin” parallel plate capacitor changeslinearly with distance between the plates.

Further, you should find that application of theequation E = −∇V (Section 5.14) to the solutionabove yields the expected result for the electricfield intensity: E ≈ −zVC/d. This is preciselythe result that we arrived at (without the aid ofLaplace’s Equation) in Section 2.2.

A reasonable question to ask at this point wouldbe: What about the potential field close to the

edge of the plates, or, for that matter, beyondthe plates? The field in this region is referred toas a fringing field. For the fringing field, ∂V/∂ρis no longer negligible and must be taken intoaccount. In addition it is necessary to modify theboundary conditions to account for the outsidesurfaces of the plates (that is, the sides of theplates that face away from the dielectric) and toaccount for the effect of the boundary betweenthe spacer material and free space. These issuesmake the problem much more difficult. When anaccurate calculation of a fringing field isnecessary, it is common to resort to a numericalsolution of Laplace’s Equation. Fortunately,accurate calculation of fringing fields is usuallynot required in practical engineering applications.

5.17 Boundary Conditionson the Electric FieldIntensity

[m0020]

In homogeneous media, electromagneticquantities vary smoothly and continuously. At aninterface between dissimilar media, however, it ispossible for electromagnetic quantities to bediscontinuous. These discontinuities can bedescribed mathematically as boundaryconditions, and used to to constrain solutions forthe associated electromagnetic quantities. In thissection we derive boundary conditions on theelectric field intensity E.

To begin, consider a region consisting of only twomedia which meet at an interface defined by themathematical surface S, as shown in Figure 5.7.If either one of the materials is a perfect electricalconductor (PEC), then S is an equipotentialsurface; i.e., the electric potential V is constanteverywhere on S. Since E is proportional to thespatial rate of change of potential (recallE = −∇V ; Section 5.14), it must be that

The component of E which is tangent to aperfectly-conducting surface is zero.

5.17. BOUNDARY CONDITIONS ON THE ELECTRIC FIELD INTENSITY 111

PEC

E

E EE

(equipotential)


Figure 5.7: At the surface of a perfectly-conducting region, E may be perpendicular to thesurface (two leftmost possibilities), but may notexhibit a component which is tangent to the sur-face (two rightmost possibilities).

This is sometimes written or spoken informallyas follows:

Etan = 0 for PEC (5.142)

where “Etan” is understood to be the componentof E which is tangent to S. Since the tangentialcomponent of E on the surface of a perfectconductor is zero, the electric field at the surfacemust be oriented entirely in the directionperpendicular to the surface, as shown inFigure 5.7.

The following equation expresses precisely thesame idea, but includes the calculation of thetangential component as part of the statement:

E× n = 0 (on PEC surface) (5.143)

where n is either normal (i.e., unit vectorperpendicular to the surface) to each point on S.This expression works because the cross productof any two vectors is perpendicular to eithervector (Section 4.1), and any vector which isperpendicular to either definition of n is tangentto S. For this same reason, it is not importantwhether n in Equation 5.143 points into or awayfrom the PEC region.

Now let us consider whether there is a moregeneral boundary condition that applies evenwhen neither of the media bordering S is aperfect conductor. The desired boundarycondition can be obtained directly from

BA w

t


Figure 5.8: Use of KVL to determine the bound-ary condition on E.

Kirchoff’s Voltage Law (KVL; Section 5.10):∮

C

E · dl = 0 (5.144)

Let the closed path of integration take the formof a rectangle centered on S, as shown inFigure 5.8. Let the sides A, B, C, and D beperpendicular or parallel to the surface,respectively. Let the length of the perpendicularsides be w, and let the length of the parallel sidesbe l. From KVL we have

∮

C

E · dl =∫

A

E · dl

+

∫

B

E · dl

+

∫

C

E · dl

+

∫

D

E · dl = 0 (5.145)

Now we let w and l become vanishingly smallwhile (1) maintaining the ratio l/w and (2)keeping C centered on S.

Now let us reduce w and l together while (1)maintaining a constant ratio w/l ≪ 1 and (2)keeping C centered on S. In this process thecontributions from the B and D segmentsbecome equal in magnitude but opposite in sign;i.e., ∫

B

E · dl+∫

D

E · dl → 0 (5.146)

This leaves∮

C

E · dl →∫

A

E · dl+∫

C

E · dl → 0 (5.147)




Let us define the unit vector t (“tangent”) asshown in Figure 5.8. When the lengths of sides Aand C become sufficiently small, we can writethe above expression as follows:

E1 · t∆l −E2 · t∆l → 0 (5.148)

where E1 and E2 are the fields evaluated on thetwo sides of the boundary and ∆l → 0 is thelength of sides A and C while this is happening.Note that the only way Equation 5.148 can betrue is if the tangential components of E1 and E2

are equal. In other words:

The tangential component of E must be con-tinuous across an interface between dissimilarmedia.

Note that this is a generalization of the result weobtained earlier for the case in which one of themedia was a PEC – in that case the tangentcomponent of E on the other side of the interfacemust be zero because it is zero in the PECmedium.

As before, we can express this idea in compactmathematical notation. Using the same ideaused to obtain Equation 5.143, we have found

E1 × n = E2 × n on S (5.149)

or, as it is more commonly written:

n× (E1 −E2) = 0 on S (5.150)

We conclude this section with a note about thebroader applicability of this boundary condition:

Equation 5.150 is the boundary conditionthat applies to E for both the electrostaticand the general (time-varying) case.

Although a complete explanation is not possiblewithout the use of the Maxwell-Faraday Equaion(Section 8.8), the reason why this boundarycondition applies in the time-varying case can bedisclosed here. In the presence of time-varyingmagnetic fields, the right-hand side of

PEC

D=nρs


Figure 5.9: The component of D which is perpen-dicular to a perfectly-conducting surface is equalto the charge density on the surface.

Equation 5.144 may become non-zero and isproportional to the area defined by the closedloop. However the above derivation requires thearea of this loop to approach zero, in which casethe possible difference from Equation 5.144 alsoconverges to zero. Therefore the boundarycondition expressed in Equation 5.150 appliesgenerally.

5.18 Boundary Conditionson the Electric FluxDensity

[m0021]

In this section we derive boundary conditions onthe electric flux density D. The considerationsare quite similar to those encountered in thedevelopment of boundary conditions on theelectric field intensity (E) in Section 5.17, so thereader may find it useful to review that sectionbefore attempting this section. This section alsoassumes familiarity with the concepts of electricflux, electric flux density, and Gauss’ Law; for arefresher Sections 2.4 and 5.5 are suggested.

To begin, consider a region at which twootherwise-homogeneous media meet at aninterface defined by the mathematical surface S,as shown in Figure 5.9. Let one of these regionsbe a perfect electrical conductor (PEC). InSection 5.17 we established that the tangentialcomponent of the electric field must be zero, and


5.18. BOUNDARY CONDITIONS ON THE ELECTRIC FLUX DENSITY 113

therefore the electric field is directed entirely inthe direction perpendicular to the surface. Wefurther know that the electric field within theconductor is identically zero. Therefore D at anypoint on S is entirely in the directionperpendicular to the surface, and pointing intothe non-conducting medium. However it is alsopossible to determine the magnitude of D. Weshall demonstrate in this section that

At the surface of a perfect conductor, themagnitude of D is equal to the surface chargedensity ρs (units of C/m2) at that point.

The following equation expresses precisely thesame idea, but includes the calculation of theperpendicular component as part of thestatement:

D · n = ρs (on PEC surface) (5.151)

where n is the normal to S pointing into thenon-conducting region. (Note that theorientation of n is now important: We haveassumed n points into region 1, and we must nowstick with this choice.) Before proceeding withthe derivation, it may be useful to note that thisresult is not surprising: The very definition ofelectric flux (Section 2.4) indicates that D shouldbe equal to a surface charge density. But we canshow this rigorously, and in the process we cangeneralize this result to the more-general case inwhich neither of the two materials are PEC.

The desired more-general boundary conditionmay be obtained from the integral form of Gauss’Law (Section 5.5), as illustrated in Figure 5.10.Let the surface of integration S ′ take the form ofclosed cylinder centered at a point on theinterface, and for which the flat ends are parallelto the surface and perpendicular to n. Let theradius of this cylinder be a, and let the length ofthe cylinder be 2h. From Gauss’ Law we have∮

S′

D · ds =∫

top

D · ds

+

∫

side

D · ds

+

∫

bottom

D · ds = Qencl (5.152)

rqstuw y

rqstuw

n

a

|

|


Figure 5.10: Use of Gauss’ Law to determine theboundary condition on D.

where the “top” and “bottom” are in Regions 1and 2, respectively, and Qencl is the chargeenclosed by S ′. Now let us reduce h and atogether while (1) maintaining a constant ratioh/a≪ 1 and (2) keeping S ′ centered on S.Because h≪ a, the area of the side can be madenegligible relative to the area of the top andbottom. Then as h→ 0 we are left with

∫

top

D · ds+∫

bottom

D · ds → Qencl (5.153)

As the area of the top and bottom sides becomeinfinitesimal the variation in D over these areasbecomes negligible. Now we have simply:

D1 · n∆A+D2 · (−n)∆A→ Qencl (5.154)

where D1 and D2 are the electric flux densityvectors in medium 1 and medium 2, respectively,and ∆A is the area of the top and bottom sides.The above expression can be rewritten

Qencl∆A

→ n · (D1 −D2) (5.155)

Note that the left side of the equation mustrepresent a actual, physical surface charge; this isapparent from dimensional analysis and the factthat h is now infinitesimally small. Therefore:

n · (D1 −D2) = ρs (5.156)

where, as noted above, n points into region 1.Summarizing:



Any discontinuity in the normal componentof the electric flux density across the bound-ary between two material regions is equal tothe surface charge.

Now let us verify that this is consistent with ourpreliminary finding, in which we assumedRegion 2 was a PEC. In that case D2 = 0, so wesee that Equation 5.151 is satisfied, as expected.If neither Region 1 nor Region 2 is PEC andthere is no surface charge on the interface, thenwe find n · (D1 −D2) = 0; i.e.,

In the absence of surface charge, the normalcomponent of the electric flux density mustbe continuous across the boundary.

Finally, we note that since D = ǫE,Equation 5.156 implies the following boundarycondition on E:

n · (ǫ1E1 − ǫ2E2) = ρs (5.157)

where ǫ1 and ǫ2 are the permittivities inRegions 1 and 2, respectively. The aboveequation illustrates one reason why we sometimesprefer the “flux” interpretation of the electricfield to the “field intensity” interpretation of theelectric field: Comparing Equations 5.156 and5.157, we see that former results in a simplerboundary condition with no loss of information.

5.19 Charge and ElectricField for a PerfectlyConducting Region

[m0025]

In this section we consider the behavior of chargeand the electric field in the vicinity of a perfectelectrical conductor (PEC).

First, note that the electric field – both theelectric field intensity E and electric flux densityD – throughout a PEC region is zero. This isbecause the electrical potential throughout a


Figure 5.11: An infinite flat slab of PEC in thepresence of an applied electric field.

PEC region must be constant.3 Recall that theelectric field is proportional to the spatial rate ofchange of electrical potential (i.e., E = −∇V ;Section 5.14). Thus, the electric field must bezero throughout a PEC region.

Second, the electric field is oriented directly awayfrom (i.e., perpendicular to) the PEC surface,and the magnitude of D is equal to the surfacecharge density ρs (C/m2) (Section 5.18).

Now we address the question of chargedistribution; i.e., the location and density ofcharge. Consider the scenario shown inFigure 5.11. Here, a flat slab of PEC material isembedded in dielectric material.4 The thicknessof the slab is finite, whereas the length and widthof the slab is infinite. The region above the slabis defined as Region 1 and has permittivity ǫ1.The region below the slab is defined as Region 2and has permittivity ǫ2. Electric fields E1 andE2 are present in Regions 1 and 2, respectively,as shown in Figure 5.11. To begin, let us assumethat these fields are the result of some externalstimulus that results in the direction of thesefields being generally upward, as shown inFigure 5.11.

3This idea is explored in some depth in Section 6.3.4For the purposes of this section, it suffices to inter-

pret “dielectric” as a “nonconducting and well character-ized entirely in terms of its permittivity”. For more, seeSection 5.20.


5.19. CHARGE AND ELECTRIC FIELD FOR A PERFECTLY CONDUCTING REGION 115

Now: What do we know about E1 and E2? First,both fields must satisfy the relevant boundaryconditions. That is, the component of E1 whichis tangent to the upper PEC surface is zero, sothat E1 is directed entirely in a direction normalto the surface. Similarly, the component of E2

which is tangent to the lower PEC surface iszero, so that E2 is directed entirely in a directionnormal to the surface. At this point we have notdetermined the magnitudes or signs of E1 andE2; we have established only that there are nonon-zero components tangential to (i.e., parallelto) the PEC surfaces.

Next, recall that the electric field must be zerowithin the slab. Therefore charge is uniformlydistributed on the upper and lower surfaces withequal surface charge density. This can beexplained as follows. First, the charge on theupper surface must be positive, since E1 isdirected away from the slab; and the charge onthe lower surface must be negative, since E2 isdirected toward the slab. Next, recall that theelectric field associated with an infinite sheet ofcharge is uniform and directed either toward oraway from the charge depending on sign(Section 5.4). Therefore within the slab theelectric field associated with the positive chargedistribution on the top surface of the slab isequal and opposite the electric field associatedwith the negative charge distribution on thebottom surface of the slab. As a result, thesefields cancel and the total field is zero, as isrequired by the equipotential condition within aPEC. Any other distribution of charge will resultin a non-zero electric field within the slab, whichwould violate the equipotential property of thePEC. We come to the following conclusions thatseem to apply in general:

There can be no static charge within a PEC.

and it follows that

Charge associated with a PEC lies on the sur-face.

The same argument cannot be used to implythat the fields outside the slab are zero, however.

−

− − −−−−−−

+

++ + + +

+ +

−−−

−

+

+

+

++

−−−− − − −

++ + + +

+

+

+

by C. Burks (modified)

Figure 5.12: Electric field lines due to a pointcharge in the vicinity of PEC regions (shaded) ofvarious shapes.

This is because the boundary conditions on D1

and D2 in the dielectric regions require thesefields to be non-zero if the surface charge densityon the PEC is non-zero. In other words, thesurface charge supports the discontinuity in thenormal component of the electric fields. In fact,we can now determine those values: D1 and D2

have magnitudes + |ρs| and − |ρs| respectively,because the surface charge density on both sidesof the slab have equal magnitude. However theelectric field intensity E1 = D1/ǫ1, whereasE2 = D2/ǫ2; i.e., these are different. That is, theelectric field intensities are unequal unless thepermittivities in each dielectric region are equal.

Finally, let us consider the structure of theelectric field in more general cases. Figure 5.12shows field lines in a homogeneous dielectricmaterial in which a point charge and PECregions of various shapes are embedded. Notethat electric field lines now bend in the dielectricso as to satisfy the requirement that thetangential component of the electric field be zeroon PEC surfaces. Also note that the chargedistribution arranges itself on the PEC surfacesso as to maintain zero electric field and constantpotential within the cube.


5.20 Dielectric Media

[m0107]

Dielectric is particular category of materialswhich exhibit low conductivity5 because theirconstituent molecules remain intact whenexposed to an electric field, as opposed toshedding electrons as is the case in goodconductors. Subsequently dielectrics do noteffectively pass current, and are thereforeconsidered “‘good insulators” as well as “poorconductors”. An important application ofdielectrics in electrical engineering is as a spacermaterial in printed circuit boards (PCBs),coaxial cables, and capacitors.

Examples of dielectrics include air, glass, teflon,and fiberglass epoxy (the material used incommon “FR4” printed circuit boards). Theseand other dielectrics are listed along with valuesof their constitutive parameters in Section A.1.

The primary electromagnetic characterization ofdielectric materials is relative permittivity ǫr(Section 2.3), which ranges from very close to 1upward to roughly 50, and less than 6 or so formost common dry materials. The permeability ofdielectric materials is approximately equal to thefree-space value (i.e., µ ≈ µ0), and so thesematerials are said to be “non-magnetic”.

Additional Reading:

• “Dielectric” on Wikipedia.

5.21 Dielectric Breakdown

[m0109]

The permittivity of an ideal dielectric isindependent of the magnitude of an appliedelectric field; the material is said to be “linear”.6

However all practical dielectrics begin to fail in

5See Section 2.8 for a refresher on parameters describingproperties of materials.

6See Section 2.8 for a review of this concept.

this respect with sufficiently high electric field.Typically the failure is abrupt, and is observed asa sudden, dramatic increase in conductivity,signaling that electrons are being successfullydislodged from their host molecules. Thethreshold value of the electric field intensity atwhich this occurs is known as the dielectricstrength, and the sudden change in behaviorobserved in the presence of an electric fieldgreater than this threshold value is known asdielectric breakdown.

Dielectric strength varies from about 3 MV/mfor air to about 200 MV/m in mica (a dielectriccommonly used in capacitors).

Dielectric breakdown is typically accompanied by“arcing”, which is a sudden flow of currentbetween reservoirs of charge. A well knownexample of this phenomenon is lightning, whichoccurs when charge is exchanged between skyand ground when air (a dielectric) exhibitsbreakdown. Dielectric breakdown in solids istypically catastrophic to the material.

Additional Reading:

• “Electrical Breakdown” on Wikipedia.

5.22 Capacitance

[m0112]

When regions of positive and negative chargecoexist in a region of free space, Coulomb forces(Section 5.1) will attempt to decrease theseparation between the charges. As noted inSection 5.8, this can be interpreted as a tendencyof a system to reduce it’s potential energy. If thecharges are fixed in place, then the potentialenergy remains constant. This potential energy isproportional to the Coulomb force. Referringback to Section 5.1, the Coulomb force is:

• Proportional to quantity of positive chargesquared,

• Inversely proportional to the separation

https://en.wikipedia.org/wiki/Dielectric

https://en.wikipedia.org/wiki/Electrical_breakdown

5.22. CAPACITANCE 117

between the positive and negative charges,and

• Inversely proportional to the permittivity ofthe material separating the charges.

Therefore the potential energy of the system issimilarly dependent on charge, separation, andpermittivity. Furthermore, we see that theability of a system to store energy in this mannerdepends on the geometry of the structure andthe permittivity of the intervening material.

Now recall that the electric field intensity E isessentially defined in terms of the Coulomb force;i.e., F = qE (Section 2.2). So, rather thanthinking of the potential energy of the system asbeing associated with the Coulomb force, it isequally valid to think of the potential energy asbeing stored in the electric field associated withthe charge distribution. It follows from findingsof the previous paragraph that the energy storedin the electric field depends on the geometry andpermittivity of the structure under consideration.This relationship is what we mean bycapacitance. We may fairly summarize thisinsight as follows:

Capacitance is the ability of a structure tostore energy in an electric field.

The capacitance of a structure depends on itsgeometry and the permittivity of the mediumseparating regions of positive and negativecharge.

Note that capacitance does not depend oncharge, which we view as either a stimulus orresponse from this point of view. Thecorresponding response or stimulus, respectively,is the net potential associated with this charge.This leads to the following definition:

C ,Q+

V(5.158)

where Q+ (units of C) is the total positivecharge, V (units of V) is the potential associatedwith this charge (defined such that it is positive),and C (units of F) is the associated capacitance.So:

-

~~~ ~~~~

-- -- -

EV

ge

Q+

ge

Q+-

PEC

Figure 5.13: Electrostatic interpretation of ca-pacitance.

In practice, capacitance is defined as the ratioof charge present on one conductor of a two-conductor system, to the potential differencebetween the conductors (Equation 5.158).

In other words, a structure is said to have greatercapacitance if it stores more charge – andtherefore stores more energy – in response to agiven potential difference.

Figure 5.13 shows the relevant features of thisdefinition. Here a battery imposes the potentialdifference V between two regions ofperfectly-conducting material. Q+ is the totalcharge on the surface of the PEC region attachedto the positive terminal of the battery. An equalamount of negative charge appears on the surfaceof the PEC region attached to the negativeterminal of the battery (Section 5.19). Thischarge distribution gives rise to the electric fieldshown in Figure 5.13. Assuming the two PECregions are fixed in place, Q+ will increaselinearly with increasing V , at a rate determinedby the capacitance C of the structure.

A capacitor is a device which is designed toexhibit a specified capacitance. We can nowmake the connection to the concept of thecapacitor as it appears in elementary circuittheory. In circuit theory, the behavior of devicesis characterized in terms of terminal voltage VTin response to terminal current IT , and viceversa. First, note that current does not normally


flow through a capacitor,7 so when we refer to“terminal current” for a capacitor, what we reallymean is the flow of charge arriving or departingfrom one of the conductors via the circuit, whichis equal to the flow of charge departing orarriving (respectively) at the other conductor.This gives the appearance of current flow throughthe capacitor when the current is examined fromoutside the capacitor. With that settled, weproceed as follows: Using Equation 5.158 weexpress the voltage VT across the terminals of acapacitor having capacitance C:

VT =Q+

C(5.159)

We seek a relationship between VT and IT .Current is charge per unit time, so the charge oneither conductor is the integral of IT over time;i.e., amps integrated over time is charge. If wedefine IT as being positive in the direction of theflow of positive charge as is the usual convention,then we have:

VT (t) =1

C

∫ t

t0

IT (τ) dτ +1

CQ+(t0) (5.160)

where t0 is an arbitrary start time. Againapplying Equation 5.158 we see that the secondterm is simply VT (t0). This is the expectedrelationship from elementary circuit theory.

Additional Reading:

• “Capacitance” on Wikipedia.

• “Capacitor” on Wikipedia.

5.23 The Thin Parallel PlateCapacitor

[m0070]

Let us now determine the capacitance of acommon type of capacitor known as the thinparallel plate capacitor, shown in Figure 5.14.

7If it does, it’s probably experiencing dielectric break-

down; see Section 5.21.

d

z

ρ

Figure 5.14: Thin parallel plate capacitor.

This capacitor consists of two flat plates, eachhaving area A, separated by distance d. Tofacilitate discussion, let us place the origin of thecoordinate system at the center of the lowerplate, with the +z axis directed toward theupper plate such that the upper plate lies in thez = +d plane.

Below we shall find the capacitance by assuminga particular charge on one plate, using theboundary condition on the electric flux density Dto relate this charge density to the internalelectric field, and then integrating over theelectric field to obtain the voltage. Thencapacitance is the ratio of the assumed charge tothe resulting potential difference.

The principal difficulty in this approach is findingthe electric field. To appreciate the problem, firstconsider that if the area of the plates was infinite,then the electric field would be very simple: Itwould begin at the positively-charged plate andextend in a perpendicular direction toward thenegatively-charged plate (Section 5.19).Furthermore, the field would be constanteverywhere between the plates. This much isapparent from symmetry alone. However whenthe plate area is finite, then we expect a fringingfield to emerge. “Fringing field” is simply a termapplied to the non-uniform field that appears atthe edge of the plates. The field is non-uniformin this region because the boundary conditionson the outside (outward-facing) surfaces of theplates have a significant effect in this region. Inthe central region of the capacitor, however, thefield is not much different from the field thatexists in the case of infinite plate area.

In any parallel plate capacitor of finite size, some

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https://en.wikipedia.org/wiki/Capacitor

5.23. THE THIN PARALLEL PLATE CAPACITOR 119

fraction of the energy will be stored by theapproximately uniform field of the central region,and the rest will be stored in the fringing field.We can make the latter negligible relative to theformer by making the capacitor very “thin”, inthe sense that the smallest identifiable dimensionof the plate is much greater than d. Under thiscondition, we may obtain a good approximationof the capacitance by simply neglecting thefringing field, since an insignificant fraction ofthe energy is stored there.

Imposing the “thin” condition leads to threeadditional simplifications. First, the surfacecharge distribution may be assumed to beapproximately uniform over the plate, whichgreatly simplifies the analysis. Second, the shapeof the plates becomes irrelevant: They might becircular, square, triangular, etc.: Whencomputing capacitance in the “thin” case, onlythe plate area A is important. Third, thethickness of each of the plates becomes irrelevant.

We are now ready to determine the capacitanceof the thin parallel plate capacitor. Here are thesteps:

1. Assume a total positive charge Q+ on theupper plate.

2. Invoking the “thin” condition, we assumethe charge density on the plates is uniform.Thus the surface charge density on bottomside of the upper plate is ρs,+ = Q+/A(C/m2).

3. From the boundary condition on the bottomsurface of the upper plate, D on this surfaceis −zρs,+.

4. The total charge on the lower plate, Q−,must be equal and opposite the total chargeon the upper plate; i.e, Q− = −Q+.Similarly the surface charge density on theupper surface of the lower plate, ρs,−, mustbe −ρs,+

5. From the boundary condition on the topsurface of the lower plate (Section 5.18), Don this surface is +zρs,−. Since

+zρs,− = −zρs,+, D on the facing sides ofthe plates is equal.

6. Again invoking the “thin” condition, weassume D between the plates hasapproximately the same structure as wewould see if the plate area was infinite.Therefore we are justified in assumingD ≈ −zρs,+ everywhere between the plates.(You might also see that this is self-evidentfrom the definition of D as the flux densityof electric charge (Section 2.4).)

7. With an expression for the electric field inhand, we may now compute the potentialdifference V between the plates as follows(Section 5.8):

V = −∫

C

E · dl

= −∫ d

0

(1

ǫD

)· (zdz)

= −∫ d

0

(−z

ρs,+ǫ

)· (zdz)

= +ρs,+ d

ǫ(5.161)

8. Finally,

C =Q+

V=

ρs,+ A

ρs,+ d/ǫ=ǫA

d(5.162)

Summarizing:

C ≈ ǫA

d(5.163)

The capacitance of a parallel plate capacitorhaving plate separation much less than thesize of the plate is given by Equation 5.163.This is an approximation because the fringingfield is neglected.

It’s worth noting that this is dimensionallycorrect; i.e., F/m times m2 divided by m yieldsF. It’s also worth noting the effect of the variousparameters:


Capacitance increases in proportion to per-mittiviy and plate area, and decreases in pro-portion to distance between the plates.

Example 5.9. Printed circuit boardcapacitance.

Printed circuit boards commonly include a“ground plane”, which serves as the voltagedatum for the board, and at least one“power plane”, which is used to distribute aDC supply voltage (See “AdditionalReading” at the end of this section). Theseplanes are separated by a dielectric material,and so the area in common between theplanes exhibits capacitance. Thiscapacitance may be viewed as an equivalentdiscrete capacitor in parallel with the powersupply, which may be either negligible,significant and beneficial, or significant andharmful. So, it is useful to know the value ofthis equivalent capacitor.

For a common type of circuit board, thedielectric thickness is about 1.6 mm and therelative permittivity of the material is about4.5. If the area in common between theground and power planes is 25 cm2, what isthe value of the equivalent bypass capacitor?

Solution. From the problem statement,ǫ ∼= 4.5ǫ0, A ∼= 25 cm2 = 2.5 × 10−3 m2,and d ∼= 1.6 mm. Using Equation 5.163, thevalue of the equivalent capacitor is 62.3 pF.

Additional Reading:

• “Printed circuit board” on Wikipedia.

5.24 Capacitance of aCoaxial Structure

[m0113]

Let us now determine the capacitance ofcoaxially-arranged conductors, shown in

b

-

az

ρ

l

V


Figure 5.15: Determining the capacitance ofcoaxial line.

Figure 5.15. The capacitance of this structure isneeded to calculate the characteristic impedanceof coaxial transmission line, as addressed inSections 3.4 and 3.10.

For our present purposes we may model thestructure as consisting of two concentricperfectly-conducting cylinders of radii a and b,separated by an ideal dielectric havingpermittivity ǫs. We place the +z axis along thecommon axis of the concentric cylinders so thatthe cylinders may be described asconstant-coordinate surfaces ρ = a and ρ = b.

In this section we shall find the capacitance byassuming a total charge Q+ on the innerconductor and integrating over the associatedelectric field to obtain the voltage between theconductors. Then capacitance is computed as theratio of the assumed charge to the resultingpotential difference. This strategy is the similarto that employed in Section 5.23 for the parallelplate capacitor, so it may be useful to reviewthat section before attempting this derivation.

The first step is to find the electric field insidethe structure. This is relatively simple if weassume that the structure has infinite length(i.e., l → ∞), since then there are no fringingfields and the internal field will be utterlyconstant with respect to z. In the central regionof a finite-length capacitor, however, the field isnot much different from the field that exists in

https://en.wikipedia.org/wiki/Printed_circuit_board


5.24. CAPACITANCE OF A COAXIAL STRUCTURE 121

the case of infinite length, and if the energystorage in fringing fields is negligible compared tothe energy storage in this central region thenthere is no harm in assuming the internal field isconstant with z. Alternatively we may think ofthe length l as pertaining to one short section ofa much longer structure, and thereby obtain thecapacitance per length as opposed to the totalcapacitance. Note that the latter is exactly whatwe need for the transmission line lumped-elementequivalent circuit model (Section 3.4).

To determine the capacitance, we invoke thedefinition (Section 5.22):

C ,Q+

V(5.164)

where Q+ is the charge on the positively-chargedconductor and V is the potential measured fromthe negative conductor to the positive conductor.The charge on the inner conductor isuniformly-distributed with density

ρl =Q+

l(5.165)

which has units of C/m. Now we will determinethe electric field intensity E, integrate E over thepath between conductors to get V , and thenapply Equation 5.164 to obtain the capacitance.

The electric field intensity for this scenario wasdetermined in Section 5.6, “Electric Field Due toan Infinite Line Charge using Gauss’ Law”,where we found

E = ρρl

2πǫsρ(5.166)

The reader should note that in that section wewere considering merely a line of charge; not acoaxial structure. So on what basis do we claimthe field is the same? This is a consequence ofGauss’ Law (Section 5.5)

∮

S

D · ds = Qencl (5.167)

which we used in Section 5.6 to find the field. Ifin this new problem we specify the samecylindrical surface S with radius ρ < b, then theenclosed charge is the same. Furthermore the

presence of the outer conductor does not changethe radial symmetry of the problem, and nothingelse remains that can change the outcome. Thisis worth noting for future reference:

The electric field inside a coaxial structurecomprised of concentric conductors and hav-ing uniform charge density on the inner con-ductor is identical to the electric field of a linecharge in free space having the same chargedensity.

Next we get V using (Section 5.8)

V = −∫

C

E · dl (5.168)

where C is any path from the negatively-chargedouter conductor to the positively-charged innerconductor. Since this can be any such path(Section 5.9), we may as well choose the simplestone. This path is the one which traverses a radialof constant φ and z. Thus:

V = −∫ a

ρ=b

(ρ

ρl2πǫsρ

)· (ρdρ)

= − ρl2πǫs

∫ a

ρ=b

dρ

ρ

= +ρl

2πǫs

∫ b

ρ=a

dρ

ρ

= +ρl

2πǫsln

(b

a

)(5.169)

Wrapping up:

C ,Q+

V=

ρll

(ρl/2πǫs) ln (b/a)(5.170)

Note that factors of ρl in the numerator anddenominator cancel out, leaving:

C =2πǫsl

ln (b/a)(5.171)

Note that this expression is dimensionallycorrect, having units of F. Also note that theexpression depends only on materials (throughǫs) and geometry (through l, a, and b). The


expression does not depend on charge or voltage,which would imply non-linear behavior.

To make the connection back to lumped-elementtransmission line model parameters (Sections 3.4and 3.10), we simply divide by l to get theper-unit length parameter:

C ′ =2πǫs

ln (b/a)(5.172)

Example 5.10. Capacitance of RG-59coaxial cable.

RG-59 coaxial cable consists of an innerconductor having radius 0.292 mm, an outerconductor having radius 1.855 mm, and apolyethylene spacing material havingrelative permittivity 2.25. Estimate thecapacitance per length of RG-59.

Solution. From the problem statement,a = 0.292 mm, b = 1.855 mm, andǫs = 2.25ǫ0. Using Equation 5.172 we findC ′ = 67.7 pF/m.

5.25 Electrostatic Energy

[m0114]

Consider a structure consisting of two perfectconductors, both fixed in position and separatedby an ideal dielectric. This could be a capacitor,or it could be one of a variety of capacitivestructures which are not explicitly intended to bea capacitor; for example, a printed circuit board.When a potential difference is applied betweenthe two conducting regions, a positive charge Q+

will appear on the surface of the conductor at thehigher potential, and a negative chargeQ− = −Q+ will appear on the surface of theconductor at the lower potential (Section 5.19).Since the conductors are not free to move,potential energy is stored in the electric fieldassociated with the surface charges(Section 5.22).

We now ask the question: What is the energystored in this field? The answer to this questionhas relevance in several engineering applications.For example, when capacitors are used asbatteries, it is useful to know to amount ofenergy that can be stored. Also, any system thatincludes capacitors or has unintendedcapacitance is using some fraction of the energydelivered by the power supply to charge theassociated structures. In many electronic systems– and in digital systems in particular – capacitorsare periodically charged and subsequentlydischarged at a regular rate. Since power isenergy per unit time, this cyclic charging anddischarging of capacitors consumes power.Therefore energy storage in capacitorscontributes to the power consumption of modernelectronic systems – We’ll delve into that topic inmore detail in an upcoming example.

Since capacitance C relates the charge Q+ to thepotential difference V between the conductors,this is the natural place to start. From thedefinition of capacitance (Section 5.22):

V =Q+

C(5.173)

From Section 5.8, electric potential is defined asthe work done (i.e., energy injected) by moving acharged particle, per unit of charge; i.e.,

V =We

q(5.174)

where q is the charge borne by the particle andWe (units of J) is the work done by moving thisparticle across the potential difference V . Sincewe are dealing with charge distributions asopposed to charged particles, it is useful toexpress this in terms of the contribution ∆We

made to We by a small charge ∆q. Letting ∆qapproach zero we have

dWe = V dq (5.175)

Now consider what must happen to transitionthe system from having zero charge (q = 0) tothe fully-charged static condition (q = Q+). Thisrequires moving the differential amount of chargedq across the potential difference V = q/C,beginning with q = 0 and continuing until

5.25. ELECTROSTATIC ENERGY 123

q = Q+. Therefore the total amount of workdone in this process is:

We =

∫ Q+

q=0

dWe

=

∫ Q+

0

V dq

=

∫ Q+

0

q

Cdq

=1

2

Q2+

C(5.176)

Equation 5.176 can be expressed entirely interms of electrical potential by noting again thatC = Q+/V , so

We =1

2CV 2 (5.177)

Since there are no other processes to account forthe injected energy, the energy stored in theelectric field is equal to We. Summarizing:

The energy stored in the electric field of acapacitor (or a capacitive structure) is givenby Equation 5.177.

Example 5.11. Why multicore computingis power-neutral.

Readers are likely aware that computersincreasingly use multicore processors asopposed to single-core processors. For ourpresent purposes, a “core” is defined as thesmallest combination of circuitry thatperforms computation. A multicoreprocessor consists of multiple identical coreswhich run in parallel. Since a multicoreprocessor consists of N identical processors,you might expect power consumptionincrease by N relative to a single-coreprocessor. However this is not the case. Tosee why, first realize that the powerconsumption of a modern computing core isdominated by the energy required tocontinuously charge and discharge themultitude of capacitances within the core.From Equation 5.177, the required energy is

12C0V

20 per clock cycle, where C0 is the sum

capacitance (remember, capacitors inparallel add) and V0 is the supply voltage.Power is energy per unit time, so the powerconsumption for a single core is

P0 =1

2C0V

20 f0 (5.178)

where f0 is the clock frequency. In a N -coreprocessor, the sum capacitance is increasedby N . However the frequency is decreasedby N since the same amount of computationis being distributed among the N cores.Therefore the power consumed by an N -coreprocessor is

PN =1

2(NC0)V

20

(f0N

)= P0 (5.179)

In other words, the increase in powerassociated with replication of hardware isnominally offset by the decrease in powerenabled by reducing the clock rate. In yetother words, the total energy of the N -coreprocessor is N times the energy of the singlecore processor at any given time; howeverthe multicore processor needs to rechargecapacitances 1/N times as often.

Before moving on, it should be noted thatthe usual reason for pursuing a multicoredesign is to increase the amount ofcomputation that can be done; i.e., toincrease the product f0N . Nevertheless, it isextremely helpful that power consumption isproportional to f0 only, and is independentof N .

The thin parallel plate capacitor (Section 5.23) isrepresentative of a large number of practicalapplications, so it is instructive to consider theimplications of Equation 5.177 for this structurein particular. For the thin parallel platecapacitor,

C ≈ ǫA

d(5.180)

where A is the plate area, d is the separationbetween the plates, and ǫ is the permittivity ofthe material between the plates. This is an


approximation because the fringing field isneglected; we shall proceed as if this is an exactexpression. Applying Equation 5.177:

We =1

2

(ǫA

d

)(Ed)

2(5.181)

where E is the magnitude of the electric fieldintensity between the plates. Rearrangingfactors, we obtain:

We =1

2ǫE2 (Ad) (5.182)

Recall that the electric field intensity in the thinparallel plate capacitor is approximately uniform.Therefore the density of energy stored in thecapacitor is also approximately uniform. Notingthat the product Ad is the volume of thecapacitor, we find that this energy density is

we =1

2ǫE2 (5.183)

which has units of energy per unit volume(J/m3).

The above expression provides an alternativemethod to compute the total electrostatic energy.Within a mathematical volume V, the totalelectrostatic energy is simply the integral of theenergy density over V; i.e.,

We =

∫

V

we dv (5.184)

This works even if E and ǫ varies with position,so even though we arrived at this result using theexample of the thin parallel-plate capacitor, ourfindings at this point apply generally.Substituting Equation 5.183 we obtain:

We =1

2

∫

V

ǫE2dv (5.185)

Summarizing:

The energy stored by the electric field presentwithin a volume is given by Equation 5.185.

It’s worth noting that this energy increases withthe permittivity of the medium, which makessense since capacitance is proportional topermittivity.

[m0034]

5.25. ELECTROSTATIC ENERGY 125

Image Credits

Fig. 5.1: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0102 fCoulombsLaw.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 5.2: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0104 fRing.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 5.3: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0104 fDisk.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 5.4: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:4.5 m0149 commons.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 5.5: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0064 fResistor.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 5.7: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0020 fPEC.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 5.8: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0020 fKVL.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 5.9: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0021 fPEC.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 5.10: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0021 fGL.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 5.11: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0025 fFlatSlab.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 5.12: by C. Burks, https://commons.wikimedia.org/wiki/File:Electrostatic induction.svg,in public domain. Minor modifications.

Fig. 5.15: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0113 fCapCoax.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

https://commons.wikimedia.org/wiki/File:M0102_fCoulombsLaw.svg


https://commons.wikimedia.org/wiki/File:M0104_fRing.svg


https://commons.wikimedia.org/wiki/File:M0104_fDisk.svg


https://commons.wikimedia.org/wiki/File:4.5_m0149_commons.svg


https://commons.wikimedia.org/wiki/File:M0064_fResistor.svg


https://commons.wikimedia.org/wiki/File:M0020_fPEC.svg


https://commons.wikimedia.org/wiki/File:M0020_fKVL.svg


https://commons.wikimedia.org/wiki/File:M0021_fPEC.svg


https://commons.wikimedia.org/wiki/File:M0021_fGL.svg


https://commons.wikimedia.org/wiki/File:M0025_fFlatSlab.svg


https://commons.wikimedia.org/wiki/File:Electrostatic_induction.svg

https://commons.wikimedia.org/wiki/File:M0113_fCapCoax.svg


Chapter 6

Steady Current and Conductivity

6.1 Convection andConduction Currents

[m0110]

In practice, we deal with two physicalmechanisms for current: convection andconduction. The distinction between these typesof current is important in electromagneticanalysis.

Convection current consists of chargedparticles traversing space in response tomechanical forces, as opposed to being guided bythe Coulomb force (Sections 2.2 and/or 5.1)exerted on particles by an electric field. Anexample of a convection current is a cloudbearing free electrons which moves through theatmosphere driven by wind.

Conduction current consists of chargedparticles moving in response to Coulomb forces,and not merely being carried by motion of thesurrounding material. Specifically, the electricfield dislodges weakly-bound electrons fromatoms, which then subsequently travel somedistance before reassociating with other atoms.Thus, the individual electrons in a conductioncurrent do not necessarily travel the full distanceover which the current is perceived to exist.

The distinction between convection andconduction is important because Ohm’s Law(Section 6.3) – which specifies the relationshipbetween electric field intensity and current –applies only to conduction current.

Additional Reading:

• “Electric current” on Wikipedia.

6.2 Current Distributions

[m0101]

In elementary electric circuit theory, current isthe rate at which electric charge passes aparticular point in a circuit: For example, 1 A is1 C per second. In this view current is a scalarquantity, and there are only two possibledirections because charge is constrained to flowalong defined channels. Direction is identified bythe sign of the current with respect to a referencedirection, which is in turn defined with respect toa reference voltage polarity. The convention inelectrical engineering defines positive current asthe flow of positive charge into the positivevoltage terminal of a passive device such as aresistor, capacitor, or inductor. For an activedevice such as a battery, positive currentcorresponds to the flow of positive charge out ofthe positive voltage terminal.

However this kind of thinking only holds upwhen the systems being considered arewell-described as “lumped” devices connected byinfinitesimally thin, filament-like connectionsrepresenting wires and circuit board traces.Many important problems in electricalengineering concern situations in which currentfails to do this: Examples include wire- and


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6.2. CURRENT DISTRIBUTIONS 127

pin-type interconnects at radio frequencies,circuit board and enclosure grounding, andphysical phenomena such as lightning. Toaccommodate this more general class of problemswe must define current as a vector quantity.Furthermore current in these problems spreadsout over surfaces and within volumes, so we mustalso consider spatial distributions of current.

Line Current. As noted above, if a current I isconstrained to follow a particular path, then theonly other consideration is direction. Thus, a linecurrent is specified mathematically as lI, wherethe direction l may vary with position along thepath. For example, in a straight wire l isconstant, whereas in a coil l varies with positionalong the coil.

Surface Current Distribution. In some cases,current may be distributed over a surface. Forexample, the radio-frequency current on a wire ofradius a made from a metal with sufficiently highconductivity can be modeled as a uniform surfacecurrent existing on the wire surface. In this casethe current is best described as a surface currentdensity Js, which is the total current I on thewire divided by the circumference 2πa of thewire:

Js = uI

2πa(units of A/m) (6.1)

where u is the direction of current flow.

Volume Current Distribution. Imagine thatcurrent is distributed within a volume. Let u∆ibe the current passing through a small openplanar surface defined within this volume, and let∆s be the area of this surface. The volumecurrent density J at any point in the volume isdefined as

J , lim∆s→0

u∆i

∆s= u

di

ds(units of A/m2) (6.2)

In general J is a function of position within thisvolume. Subsequently the total current passingthrough a surface S is

I =

∫

S

J · ds (units of A) (6.3)

In other words, volume current densityintegrated over a surface yields total current

through that surface. You might recognize thisas a calculation of flux, and it is.1

Example 6.1. Current and current densityin a wire of circular cross-section.

Figure 6.1 shows a straight wire havingcross-sectional radius a = 3 mm. A batteryis connected across the two ends of the wireresulting in a volume current densityJ = z8 A/m2 which is uniform throughoutthe wire. Find the net current I through thewire.

Solution. The net current is

I =

∫

S

J · ds

We choose S to be the cross-sectionperpendicular to the axis of the wire. Also,we choose ds to point such that I is positivewith respect to the sign convention shown inFigure 6.1 , which is the usual choice inelectric circuit analysis. With these choices,we have

I =

∫ a

ρ=0

∫ 2π

φ=0

(z8 A/m

2)· (zρ dρ dφ)

=(8 A/m

2)∫ a

ρ=0

∫ 2π

φ=0

ρ dρ dφ

=(8 A/m

2)·(πa2

)

= 226 µA

This answer is independent of thecross-sectional surface S used to do thecalculation. For example, you could use analternative surface which is tilted 45 awayfrom the axis of the wire: In that case thecross-sectional area would increase, but thedot product of J and ds would beproportionally less, and the outcome wouldbe the same. Since the choice of S isarbitrary – any surface with edges at theperimeter of the wire will do – you should

1However it is not common to refer to net current as aflux; go figure!

128 CHAPTER 6. STEADY CURRENT AND CONDUCTIVITY

J

I

a

z

-

ds


Figure 6.1: Net current I and current density Jin a wire of circular cross-section.

make the choice which makes the problem assimple as possible, as we have done above.

Additional Reading:

• “Electric current” on Wikipedia.

• “Current density” on Wikipedia.

6.3 Conductivity

[m0010]

Conductivity is one of the three primary“constitutive parameters” that is commonly usedto characterize the electromagnetic properties ofmaterials (Section 2.8). The key idea is this:

Conductivity is a property of materials thatdetermines conduction current density in re-sponse to an applied electric field.

Recall that conduction current is the flow ofcharge in response to an electric field, via theCoulomb force (Section 6.1). Although theCoulomb force is straightforward to calculate(Section 5.1), the result is merely the forceapplied, not the speed at which charge moves inresponse. The latter is determined by themobility of charge, which is in turn determined

by the atomic and molecular structure of thematerial. Conductivity relates current density tothe applied field directly, without requiring oneto grapple separately with the issues of appliedforce and charge mobility.

In the absence of material – that is, in a true,perfect vacuum – conductivity is zero becausethere is no charge available to form current, andtherefore the current is zero no matter whatelectric field is applied. At the other extreme, agood conductor is a material that contains asupply of charge that is able to move freelywithin the material. When an electric field isapplied to a good conductor, charge-bearingmaterial constituents eagerly move in thedirection determined by the electric field,creating current flow in that direction. Thisrelationship is summarized by Ohm’s Law forElectromagnetics :

J = σE (6.4)

where E is electric field intensity (V/m); J is thevolume current density, a vector describing thecurrent flow, having units of A/m2 (seeSection 6.2); and σ is conductivity. Since E hasunits of V/m, we see σ has units of Ω−1m−1,which is more commonly expressed as S/m,where 1 S (“siemens”) is defined as 1 Ω−1.Section A.3 provides values of conductivity for arepresentative set of materials.

Conductivity σ is expressed in units of S/m,where 1 S = 1 Ω−1.

It is important to note that the current beingaddressed here is conduction current, and notconvection current, displacement current, orsome other form of current – see Section 6.2 forelaboration. Summarizing:

Ohm’s Law for Electromagnetics (Equa-tion 6.4) states that volume density of con-duction current (A/m2) equals conductivity(S/m) times electric field intensity (V/m).

The reader is likely aware that there is also an“Ohm’s Law” in electric circuit theory, which


https://en.wikipedia.org/wiki/Electric_current

https://en.wikipedia.org/wiki/Current_density

6.3. CONDUCTIVITY 129

states that current I (units of A) is voltage V(units of V) divided by resistance R (units of Ω);i.e., I = V/R. This is in fact a special case ofEquation 6.4; see Section 6.4 for more about this.

As mentioned above, σ depends on both theavailability and mobility of charge within thematerial. At the two extremes we have perfectinsulators, for which σ = 0; and perfectconductors, for which σ → ∞. Some materialsapproach these extremes, whereas others fallmidway between these conditions. Here are a fewclasses of materials that are frequentlyencountered:

• A perfect vacuum – “free space” – containsno charge and therefore is a perfect insulatorwith σ = 0.

• Good insulators typically have conductivities≪ 10−10 S/m, which is sufficiently low thatthe resulting currents can usually be ignored.The most important example is air, whichhas a conductivity only slightly greater thanthat of free space. An important class ofgood insulators is lossless dielectrics,2 whichare well-characterized in terms ofpermittivity (ǫ) alone, and for which µ = µ0

and σ = 0 may usually be assumed.

• Poor insulators have conductivities whichare low, but nevertheless sufficiently highthat the resulting currents cannot beignored. For example, the dielectric materialwhich is used to separate the conductors in atransmission line must be considered a poorinsulator as opposed to a good (effectivelylossless) insulator in order to characterizeloss per length along the transmission line,which can be significant.3 These lossydielectrics are well-characterized in terms ofǫ and σ, and typically µ = µ0 can beassumed.

• Semiconductors such as those materials usedin integrated circuits have intermediateconductivities; typically in the range 10−4 to101 S/m.

2See Section 5.20 for a discussion of dielectric materials.3Review Sections 3.4, 3.9, and associated sections for a

refresher on this issue.

• Good conductors are materials with veryhigh conductivities; typically greater than105 S/m. An important category of goodconductors includes metals, with certainmetals including alloys of aluminum, copper,and gold reaching conductivities on theorder of 108 S/m. In such materialsminuscule electric fields give rise to largecurrents. There is no significant storage ofenergy in such materials, and so the conceptof permittivity is not relevant for goodconductors.

The reader should take care to note that termssuch as good conductor and poor insulator arequalitative and subject to context. What may beconsidered a “good insulator” in one applicationmay be considered to be a “poor insulator” inanother.

One relevant category of material was notincluded in the above list: perfect conductors. Aperfect conductor is a material in which σ → ∞.It is tempting to interpret this as meaning thatany electric field gives rise to infinite currentdensity; however this is not plausible even in theideal limit. Instead, this condition is interpretedas meaning that E = J/σ → 0 throughout thematerial; i.e., E is zero independently of anycurrent flow in the material. An importantconsequence is that the potential field V is equalto a constant value throughout the material.(Recall E = −∇V (Section 5.14), so constant Vmeans E = 0.) We refer to the volume occupiedby such a material as an equipotential volume.This concept is useful as an approximation of thebehavior of good conductors; for example, metalsare often modeled as perfectly-conductingequipotential volumes in order to simplifyanalysis.

A perfect conductor is a material for whichσ → ∞, E → 0, and subsequently V (theelectric potential) is constant.

One final note: It is important to remain awareof the assumptions we have made aboutmaterials in this book, which are summarized inSection 2.8. In particular, we continue to assume


that materials are linear unless otherwiseindicated. For example: Whereas air is normallyconsidered to be an insulator and therefore apoor conductor, anyone who has ever witnessedlightning has seen a demonstration that underthe right conditions – i.e., sufficiently largepotential difference between earth and sky – aircan be come a very good conductor. Thisparticular situation is known as dielectricbreakdown (see Section 5.21). The non-linearityof materials can become evident before reachingthe point of dielectric breakdown, so one must becareful to consider this possibility when dealingwith strong electric fields.

Additional Reading:

• “Electrical Resistivity and Conductivity” onWikipedia.

• “Ohm’s Law” on Wikipedia.

6.4 Resistance

[m0071]

The concept of resistance is most likely familiarto readers via Ohm’s Law for Devices ; i.e.,V = IR where V is the potential differenceassociated with a current I. This is correct, butit is not the whole story. Let’s begin with astatement of intent:

Resistance R (Ω) is a characterization of theconductivity of a device (as opposed to a ma-terial) in terms of Ohms Law for Devices; i.e.,V = IR.

Resistance is a property of devices such asresistors, which are intended to provideresistance; as well as being a property of mostpractical electronic devices, whether it is desiredor not.

Resistance is a manifestation of the conductivityof the materials comprising the device, whichsubsequently leads to the “V = IR” relationship.This brings us to a very important point and a

common source of confusion: Resistance is notnecessarily the real part of impedance. Let’s takea moment to elaborate. Impedance (Z) is definedas the ratio of voltage to current; i.e., V/I; or

equivalently in the phasor domain as V /I.Essentially any device – not just devicesexhibiting resistance – can be characterized interms of this ratio. Consider for example theinput impedance of a terminated transmissionline (Section 3.15). This impedance may have anon-zero real-valued component even when thetransmission line and the terminating load arecomprised of perfect conductors. Summarizing:

Resistance results in a real-valued impedance;but not all devices exhibiting a real-valuedimpedance exhibit resistance, and the realcomponent of a complex-valued impedancedoes not necessarily represent resistance.

Restating the main point in yet other words:Resistance pertains to limited conductivity, notsimply to voltage-current ratio.

Also important to realize is that whereasconductivity σ (units of S/m) is a property ofmaterials, resistance depends on bothconductivity and the geometry of the device. Inthis section we address the question of how theresistance of a device can be is determined. Thefollowing example serves this purpose.

Figure 6.2 shows a straight wire of length lcentered on the z axis, forming a cylinder of amaterial having conductivity σ andcross-sectional radius a. The ends of the cylinderare covered by perfectly-conducting plates towhich the terminals are attached. A battery isconnected to the terminals, resulting in a uniforminternal electric field E between the plates.

To calculate R for this device, let us firstcalculate V , then I, and finally R = V/I. Herewe go: First, we compute the potential differenceusing the following result from Section 5.8:

V = −∫

C

E · dl (6.5)

Here we can view V as a “given” (being thevoltage of the battery), but wish to evaluate the

https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity

https://en.wikipedia.org/wiki/Ohm's_law

6.4. RESISTANCE 131

E

I

a

z

-

z z=

σ


Figure 6.2: Analysis of the resistance of straightwire of circular cross-section.

right hand side so as to learn something aboutthe effect of the conductivity and geometry ofthe wire. The appropriate choice of C begins atz = 0 and ends at z = l, following the axis of thecylinder. (Remember: C defines the direction ofincreasing potential, and the resulting potentialdifference will be the node voltage at the endpoint minus the node voltage at the start point.)Thus, dl = zdz and we have

V = −∫ l

z=0

E · (zdz)

We do not yet know E; however we know it isconstant throughout the device and points in the−z direction since this is the direction of currentflow and Ohm’s Law for Electromagnetics(Section 6.3) requires the electric field to point inthe same direction. Thus we may writeE = −zEz where Ez is a constant. We now find:

V = −∫ l

z=0

(−zEz) · (zdz)

= +Ez

∫ l

z=0

dz

= +Ezl

The current I is given by (Section 6.2)

I =

∫

S

J · ds

We choose the surface S to be the cross-sectionperpendicular to the axis of the wire. Also, wechoose ds to point such that I is positive withrespect to the sign convention shown inFigure 6.2. With these choices, we have

I =

∫ a

ρ=0

∫ 2π

φ=0

J · (−z ρ dρ dφ)

From Ohm’s Law for Electromagnetics, we have

J = σE = −zσEz (6.6)

So now

I =

∫ a

ρ=0

∫ 2π

φ=0

(−zσEz) · (−zρ dρ dφ)

= σEz

∫ a

ρ=0

∫ 2π

φ=0

ρ dρ dφ

= σEz(πa2

)

Finally:

R =V

I=

Ezl

σEz (πa2)=

l

σ (πa2)

This is a good-enough answer for the problemposed, but it is easily generalized a bit further.Noting that πa2 in the denominator is thecross-sectional area A of the wire, we may write

R =l

σA(6.7)

i.e., the resistance of a wire having cross-sectionalarea A – regardless of the shape of thecross-section, is given by the above equation.

The resistance of a right cylinder of mate-rial, given by Equation 6.7, is proportionalto length and inversely proportional to cross-sectional area and conductivity.

It is important to remember that Equation 6.7presumes that the volume current density J isuniform over the cross-section of the wire. This isan excellent approximation for thin wires at“low” frequencies including, of course, DC. Athigher frequencies it may not be a goodassumption that J is uniformly-distributed over



the cross-section of the wire, and at sufficientlyhigh frequencies one finds instead that thecurrent is effectively limited to the exteriorsurface of the wire. In the “high frequency” case,A in Equation 6.7 is reduced from the physicalarea to a smaller value corresponding to thereduced area through which almost all of thecurrent flows. Therefore R increases withincreasing frequency.

In order to quantify the high frequency behaviorof R (including determination of whatconstitutes “high frequency” in this context)requires concepts beyond the theory ofelectrostatics, and so is addressed elsewhere.

Example 6.2. Resistance of 22 AWGhookup wire.

A common type of wire found in DCapplications is 22AWG (“American WireGauge”; see “Additional Resources” at theend of this section) copper solid-conductorhookup wire. This type of wire has circularcross-section with diameter 0.644 mm.What is the resistance of 3 m of this wire?Assume copper conductivity of 58 MS/m.

Solution. From the problem statement, thediameter 2a = 0.644 mm, σ = 58× 106 S/m,and l = 3 m. The cross-sectional area isA = πa2 ∼= 3.26× 10−7 m2. UsingEquation 6.7 we obtain R = 159 mΩ.

Example 6.3. Resistance of steel pipe.

A pipe is 3 m long and has inner and outerradii of 5 mm and 7 mm respectively. It ismade from steel having conductivity4 MS/m. What is the DC resistance of thispipe?

Solution. We can use Equation 6.7 if wecan determine the cross-sectional area Athrough which the current flows. This areais simply the area defined by the outerradius, πb2, minus the area defined by theinner radius πa2. Thus

A = πb2 − πa2 ∼= 7.54× 10−5 m2. From theproblem statement we also determine thatσ = 4× 106 S/m and l = 3 m. UsingEquation 6.7 we obtain R ∼= 9.95 mΩ.

Additional Reading:

• “Resistor” on Wikipedia.

• “Ohm’s Law” on Wikipedia.

• “American wire gauge” on Wikipedia.

6.5 Conductance

[m0105]

Conductance, like resistance (Section 6.4), is aproperty of devices. Specifically:

Conductance G (Ω−1 or S) is the reciprocalof resistance R.

Therefore conductance depends on both theconductivity of the materials used in the deviceas well as the geometry of the device.

A natural question to ask is: Why do we requirethe concept of conductance, if it simply thereciprocal of resistance? The short answer is thatthe concept of conductance is not required; or,rather, we need only resistance or conductance,and not both. The concept appears inengineering analysis for two reasons:

• Conductance is sometimes considered to be amore intuitive description of the underlyingphysics in cases where the applied voltage isconsidered to be the independent “stimulus”and current is considered to be the response.This is why conductance appears in thelumped element model for transmission lines(Section 3.4), for example.

• Characterization in terms of conductancemay be preferred when considering thebehavior of devices in parallel, since the

https://en.wikipedia.org/wiki/Resistor

https://en.wikipedia.org/wiki/Ohm's_law

https://en.wikipedia.org/wiki/American_wire_gauge

6.5. CONDUCTANCE 133

conductance of a parallel combination issimply the sum of the conductances of thedevices.

Example 6.4. Conductance of a coaxialstructure.

Let us now determine the conductance of astructure consisting of coaxially-arrangedconductors separated by a lossy dielectric, asshown in Figure 6.3. The conductance ofthis structure is of interest in determiningthe characteristic impedance of coaxialtransmission line, as addressed inSections 3.4 and 3.10.

For our present purposes we may model thestructure as two concentricperfectly-conducting cylinders of radii a andb, separated by a lossy dielectric havingconductivity σs. We place the +z axis alongthe common axis of the concentric cylindersso that the cylinders may be described asconstant-coordinate surfaces ρ = a andρ = b.

There are at least 2 ways to solve thisproblem. One method is to follow theprocedure that was used to find thecapacitance of this structure in Section 5.24.Adapting that approach to the presentproblem, one would assume a potentialdifference V between the conductors, fromthat determine the resulting electric fieldintensity E, and then using Ohm’s Law forElectromagnetics (Section 6.3) determinethe density J = σsE of the current thatleaks directly between conductors. From thisone is able to determine the total leakagecurrent I, and subsequently the conductanceG , I/V . Although highly recommended asan exercise for the student, in this sectionwe take an alternative approach so as todemonstrate that there are a variety ofapproaches available for such problems.

The method we shall use below is as follows:(1) Assume a leakage current I between the

conductors; (2) Determine the associatedcurrent density J, which is possible usingonly geometrical considerations; (3)Determine the associated electric fieldintensity E using J/σs; (4) Integrate E overa path between the conductors to get V .Then as before conductance G , I/V .

The current I is defined as shown inFigure 6.3, with reference directionaccording to the engineering convention thatpositive current flows out of the positiveterminal of a source. (It’s worth noting thatthe opposite direction may be used for thereference direction as well, with theappropriate changes in sign throughout therest of the work.) The associated currentdensity must flow in the same direction, andthe circular symmetry of the problemtherefore constrains J to have the form

J = ρI

A(6.8)

where A is the area through which I flows.In other words, current flows radiallyoutward from the inner conductor to theouter conductor, with density thatdiminishes inversely with the area throughwhich the total current flows. (It may behelpful to view J as a flux density and I as aflux, as noted in Section 6.2.) This area issimply circumference 2πρ times length l, so

J = ρI

2πρl(6.9)

which exhibits the correct units of A/m2.

Now from Ohm’s Law for Electromagneticswe find the electric field within the structureis

E =J

σs= ρ

I

2πρlσs(6.10)

Next we get V using (Section 5.8)

V = −∫

C

E · dl (6.11)


where C is any path from thenegatively-charged outer conductor to thepositively-charged inner conductor. Sincethis can be any such path (Section 5.9), weshould choose the simplest one. Thesimplest path is the one which traverses aradial of constant φ and z. Thus:

V = −∫ a

ρ=b

(ρ

I

2πρlσs

)· (ρdρ)

= − I

2πlσs

∫ a

ρ=b

dρ

ρ

= +I

2πlσs

∫ b

ρ=a

dρ

ρ

= +I

2πlσsln

(b

a

)(6.12)

Wrapping up:

G ,I

V=

I

(I/2πlσs) ln (b/a)(6.13)

Note that factors of I in the numerator anddenominator cancel out, leaving:

G =2πlσsln (b/a)

(6.14)

Note that Equation 6.14 is dimensionallycorrect, having units of S = Ω−1. Also notethat this is expression depends only onmaterials (through σs) and geometry(through l, a, and b). Notably it does notdepend on current or voltage, which wouldimply non-linear behavior.

To make the connection back tolumped-element transmission line modelparameters (Sections 3.4 and 3.10), wesimply divide by l to get the per-lengthparameter:

G′ =2πσs

ln (b/a)(6.15)

b

a

z

ρ

l

I

s

J

Figure 6.3: Determining the conductance of astructure consisting of coaxially-arranged conduc-tors separated by a lossy dielectric.

Example 6.5. Conductance of RG-59coaxial cable.

RG-59 coaxial cable consists of an innerconductor having radius 0.292 mm, an outerconductor having radius 1.855 mm, and apolyethylene spacing material exhibitingconductivity of about 5.9× 10−5 S/m.Estimate the conductance per length ofRG-59.

Solution. From the problem statement,a = 0.292 mm, b = 1.855 mm, andσs ∼= 5.9× 10−5 S/m. Using Equation 6.15we find G′ ∼= 200 µS/m.

6.6 Power Dissipation inConducting Media

[m0106]

The displacement of charge in response to theforce exerted by an electric field constitutes areduction in the potential energy of the system(Section 5.8). If the charge is part of a steadycurrent, there must be an associated loss ofenergy that occurs at a steady rate since thecurrent itself is steady. Since power is energy perunit time, the loss of energy associated withcurrent is expressible as power dissipation. Inthis section we address two questions: (1) How

6.6. POWER DISSIPATION IN CONDUCTING MEDIA 135

much power is dissipated in this manner, and (2)What happens to the lost energy?

First, recall that work is force times distancetraversed in response to that force (Section 5.8).Stated mathematically:

∆W = +F ·∆l (6.16)

where the vector F is the force (units of N)exerted by the electric field, the vector ∆l is thedirection and distance (units of m) traversed,and ∆W is the work done (units of J) as a result.Note that a “+” has been explicitly indicated;this is to emphasize the distinction from thework that being considered in Section 5.8. Inthat section, the work “∆W” represented energyfrom an external source that was being used toincrease the potential energy of the system bymoving charge “upstream” relative to the electricfield. Now, ∆W represents this internal energyas it is escaping from the system in the form ofkinetic energy; therefore positive ∆W now meansa reduction in potential energy; hence the signchange.4

The associated power ∆P (units of W) is ∆Wdivided by the time ∆t (units of s) required forthe distance ∆l to be traversed:

∆P =∆W

∆t= F · ∆l

∆t(6.17)

Now we’d like to express force in terms of theelectric field exerting this force. Recall that theforce exerted by an electric field intensity E(units of V/m) on a particle bearing charge q(units of C) is qE (Section 2.2). However we’dlike to express this force in terms of a current, asopposed to a charge. An expression in terms ofcurrent can be constructed as follows. First notethat the total charge in a small volume “cell” isthe volume charge density ρv (units of C/m3)times the volume ∆v of the cell; i.e., q = ρv∆v(Section 5.3). Therefore:

F = qE = ρv ∆v E (6.18)

4It could be argued that it is bad form to use the samevariable to represent both tallies; nevertheless it is com-mon practice and so we simply remind the reader that it isimportant to be aware of the definitions of variables eachtime they are (re)introduced.

and subsequently

∆P = ρv ∆v E · ∆l

∆t= E ·

(ρv

∆l

∆t

)∆v (6.19)

The quantity in parentheses has units of C/m3 ·m · s−1, which is A/m2. Apparently this quantityis the volume current density J, so we have

∆P = E · J ∆v (6.20)

In the limit as ∆v → 0 we have

dP = E · J dv (6.21)

and integrating over the volume V of interest weobtain

P =

∫

V

dP =

∫

V

E · J dv (6.22)

The above expression is commonly known asJoule’s Law. In our situation it is convenient touse Ohm’s Law for Electromagnetics (J = σE;Section 6.3) to get everything in terms ofmaterials properties (σ), geometry (V), and theelectric field:

P =

∫

V

E · (σE) dv (6.23)

which is simply

P =

∫

V

σ |E|2 dv (6.24)

Thus:

The power dissipation associated with cur-rent is given by Equation 6.24. This poweris proportional to conductivity and propor-tional to electric field intensity squared.

This result facilitates the analysis of powerdissipation in materials exhibiting loss; i.e.,having finite conductivity. But what is the powerdissipation in a perfectly conducting material?For such a material, σ → ∞ and E → 0 nomatter how much current is applied(Section 6.3). In this case, Equation 6.24 is notvery helpful. However, as we just noted, being aperfect conductor means E → 0 no matter howmuch current is applied, so the power dissipatedin a perfect conductor must be zero.


When conductivity is finite, Equation 6.24 servesas a more-general version of a concept fromelementary circuit theory, as we shall nowdemonstrate. Let E = zEz, so |E|2 = E2

z . ThenEquation 6.24 becomes:

P =

∫

V

σE2z dv = σE2

z

∫

V

dv (6.25)

The second integral in Equation 6.25 is clearly acalculation of volume. Let’s assume the volume isa cylinder aligned along the z axis. The volumeof this cylinder is the cross-sectional area A timesthe length l. Then the above equation becomes:

P = σE2z A l (6.26)

For reasons that will become apparent veryshortly, let’s reorganize the above expression asfollows:

P = (σEzA) (Ezl) (6.27)

Note that σEz is the current density in A/m2,which when multiplied by A gives the totalcurrent. Therefore the quantity in the first set ofparentheses is simply the current I. Also notethat Ezl is the potential difference over thelength l, which is simply the node-to-nodevoltage V (Section 5.8). Therefore we havefound:

P = IV (6.28)

as expected from elementary circuit theory.

Now: What happens to the energy associatedwith this dissipation of power? The displacementof charge carriers in the material is limited bythe conductivity, which itself is finite because,simply put, other constituents of the material getin the way. If charge is being displaced asdescribed in this section, then energy is beingused to displace the charge-bearing particles.The motion of constituent particles is observedas heat – in fact, this is essentially the definitionof heat. Therefore:

The power dissipation associated with theflow of current in any material which is not aperfect conductor manifests as heat.

This phenomenon is known as joule heating,ohmic heating, and by other names. This

conversion of electrical energy to heat is themethod of operation for toasters, electric spaceheaters, and many other devices that generateheat. It is of course also the reason thatelectronic devices become hot.

Additional Reading:

• “Joule Heating” on Wikipedia.

[m0057]

https://en.wikipedia.org/wiki/Joule_heating

6.6. POWER DISSIPATION IN CONDUCTING MEDIA 137

Image Credits

Fig. 6.1: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0101 fMatEx1.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 6.2: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0071 fMatEx1.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

https://commons.wikimedia.org/wiki/File:M0101_fMatEx1.svg


https://commons.wikimedia.org/wiki/File:M0071_fMatEx1.svg


Chapter 7

Magnetostatics

[m0117]

Magnetostatics is the theory of the magnetic fieldin conditions in which it’s behavior isindependent of electric fields, including

• The magnetic field associated with variousspatial distributions of steady current;

• The energy associated with the magneticfield; and

• Inductance, which is the ability of astructure to store energy in a magnetic field.

The word ending “-statics” refers to the fact thatthese aspects of electromagnetic theory can bedeveloped by assuming that the sources of themagnetic field are time-invariant; we might saythat magnetostatics is the study of the magneticfield at DC. However many aspects ofmagnetostatics are applicable at “AC” as well.

7.1 Comparison ofElectrostatics andMagnetostatics

[m0115]

Students encountering magnetostatics for thefirst time have usually been exposed toelectrostatics already. Electrostatics andmagnetostatics exhibit many similarities. These

are summarized in Table 7.1. The elements ofmagnetostatics presented in this table are allformally introduced in other sections; the solepurpose of this table is to point out thesimilarities.

The technical term for these similarities isduality. Duality also exists between voltage andcurrent in electrical circuit theory. For moreabout the concept of duality, see “AdditionalReading” at the end of this section.

Additional Reading:

• Duality (electricity and magnetism) onWikipedia.

• Duality (electrical circuits) on Wikipedia.

7.2 Gauss’ Law for MagneticFields: Integral Form

[m0018]

Gauss’ Law for Magnetic Fields (GLM) is one ofthe four fundamental laws of classicalelectromagnetics, collectively known as Maxwell’sEquations. Before diving in, the reader isstrongly encouraged to review Section 2.5. Inthat section GLM emerges from the “fluxdensity” interpretation of the magnetic field.GLM is not identified in that section, but now weare ready for an explicit statement:


https://en.wikipedia.org/wiki/Duality_(electricity_and_magnetism)

https://en.wikipedia.org/wiki/Duality_(electrical_circuits)


7.2. GAUSS’ LAW FOR MAGNETIC FIELDS: INTEGRAL FORM 139

electrostatics magnetostatics

Sources static charge steady current,magnetizable material

Field intensity E (V/m) H (A/m)Flux density D (C/m2) B (Wb/m2=T)Material relations D = ǫE B = µH

J = σEForce on charge q F = qE F = qv ×BMaxwell’s Eqs.

∮SD · ds = Qencl

∮SB · ds = 0

(integral)∮CE · dl = 0

∮CH · ds = Iencl

Maxwell’s Eqs. ∇ ·D = ρv ∇ ·B = 0(differential) ∇×E = 0 ∇×H = J

Boundary Conditions n× [E1 −E2] = 0 n× [H1 −H2] = Jsn · [D1 −D2] = ρs n · [B1 −B2] = 0

Energy storage Capacitance InductanceEnergy density we =

12ǫE

2 wm = 12µH

2

Energy dissipation Resistance

Table 7.1: A summary of the duality between electrostatics and magnetostatics.

Gauss’ Law for Magnetic Fields (Equa-tion 7.1) states that the flux of the magneticfield through a closed surface is zero.

This is expressed mathematically as follows:

∮

S

B · ds = 0 (7.1)

where B is magnetic flux density and S is aclosed surface with outward-pointing differentialsurface normal ds. It may be useful to considerthe units: B has units of Wb/m2, thereforeintegrating B over a surface gives a quantitywith units of Wb, which is magnetic flux, asindicated above.

GLM can also be interpreted in terms ofmagnetic field lines. For the magnetic fluxthrough a closed surface to be zero, every fieldline entering the volume enclosed by S must alsoexit this volume: Field lines may not begin orend within the volume. The only way this can betrue for every possible surface S is if magneticfield lines always form closed loops. This is infact what we find in practice, as shown inFigure 7.1.

c© Youming / K. Kikkeri CC BY SA 4.0

Figure 7.1: Gauss’ Law for Magnetostatics ap-plied to a two-dimensional bar magnet: For thesurface S = SA, every field line entering S alsoleaves S, so the flux through S is zero. For thesurface S = SB , every field line within S remainsin S, so the flux through S is again zero.


140 CHAPTER 7. MAGNETOSTATICS

Following this argument one step further, notethat GLM implies that there can be noparticular particle or structure that can be thesource of the magnetic field, since if there were,that would be a start point for field lines. This isone way in which the magnetic field is verydifferent from the electric field, for which everyfield line begins at a charged particle. So, whenwe say that current (for example) is the source ofthe magnetic field, we mean only that the fieldcoexists with current, and not that the magneticfield is somehow attached to the current.Summarizing, there is no “localizable” quantity,analogous to charge for electric fields, associatedwith magnetic fields. This is just another way inwhich magnetic fields are weird!

Summarizing:

Gauss’ Law for Magnetic Fields requires thatmagnetic field lines form closed loops. Fur-thermore, there is no particle that can beidentified as the source of the magnetic field.

Additional Reading:

• “Gauss’ Law for Magnetism” on Wikipedia.


7.3 Gauss’ Law forMagnetism: DifferentialForm

[m0047]

The integral form of Gauss’ Law (Section 7.2)states that the magnetic flux through a closedsurface is zero. In mathematical form:

∮

S

B · ds = 0 (7.2)

where B is magnetic flux density and S is theenclosing surface. Just as Gauss’s Law forelectrostatics has both integral (Sections 5.5) anddifferential (Section 5.7) forms, so too does

Gauss’ Law for Magnetic Fields. Here we areinterested in the differential form for the samereason: Given a differential equation and theboundary conditions imposed by structure andmaterials, we may then solve for the magneticfield in very complicated scenarios.

The equation we seek may be obtained fromEquation 7.2 using the Divergence Theorem(Section 4.7), which in the present case may bewritten:

∫

V

(∇ ·B) dv =

∮

S

B · ds (7.3)

Where V is the mathematical volume boundedby the closed surface S. From Equation 7.2 wesee that the right hand side of the equation iszero, leaving:

∫

V

(∇ ·B) dv = 0 (7.4)

The above relationship must hold regardless ofthe specific location or shape of V. The only waythis is possible is if the integrand is everywhereequal to zero. We conclude:

∇ ·B = 0 (7.5)

In other words, the flux per unit volume of themagnetic field is always zero. This is anotherway of saying that there is no point in space thatcan be considered to be the source of themagnetic field, for if it were, then the total fluxthrough a bounding surface would be greaterthan zero. Said yet another way: The source ofthe magnetic field is not localizable.

The differential (“point”) form of Gauss’ Lawfor Magnetic Fields (Equation 7.5) states thatthe flux per unit volume of the magnetic fieldis always zero.

Additional Reading:

• “Gauss’ Law for Magnetism” on Wikipedia.

https://en.wikipedia.org/wiki/Gauss'_law_for_magnetism


https://en.wikipedia.org/wiki/Gauss'_law_for_magnetism

7.4. AMPERE’S CIRCUITAL LAW (MAGNETOSTATICS): INTEGRAL FORM 141

Iencl


Figure 7.2: Reference directions for Ampere’sCircuital Law (Equation 7.6).

7.4 Ampere’s Circuital Law(Magnetostatics):Integral Form

[m0019]

Ampere’s circuital law (ACL) relates current tothe magnetic field associated with the current. Inthe magnetostatic regime the law is (see alsoFigure 7.2):

∮

C

H · dl = Iencl (7.6)

That is, the integral of the magnetic fieldintensity H over a closed path C is equal to thecurrent enclosed by that path, Iencl. Beforeproceeding to interpret this law, it is useful to seethat it is dimensionally correct: That is, H (unitsof A/m) integrated over a distance (units of m)yields a quantity with units of current (i.e., A).

In general Iencl may be either positive ornegative. The direction corresponding to positivecurrent flow must be correctly associated with C.The relationship follows the right-hand rule ofStokes’ Theorem (Section 4.9) summarized asfollows: The direction of positive Iencl is thedirection in which the fingers of the right handintersect any surface S bordered by C when thethumb of the right hand points in the direction ofintegration. The connection to Stokes’ Theoremis not a coincidence: See Section 7.9 for more

about this.

Also note that S can be any surface which isbounded by C – not just the taut surface impliedin Figure 7.2.

The integral form of Ampere’s Circuital Lawfor magnetostatics (Equation 7.6) relates themagnetic field along a closed path to the totalcurrent flowing through any surface boundedby that path.

ACL plays a role in magnetostatics that is verysimilar to the role played by the integral form ofGauss’ Law (Section 5.5) in electrostatics. Thatis, ACL provides a means calculate the magneticfield given the source current. ACL also has asimilar limitation: Generally symmetry isrequired to simplify the problem sufficiently sothat the integral equation may be solved.Fortunately a number of important problems fallin this category. Examples include the problemsaddressed in Sections 7.5 (magnetic field of a linecurrent), 7.6 (magnetic field inside a straightcoil), 7.7 (magnetic field of a toroidal coil), 7.8(magnetic field of a current sheet), and 7.11(boundary conditions on the magnetic fieldintensity). For problems in which the necessarysymmetry is not available, the differential form ofACL may be required (Section 7.9).

Finally, be aware that the form of ACL addressedhere applies to magnetostatics only. In thepresence of a time-varying electric field, the rightside of ACL includes an additional componentknown as the displacement current (Section 8.9).

Additional Reading:


• “Ampere’s Circuital Law” on Wikipedia.

7.5 Magnetic Field of anInfinitely-Long StraightCurrent-Bearing Wire



https://en.wikipedia.org/wiki/Ampere's_circuital_law


z

y

x

a

I

ρ


Figure 7.3: Determination of the magnetic fielddue to steady current in an infinitely-long straightwire.

[m0119]

In this section we use the magnetostatic form ofAmpere’s Circuital Law (ACL) (Section 7.4) todetermine the magnetic field due to a steadycurrent I (units of A) in an infinitely-longstraight wire. The problem is illustrated inFigure 7.3. The wire is an electrically-conductingcircular cylinder of radius a. Since the wire is acylinder, the problem is easiest to work incylindrical coordinates with the wire alignedalong the z axis.

Here’s the relevant form of ACL:

∮

C

H · dl = Iencl (7.7)

where Iencl is the current enclosed by the closedpath C. ACL works for any closed path, so toexploit the symmetry of the cylindricalcoordinate system we choose a circular path ofradius ρ in the z = 0 plane, centered at the

origin. With this choice we have

Iencl = I for ρ ≥ a (7.8)

For ρ < a, Iencl < I. The current will bedistributed uniformly throughout the wire(Section 6.4). Since the current is uniformlydistributed over the cross section, Iencl is lessthan the total current I by the same factor thatthe area enclosed by C is less than πa2, thecross-sectional area of the wire. The areaenclosed by C is simply πρ2, so we have

Iencl = Iπρ2

πa2= I

ρ2

a2for ρ < a (7.9)

For the choice of C made above, Equation 7.7becomes

∫ 2π

φ=0

H ·(φ ρ dφ

)= Iencl (7.10)

Note that we have chosen to integrate in the +φdirection. Therefore the right hand rule specifiesthat positive Iencl corresponds to current flowingin the +z direction, which is consistent with thedirection indicated in Figure 7.3. (Here’s anexcellent exercise to test your understanding:Change the direction of the path of integrationand confirm that you get the same resultobtained at the end of this section. Changing thedirection of integration should not change themagnetic field associated with the current!)

The simplest way to solve for H fromEquation 7.10 is to use a symmetry argument,which proceeds as follows:

• Since the distribution of current is uniformand infinite in the z-dimension, H can’tdepend on z, and so H · z must be zeroeverywhere.

• The problem is identical after any amount ofrotation in φ; therefore the magnitude of Hcannot depend on φ. This is a form of radialsymmetry. Since we determined above thatH can’t depend on z either, it must be thatthe magnitude of H can depend only on ρ.


7.6. MAGNETIC FIELD INSIDE A STRAIGHT COIL 143

• The radial symmetry of the problem alsorequires that H · ρ be equal to zero. If thiswere not the case, then the field would notbe radially symmetric. Since we determinedabove that H · z is also zero, it must be thatH is entirely ±φ-directed.

From the above considerations, the most generalform of the magnetic field intensity can bewritten H = φH(ρ). Substituting this intoEquation 7.10, we obtain

Iencl =

∫ 2π

φ=0

[φH(ρ)

]·(φ ρ dφ

)

= ρH(ρ)

∫ 2π

φ=0

dφ

= 2πρH(ρ) (7.11)

Therefore H(ρ) = Iencl/2πρ. Reassociating theknown direction, we obtain:

H = φIencl2πρ

(7.12)

Therefore the field outside of the wire is:

H = φI

2πρfor ρ ≥ a (7.13)

whereas the field inside the wire is:

H = φIρ

2πa2for ρ < a (7.14)

(By the way, this is a good time for a unitscheck.)

Note that as ρ increases from zero to a (i.e.,inside the wire), the magnetic field isproportional to ρ, and therefore increases.However as ρ continues to increase beyond a (i.e.,outside the wire), the magnetic field isproportional to ρ−1, and therefore decreases.

If desired, the associated magnetic flux densitycan be obtained using B = µH.

Summarizing:

I

H

c© Jfmelero CC BY SA 4.0 (modified)

Figure 7.4: Right hand rule for the relationshipbetween the direction of current and the directionof the magnetic field.

The magnetic field due to current in an infi-nite straight wire is given by Equations 7.13(outside the wire) and 7.14 (inside the wire).

The magnetic field is +φ-directed for currentflowing in the +z direction, so the magneticfield lines form concentric circles perpendicu-lar to and centered on the wire.

Finally, we point out another “right hand rule”that emerges from this solution, shown inFigure 7.4 and summarized below:

The magnetic field due to current in an in-finite straight wire points in the direction ofthe fingers of the right hand when the thumbof the right hand is aligned in the directionof current flow.

This simple rule turns out to be handy in quicklydetermining the relationship between thedirections of the magnetic field and current flowin many other problems, and so is well worthcommitting to memory.

7.6 Magnetic Field Inside aStraight Coil

[m0120]

In this section we use the magnetostatic integral



by Zureks (public domain)

Figure 7.5: A straight coil.

Figure 7.6: Determination of the magnetic fielddue to DC current in a coil.

form of Ampere’s Circuital Law (ACL)(Section 7.4) to determine the magnetic fieldinside a straight coil of the type shown inFigure 7.5 in response to a steady (i.e., DC)current. The result has a number of applications,including the analysis and design of inductors,solenoids (coils which are used as magnets,typically as part of an actuator), and as abuilding block and source of insight for morecomplex problems.

The present problem is illustrated in Figure 7.6.The coil is circular with radius a and length l,and consists of N turns (windings) of wire woundwith uniform winding density. Since the coilforms a cylinder, the problem is easiest to workin cylindrical coordinates with the axis of the coilaligned along the z axis.

To begin, let’s take stock of what we alreadyknow about the answer, which is actually quite abit. The magnetic field deep inside the coil isgenerally aligned with axis of the coil as shown inFigure 7.7. This can be explained using theresult for the magnetic field due to a straight linecurrent (Section 7.5), in which we found that themagnetic field follows a “right hand rule”: The

c© Geek3 CC BY SA 3.0

Figure 7.7: Magnetic field lines inside a straightcoil with closely-spaced windings. (Dotted circlesrepresent current flowing up/out from the page;crossed circles represent current flowing down/intothe page.)

I

c© Chetvorno CC0 1.0 (modified)

Figure 7.8: Magnetic field due to a single loop.

magnetic field points in the direction of thefingers of the right hand when the thumb of theright hand is aligned in the direction of currentflow. The wire comprising the coil is obviouslynot straight, but we can consider just one shortsegment of one turn and then sum the results forall such segments. When we consider this for asingle turn of the coil, the situation is as shownin Figure 7.8. Summing the results for manyloops, we see that the direction of the magneticfield inside the coil must be generally in the +zdirection when the current I is applied as shownin Figures 7.6 and 7.7. One caveat however: Thewindings must be sufficiently closely-spaced thatthe magnetic field lines can only pass through theopenings at the end of the coil, and do not takeany “shortcuts” between individual windings.


https://creativecommons.org/publicdomain/zero/1.0/deed.en

7.6. MAGNETIC FIELD INSIDE A STRAIGHT COIL 145

Figure 7.7 also indicates that the magnetic fieldlines near the ends of the coil diverge from theaxis of the coil. This is understandable given theresult from Section 7.2 that magnetic field linesform closed loops. That is, each field line isapproximately +z directed inside the coil, butdiverges significantly from the z axis beforeentering and after emerging from the ends of thecoil. The relatively complex structure of themagnetic fields near the ends of the coil (the“fringing fields”) and outside of the coil makethem relatively difficult to analyze. Thereforehere we shall restrict our attention to themagnetic field deep inside the coil. Thisrestriction turns out to be of little consequence inthe engineering applications in which our resultis used.

Also, it is apparent from the radial symmetry ofthe coil that the magnitude of the magnetic fieldcannot depend on φ. Putting these findingstogether, we find that the most general form forthe magnetic field intensity deep inside the coil isH ≈ zH(ρ). That is, the direction of H is ±zand the magnitude of H depends, at most, on ρ.In fact, we will soon find with the assistance ofACL that the magnitude of H doesn’t depend onρ either.

Here’s the relevant form of ACL:∮

C

H · dl = Iencl (7.15)

where Iencl is the current enclosed by the closedpath C. ACL works for any closed path thatencloses the current if interest. Also, forsimplicity, we prefer a path that lies on aconstant-coordinate surface. The selected path isshown in Figure 7.9. The benefits of thisparticular path will soon become apparent.However, note for now that this particular choiceis consistent with the right-hand rule relating thedirection of C to the direction of positive I. Thatis, when I is positive, the current in the turns ofthe coil pass through the surface bounded by Cin the same direction as the fingers of the righthand when the thumb is aligned in the indicateddirection of C.

Let’s define N to be the number of windings

z

l'

ρ2

ρ1

A

D

C

B

z1 z2

cylindrical form of coil


Figure 7.9: Selected path of integration.

(equivalently, “turns” or “loops”) in the coil.Then the winding density of the coil is N/l(turns/m). Let the path length in the z directionbe l′, as indicated in Figure 7.9. Then theenclosed current is

Iencl =N

ll′I (7.16)

That is, the number of turns per unit lengthtimes length gives number of turns, and thisquantity times the current through the wire isthe total amount of current crossing the surfacebounded by C.

For the choice of C made above, and taking ourapproximation for the form of H as exact,Equation 7.15 becomes

∮

C

[zH(ρ)] · dl = N

ll′I (7.17)

The integral consists of segments A, B, C, andD, as shown in Figure 7.9. Let us consider theresult for each of these segments individually:

• The integral over segments B and D is zerobecause dl = ρdρ for these segments, and soH · dl = 0 for these segments.

• It is also possible to make the contributionfrom Segment C go to zero simply by letting



ρ2 → ∞. The argument is as follows: Themagnitude of H outside the coil mustdecrease with distance from the coil, so for ρsufficiently large, H(ρ) becomes negligible.If that’s the case, then the integral overSegment C also becomes negligible.

With ρ2 → ∞, only Segment A contributessignificantly to the integral over C andEquation 7.17 becomes:

N

ll′I =

∫ z2

z1

[zH(ρ1)] · (zdz)

= H(ρ1)

∫ z2

z1

dz

= H(ρ1) [z2 − z1] (7.18)

Note z2 − z1 is simply l′. Also, we have foundthat the result is independent of ρ1, asanticipated earlier. Summarizing:

H ≈ zNI

linside coil (7.19)

Let’s take a moment to consider the implicationsof this remarkably simple result.

• Note that it is dimensionally correct; that is,current divided by length gives units ofA/m, which are the units of H.

• We have found that the magnetic field issimply winding density (N/l) times current.To increase the magnetic field you can eitheruse more turns per unit length, or increasethe current.

• We have found that the magnetic field isuniform inside the coil; that is, the magneticfield along the axis is equal to the magneticfield close to the cylinder wall formed by thecoil. However this does not apply close toends of the coil, since we have neglected thefringing field.

These findings have useful applications in morecomplicated and practical problems, so it isworthwhile taking note of these now.Summarizing:

c© Slick CC0 1.0

Figure 7.10: A toroidal coil used as a large-valueinductor in the power supply of a wireless router.

The magnetic field deep inside a straight coil(Equation 7.19) is uniform and proportionalto winding density and current.

Additional Reading:

• “Electromagnetic Coil” on Wikipedia.

• “Solenoid” on Wikipedia.

7.7 Magnetic Field of aToroidal Coil

[m0049]

A toroid is a cylinder in which the ends arejoined to form a closed loop. An example of atoroidal coil is shown in Figure 7.10. Toroidalcoils are commonly used to form inductors andtransformers. The principal advantage of toroidalcoils over straight coils in these applications ismagnetic field containment : As we shall see inthis section, the magnetic field outside of atoroidal coil can be made negligibly small. Thisreduces concern about interactions between thisfield and other fields and structures in thevicinity.

In this section we use the magnetostatic form ofAmpere’s Circuital Law (ACL) (Section 7.4) todetermine the magnetic field due to a steady(DC) current flowing through a toroidal coil.


https://en.wikipedia.org/wiki/Electromagnetic_coil

https://en.wikipedia.org/wiki/Solenoid

7.7. MAGNETIC FIELD OF A TOROIDAL COIL 147

a

b

I

H

Figure 7.11: Geometry of a toroidal coil.

The present problem is illustrated in Figure 7.11.The toroid is circular with inner and outer radiia and b, respectively. The coil consists of Nturns (windings) of wire wound with uniformwinding density. This problem is easiest to workin cylindrical coordinates with the toroidcentered on the origin in z = 0 plane.

To begin, let’s take stock of what we alreadyknow about the answer, which is actually quite abit. First, a review of Section 7.6 (“MagneticField Inside a Straight Coil”) is recommended.There it is shown that the magnetic field deepinside a straight coil is aligned with axis of thecoil. This can be explained using the result forthe magnetic field due to a straight line current(Section 7.5), in which we found that themagnetic field follows a “right hand rule”: Themagnetic field points in the direction of thefingers of the right hand when the thumb of theright hand is aligned in the direction of currentflow. The wire comprising the coil is obviouslynot straight, but we can consider just one shortsegment of one turn and then sum the results forall such segments. When we do this, we see thatthe direction of the magnetic field inside the coilmust be in the +φ direction when the current Iis applied as shown in Figure 7.11. Also: Becausethe problem is identical after any amount of

rotation around the z axis, the magnitude of themagnetic field cannot depend on φ. Puttingthese findings together, we find that the mostgeneral form for the magnetic field intensityinside or outside the coil is H = φH(ρ, z).

Here’s the relevant form of ACL:∮

C

H · dl = Iencl (7.20)

where Iencl is the current enclosed by the closedpath C. ACL works for any closed path, but weneed one which encloses some current so as toobtain a relationship between I and H. Also, forsimplicity, we prefer a path that lies on aconstant-coordinate surface. The selected path isa circle of radius ρ centered on the origin in thez = z0 plane, as shown in Figure 7.12. Wefurther require C to lie entirely inside the coil,which ensures that the enclosed current includesthe current of all the wires as they pass throughthe hole at the center of the toroid. We choosethe direction of integration to be in the +φdirection, which is consistent with the indicateddirection of positive I according to the righthand rule. That is, when I is positive, thecurrent in the turns of the coil pass through thesurface bounded by C in the same direction asthe fingers of the right hand when the thumb isaligned in the indicated direction of C.

In terms of the variables we have defined, theenclosed current is simply

Iencl = NI (7.21)

Equation 7.20 becomes

∮

C

[φH(ρ, z0)

]· dl = NI (7.22)

Now evaluating the integral:

NI =

∫ 2π

0

[φH(ρ, z0)

]·(φ ρ dφ

)

= ρH(ρ, z0)

∫ 2π

0

dφ

= 2πρH(ρ, z0) (7.23)

It is now clear that the result is independent of


x

y

a

b

I

ρ

C

Figure 7.12: Selected path of integration.

z0. Summarizing:

H = φNI

2πρinside coil (7.24)

Let’s take a moment to consider the implicationsof this result.

• Note that it is dimensionally correct; that is,current divided by the circumference of C(2πρ) gives units of A/m, which are theunits of H.

• We have found that the magnetic field isproportional to winding density (i.e.,number of turns divided by circumference)times current. To increase the magnetic fieldyou can either use more turns, or increasethe current.

• Remarkably, we have found that themagnitude of the magnetic field inside thecoil depends only on ρ; i.e., the distancefrom the central (here, z) axis. It isindependent of z.

Summarizing:

The magnetic field inside a toroidal coil(Equation 7.24) depends only on distancefrom the central axis and is proportional towinding density and current.

Now let us consider what happens outside thecoil. For this, we consider any path of integration(C) that lies completely outside the coil. Notethat any such path encloses no current andtherefore Iencl = 0 for any such path. In this casewe have: ∮

C

H · dl = 0 (7.25)

There are two ways this could be true: Either Hcould be zero everywhere along the path, or Hcould be non-zero along the path in such a waythat the integral turns out to be zero. The radialsymmetry of the problem rules out the secondpossibility: If H is radially symmetric and C isradially symmetric, then the sign of H · dl shouldnot change over C. Therefore:

The magnetic field everywhere outside anideal toroidal coil is zero.

Note the caveat signalled by the use of theadjective “ideal”. In a non-ideal toroidal coil –i.e., any practical device – we expect there willbe some leakage of magnetic flux between thewindings. In practice, this leakage can be madenegligibly small by using a sufficiently highwinding density, and winding the wire onmaterial on a toroidal form (a “core”) havingsufficiently large permeability. The use of ahigh-permeability core, in particular, willdramatically improve the already pretty-goodcontainment. In fact, the use of such a coreallows the spacing between windings to becomequite large before leakage becomes significant.

One final observation about toroidal coils: At nopoint in the derivation of the magnetic field didwe need to consider the cross-sectional shape ofthe coil: We merely needed to know whether Cwas inside or outside the coil. Therefore:

The magnetic field inside an ideal toroidal coil

7.8. MAGNETIC FIELD OF AN INFINITE CURRENT SHEET 149

z

C

JsLz

Ly

y

Figure 7.13: Analysis of the magnetic field dueto an infinite thin sheet of current.

does not depend on the cross-sectional shapeof the coil.

Additional Reading:

• “Toroidal Inductors and Transformers” onWikipedia.

———————-

7.8 Magnetic Field of anInfinite Current Sheet

[m0121]

We now consider the magnetic field due to aninfinite sheet of current, shown in Figure 7.13.The solution to this problem is useful as abuilding block and source of insight in morecomplex problems, as well as being a usefulapproximation to some practical problemsinvolving current sheets of finite extent including,for example, microstrip transmission line andground plane currents in printed circuit boards.

The current sheet in Figure 7.13 lies in the z = 0plane and the current density is Js = xJs (unitsof A/m); i.e., the current is uniformly distributedsuch that the total current crossing any segmentof width ∆y along the y direction is Js∆y.

To begin, let’s take stock of what we alreadyknow about the answer, which is actually quite abit. For example, imagine the current sheet as acontinuum of strips parallel to the x axis andvery thin in the y dimension. Each of these stripsindividually behaves like a straight line currentI = Js∆y (units of A), where ∆y is the width ofa strip. The magnetic field due to each of thesestrips is determined by a “right hand rule”: Themagnetic field points in the direction of thefingers of the right hand when the thumb of theright hand is aligned in the direction of currentflow. (Section 7.5). It is apparent from this muchthat H can have no y component, since the fieldof each individual strip has no y component.When the magnetic field due to each strip isadded to that of all the other strips, the zcomponent of the sum field must be zero due tosymmetry. It is also clear from symmetryconsiderations that the magnitude of H cannotdepend on x or y. Summarizing, we havedetermined that the most general form for H isyH(z), and furthermore the sign of H(z) mustbe positive for z < 0 and negative for z > 0.

It’s possible to solve this problem by actuallysumming over the continuum of current strips asimagined above.1 However it’s far easier to useAmpere’s Circuital Law (ACL; Section 7.4):Here’s the relevant form of ACL:

∮

C

H · dl = Iencl (7.26)

where Iencl is the current enclosed by the path C.ACL works for any closed path, but we need onewhich encloses some current so as to obtain arelationship between Js and H. Also, forsimplicity, we prefer a path that lies on aconstant-coordinate surface. A convenient pathin this problem is a rectangle lying in the x = 0plane and centered on the origin, as shown inFigure 7.13. We choose the direction ofintegration to be counter-clockwise from theperspective shown in Figure 7.13, which isconsistent with the indicated direction of positiveJs according to the right-hand rule. That is,when Js is positive (current flowing in the +x

1In fact, this is pretty good thing to try, if for no otherreason than to see how much simpler it is to use ACL in-stead.

https://en.wikipedia.org/wiki/Toroidal_inductors_and_transformers


direction), the current passes through the surfacebounded by C in the same direction as the fingersof the right hand when the thumb is aligned inthe indicated direction of C.

Let us define Ly to be the width of therectangular path of integration in the ydimension, and Lz to be the width in the zdimension. In terms of the variables we havedefined, the enclosed current is simply

Iencl = JsLy (7.27)

Equation 7.26 becomes∮

C

[yH(z)] · dl = JsLy (7.28)

Note that H · dl = 0 for the vertical sides of thepath, since H is y-directed and dl = zdz on thosesides. Therefore only the horizontal sidescontribute to the integral and we have:

∫ +Ly/2

−Ly/2

[yH

(−Lz

2

)]· (ydy)

+

∫ −Ly/2

+Ly/2

[yH

(+Lz2

)]· (ydy) = JsLy (7.29)

Now evaluating the integrals:

H

(−Lz

2

)Ly −H

(+Lz2

)Ly = JsLy (7.30)

Note that all factors of Ly cancel in the aboveequation. Furthermore,H(−Lz/2) = −H(+Lz/2) due to (1) symmetrybetween the upper and lower half-spaces and (2)the change in sign between these half-spaces,noted earlier. We use this to eliminateH(+Lz/2) and solve for H(−Lz/2) as follows:

2H(−Lz/2) = Js (7.31)

yielding

H(−Lz/2) = +Js2

(7.32)

and therefore

H(+Lz/2) = −Js2

(7.33)

Furthermore, note that H is independent of Lz;for example, the result we just found indicates

the same value of H(+Lz/2) regardless of thevalue of Lz. Therefore H is uniform throughoutall space, except for the change of signcorresponding for the field above vs. below thesheet. Summarizing:

H = ±yJs2

for z ≶ 0 (7.34)

The magnetic field intensity due to an infi-nite sheet of current (Equation 7.34) is spa-tially uniform except for a change of sign cor-responding for the field above vs. below thesheet.

7.9 Ampere’s Law(Magnetostatics):Differential Form

[m0118]

The integral form of Amperes’ Circuital Law(ACL; Section 7.4) for magnetostatics relates themagnetic field along a closed path to the totalcurrent flowing through any surface bounded bythat path. In mathematical form:

∮

C

H · dl = Iencl (7.35)

where H is magnetic field intensity, C is theclosed curve, and Iencl is the total currentflowing through any surface bounded by C. Inthis section we derive the differential form of thisequation. In some applications this differentialequation, combined with boundary conditionsassociated with discontinuities in structure andmaterials, can be used to solve for the magneticfield in arbitrarily complicated scenarios. A moredirect reason for seeking out this differentialequation is that we gain a little more insight intothe relationship between current and themagnetic field, disclosed at the end of thissection.

The equation we seek may be obtained usingStokes’ Theorem (Section 4.9), which in the

7.10. BOUNDARY CONDITIONS ON THE MAGNETIC FLUX DENSITY (B) 151

present case may be written:

∫

S

(∇×H) · ds =∮

C

H · dl (7.36)

where S is any surface bounded by C, and ds isthe differential surface area combined with unitvector that is perpendicular to that surface inthe direction determined by the right-hand rule.ACL tells us that the right side of the aboveequation is simply Iencl. We may express Iencl asthe integral of the volume current density J(units of A/m2; Section 6.2) as follows:

Iencl =

∫

S

J · ds (7.37)

so we may rewrite Equation 7.36 as follows:

∫

S

(∇×H) · ds =∫

S

J · ds (7.38)

The above relationship must hold regardless ofthe specific location or shape of S. The only waythis is possible for all possible surfaces in allapplicable scenarios is if the integrands are equal.Thus we obtain the desired expression:

∇×H = J (7.39)

that is, the curl of the magnetic field intensity ata point is equal to the volume current density atthat point. Recalling the properties of the curloperator (Section 4.8) – in particular, that curlinvolves derivatives with respect to direction –we conclude:

The differential form of Ampere’s CircuitalLaw for magnetostatics (Equation 7.39) indi-cates that the volume current density at anypoint in space is proportional to the spatialrate of change of the magnetic field, and isperpendicular to the magnetic field at thatpoint.

Additional Reading:

• “Ampere’s circuital law” on Wikipedia.


r

region 2

n

a


Figure 7.14: Determination of the boundary con-dition on B at the interface between material re-gions.

7.10 Boundary Conditionson the Magnetic FluxDensity (B)

[m0022]

In homogeneous media, electromagneticquantities vary smoothly and continuously. At aninterface between dissimilar media, however, it ispossible for electromagnetic quantities to bediscontinuous. Continuities and discontinuities infields can be described mathematically byboundary conditions, and used to to constrainsolutions for fields away from these interfaces.

In this section we derive the boundary conditionon the magnetic flux density B at a smoothinterface between two material regions, as shownin Figure 7.14.2 The desired boundary conditionmay be obtained from Gauss’ Law for MagneticFields (GLM; Section 7.2):

∮

S

B · ds = 0 (7.40)

where S is any closed surface. Let S take theform of cylinder centered at a point on theinterface, and for which the flat ends are parallelto the surface and perpendicular to n, as shown

2It may be helpful to note the similarity (duality, in fact)between this derivation and the derivation of the associatedboundary condition on D presented in Section 5.18.





in Figure 7.14. Let the radius of this cylinder bea, and let the length of the cylinder be 2h. FromGLM we have

∮

S

B · ds =∫

top

B · ds

+

∫

side

B · ds

+

∫

bottom

B · ds = 0 (7.41)

Now let us reduce h and a together while (1)maintaining a constant ratio h/a≪ 1 and (2)keeping S centered on the interface. Becauseh≪ a, the area of the side can be madenegligible relative to the area of the top andbottom. Then as h→ 0 we are left with

∫

top

B · ds+∫

bottom

B · ds → 0 (7.42)

As the area of the top and bottom sides becomeinfinitessmal the variation in B over these areasbecomes negligible. Now we have simply:

B1 · n∆A+B2 · (−n)∆A→ 0 (7.43)

where B1 and B2 are the magnetic flux densitiesat the interface but in regions 1 and 2,respectively, and ∆A is the area of the top andbottom sides. Note that the orientation of n isimportant: We have assumed n points intoregion 1, and we must now stick with this choice.Thus, we obtain

n · (B1 −B2) = 0 (7.44)

where, as noted above, n points into region 1.

Summarizing:

The normal (perpendicular) component of Bacross the boundary between two material re-gions is continuous.

It is worth noting what this means for themagnetic field intensity H: Since B = µH, itmust be that

r

BA w

tn

Figure 7.15: Determining the boundary conditiononH at the smooth interface between two materialregions.

The normal (perpendicular) component of Hacross the boundary between two material re-gions is discontinuous if the permeabilities areunequal.

7.11 Boundary Conditionson the Magnetic FieldIntensity (H)

[m0023]

In homogeneous media, electromagneticquantities vary smoothly and continuously. At aninterface between dissimilar media, however, it ispossible for electromagnetic quantities to bediscontinuous. Continuities and discontinuities infields can be described mathematically byboundary conditions, and used to to constrainsolutions for fields away from these interfaces. Inthis section we derive boundary conditions on themagnetic field intensity H.

To begin, consider a region consisting of only twomedia which meet at a smooth interface asshown in Figure 7.15. The desired boundarycondition can be obtained directly fromAmpere’s Circuital Law (ACL; Section 7.4):

∮

C

H · dl = Iencl (7.45)

where C is any closed path and Iencl is thecurrent that flows through the surface bounded

7.12. INDUCTANCE 153

by that path. Let C take the form of a rectanglecentered on a point on the interface as shown inFigure 7.15, perpendicular to the direction ofcurrent flow at that location. Let the sides A, B,C, and D be perpendicular and parallel to theinterface. Let the length of the parallel sides be l,and let the length of the perpendicular sides bew. From ACL we have

∮H · dl =

∫

A

H · dl

+

∫

B

H · dl

+

∫

C

H · dl

+

∫

D

H · dl = Iencl (7.46)

Now we let w and l become vanishingly smallwhile (1) maintaining the ratio l/w and (2)keeping C centered on the interface. In thisprocess the contributions from the B and Dsegments become equal in magnitude butopposite in sign; i.e.,

∫

B

H · dl+∫

D

H · dl → 0 (7.47)

This leaves∫

A

H · dl+∫

C

H · dl → Iencl (7.48)

Let us define the unit vector t (“tangent”) asshown in Figure 7.15. Now we have simply:

H1 · t∆l −H2 · t∆l = Iencl (7.49)

where H1 and H2 are the fields evaluated on thetwo sides of the boundary, and ∆l → 0 is thelength of sides A and C while this is happening.Note that Iencl in this situation is

Iencl = Js ·(t× n

)∆l (7.50)

where n is the normal to the surface, pointinginto Region 1. Now Equation 7.49 can bewritten:

H1 · t∆l −H2 · t∆l = Js ·(t× n

)∆l (7.51)

Eliminating the common factor of ∆l andarranging terms on the left:

(H1 −H2) · t = Js ·(t× n

)(7.52)

The left side of the above equation is thecomponent of H1 −H2 which lies in the tdirection. Another way to compute this verysame quantity is to first take the cross productwith n, yielding a vector which is perpendicularto both H and n; and then taking the dotproduct with t× n to obtain the desired scalar.Here’s that idea expressed mathematically:

(H1 −H2) · t = (H1 −H2)× n ·(t× n

)(7.53)

so we may rewrite Equation 7.52 as follows:

[(H1 −H2)× n] ·(t× n

)= Js ·

(t× n

)(7.54)

This must be true for all possible values of t× n;i.e. for all possible directions of current flow onthe surface. Therefore

(H1 −H2)× n = Js (7.55)

This is the desired boundary condition. The leftside of the above equation is the magnitude of thecomponent of H1 −H2 which is tangent to thesurface. Therefore we have found the following:

A discontinuity in the tangential componentof H across a material interface must be sup-ported by a surface current in a perpendiculardirection.

Since we frequently encounter situations in whichthere is no current on the interface, it is useful tointerpret the boundary condition for the case inwhich Js = 0. Here it is:

In the absence of surface current, the tan-gential component of H must be continuousacross the interface.

7.12 Inductance

[m0123]

Current creates a magnetic field, whichsubsequently exerts force on othercurrent-bearing structures. For example, thecurrent in each turn of a coil exerts a force on


the current in every other turn of the coil. If thewindings are fixed in place, then this force isunable to do work (i.e., move the windings), soinstead the coil stores potential energy. Thispotential energy can be released by turning offthe external source: When this happens, theremaining charge continues to flow, nowpropelled by the magnetic force. As this supplyof charge diminishes, so to does the magneticfield, until both dwindle to zero. At this pointpotential energy is expended. To restorepotential energy, the external source must beturned on, restoring the supply of charge, whichflows and thereby regenerates the magnetic field.

Now recall that the magnetic field is essentiallydefined in terms of the force associated with thispotential energy; i.e., F = qv ×B where q is thecharge of a particle comprising the current, v isthe velocity of the particle, and B is magneticflux density (Section 2.5). So, rather thanthinking of the potential energy of the system asbeing associated with the magnetic force appliedto charge, it is equally valid to think of thepotential energy as being stored in the magneticfield associated with the current distribution.The energy stored in the magnetic field dependson the geometry of the current-bearing structureand the permeability of the intervening material,because the magnetic field depends on theseparameters.

The relationship between current applied to astructure and the energy stored in the associatedmagnetic field is what we mean by inductance.We may fairly summarize this insight as follows:

Inductance is the ability of a structure tostore energy in a magnetic field.

The inductance of a structure depends onthe geometry of its current-bearing struc-tures and the permeability of the interveningmedium.

Note that inductance does not depend oncurrent, which we view as either a stimulus orresponse from this point of view. The

corresponding response or stimulus, respectively,is the magnetic flux associated with this current.This leads to the following definition:

L =Φ

I(single linkage) (7.56)

where Φ (units of Wb) is magnetic flux, I (unitsof A) is the current responsible for this flux, andL (units of H) is the associated inductance. (The“single linkage” caveat will be explained below.)In other words, a device with high inductancegenerates a large magnetic flux in response to agiven current, and therefore stores more energyfor a given current than a device with lowerinductance.

To use Equation 7.56 we must carefully definewhat we mean by “magnetic flux” in this case.Generally, magnetic flux is magnetic flux density(again, B, units of Wb/m2) integrated over aspecified surface S, so

Φ =

∫

S

B · ds (7.57)

where ds is the differential surface area vector,with direction normal to S. However, this leavesunanswered the following questions: Which S,and which of the two possible normal directionsof ds? For a meaningful answer, S must uniquelyassociate the magnetic flux to the associatedcurrent. Such an association exists if we requirethe current to form a closed loop. This is shownin Figure 7.16. Here C is the closed loop alongwhich the current flows, S is a surface boundedby C, and the direction of ds is defined accordingto the right-hand rule of Stokes’ Theorem(Section 4.9). Note that C can be a closed loop ofany shape; i.e., not just circular, and notrestricted to lying in a plane. Further note thatS used in the calculation of Φ can be any surfacebounded by C: This is because magnetic fieldlines form closed loops, so any one magnetic fieldline intersects any open surface bounded by Cexactly once. Such an intersection is sometimescalled a “linkage”. So there we have it: Werequire the current I to form a closed loop, wemeasure the magnetic flux through this loopusing the sign convention of the right-hand rule,and the ratio is the inductance.

7.12. INDUCTANCE 155

I

B

ds


Figure 7.16: Association between a closed loopof current and the associated magnetic flux.

Many structures consist of multiple such loops –the coil is of course one of these. In a coil, eachwinding carries the same current, and themagnetic fields of the windings add to create amagnetic field which grows in proportion to thenumber of windings (Section 7.6). The magneticflux density inside a coil is proportional to thenumber of turns, N , so the flux Φ inEquation 7.56 should properly be indicated asNΦ. Another way to look at this is that we arecounting the number of times the same current isable to generate a unique set of magnetic fieldlines which intersect S. For this reason it iscommon to refer to the quantity NΦ as “fluxlinkage”.

Summarizing, our complete definition forinductance is

L =NΦ

I(identical linkages) (7.58)

An engineering definition of inductance isEquation 7.58, with the magnetic flux definedto be that associated with a single closed loopof current with sign convention as indicatedin Figure 7.16, and N defined to be the num-ber of times the same current I is able tocreate that flux.

What happens if the loops have different shapes?

For example, what if the coil is not a cylinder,but rather cone-shaped? (Yes, there is such athing: See “Additional Reading” at the end ofthis section.) In this case one needs a better wayto determine the flux linkage, since the fluxassociated with each loop of current will bedifferent. However this is beyond the scope ofthis section.

An inductor is a device which is designed toexhibit a specified inductance. We can now makethe connection to the concept of the inductor asit appears in elementary circuit theory. First werewrite Equation 7.58 as follows:

I =NΦ

L(7.59)

Taking the derivative of both sides of thisequation with respect to time, we obtain:

d

dtI =

N

L

d

dtΦ (7.60)

Now we need to reach beyond the realm ofmagnetostatics for just a moment. It is shownSection 8.3 (“Faraday’s Law”) that the change inΦ associated with a change in current results inthe creation of an electrical potential equal to−NdΦ/dt realized over the loop C. In otherwords, the terminal voltage V is +NdΦ/dt, withthe change of sign intended to keep the resultconsistent with the sign convention relatingcurrent and voltage in passive devices. ThereforedΦ/dt in Equation 7.60 is equal to V/N . Makingthe substitution we find:

V = Ld

dtI (7.61)

This is the expected relationship fromelementary circuit theory.

Another circuit theory concept related toinductance is mutual inductance. Whereasinductance relates changes in current toinstantaneous voltage in the same device(Equation 7.61), mutual inductance relateschanges in current in one device to instantaneousvoltage in a different device. This can occurwhen the two devices are coupled (“linked”) bythe same magnetic field. For example,transformers (Section 8.5) typically consist of



separate coils which are linked by the samemagnetic field lines. The voltage across one coilmay be computed as the time-derivative ofcurrent on the other coil times the mutualinductance.

Let us conclude this section by taking a momentto dispel a common misconception aboutinductance. The misconception pertains to thefollowing question: If the current does not form aclosed loop, what is the inductance? Forexample, engineers sometimes refer to theinductance of a pin or lead of an electroniccomponent. A pin or lead is not a closed loop, sothe formal definition of inductance given above –ratio of magnetic flux to current – does notapply. The broader definition of inductance – theability to store energy in a magnetic field – doesapply, but this is not what is meant by “pininductance” or “lead inductance”. What isactually meant is the imaginary part of theimpedance of the pin or lead – i.e., the reactance– expressed as an equivalent inductance. In otherwords: The reactance of an inductive device ispositive, so any device which also exhibits apositive reactance can be viewed from a circuittheory perspective as an equivalent inductance.This is not referring to the storage of energy in amagnetic field; it merely means that the devicecan be modeled as an inductance in a circuitdiagram. In the case of “pin inductance” theculprit is not actually inductance, but rather skineffect (see “Additional References” at the end ofthis section). Summarizing:

Inductance implies positive reactance, butpositive reactance does not necessarily implyinductance.

Additional Reading:

• “Inductance” on Wikipedia.

• “Inductor” on Wikipedia.

• T.A. Winslow,“Conical Inductors for Broadband Applications”,IEEE Microwave Mag., Vol. 6, No. 1, Mar2005, pp. 68–72.

• “Skin Effect” on Wikipedia.

Figure 7.17: Determination of the inductance ofa straight coil.

7.13 Inductance of a StraightCoil

[m0124]

In this section we determine the inductance of astraight coil, as shown in Figure 7.17. The coil iscircular with radius a and length l, and consistsof N turns (windings) of wire wound withuniform winding density. Since the coil forms acylinder, the problem is easiest to work incylindrical coordinates with the axis of the coilaligned along the z axis.

Inductance L in this case is given by(Section 7.12)

L =NΦ

I(7.62)

where I is current and Φ is the magnetic fluxassociated with one winding of the coil. Magneticflux in this case is given by

Φ =

∫

S

B · ds (7.63)

where B is the magnetic flux density (units of T= Wb/m2), S is the surface bounded by a singlecurrent loop, and ds points in the directiondetermined by the right hand rule with respectto the direction of positive current flow.

First, let’s determine the magnetic field. Themagnetic flux density deep inside the coil is(Section 7.6):

B ≈ zµNI

l(7.64)

https://en.wikipedia.org/wiki/Inductance

https://en.wikipedia.org/wiki/Inductor

https://doi.org/10.1109/MMW.2005.1418000

https://en.wikipedia.org/wiki/Skin_effect

7.14. INDUCTANCE OF A COAXIAL STRUCTURE 157

Is it reasonable to use this approximation here?Since inductance pertains to energy storage, thequestion is really this: What fraction of theenergy is stored in a field which is well-describedby this approximation, as opposed to energystored in the “fringing field” close to the ends ofthe coil. If we make l sufficiently large relative toa, then presumably energy storage in the fringingfield can be sufficiently small to be negligible incomparison. Since the alternative leads to amuch more complicated problem, we shallassume this.

Next, we determine Φ. In this case a naturalchoice for S is the interior cross-section of thecoil in a plane perpendicular to the axis. Thedirection of ds must be z since this is thedirection in which the fingers of the right handpoint when the current flows in the directionindicated in Figure 7.17. Thus we have

Φ ≈∫

S

(zµNI

l

)· (zds)

=µNI

l

∫

S

ds

=µNI

lA (7.65)

where A is the cross-sectional area of the coil.

Finally from Equation 7.62 we obtain

L ≈ µN2A

l(l ≫ a) (7.66)

Note that this is dimensionally correct; that is,permeability (units of H/m) times area (units ofm2) divided by length (units of m) gives units ofH, as expected. Also, it is worth noting thatinductance is proportional to permeability andcross-sectional area, and inversely proportional tolength. Interestingly the inductance isproportional to N2 as opposed to N ; this isbecause field strength increases with N , andindependently there are N flux linkages. Finally,we note that the inductance does not depend onthe shape of the coil cross-section, but only thearea of the cross-section.

The inductance of a long straight coil is givenapproximately by Equation 7.66.

Again, this result is approximate because itneglects the non-uniform “fringing field” near theends of the coil. Nevertheless, this resultfacilitates useful engineering analysis and design.

Additional Reading:

• “Inductance” on Wikipedia.

7.14 Inductance of a CoaxialStructure

[m0125]

Let us now determine the inductance of coaxialstructure, shown in Figure 7.18. The inductanceof this structure is of interest for a number ofreasons; in particular, for determining thecharacteristic impedance of coaxial transmissionline, as addressed in Sections 3.4 and 3.10.

For our present purposes we may model thestructure as consisting of two concentricperfectly-conducting cylinders of radii a and b,separated by a homogeneous material havingpermeability µ. To facilitate discussion, let usplace the +z axis along the common axis of theconcentric cylinders, so that the cylinders may bedescribed as the surfaces ρ = a and ρ = b.

Below we shall find the inductance by assuming acurrent I on the inner conductor and integratingover the resulting magnetic field to obtain themagnetic flux Φ between the conductors. Theninductance can be determined as the ratio of theresponse flux to the source current.

A note before we get started: The derivation weare about to do has a number of remarkablesimilarities to the derivation of the capacitance ofa coaxial structure, addressed in Section 5.24.The reader may benefit from a review of thatsection before attempting this derivation,although a review is certainly not required.

https://en.wikipedia.org/wiki/Inductance


b

a

z

ρ

z=l

I

μ

z=0

H

Figure 7.18: Determining the inductance of coax-ial line.

The first step is to find the magnetic field insidethe structure. This is relatively simple if we mayassume that the structure has infinite length(i.e., l → ∞), since then there are no fringingfields and the internal field will be utterlyconstant with respect to z. In the central regionof a finite-length inductor, however, the field isnot much different from the field that exists inthe case of infinite length, and if the energystorage in fringing fields is negligible compared tothe energy storage in this central region thenthere is no harm in assuming the internal field isconstant with z. Alternatively we may think ofthe length l as pertaining to one short section ofa much longer structure, and thereby obtain theinductance per length as opposed to the totalinductance for the structure. Note that the latteris exactly what we need for the transmission linelumped-element equivalent circuit model(Section 3.4).

To determine the inductance, we invoke thedefinition (Section 7.12):

L ,Φ

I(7.67)

A current I flowing in the +z direction on theinner conductor gives rise to a magnetic fieldinside the coaxial structure. The magnetic fieldintensity for this scenario was determined inSection 7.5, “Magnetic Field of an Infinitely-LongStraight Current-Bearing Wire”, where we found

H = φI

2πρ, a ≤ ρ ≤ b (7.68)

The reader should note that in that section wewere considering merely a line of current; not acoaxial structure. So on what basis do we claimthe field for inside the coaxial structure is thesame? This is a consequence of Amperes Law(Section 7.4)

∮

C

H · dl = Iencl (7.69)

If in this new problem we specify the samecircular path C with radius greater than a andless than b, then the enclosed current is thesame. Furthermore the presence of the outerconductor does not change the radial symmetryof the problem, and nothing else remains thatcan change the outcome. This is worth noting forfuture reference:

The magnetic field inside a coaxial structurecomprised of concentric conductors bearingcurrent I is identical to the magnetic field ofthe line current I in free space.

We’re going to need magnetic flux density (B) asopposed to H in order to get the magnetic flux.This is simple since they are related by thepermeability of the medium; i.e., B = µH. Thus:

B = φµI

2πρ, a ≤ ρ ≤ b (7.70)

Next we get Φ by integrating over the magneticflux density

Φ =

∫

S

B · ds (7.71)

where S is any open surface through which allmagnetic field lines within the structure mustpass. Since this can be any such surface, we mayas well choose the simplest one. The simplestsuch surface is a plane of constant φ, since such aplane is constant-coordinate surface andperpendicular to the magnetic field lines. Using

7.15. MAGNETIC ENERGY 159

this surface we find:

Φ =

∫ b

ρ=a

∫ l

z=0

(φµI

2πρ

)·(φdρdz

)

=µI

2π

∫ b

ρ=a

∫ l

z=0

dρ

ρdz

=µI

2π

(∫ l

z=0

dz

)(∫ b

ρ=a

dρ

ρ

)

=µIl

2πln

(b

a

)(7.72)

Wrapping up:

L ,Φ

I=

(µIl/2π) ln (b/a)

I(7.73)

Note that factors of I in the numerator anddenominator cancel out, leaving:

L =µl

2πln

(b

a

)(7.74)

Note that this is dimensionally correct, havingunits of H. Also note that this is expressiondepends only on materials (through µ) andgeometry (through l, a, and b). Notably it doesnot depend on current, which would implynon-linear behavior.

To make the connection back to lumped-elementtransmission line model parameters (Sections 3.4and 3.10), we simply divide by l to get theper-unit length parameter:

L′ =µ

2πln

(b

a

)(7.75)

which has units of H/m.

Example 7.1. Inductance of RG-59 coaxialcable.

RG-59 coaxial cable consists of an innerconductor having radius 0.292 mm, an outerconductor having radius 1.855 mm, andpolyethylene (a non-magnetic dielectric)spacing material. Estimate the inductanceper length of RG-59.

Solution. From the problem statement,a = 0.292 mm, b = 1.855 mm, and µ ∼= µ0

since the spacing material is non-magnetic.Using Equation 7.75 we find L′ ∼= 370 nH/m.

7.15 Magnetic Energy

[m0127]

Consider a structure exhibiting inductance; i.e.,one which is able to store energy in a magneticfield in response to an applied current. Thisstructure could be a coil, or it could be one of avariety of inductive structures which are notexplicitly intended to be an inductor; forexample, a coaxial transmission line. Whencurrent is applied, the current-bearing elementsof the structure exert forces on each other. Sincethese elements are not normally free to move, wemay interpret this force as potential energystored in the magnetic field associated with thecurrent (Section 7.12).

We now ask the question: How much energy isstored in this field? The answer to this questionhas relevance in several engineering applications.One issue is that any system that includesinductance is using some fraction of the energydelivered by the power supply to energize thisinductance. In many electronic systems – inpower systems in particular – inductors areperiodically energized and de-energized at aregular rate. Since power is energy per unit time,this consumes power. Therefore energy storage ininductors contributes to the power consumptionof electrical systems.

The stored energy is most easily determinedusing circuit theory concepts. First, we note thatthe electrical potential difference v(t) (units ofV) across an inductor is related to the currenti(t) (units of A) through the inductor as follows(Section 7.12):

v(t) = Ld

dti(t) (7.76)

where L (units of H) is the inductance. The


instantaneous power associated with the device is

p(t) = v(t)i(t) (7.77)

Energy (units of J) is power (units of J/s)integrated over time. Let Wm be the energystored in the inductor. At some time t0 in thepast, i(t0) = 0 and Wm = 0. As current isapplied, Wm increases monotonically. At thepresent time t, i(t) = I. Thus the present valueof the magnetic energy is:

Wm =

∫ t0+t

t0

p(τ)dτ (7.78)

Now evaluating this integral using therelationships established above:

Wm =

∫ t+t0

t0

v(τ)i(τ)dτ

=

∫ t+t0

t0

[Ld

dτi(τ)

]i(τ)dτ

= L

∫ t+t0

t0

[d

dτi(τ)

]i(τ)dτ (7.79)

Changing the variable of integration from τ (anddτ) to i (and di) we have

Wm = L

∫ t+t0

t0

di

dτi dτ

= L

∫ I

0

i di (7.80)

Evaluating the integral we obtain the desiredexpression

Wm =1

2LI2 (7.81)

The energy stored in an inductor in responseto a steady current I is Equation 7.81. Thisenergy increases in proportion to inductanceand in proportion to the square of current.

The long straight coil (Section 7.13) isrepresentative of a large number of practicalapplications, so it is useful to interpret the abovefindings in terms of this structure in particular.For this structure we found

L =µN2A

l(7.82)

where µ is the permeabity, N is the number ofturns, A is cross-sectional area, and l is length.The magnetic field intensity inside this structureis related to I by (Section 7.6):

H =NI

l(7.83)

Substituting these expressions intoEquation 7.81, we obtain

Wm =1

2

[µN2A

l

] [Hl

N

]2

=1

2µH2Al (7.84)

Recall that the magnetic field inside a long coil isapproximately uniform. Therefore the density ofenergy stored inside the coil is approximatelyuniform. Noting that the product Al is thevolume inside the coil, we find that this energydensity is

wm =1

2µH2 (7.85)

which has units of energy per unit volume(J/m3).

The above expression provides an alternativemethod to compute the total magnetostaticenergy in any structure. Within a mathematicalvolume V, the total electrostatic energy is simplythe integral of the energy density over V; i.e.,

Wm =

∫

V

wm dv (7.86)

This works even if the magnetic field and thepermeability vary with position. SubstitutingEquation 7.85 we obtain:

Wm =1

2

∫

V

µH2dv (7.87)

Summarizing:

The energy stored by the magnetic fieldpresent within any defined volume is givenby Equation 7.87.

It’s worth noting that this energy increases withthe permeability of the medium, which makes

7.16. MAGNETIC MATERIALS 161

sense since inductance is proportional topermeability.

Finally we reiterate that although we arrived atthis result using the example of the long straightcoil, Equation 7.87 is completely general.

7.16 Magnetic Materials

[m0058]

As noted in Section 2.5, magnetic fields arise inthe presence of moving charge (i.e., current) andin the presence of certain materials. In thissection we address these “magnetic materials”.

A magnetic material may be defined as asubstance which is exhibits permeability µ(Section 2.6) that is different from thepermeability of free space µ0. Since the magneticflux density B is related to the magnetic fieldintensity H via B = µH, magnetic materialsexhibit magnetic flux density in response to agiven magnetic field intensity that is significantlygreater than that of other materials. Magneticmaterials are also said to be “magnetizable”,meaning that the application of a magnetic fieldto such a material causes the material itself tobecome a source of the magnetic field.

Magnetic media are typically metals orsemiconductors; or heterogeneous mediacontaining such materials, including ferrite,which consists of iron particles suspended in aceramic. Magnetic media are commonly classifiedaccording to the physical mechanism responsiblefor their magnetizability. These mechanismsinclude paramagnetism, diamagnetism, andferromagnetism. All three of these mechanismsinvolve quantum mechanical processes operatingat the atomic and subatomic level, and are notwell-explained by classical physics. Theseprocesses are beyond the scope of this book (butinformation is available via “AdditionalReferences” at the end of this section). Howeverit is possible however to identify somereadily-observable differences between thesecategories of magnetic media.

Paramagnetic and diamagnetic materialsexhibit permeability which is only very slightlydifferent than µ0; and typically by much lessthan 0.01%. These materials exhibit very weakand temporary magnetization. The principaldistinction between paramagnetic anddiamagnetic media is in the persistence andorientation of induced magnetic fields.Paramagnetic materials – including aluminum,magnesium, and platinum – exhibit a very weakpersistent magnetic field, and the magnetic fieldinduced in the material is aligned in the samedirection as the impressed (external) magneticfield. Diamagnetic materials – including copper,gold, and silicon – do not exhibit a persistentmagnetic field, and the magnetic field induced inthe material is (counter to intuition!) aligned inthe opposite direction as the impressed magneticfield. The magnetization of paramagnetic anddiamagnetic media is typically so weak that it isnot often a consideration in engineering analysisand design.

Paramagnetic and diamagnetic media exhibitpermeability only very slightly different thanthat of free space, with little or no magneti-zation.

Ferromagnetic materials, on the other hand,exhibit permeability which can be many orders ofmagnitude greater than µ0. (See Appendix A.2for some example values.) These materials canbe readily and indefinitely magnetized; thuspermanent magnets are typically comprised offerromagnetic materials. Commonly-encounteredferromagnetic materials include iron, nickel, andcobalt.

Ferromagnetic materials are significantlynon-linear (Section 2.8), exhibiting saturationand hysteresis. This is illustrated in Figure 7.19.In this plot, the origin represents a ferromagneticmaterial which is unmagnetized and in a regionfree of an external magnetic field. The externalmagnetic field is quantified in terms of H, plottedalong the horizontal axis. As the external field isincreased, so to is B in the material, according tothe relationship B = µH. Right away we see thematerial is non-linear, since the slope of thecurve – and hence µ – is not constant.


B

saturation

H

c© Ndthe (modified) CC BY SA 3.0

Figure 7.19: Non-linearity in a ferromagnetic ma-terial manifesting as saturation and hysteresis.

Once the external magnetizing field H exceeds acertain value, the response field B no longersignificantly increases. This is saturation. Oncesaturated, further increases in the external fieldresult do not significantly increase themagnetization of the material, and so there is nosignificant increase in B.

From this state of saturation, let us now reducethe external field. We find that the rate ofdecrease in B with respect to H is significantlyslower than the rate that B originally increasedwith respect to H. In fact, B is still greater thanzero even when H has been reduced to zero. Atthis point the magnetization of the material isapparent, and a device comprised of this materialcould be used as a magnet.

If we now apply an external field in the reversedirection, we find that we are eventually able tozero and then redirect the response field. As wecontinue to decrease H (that is, increase themagnitude in the reverse direction), we onceagain reach saturation.

The same behavior is observed when we onceagain increase H: The material is eventuallydemagnetized, remagnetized in the oppositedirection and then saturated in that direction.At this point is apparent that a return to the

start condition (H = B = 0; i.e., demagnetizedwhen there is no external field) is not possible.

Hysteresis is the name that we apply to thisparticular form of non-linear behavior. Hysteresishas important implications in engineeringapplications. First, as identified above, it is afundamental principle in the analysis and designof magnets. In applications where aferromagnetic material is being used becausehigh permeability is desired – e.g., in inductors(Section 7.12) and transformers (Section 8.5) –hysteresis complicates the design and imposeslimits on the performance of the device.

Hysteresis may also be exploited as a form ofmemory. This is apparent from Figure 7.19: IfB > 0, then recent values of H must have beenrelatively large and positive. Similarly, If B < 0,then recent values of H must have been relativelylarge and negative. Furthermore, the most recentsign of H can be inferred even if the currentvalue of H is zero. In this sense the material“remembers” the past history of itsmagnetization, and may thereby exhibitsmemory. This is the enabling principle for anumber of digital data storage devices, includinghard drives (see “Additional Reading” at the endof this section).

Ferromagnetic media exhibit permeability µthat is orders of magnitude greater than thatof free space, and are readily magnetizable.These materials are also nonlinear in µ, whichmanifests as saturation and hysteresis.

Additional Reading:

• Section A.2 (“Permeability of SomeCommon Materials”)

• “Magnetism” on Wikipedia.

• “Ferrite (magnet)” on Wikipedia.

• “Paramagnetism” on Wikipedia.

• “Diamagnetism” on Wikipedia.

• “Ferromagnetism” on Wikipedia.


https://en.wikipedia.org/wiki/Magnetism

https://en.wikipedia.org/wiki/Ferrite_(magnet)

https://en.wikipedia.org/wiki/Paramagnetism

https://en.wikipedia.org/wiki/Diamagnetism

https://en.wikipedia.org/wiki/Ferromagnetism

7.16. MAGNETIC MATERIALS 163

• “Magnetic hysteresis” on Wikipedia.

• “Magnetic storage” on Wikipedia.

• “Hard disk drive” on Wikipedia.

[m0150]

https://en.wikipedia.org/wiki/Magnetic_hysteresis

https://en.wikipedia.org/wiki/Magnetic_storage

https://en.wikipedia.org/wiki/Hard_disk_drive


Image Credits

Fig. 7.1: c© Youming / K. Kikkeri,https://commons.wikimedia.org/wiki/File:M0018 fGLMBarMagnet (2).svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 7.2: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0019 fACL.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 7.3: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0119 fLSW.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 7.4: c© Jfmelero, https://et.wikipedia.org/wiki/Fail:Manoderecha.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/), modified.

Fig. 7.5: Zureks, https://en.wikipedia.org/wiki/File:Solenoid-1.png,public domain.

Fig. 7.7: c© Geek3, https://commons.wikimedia.org/wiki/File:VFPt Solenoid correct.svg,CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/).

Fig. 7.8: Chetvorno, https://en.wikipedia.org/wiki/File:Magnetic field of loop 3.svg,public domain via CC0 1.0, modified.

Fig. 7.9: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0120 fPath.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 7.10: Slick, https://en.wikipedia.org/wiki/File:3Com OfficeConnect ADSL Wireless 11g Firewall Router 2012-10-28-0869.jpg,public domain via CC0 1.0 (https://creativecommons.org/publicdomain/zero/1.0/deed.en).

Fig. 7.14: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0022 fBoundary.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 7.16: c© K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0123 fFluxCurrent.svg,CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).

Fig. 7.19: c© Ndthe, https://commons.wikimedia.org/wiki/File:HysteresisSVG.svg,CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/deed.en), modified by author.

https://commons.wikimedia.org/wiki/File:M0018_fGLMBarMagnet_(2).svg


https://commons.wikimedia.org/wiki/File:M0019_fACL.svg


https://commons.wikimedia.org/wiki/File:M0119_fLSW.svg


https://et.wikipedia.org/wiki/Fail:Manoderecha.svg


https://en.wikipedia.org/wiki/File:Solenoid-1.png

https://commons.wikimedia.org/wiki/File:VFPt_Solenoid_correct.svg


https://en.wikipedia.org/wiki/File:Magnetic_field_of_loop_3.svg


https://commons.wikimedia.org/wiki/File:M0120_fPath.svg



https://commons.wikimedia.org/wiki/File:M0022_fBoundary.svg


https://commons.wikimedia.org/wiki/File:M0123_fFluxCurrent.svg


https://commons.wikimedia.org/wiki/File:HysteresisSVG.svg


Chapter 8

Time-Varying Fields

8.1 Comparison of Staticand Time-VaryingElectromagnetics

[m0013]

Students encountering time-varyingelectromagnetic fields for the first time haveusually been exposed to electrostatics andmagnetostatics already. These disciplines exhibitmany similarities as summarized in Table 8.1.The principles of time-varying electromagneticspresented in this table are all formally introducedin other sections; the sole purpose of this table isto point out the differences. We can summarizethe differences as follows:

Maxwell’s Equations in the general (time-varying) case include extra terms that do notappear in the equations describing electro-statics and magnetostatics. These terms in-volve time derivatives of fields, and describecoupling between electric and magnetic fieldsthat does not exist in electro- and magneto-static conditions.

The coupling between electric and magneticfields in the time-varying case has one ratherprofound consequence in particular: It becomespossible for fields to continue to exist even aftertheir sources – i.e., charges and currents – areturned off. What kind of field can continue toexist in the absence of a source? Such a field iscommonly called a wave. Waves representingsignals in transmission lines and signals

propagating away from an antenna are examples.

Additional Reading:


8.2 ElectromagneticInduction

[m0129]

When an electrically-conducting structure isexposed to a time-varying magnetic field, anelectrical potential difference is induced acrossthe structure. This phenomenon is known aselectromagnetic induction. A convenientintroduction to electromagnetic induction isprovided by Lenz’s Law. This section explainselectromagnetic induction in the context ofLenz’s Law and provides two examples.

Let us begin with the example depicted inFigure 8.1, involving a cylindrical coil. Attachedto the terminals of the coil is a resistor for whichwe may identify an electric potential difference Vand current I. The sign conventions indicated forV and I are arbitrary, but it is important to beconsistent once they are established.

Now let us introduce a bar magnet as shown inFigure 8.1. The magnet is centered along theaxis of coil, to the right of the coil, and with itsnorth pole facing toward the coil. The magnet isresponsible for the magnetic flux density Bimp.




166 CHAPTER 8. TIME-VARYING FIELDS

Electrostatics / Time-VaryingMagnetostatics (Dynamic)

Electric & magnetic independent possibly coupledfields are...Maxwell’s Eqns.

∮SD · ds = Qencl

∮SD · ds = Qencl

(integral)∮CE · dl = 0

∮CE · dl = − ∂

∂t

∫SB · ds∮

SB · ds = 0

∮SB · ds = 0∮

CH · ds = Iencl

∮CH · ds = Iencl+

∫S∂∂tD · ds

Maxwell’s Eqns. ∇ ·D = ρv ∇ ·D = ρv(differential) ∇×E = 0 ∇×E = − ∂

∂tB∇ ·B = 0 ∇ ·B = 0∇×H = J ∇×H = J+ ∂

∂tD

Table 8.1: Comparison of principles governing static and time-varying electromagnetic fields. Differ-ences in the time-varying case relative to the static case are highlighted in blue.

Y. Qin (modified) CC BY 4.0

Figure 8.1: An experiment demonstrating elec-tromagnetic induction and Lenz’s Law.

We refer to Bimp as an impressed magnetic fieldbecause this field exists independently of anyresponse that may be induced by interactionwith the coil. Note that Bimp points to the leftinside the coil.

The experiment consists of three tests. We willfind in two of these tests that current flows (i.e.,|I| > 0) and subsequently there is an inducedmagnetic field Bind due to this current. It is thedirection of the current and subsequently thedirection of Bind inside the coil that we wish toobserve. The findings are summarized below andin Table 8.2.

• When the magnet is motionless, we have theunsurprising result that there is no currentin the coil, Therefore no magnetic field is

induced, and the total magnetic field issimply equal to Bimp.

• When the magnet moves toward the coil, weobserve current which is positive withrespect to the reference direction indicatedin Figure 8.1. This current creates aninduced magnetic field Bind which points tothe right, as predicted by magnetostaticconsiderations from the right-hand rule.Since Bimp points to the left, it appears thatthe induced current is opposing the increasein the magnitude of the total magnetic field.

• When the magnet moves away from the coil,we observe current which is negative withrespect to the reference direction indicatedin Figure 8.1. This current yields Bind

which points to the left. Since Bimp pointsto the left, it appears that the inducedcurrent is opposing the decrease in themagnitude of the total magnetic field.

The first conclusion one may draw from thisexperiment is that changes in the magnetic fieldcan induce current. This was the claim made inthe first paragraph of this section and is aconsequence of Faraday’s Law, which is tackledin detail in Section 8.3. The second conclusion –also associated with Faraday’s Law – is the pointof this section, and goes like this: The inducedmagnetic field – that is, the one due to the


8.2. ELECTROMAGNETIC INDUCTION 167

Magnet is ... |Bimp| in coil is ... Circuit Response Bind inside coilMotionless constant V = 0, I = 0 noneMoving toward coil increasing V > 0, I > 0 Pointing rightMoving away from coil decreasing V < 0, I < 0 Pointing left

Table 8.2: Results of the experiment associated with Figure 8.1.

current induced in the coil – always opposes thechange in the impressed magnetic field.Generalizing:

Lenz’s Law states that the current that isinduced by a change in an impressed mag-netic field creates an induced magnetic fieldthat opposes (acts to reduce the effect of) thechange in the total magnetic field.

When the magnet moves, three things happen:(1) A current is induced, (2) A magnetic field isinduced (which adds to the impressed magneticfield), and (3) the value of V becomes non-zero.Lenz’s Law does not address the followingquestion: Which of these are responding directlyto the change in the impressed magnetic field,and which of these are simply responding tochanges in the other quantities? Lenz’s Law mayleave you with the incorrect impression that it isI that is induced, and that Bind and V aresimply responding to this current. In truth, thequantity which is induced is actually V . This canbe verified in the above experiment by replacingthe resistor with a high-impedance voltmeter,which will indicate that V is changing eventhough no current is flowing. Current flow is aresponse to this potential. Nevertheless, we say Iis induced, even if indirectly through V .

So, if Lenz’s Law is not entirely forthright as anexplanation of the underlying physics, then whatis it good for? Lenz’s Law is often useful forquickly determining the direction of current flowin practical electromagnetic induction problemswithout resorting to the mathematics associatedwith Faraday’s Law. Here’s an example:

Example 8.1. Electromagnetic inductionthrough a transformer.

Figure 8.2 shows a rudimentary circuitconsisting of a battery and a switch on theleft, a voltmeter on the right, and atransformer linking the two. It is notnecessary to be familiar with transformersto follow this example; suffice it to say thatthe transformer considered here consists oftwo coils wound around a common toroidalcore, which serves to contain magnetic flux.In this way the flux generated by either coilis delivered to the other coil with negligibleloss.

The experiment begins with the switch onthe left in the open state. Thus there is nocurrent and no magnetic field apparent inthe coil on the left. The voltmeter reads 0 V.When the switch is closed, what happens?

Solution. Closing the switch creates acurrent in the coil on the left. Given theindicated polarity of the battery, thiscurrent flows counter-clockwise through thecircuit on the left, with current arriving inthe left coil through the bottom terminal.Given the indicated direction of the windingin the left coil, the impressed magnetic fieldBimp is oriented counter-clockwise throughthe toroidal core. The coil on the right“sees” Bimp increase from zero to somelarger value. Since the voltmeterpresumably has input high impedance, nocurrent flows. However if current were ableto flow, Lenz’s Law dictates that it would beinduced to flow in a counter-clockwisedirection around the circuit on the right,since the induced magnetic field Bind wouldthen be clockwise-directed so as to opposethe increase in Bimp. Therefore the


- - -

-

E. Bach CC0 1.0 (modified)

Figure 8.2: Electromagnetic induction through atransformer.

potential measured at the bottom of theright coil would be higher than the potentialat the top of the right coil. The figureindicates that the voltmeter measures thepotential at its right terminal relative to itsleft terminal, so the needle will deflect to theright. This deflection will be temporary,since the current provided by the batterybecomes constant at a new non-zero valueand Bind responds only to the change inBimp. The voltmeter reading will remain atzero for as long as the switch remains closedand the current remains steady.

Here are some follow up exercises to testyour understanding of what is going on: (1)Now open the switch. What happens? (2)Repeat the original experiment, but beforestarting swap the terminals on the battery.

Additional Reading:

• “Electromagnetic Induction” on Wikipedia.

• “Lenz’s Law” on Wikipedia.

8.3 Faraday’s Law

[m0055]

Faraday’s Law describes the generation ofelectric potential by a time-varying magneticflux. This is a form of electromagnetic induction,as discussed in Section 8.2.

VT

-

R

I

B

Figure 8.3: A single loop of wire in the presenceof an impressed magnetic field.

To begin, consider the scenario shown inFigure 8.3: A single loop of wire in the presenceof an impressed magnetic field B. For reasonsexplained later, we introduce a small gap anddefine VT to be the potential difference measuredacross the gap according to the sign conventionindicated. The resistance R may be any valuegreater than zero, including infinity; i.e., a literalgap.

As long as R is not infinite, we know from Lenz’sLaw (Section 8.2) to expect that a time-varyingmagnetic field will cause a current to flow in thewire. Lenz’s Law also tells us the direction inwhich the current will flow. However Lenz’s Lawdoes not tell us the magnitude of the current,and sidesteps some important physics that hasprofound implications for the analysis and designof electrical devices including generators andtransformers.

The more complete picture is given by Faraday’sLaw. In terms of the scenario of Figure 8.3,Faraday’s Law relates the potential VT inducedby the time variation of B. VT then gives rise tothe current identified in Lenz’s Law; themagnitude of this current is simply VT /R.Without further ado, here’s Faraday’s Law forthis single loop scenario:

VT = − ∂

∂tΦ (single loop) (8.1)

Here Φ is the magnetic flux (units of Wb)associated with any open surface S bounded by


https://en.wikipedia.org/wiki/Electromagnetic_induction

https://en.wikipedia.org/wiki/Lenz's_law

8.3. FARADAY’S LAW 169

the loop:

Φ =

∫

S

B · ds (8.2)

where B is magnetic flux density (units of T orWb/m2) and ds is the differential surface areavector.

To make headway with Faraday’s Law, one mustbe clear about the meanings of S and ds. If thewire loop in the present scenario lies in a plane,then a good choice for S is the simply the planararea bounded by the loop. However any surfacethat is bounded by the loop will work, includingnon-planar surfaces that extend above and/orbelow the plane of the loop: All that is requires isthat every magnetic field line that passes throughthe loop also passes through S. In other words itis only necessary that the curve C defining theedge of the open surface S correspond to theloop. Subsequently the magnitude of ds is thedifferential surface element (ds) and the directionof ds is a unit vector n which is perpendicular toeach point on S, so ds = nds.

This leaves just one issue remaining: Theorientation of n. This is sorted out in Figure 8.4.There are two possible ways for a vector to beperpendicular to a surface, and the directionchosen for n will affect the sign of VT . Thereforen must be somehow related to the polaritychosen for VT . Here’s the relationship: Let C bethe curve that begins at the “−” terminal of VTand follows the entire perimeter of the loop,ending at at the “+” terminal. Then n isdetermined by the following “right-hand rule”: npoints in the direction of the fingers of the righthand when the thumb of the right hand is alignedin the direction of C. It’s worth noting that thisconvention is precisely the convention used torelate n and C in Stokes’ Theorem (Section 4.9).

Now let us recap how Faraday’s Law is appliedto the single loop scenario of Figure 8.3:

1. Assign “+” and “−” terminals to the gapvoltage VT .

2. The orientation of n is determined by theright hand rule, taking the direction of C to

VT

-

RS

n

C

Figure 8.4: Relationship between the polarityof VT and orientations of C and n in the planarsingle-loop scenario.

be the perimeter of the loop beginning at“−” and ending at “+”

3. B yields a magnetic flux Φ associated withthe loop according to Equation 8.2. S is anyopen surface which intersects all themagnetic field lines which pass through theloop (so you might as well choose S in a waythat results in the simplest possibleintegration).

4. VT is the time derivative of Φ with a changeof sign (Equation 8.1).

5. The current I flowing in the loop is VT /R,with the reference direction (i.e., direction ofpositive current) being from “+” to “−”through the resistor. Think of the loop as avoltage source, and you’ll get the correctreference direction for I.

At this point let us reiterate that theelectromagnetic induction described by Faraday’sLaw induces potential (in this case, VT ), and notcurrent. The current in the loop is simply theinduced voltage divided by the resistance of theloop. This point is easily lost, especially in lightof Lenz’s Law, which seems to imply thatcurrent, as opposed to potential, is the thingbeing induced.

Wondering about the significance of the minussign in Equation 8.1? That is specifically Lenz’sLaw: The current I that ends up circulating inthe loop generates its own magnetic field (“Bind”


in Section 8.2) which is distinct from theimpressed magnetic field B and which tends tooppose change in B. Thus we see that Faraday’sLaw subsumes Lenz’s Law.

Frequently one is interested in a structure whichconsists of multiple identical loops. We have inmind here something like a coil with N ≥ 1windings tightly packed together. In this caseFaraday’s Law is

VT = −N ∂

∂tΦ (8.3)

Note that the difference is simply that the gappotential VT is greater by N . In other words,each winding of the coil contributes a potentialgiven by Equation 8.1, and these potentials addin series.

Faraday’s Law, given in general by Equa-tion 8.3, states that the potential induced ina coil is proportional to the time derivative ofthe magnetic flux through the coil.

The induced potential VT is often referred to as“emf”, which is a contraction of the termelectromotive force – a misnomer to be sure, sinceno actual force is implied; only potential. Theterm “emf” is nevertheless frequently used in thecontext of Faraday’s Law applications forhistorical reasons. Above we considered thegeneration of emf by time variation of B.However Equation 8.1 indicates that whatactually happens is that the emf is the result oftime variation of the magnetic flux, Φ. Magneticflux is magnetic flux density integrated over area,so it appears that emf can also be generatedsimply by varying S, independently of any timevariation of B. In other words, emf may begenerated even when B is constant, by insteadvarying the shape or orientation of the coil. So,it appears we have a variety of schemes by whichwe can generate emf. Here they are:

1. Time-varying B (as we considered above).For example, B might be due to apermanent magnet which is moved (e.g.,translated or rotated) in the vicinity of the

coil, as described in Section 8.2. Or, morecommonly, B might be due to a different coilthat bears a time-varying current. Thesemechanisms are collectively referred to astransformer emf, and is the underlyingprinciple of operation of transformers(Section 8.5).

2. The perimeter C – and thus the surface Sover which Φ is determined – can betime-varying. For example, a wire loopmight be rotated or changed in shape in thepresence of a constant magnetic field. Thismechanism is referred to as motional emf,and is the underlying principle of operationof generators (Section 8.7).

3. Transformer and motional emf can exist invarious combinations. From the perspectiveof Faraday’s Law, transformer and motionalemf are the same in the sense in either caseΦ is time-varying, which is all that isrequired to generate emf.

Finally, a comment on the generality of Faraday’sLaw. Above we have introduced Faraday’s Lawas if it were specific to loops and coils of wire.However the truth of the matter is that Faraday’sLaw is fundamental physics: If you can define aclosed path – current-bearing or not – then youcan compute the potential difference achieved bytraversing that path using Faraday’s Law. Thevalue you compute is the potential associatedwith electromagnetic induction, and existsindependently and in addition to the potentialdifference associated with the static electric field(e.g., Section 5.12). In other words:

Faraday’s Law indicates the contribution ofelectromagnetic induction (the generation ofemf by a time-varying magnetic flux) to thepotential difference achieved by traversing aclosed path.

In Section 8.8 this insight is used to transformthe static form of Kirchoff’s Voltage Law(Section 5.10) – which gives the potentialdifference associated with electric field only –into the Maxwell-Faraday Equation (Section 8.8),

8.4. INDUCTION IN A MOTIONLESS LOOP 171

which is a general statement about therelationship between the instantaneous value ofelectric fields and the time derivative of magneticfields.

Additional Reading:

• “Faraday’s law of induction” on Wikipedia.

8.4 Induction in aMotionless Loop

[m0056]

In this section we consider the problem depictedin Figure 8.5: A single motionless loop of wire inthe presence of a spatially-uniform buttime-varying magnetic field. A small gap isintroduced in the loop, allowing us to measurethe induced potential VT . Additionally aresistance R is connected across VT in order toallow a current to flow. This problem wasconsidered in Section 8.3 as an introduction toFaraday’s Law; in this section we shall actuallywork the problem and calculate some values.This is intended to serve as an example of theapplication of Faraday’s Law, a demonstration oftransformer emf, and will serve as a first steptoward an understanding of transformers asdevices.

In the present problem, the loop is centered inthe z = 0 plane. The magnetic flux density isB = bB(t); i.e., time-varying magnitude B(t)

with respect to a constant direction b. Becausethis magnetic field is spatially uniform (i.e., thesame everywhere), we will find that only the areaof the loop is important, and not it’s specificshape. For this reason, it will not be necessary tospecify the radius of the loop or even require thatit be a circular loop. Our task is to findexpressions for the induced potential VT and theresponse current I.

To begin, remember that Faraday’s Law is acalculation of electric potential, and not current.So, the approach is to first find VT , and then find

VT

-

R

I

B

Figure 8.5: A single loop of wire in the presence ofan impressed spatially-uniform but time-varyingmagnetic field.

the current I that flows through the gapresistance in response.

The sign convention for VT is arbitrary; here wehave selected “+” and “−” terminals asindicated in Figure 8.5.1 Following the standardconvention for the reference direction of currentthrough a passive device, I should be directed asshown in Figure 8.5. It is worth repeating thatthese conventions for the signs of VT and I aremerely references ; for example we may well findthat I is negative, which means that currentflows in a clockwise direction in the loop.

We now invoke Faraday’s Law:

VT = −N ∂

∂tΦ (8.4)

The number of windings N in the loop is 1, andΦ is the magnetic flux through the loop. Thus:

VT = − ∂

∂t

∫

S

B · ds (8.5)

where S is any open surface that intersects all ofthe magnetic field lines that pass through theloop. The simplest such surface is simply theplanar surface defined by the perimeter of theloop. Then ds = nds, where ds is the differentialsurface element and n is the normal to the plane

1A good exercise for the student is to repeat this prob-lem with the terminal polarity reversed; one should obtainthe same answer.

https://en.wikipedia.org/wiki/Faraday's_law_of_induction


of the loop. Which of the two possible normals tothe loop? This is determined by the right-handrule of Stokes’ Theorem: From the “−” terminal,we point the thumb of the right hand in thedirection that leads to the “+” terminal bytraversing the perimeter of the loop. When wedo this, we find the fingers of the right handintersect S in the manner as the n vector. Tomaintain the generality of results derived below,we shall not make the substitution n = +z;nevertheless we see this is the case for a loopparallel to the z = 0 plane with the polarity ofVT indicated in Figure 8.5.

Taking this all into account, we have

VT = − ∂

∂t

∫

S

(bB(t)

)· (nds)

= −(b · n

) ∂

∂t

∫

S

B(t) ds

= −(b · n

)( ∂

∂tB(t)

)∫

S

ds (8.6)

The integral in the above expression is simplythe area of the loop, which is a constant; let thesymbol A represent this area. We obtain

VT = −(b · nA

) ∂

∂tB(t) (8.7)

which is the expression we seek. Note that thequantity b · nA is the projected area of the loop.The projected area is equal to A when the themagnetic field lines are perpendicular to the loop(i.e., b = n), and decreases to zero as b · n → 0.Summarizing:

The magnitude of the transformer emf in-duced by a spatially-uniform magnetic fieldis equal to the projected area times the timederivative of the magnetic flux density, witha change of sign. (Equation 8.7).

A few observations about this result:

• As promised earlier, we have found that theshape of the loop is irrelevant; i.e., a squareloop having the same area and planarorientation would result in the same VT .

This is because the magnetic field isspatially uniform, and because it is themagnetic flux (Φ) and not the magnetic fieldor shape of the loop alone that determinesthe induced potential.

• The induced potential is proportional to A;i.e., VT can be increased by increasing thearea of the loop.

• The peak magnitude of the induced potentialis maximized when the plane of the loop isperpendicular to the magnetic field lines.

• The induced potential goes to zero when theplane of the loop is parallel to the magneticfield lines. Said another way: There is noinduction unless magnetic field lines passthrough the loop.

• The induced potential is proportional to therate of change of B. If B is constant in time,then there is no induction.

Finally, the current in the loop is simply

I =VTR

(8.8)

Again, electromagnetic induction inducespotential, and the current flows only in responseto the induced potential as determined by Ohm’sLaw. In particular: If the resistor is removed,then R→ ∞ and I → 0, but VT is unchanged.

One final comment: Even though the current I isnot a direct result of electromagnetic induction,we can use I as a check of the result using Lenz’sLaw (Section 8.2). We’ll demonstrate this in theexample below.

Example 8.2. Induction in a motionlesscircular loop by a linearly-increasingmagnetic field

Let the loop be planar in the z = 0 plane,and circular with radius a = 10 cm. Let the

8.4. INDUCTION IN A MOTIONLESS LOOP 173

magnetic field be zB(t) where

B(t) = 0 , t < 0

= B0t/t0 , 0 ≤ t ≤ t0

= B0 , t > t0 (8.9)

i.e., B(t) begins at zero and increaseslinearly to B0 at time t0, after which itremains constant at B0. Let B0 = 0.2 T,t0 = 3 s, and let the loop be closed by aresistor R = 1 kΩ. What current I flows inthe loop?

Solution. Adopting the sign conventions ofFigure 8.5 we first note that n = +z; this isdetermined by the right-hand rule withrespect to the indicated polarity of VT .Thus Equation 8.7 becomes

VT = −(b · zA

) ∂

∂tB(t)

Note b · zA = A since b = z; i.e., becausethe plane of the loop is perpendicular to themagnetic field lines. Since the loop iscircular, A = πa2. Also

∂

∂tB(t) = 0 , t < 0

= B0/t0 , 0 ≤ t ≤ t0

= 0 , t > t0

Putting this all together:

VT = −πa2B0

t0= −2.09 mV , 0 ≤ t ≤ t0

and VT = 0 before and after this time, sinceB is constant during those times.Subsequently the induced current is

I =VTR

= −2.09 µA , 0 ≤ t ≤ t0

and I = 0 before and after this time. Wehave found that the induced current is aconstant clockwise flow that exists onlywhile B is increasing.

Finally, let’s see if the result is consistentwith Lenz’s Law. The current induced while

B is changing gives rise to an inducedmagnetic field Bind. From the “right handrule” that relates the direction of I to thedirection of Bind (Section 7.5), the directionof Bind is −z inside the loop. In otherwords: The magnetic field associated withthe induced current opposes the increasingimpressed magnetic field that induced thecurrent, in accordance with Lenz’s Law.

Example 8.3. Induction in a motionlesscircular loop by a sinusoidally-varyingmagnetic field.

Let us repeat the previous example, but nowwith

B(t) = B0 sin 2πf0t

with f0 = 1 kHz.

Solution. Now

∂

∂tB(t) = 2πf0B0 cos 2πf0t

SoVT = −2π2f0a

2B0 cos 2πf0t

Subsequently

I =VTR

= −2π2f0a2B0

Rcos 2πf0t

Substituting values, we have:

I = −(395 mA) cos [(6.28 krad/s)t]

It should be no surprise that VT and I varysinusoidally, since the source (B) variessinusoidally. A bit of useful trivia here isthat VT and I are 90 out of phase with thesource. It is also worth noting what happenswhen B = 0. This occurs twice per period,at t = nπ/2 where n is any integer,including t = 0. At these times B is zero,but VT and hence IR are decidedly non-zero;in fact, they are at their maximummagnitude. Again, it is the change in B thatinduces voltage and subsequently current,not B itself.


8.5 Transformers: Principleof Operation

[m0031]

A transformer is a device which connects twoelectrical circuits through a shared magneticfield. Transformers are used in impedancetransformation, voltage level conversion, circuitisolation, conversion between single-ended anddifferential signal modes, and other applications.2

The underlying electromagnetic principle isFaraday’s Law (Section 8.3); in particular,transformer emf.

The essential features of a transformer can bederived from the simple experiment shown inFigures 8.6 and 8.7. In this experiment, two coilsare arranged along a common axis. The windingpitch is small, so that all magnetic field lines passthrough the length of the coil, and no lines passbetween the windings. To further contain themagnetic field, we assume both coils are woundon the same core, consisting of some materialexhibiting high permeability. The upper coil hasN1 turns and the lower coil has N2 turns.

In Part I of this experiment (Figure 8.6), theupper coil is connected to a sinusoidally-varying

voltage source V(1)1 in which the subscript refers

to the coil and the superscript refers to “Part I”of this experiment. The voltage source creates acurrent in the coil, which in turn creates atime-varying magnetic field B in the core.Applying the right-hand rule that relates currentto induced magnetic field, the reference directionfor B is in the downward (−z) direction.

The lower coil has N2 turns wound in theopposite direction and is left open-circuited.Given the closely-spaced windings and use of ahigh-permeability core, we assume that themagnetic field within the lower coil is equal to Bgenerated in the upper coil. The potential

induced in the lower coil is V(1)2 with reference

polarity indicated in the figure. From Faraday’s

2See “Additional Reading” at the end of this section formore on these applications.

turns ¡(1)

+

¢

2(1)

+

¢

N1

turnsN2

Figure 8.6: Part I of an experiment demonstratingthe linking of electric circuits using a transformer.

turns£ 1(2)

+

¤

£ 2(2)

+

¤

N1

turnsN2

Figure 8.7: Part II of an experiment demonstrat-ing the linking of electric circuits using a trans-former.

8.5. TRANSFORMERS: PRINCIPLE OF OPERATION 175

Law we have

V(1)2 = −N2

∂

∂tΦ2 (8.10)

where Φ2 is the flux through a single turn of thelower coil. Thus:

V(1)2 = −N2

∂

∂t

∫

S

B · (−zds) (8.11)

Note that the direction of ds = −zds isdetermined by the polarity we have chosen for

V(1)2 .

In Part II of the experiment (Figure 8.7), wemake changes as follows: We apply a voltage

V(2)2 to the lower coil and open-circuit the upper

coil. Further, we adjust V(2)2 such that the

induced magnetic flux density is again B – thatis, equal to the field in Part I of the experiment.Now we have

V(2)1 = −N1

∂

∂tΦ1 (8.12)

which is

V(2)1 = −N1

∂

∂t

∫

S

B · (+zds) (8.13)

For reasons that will become apparent in amoment, let’s shift the leading minus sign intothe integral. We then have

V(2)1 = +N1

∂

∂t

∫

S

B · (−zds) (8.14)

Comparing this to Equations 8.10 and 8.11, wesee that we may rewrite this in terms of the fluxin the lower coil in Part I of the experiment:

V(2)1 = +N1

∂

∂tΦ2 (8.15)

In fact, we can express this in terms of thepotential in Part I of the experiment:

V(2)1 =

(−N1

N2

)(−N2

∂

∂tΦ2

)

=

(−N1

N2

)V

(1)2 (8.16)

We have found that the potential in the uppercoil in Part II is related in a simple way to the

potential in the lower coil in Part I of theexperiment. If we had done Part II first, wewould have obtained the same result but withthe superscripts swapped. Therefore it must begenerally true – regardless of the arrangement ofterminations – that

V1 = −N1

N2V2 (8.17)

This expression should be familiar fromelementary circuit theory – except possibly forthe minus sign. So what’s up that? The minussign is a consequence of the fact that the coils arewound in opposite directions. We can make theabove expression a little more general as follows:

V1V2

= pN1

N2(8.18)

where p is defined to +1 when the coils arewound in the same direction, and −1 when coilsare wound in opposite directions. (It is anexcellent exercise to confirm that this is true byrepeating the above analysis with windingdirection changed for either the upper or lowercoil, for which p will then turn out to be +1.)This is the “transformer law” of basic electriccircuit theory, from which all othercharacteristics of transformers as two-port circuitdevices can be obtained (See Section 8.6 forfollow-up on that). Summarizing:

The ratio of coil voltages in an ideal trans-former is equal to the ratio of turns with signdetermined by the relative directions of thewindings, per Equation 8.18.

A more familiar transformer design is shown inFigure 8.8: Coils wound on a toroidal core asopposed to a cylindrical core. Why do this? Thisarrangement confines the magnetic field linkingthe two coils to the core, as opposed to allowingfield lines to extend beyond the device. Thisconfinement is important in order to preventstructures outside the transformer frominterfering with the magnetic field linking thecoils. The principle of operation is in all otherrespects the same.

Additional Reading:


V1

V2

N1turns

N2turns

Φ

MagneticFlux,

TransformerCore

+

+

−

−

c© BillC (modified) CC BY SA 3.0

Figure 8.8: Transformer implemented as coilssharing a toroidal core. Here p = +1.

• “Transformer” on Wikipedia.

• “Balun” on Wikipedia.

8.6 Transformers asTwo-Port Devices

[m0032]

Section 8.5 explains the principle of operation ofthe ideal transformer. The relationship governingthe terminal voltages V1 and V2 was found to be

V1V2

= pN1

N2(8.19)

where N1 and N2 are the number of turns in theassociated coils and p is either +1 or −1depending on the relative orientation of thewindings; i.e., whether the reference direction ofthe associated fluxes is the same or opposite,respectively.

We shall now consider ratios of current andimpedance in ideal transformers, using thetwo-port model shown in Figure 8.9. By virtue ofthe reference current directions and polarities

¥2

¦2

+

§

¥1

¦1

+

§

+_ ¨2

¨1

Figure 8.9: The transformer as a two-port circuitdevice.

chosen in this figure, the power delivered by thesource V1 is V1I1, and the power absorbed by theload Z2 is −V2I2. Assuming the transformerwindings have no resistance, and assuming themagnetic flux is perfectly contained within thecore, the power absorbed by the load must equalthe power provided by the source; i.e.,V1I1 = −V2I2. Thus we have3

I1I2

= −V2V1

= −pN2

N1(8.20)

We can develop an impedance relationship forideal transformers as follows: Let Z1 be the theinput impedance of the transformer; that is, theimpedance looking into port 1 from the source.Thus we have

Z1 =V1I1

=+p (N1/N2)V2−p (N2/N1) I2

= −(N1

N2

)2(V2I2

)(8.21)

In Figure 8.9 Z2 is the the output impedance ofport 2; that is, the impedance looking out port 2into the load. Therefore Z2 = −V2/I2 (onceagain the minus sign is a result of our choice of

3 The minus signs in this equation are a result of thereference polarity and directions indicated in Figure 8.9.These are more-or-less standard in electrical two-port the-ory (see “Additional Reading” at the end of this section),but are certainly not the only reasonable choice. If yousee these expressions without the minus signs it proba-bly means that a different combination of reference direc-tions/polarities is in effect.


https://en.wikipedia.org/wiki/Transformer

https://en.wikipedia.org/wiki/Balun

8.6. TRANSFORMERS AS TWO-PORT DEVICES 177

sign conventions in Figure 8.9). Substitution ofthis result into the above equation yields

Z1 =

(N1

N2

)2

Z2 (8.22)

and therefore

Z1

Z2=

(N1

N2

)2

(8.23)

Thus we have demonstrated that a transformerscales impedance in proportion to the square ofthe turns ratio N1/N2. Remarkably, theimpedance transformation depends only on theturns ratio, and is independent of the relativedirection of the windings (p).

The relationships developed above should beviewed AC expressions, and are not normallyvalid at DC. This is because transformers exhibita fundamental limitation in their low-frequencyperformance. To see this, first recall Faraday’sLaw:

V = −N ∂

∂tΦ (8.24)

If the magnetic flux Φ isn’t time-varying, thenthere is no induced electric potential, andsubsequently no linking of the signals associatedwith the coils. At very low but non-zerofrequencies, we encounter another problem thatgets in the way: Magnetic saturation. To seethis, note we can obtain an expression for Φ fromFaraday’s Law by integrating with respect totime, yielding

Φ(t) = − 1

N

∫ t

t0

V (τ)dτ +Φ(t0) (8.25)

where V (t) is sinusoidally-varying and t0 is someearlier time at which we know the value of Φ(t0).Note the peak value of Φ after t = t0 depends onthe frequency of V (t). If the frequency of V (t) isvery low, then Φ can become very large. Sincethe associated cross-sectional areas arepresumably constant, this means that themagnetic field becomes very large. The problemwith that is that most practicalhigh-permeability materials suitable for use astransformer cores exhibit magnetic saturation;that is, the rate at which the magnetic field is

DifferentialSingle-Ended

Single-Ended

c© SpinningSpark CC BY SA 3.0 (modified)

Figure 8.10: Transformers used to convert asingle-ended (“unbalanced”) signal to a differen-tial (balanced) signal, and back.

able to increase slows with increasing magneticfield magnitude. The result of all this is that atransformer may work fine at (say) 1 MHz, butat (say) 1 Hz the transformer may exhibit anapparent loss associated with this saturation.Thus, practical transformers exhibit highpassfrequency response.

It should be noted that the highpass behavior ofpractical transistors can be useful: For example,a transformer can be used to isolate an undesiredDC offset and/or low-frequency noise in thecircuit attached to one coil from the circuitattached to the other coil.

The DC-isolating behavior of a transformer alsoallows the transformer to be used as a balun, asillustrated in Figure 8.10. A balun is a two-portdevice which transforms a single-ended(“unbalanced”) signal – that is, one having anexplicit connection to a datum (e.g., ground) –into a differential (“balanced”) signal, for whichthere is no explicit connection to a datum.Differential signals have many benefits in circuitdesign, whereas inputs and outputs to devicesmust often be in single-ended form. Thus acommon use of transformers is to effect theconversion between single-ended and differentialcircuits. Although a transformer is certainly notthe only device that can be used as a balun, ithas one very big advantage, namely bandwidth.

Additional Reading:

• “Transformer” on Wikipedia.

• “Two-port network” on Wikipedia.


https://en.wikipedia.org/wiki/Transformer

https://en.wikipedia.org/wiki/Two-port_network


• “Saturation (magnetic)” on Wikipedia.

• “Balun” on Wikipedia.

• S.W. Ellingson, “Differential Circuits”(Sec. 8.7) in Radio Systems Engineering,Cambridge Univ. Press, 2016.

8.7 The Electric Generator

[m0030]

A generator is a device which transformsmechanical energy into electrical energy, typicallyby electromagnetic induction via Faraday’s Law(Section 8.3). For example a generator mightconsist of a gasoline engine which turns acrankshaft, to which is attached a system of coilsand/or magnets. This rotation changes therelative orientations of the coils with respect tothe magnetic field in a time-varying manner,resulting in a time-varying magnetic flux andsubsequently induced electric potential. In thiscase the induced potential is said to be a form ofmotional emf, as it is due entirely to changes ingeometry as opposed to changes in themagnitude of the magnetic field. Coal- andsteam-fired generators, hydroelectric generators,wind turbines, and other energy generationdevices operate using essentially this principle.

Figure 8.11 shows a rudimentary generator whichserves as to illustrate the relevant points. Thegenerator consists of a planar loop which rotatesaround the z axis; therefore the rotation can beparameterized in φ. In this case the direction ofrotation is specified to be in the +φ direction.The frequency of rotation is f0; that is, the timerequired for the loop to make one completerevolution is 1/f0. We assume a time-invariantand spatially-uniform magnetic flux densityB = bB0 where b and B0 are both constants.For illustration purposes, the loop is indicated tobe circular. However because the magnetic fieldis time-invariant and spatially-uniform, thespecific shape of the loop is not important, as weshall see in a moment: Only the area of the loopwill matter.

z

y

x

VT-

©ª«ection ofrotation

Figure 8.11: A rudimentary single-loop genera-tor, shown at time t = 0.

The induced potential is indicated as VT withreference polarity as indicated in the figure. Thispotential is given by Faraday’s Law:

VT = − ∂

∂tΦ (8.26)

Here Φ is the magnetic flux associated with anopen surface S bounded by the loop:

Φ =

∫

S

B · ds (8.27)

As usual S can be any surface which intersects allmagnetic field lines passing through the loop; andalso as usual the simplest choice is simply theplanar area bounded by the loop. The differentialsurface element ds is nds, where n is determinedby the reference polarity of VT according to the“right hand rule” convention from Stokes’Theorem. Making substitutions, we have

Φ =

∫

S

(bB0

)· (nds)

=[b · n

]B0

∫

S

ds

=[b · n

]B0A (8.28)

https://en.wikipedia.org/wiki/Saturation_(magnetic)

https://en.wikipedia.org/wiki/Balun

8.7. THE ELECTRIC GENERATOR 179

where A is the area of the loop.

To make headway, an expression for n is needed.The principal difficulty here is that n is rotatingwith the loop, and so is time-varying. To sortthis out, first consider the situation at time t = 0,which is the moment illustrated in Figure 8.11.Beginning at the “−” terminal, we point thethumb of our right hand in the direction thatleads to the “+” terminal by traversing the loop;n is then the direction perpendicular to the planeof the loop in which the fingers of our right handpass through the loop. We see that at t = 0,n = +y. A bit later, the loop will have rotatedby one-fourth of a complete rotation, and at thattime n = −x. This occurs at t = 1/(4f0). Laterstill, the loop will have rotated by one-half of acompete rotation, and then n = −y. This occursat t = 1/(2f0). It is apparent that

n(t) = −x sin 2πf0t+ y cos 2πf0t (8.29)

as can be confirmed by checking the three specialcases identified above. Now applying Faraday’sLaw, we find

VT = − ∂

∂tΦ

= − ∂

∂t

[b · n(t)

]B0A

= −[b · ∂

∂tn(t)

]B0A (8.30)

For notational convenience we make the followingdefinition:

n′(t) , − 1

2πf0

∂

∂tn(t) (8.31)

which is simply the time derivative of n dividedby 2πf0 so as to retain a unit vector. The reasonfor including a change of sign will becomeapparent in a moment. Applying this definition,we find

n′(t) = +x cos 2πf0t+ y sin 2πf0t (8.32)

Note that this is essentially the definition of theradial basis vector ρ from the cylindricalcoordinate system (which is why we applied theminus sign in Equation 8.31). Recognizing this,we write

ρ(t) = +x cos 2πf0t+ y sin 2πf0t (8.33)

and finally

VT = +2πf0B0Ab · ρ(t) (8.34)

If the purpose of this device is to generate power,then presumably we would choose the magneticfield to be in a direction that maximizes themaximum value of b · ρ(t). Therefore, power isoptimized for B polarized entirely in somecombination of x and y, and with B · z = 0.Under that constraint, we see that VT variessinusoidally with frequency f0 and exhibits peakmagnitude

max |VT (t)| = 2πf0B0A (8.35)

It’s worth noting that the maximum voltagemagnitude is achieved when the plane of the loopis parallel to B; i.e., when b · n(t) = 0 so thatΦ(t) = 0. Why is that? Because this is whenΦ(t) is most rapidly increasing or decreasing.Conversely: When the plane of the loop isperpendicular to B (b · n(t) = 1), |Φ(t)| ismaximum but its time-derivative is zero, soVT (t) = 0 at this instant.

Example 8.4. Rudimentary electricgenerator.

The generator in Figure 8.11 consists of acircular loop of radius a = 1 cm rotating at1000 revolutions per second in a static andspatially-uniform magnetic flux density of1 mT in the +x direction. What is theinduced potential?

Solution. From the problem statement,f0 = 1 kHz, B0 = 1 mT, and b = +x.Therefore b · ρ(t) = x · ρ(t) = cos 2πf0t. Thearea of the loop is A = πa2. FromEquation 8.34 we obtain

VT (t) = (2.00 mV) cos [(6.28 krad/s) t]

Finally, we note that it is not really necessary forthe loop to rotate in the presence of a magneticfield with constant b; it works equally well forthe loop to be stationary and for b to rotate – in


fact this is essentially the same problem. In somepractical generators, both thepotential-generating coils and fields (generatedby some combination of magnets and coils)rotate.

Additional Reading:

• “Electric Generator” on Wikipedia.

8.8 The Maxwell-FaradayEquation

[m0050]

In this section we generalize Kirchoff’s VoltageLaw (KVL), previously encountered as aprinciple of electrostatics in Section 5.10. KVLstates that in the absence of a time-varyingmagnetic flux, the electric potential accumulatedby traversing a closed path C is zero. Here is thatidea in mathematic form:

V =

∮

C

E · dl = 0 (8.36)

Now recall Faraday’s Law (Section 8.3):

V = − ∂

∂tΦ = − ∂

∂t

∫

S

B · ds (8.37)

Here S is any open surface that intersects allmagnetic field lines passing through C, with therelative orientations of C and ds determined inthe usual way by the Stokes’ Theorem convention(Section 4.9). Note that Faraday’s Law agreeswith KVL in the electrostatic case: If magneticflux is constant, then Faraday’s Law says V = 0.However Faraday’s Law is very clearly notconsistent with KVL if magnetic flux istime-varying. The correction is simple enough:We can simply set these expressions to be equal.Here we go:

∮

C

E · dl = − ∂

∂t

∫

S

B · ds (8.38)

This general form is known by a variety ofnames; here we refer to it as theMaxwell-Faraday Equation (MFE).

The integral form of the Maxwell-FaradayEquation (Equation 8.38) states that the elec-tric potential associated with a closed path Cis due entirely to electromagnetic induction,via Faraday’s Law.

Despite the great significance of this expressionas one of the four Maxwell’s Equations, onemight argue that all we have done is simply towrite Faraday’s Law in a slightly more verboseway. This is true. The real power of the MFE isunleashed when it is expressed in differential, asopposed to integral form. Let us now do this.

We can transform the left-hand side ofEquation 8.38 into a integral over S using Stokes’Theorem. Applying Stokes’ theorem on the left,we obtain

∫

S

(∇×E) · ds = − ∂

∂t

∫

S

B · ds (8.39)

Now exchanging the order of integration anddifferentiation on the right hand side:

∫

S

(∇×E) · ds =∫

S

(− ∂

∂tB

)· ds (8.40)

The surface S on both sides is the same, and wehave not constrained S in any way: S can be anymathematically-valid open surface anywhere inspace, having any size and any orientation. Theonly way the above expression can be universallytrue under these conditions is if the integrandson each side are equal at every point in space.Therefore:

∇×E = − ∂

∂tB (8.41)

which is the MFE in differential form.

What does this mean? Recall that the curl of Eis a way to take a directive of E with respect toposition. Therefore the MFE constrains spatialderivatives of E to be simply related to the rateof change of B. Said plainly:

The differential form of the Maxwell-FaradayEquation (Equation 8.41) relates the changein the electric field with position to thechange in the magnetic field with time.

https://en.wikipedia.org/wiki/Electric_generator

8.9. DISPLACEMENT CURRENT AND AMPERE’S LAW 181

Now that is arguably new and usefulinformation: We now see that electric andmagnetic fields are coupled not only for lineintegrals and fluxes, but also at each point inspace.

Additional Reading:

• “Faraday’s Law of Induction” on Wikipedia.


8.9 Displacement Currentand Ampere’s Law

[m0053]

In this section we generalize Ampere’s Law,previously encountered as a principle ofmagnetostatics in Section 7.4. Ampere’s Lawstates that the conduction current Iencl flowingthrough closed path C is equal to the line integralof the magnetic field intensity H along C. Here isthat idea in mathematic form:

∮

C

H · dl = Iencl (8.42)

We shall now demonstrate that this equation isunreliable if the current is not steady; i.e., notDC.

First, consider the situation shown in Figure 8.12.Here a conduction current I flows in the wire,subsequently generating a magnetic field H thatcirculates around the wire (Section 7.5). Whenwe perform the integration in Ampere’s Lawalong any path C enclosing the wire, the result isI, as expected. In this case Ampere’s Law isworking even when I is time-varying.

Now consider the situation shown in Figure 8.13,in which we have introduced a parallel-platecapacitor. In the DC case, this situation issimple: No current flows, so there is no magneticfield and Ampere’s Law is trivially true. In theAC case, the conduction current I can benon-zero, but we must be clear about the

I

I

S1

HC

C. Burks (modified)

Figure 8.12: Ampere’s Law applied to a continu-ous line of conduction current.

physical origin of this current: What ishappening is that for one half of a period, asource elsewhere in the circuit is moving positivecharge to one side of the capacitor and negativecharge to the other side. For the otherhalf-period, the source is exchanging the chargeso that negative charge appears on the previouslypositively-charge side, and vice-versa. Note thatat no point is conduction current flowing directlyfrom one side of the capacitor to the other;instead all conduction current must flow throughthe circuit in order to arrive at the other plate.Even though there is no conduction currentbetween the plates, there is conduction current inthe wire, and therefore there is also a magneticfield associated with that current.

Now we are ready to shine a light on theproblem. Recall that from Stokes’ Theorem(Section 4.9), the line integral over C ismathematically equivalent to an integral overany open surface S that is bounded by C. Twosuch surfaces are shown in Figure 8.12 andFigure 8.13, indicated as S1 and S2. In thewire-only scenario of Figure 8.12, the choice of S

https://en.wikipedia.org/wiki/Faraday's_law_of_induction



E

I

I

S1

HC

C. Burks (modified)

Figure 8.13: Ampere’s Law applied to a parallelplate capacitor.

clearly doesn’t matter: Any valid surfaceintersects conduction current equal to I.Similarly in the scenario of Figure 8.13everything seems fine if we choose S = S1. If onthe other hand we select S2 in the parallel-platecapacitor case, then we have a problem: There isno conduction current flowing through S2, so theright side of Equation 8.42 is zero even thoughthe left side is potentially non-zero. So it appearsthat something necessary for the time-varyingcase is missing from Equation 8.42.

To resolve the problem, we postulate anadditional term in Ampere’s Law that is non-zeroin the above scenario. Specifically, we propose:

∮

C

H · dl = Ic + Id (8.43)

where Ic is the enclosed conduction current(formerly identified as Iencl) and Id is theproposed new term. If we are to accept thispostulate, then here is a list of things we knowabout Id:

• Id has units of current (A).

• Id = 0 in the DC case and is potentiallynon-zero in the AC case. This implies thatId is the time derivative of some otherquantity.

• Id must be somehow related to the electricfield.

How do we know Id must be related to theelectric field? This is because theMaxwell-Faraday Equation (Section 8.8) tells usthat spatial derivatives of E are related to timederivatives of H; i.e., E and H are coupled in thetime-varying (here, AC) case. This couplingbetween E and H must also be at work here, butwe have not yet seen E play a role. This is prettystrong evidence that Id depends on the electricfield.

Without further ado, here’s Id:

Id =

∫

S

∂D

∂t· ds (8.44)

where D is the electric flux density (units ofC/m2) and which is equal to ǫE as usual, and Sis the same open surface associated with C inAmpere’s Law. Note that this expression meetsour expectations: It is determined by the electricfield, it is zero when the electric field is constant(i.e., not time varying), and has units of current.

The quantity Id is commonly known asdisplacement current. It should be noted thatthis name is a bit misleading, since Id is not acurrent in the conventional sense. Certainly it isnot a conduction current: Conduction current isrepresented by Ic, and there is no currentconducted through an ideal capacitor. It is notunreasonable to think of Id as current in a moregeneral sense, for the following reason: At oneinstant charge is distributed one way, at anotherit is distributed in another way, so if you definecurrent as a time variation in the chargedistribution relative to S – regardless of the pathtaken by the charge – then Id is a current.However this distinction is a bit philosophical, soit may be less confusing to interpret“displacement current” instead as a separateelectromagnetic quantity that just happens tohave units of current.

8.9. DISPLACEMENT CURRENT AND AMPERE’S LAW 183

Now we are able to write the general form ofAmpere’s Law that applies even when sourcesare time-varying. Here it is:

∮

C

H · dl = Ic +

∫

S

∂D

∂t· ds (8.45)

As is the case in the Maxwell-Faraday Equation,most of the utility of Ampere’s Law is unleashedwhen expressed in differential form. To obtainthis form the first step is to write Ic as anintegral of over S; this is simply (see Section 6.2):

Ic =

∫

S

J · ds (8.46)

where J is the volume current density (units ofA/m2). So now we have

∮

C

H · dl =∫

S

J · ds+∫

S

∂D

∂t· ds

=

∫

S

(J+

∂D

∂t

)· ds (8.47)

We can transform the left side of the aboveequation into a integral over S using Stokes’Theorem (Section 4.9). We obtain

∫

S

(∇×H) · ds =∫

S

(J+

∂D

∂t

)· ds (8.48)

The surface S on both sides is the same, and wehave not constrained S in any way: S can be anymathematically-valid open surface anywhere inspace, having any size and any orientation. Theonly way the above expression can be universallytrue under these conditions is if the integrandson each side are equal at every point in space.Therefore:

∇×H = J+∂

∂tD (8.49)

which is Ampere’s Law in differential form.

What does Equation 8.49 mean? Recall that thecurl of H is a way to describe the direction andrate of change of H with position. Therefore thisequation constrains spatial derivatives of H to besimply related to J (conduction current) and thetime derivative of D (displacement current). Saidplainly:

The differential form of the general (time-varying) form of Ampere’s Law (Equa-tion 8.49) relates the change in the magneticfield with position to the change in the elec-tric field with time, plus conduction current.

As is the case in the Maxwell-Faraday Equation(Section 8.8), we see that electric and magneticfields become coupled at each point in spacewhen sources are time-varying.

Additional Reading:

• “Displacement Current” on Wikipedia.

• “Ampere’s Circuital Law” on Wikipedia.


[m0126]

https://en.wikipedia.org/wiki/Displacement_current




Image Credits

Fig. 8.1: c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0003 fCoilBarMagnet.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/), modified from the original.

Fig. 8.2: E. Bach, https://commons.wikimedia.org/wiki/File:Faraday emf experiment.svg,public domain via CC0 1.0 (https://creativecommons.org/publicdomain/zero/1.0/deed.en),modified from the original.

Fig. 8.8: c© BillC, https://commons.wikimedia.org/wiki/File:Transformer3d col3.svg,CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Modified by author.

Fig. 8.10: c© SpinningSpark, https://en.wikipedia.org/wiki/File:Transformer balance.svg,CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Modified by author.

Fig. 8.12: C. Burks, https://en.wikipedia.org/wiki/File:Displacement current in capacitor.svg,public domain. Heavily modified by author.

Fig. 8.13: C. Burks, https://en.wikipedia.org/wiki/File:Displacement current in capacitor.svg,public domain. Heavily modified by author.

https://commons.wikimedia.org/wiki/File:M0003_fCoilBarMagnet.svg


https://commons.wikimedia.org/wiki/File:Faraday_emf_experiment.svg


https://commons.wikimedia.org/wiki/File:Transformer3d_col3.svg


https://en.wikipedia.org/wiki/File:Transformer_balance.svg


https://en.wikipedia.org/wiki/File:Displacement_current_in_capacitor.svg

https://en.wikipedia.org/wiki/File:Displacement_current_in_capacitor.svg

Chapter 9

Plane Wave Propagation in LosslessMedia

9.1 Maxwell’s Equations inDifferential Phasor Form

[m0042]

In this section we derive the phasor form ofMaxwell’s Equations from the generaltime-varying form of these equations. Here weare interested exclusively in the differential(“point”) form of these equations. It is assumedthat the reader is comfortable with phasorrepresentation and its benefits; if not, a review ofSection 1.5 is recommended before attemptingthis section.

Maxwell’s Equations in differential time-domainform are Gauss’ Law (Section 5.7):

∇ ·D = ρv (9.1)

the Maxwell-Faraday Equation (MFE;Section 8.8):

∇×E = − ∂

∂tB (9.2)

Gauss’ Law for Magnetism (GSM; Section 7.3):

∇ ·B = 0 (9.3)

and Ampere’s Law (Section 8.9):

∇×H = J+∂

∂tD (9.4)

We begin with Gauss’s Law (Equation 9.1). We

define D and ρv as phasor quantities through the

usual relationship:

D = ReDejωt

(9.5)

and

ρv = Reρve

jωt

(9.6)

Substituting these expressions into Equation 9.1:

∇ ·[ReDejωt

]= Re

ρve

jωt

(9.7)

Divergence is a real-valued linear operator.Therefore we may exchange the order of the“Re” and “∇·” operations (this is “Claim 2”from Section 1.5):

Re∇ ·[Dejωt

]= Re

ρve

jωt

(9.8)

Next we note that the differentiation associatedwith the divergence operator is with respect toposition and not with respect to time, so theorder of operations may be rearranged as follows:

Re[

∇ · D]ejωt

= Re

ρve

jωt

(9.9)

Finally we note that the equality of the left andright sides of the above equation implies theequality of the associated phasors (this is“Claim 1” from Section 1.5); thus,

∇ · D = ρv (9.10)

In other words, the differential form of Gauss’Law for phasors is identical to the differential



186 CHAPTER 9. PLANE WAVE PROPAGATION IN LOSSLESS MEDIA

form of Gauss’ Law for physical time-domainquantities.

The same procedure applied to the MFE is onlya little more complicated. First we establish thephasor representations of the electric andmagnetic fields:

E = ReEejωt

(9.11)

B = ReBejωt

(9.12)

After substitution into Equation 9.2:

∇×[ReEejωt

]= − ∂

∂t

[ReBejωt

](9.13)

Both curl and time-differentiation are real-valuedlinear operations, so we are entitled to changethe order of operations as follows:

Re∇×

[Eejωt

]= −Re

∂

∂t

[Bejωt

](9.14)

On the left, we note that the time dependenceejωt can be pulled out of the argument of the curloperator, since it does not depend on position:

Re[

∇× E]ejωt

= −Re

∂

∂t

[Bejωt

]

(9.15)

On the right, we note that B is constant withrespect to time (because it is a phasor), so:

Re[

∇× E]ejωt

= −Re

B∂

∂tejωt

= −ReBjωejωt

= Re[

−jωB]ejωt

(9.16)

And so we have found:

∇× E = −jωB (9.17)

Let’s pause for a moment to consider the aboveresult. In the general time-domain version of theMFE, we must take spatial derivatives of theelectric field and time derivatives of the magneticfield. In the phasor version of the MFE, the timederivative operator has been replaced with

multiplication by jω. This is a tremendoussimplification since the equations now involvedifferentiation over position only. Furthermoreno information is lost in this simplification – fora reminder of why that is, see the discussion ofFourier Analysis at the end of Section 1.5.Without this kind of simplification, much ofwhat is now considered “basic” engineeringelectromagnetics would be intractable.

The procedure for conversion of the remainingtwo equations is very similar, yielding:

∇ · B = 0 (9.18)

∇× H = J+ jωD (9.19)

The details are left as an exercise for the reader.

The differential form of Maxwell’s Equations(Equations 9.10, 9.17, 9.18, and 9.19) involveoperations on the phasor representations ofthe physical quantities. These equations havethe advantage that differentiation with re-spect to time is replaced by multiplication byjω.

9.2 Wave Equations forSource-Free and LosslessRegions

[m0036]

Electromagnetic waves are solutions to a set ofcoupled differential simultaneous equations –namely, Maxwell’s Equations. The generalsolution to these equations includes constantswhose values are determined by the applicableelectromagnetic boundary conditions. Howeverthis direct approach can be difficult and is oftennot necessary. In unbounded homogeneousregions which are “source free” (containing nocharges or currents), a simpler approach ispossible. In this section we reduce Maxwell’sEquations to wave equations that apply to theelectric and magnetic fields in this simplerapproach. Before reading further, the reader

9.2. WAVE EQUATIONS FOR SOURCE-FREE AND LOSSLESS REGIONS 187

should consider a review of Section 1.3 (noting inparticular Equation 1.1) and Section 3.6 (waveequations for voltage and current on atransmission line). This section seeks to developthe analogous equations for electric and magneticwaves.

We can get the job done using the differential“point” phasor form of Maxwell’s Equations,developed in Section 9.1. Here they are:

∇ · D = ρv (9.20)

∇× E = −jωB (9.21)

∇ · B = 0 (9.22)

∇× H = J+ jωD (9.23)

In a source-free region there is no net charge andno current, hence ρv = 0 and J = 0 in thepresent analysis. The above equations become

∇ · D = 0 (9.24)

∇× E = −jωB (9.25)

∇ · B = 0 (9.26)

∇× H = +jωD (9.27)

Next we recall that D = ǫE and that ǫ is areal-valued constant for a medium which islossless, homogeneous, and isotropic(Section 2.8). Similarly B = µH and µ is areal-valued constant. Thus, under theseconditions, it is sufficient to consider either D orE, and either B or H. The choice is arbitrary,but in engineering applications it is customary touse E and H. Eliminating the now-redundantquantities D and B, the above equations become

∇ · E = 0 (9.28)

∇× E = −jωµH (9.29)

∇ · H = 0 (9.30)

∇× H = +jωǫE (9.31)

It is important to note that requiring the regionof interest to be source-free precludes thepossibility of loss in the medium. To see this,

let’s first be clear about what we mean by “loss”.For an electromagnetic wave, loss is observed asa reduction in the magnitude of the electric andmagnetic field with increasing distance. Thisreduction is due to the dissipation of power inthe medium. This occurs when the conductivityσ is greater than zero, because Ohm’s Law forElectromagnetics (J = σE; Section 6.3) requiresthat power in the electric field be transferred intoconduction current, and is thereby lost to thewave (Section 6.6). When we required J to bezero above, we precluded this possibility; that is,we implicitly specified σ = 0. The fact that theconstitutive parameters µ and ǫ appear inEquations 9.28–9.31, but σ does not, is furtherevidence of this.

Equations 9.28–9.31 are Maxwell’s Equationsfor a region comprised of isotropic, homo-geneous, and source-free material. Becausethere can be no conduction current in asource-free region, these equations apply onlyto material which is lossless (i.e., having neg-ligible σ).

Before moving on, one additional disclosure isappropriate: It turns out that there actually is away to use Equations 9.28–9.31 for regions inwhich loss is significant. This requires aredefinition of ǫ as a complex-valued quantity.We shall not consider this technique in thissection. We mention this simply because oneshould be aware that if permittivity appears as acomplex-valued quantity, it is because it is beingused to represent non-zero conductivity (thus,loss) as well as the physical permittivity.

To derive the wave equations we begin with theMFE, Equation 9.29. Taking the curl of bothsides of the equation we obtain

∇×(∇× E

)= ∇×

(−jωµH

)

= −jωµ(∇× H

)(9.32)

On the right we can eliminate ∇× H usingEquation 9.31:

−jωµ(∇× H

)= −jωµ

(+jωǫE

)

= +ω2µǫE (9.33)


On the left side of Equation 9.32 we apply thevector identity

∇×∇×A = ∇ (∇ ·A)−∇2A (9.34)

which in this case is

∇×∇× E = ∇(∇ · E

)−∇2E (9.35)

We may eliminate the first term on the rightusing Equation 9.28, yielding

∇×∇× E = −∇2E (9.36)

Substituting these results back intoEquation 9.32 and rearranging terms we have

∇2E+ ω2µǫE = 0 (9.37)

This is the wave equation for E. Note that it is ahomogeneous (in the mathematical sense of theword) differential equation, which is expectedsince we have derived it for a source-free region.

It is common to make the following definition

β , ω√µǫ (9.38)

so that Equation 9.37 may be written

∇2E+ β2E = 0 (9.39)

Why go the the trouble of defining β? Onereason is that β conveniently captures thecontribution of the frequency, permittivity, andpermeability all in one constant. Another reasonis to emphasize the connection to the parameterβ appearing in transmission line theory: SeeSection 3.8 for a reminder. It should be clearthat β is a phase propagation constant, havingunits of 1/m (or rad/m, if you prefer), andindicates the rate at which the phase of thepropagating wave progresses with distance.

The wave equation for H is obtained usingessentially the same procedure, which is left asan exercise for the reader. It should be clearfrom the symmetry of Equations 9.28-9.31 thatthe result will be very similar. One finds:

∇2H+ β2H = 0 (9.40)

Equations 9.39 and 9.40 are the wave equa-tions for E and H, respectively, for a regioncomprised of isotropic, homogeneous, lossless,and source-free material.

Looking ahead, note that E and H are solutionsto the same homogeneous differential equation.Consequently E and H cannot be different bymore than a constant factor and a direction. Infact we can also determine something about thefactor simply by examining the units involved:Since E has units of V/m and H has units ofA/m, this factor will be expressible in units ofthe ratio of V/m to A/m, which is Ω. Thisindicates that the factor will be an impedance.This factor is known as the wave impedance, andwill be addressed in Section 9.5. This impedanceis analogous the characteristic impedance of atransmission line (Section 3.7).

Additional Reading:

• “Wave Equation” on Wikipedia.

• “Electromagnetic Wave Equation” onWikipedia.

9.3 Types of Waves

[m0142]

Solutions to the electromagnetic wave equations(Section 9.2) exist in a variety of forms,representing different types of waves. It is usefulto identify three particular geometries forunguided waves. Each of these geometries isdefined by the shape formed by surfaces ofconstant phase, which we refer to as phasefronts.(Keep in mind the analogy betweenelectromagnetic waves and sound waves(described in Section 1.3), and to note thatsound waves also exhibit these geometries.)

A spherical wave has phasefronts that formconcentric spheres, as shown in Figure 9.1.Waves are well-modeled as spherical when thedimensions of the source of the wave are small

https://en.wikipedia.org/wiki/Wave_equation

https://en.wikipedia.org/wiki/Electromagnetic_wave_equation

9.4. UNIFORM PLANE WAVES: DERIVATION 189


Figure 9.1: The phasefronts of a spherical waveform concentric spheres.

relative to the scale at which the wave isobserved. For example, the wave radiated by anantenna having dimensions of 10 cm, whenobserved in free space over a scale of 10 km,appears to have phasefronts which are verynearly spherical. Note that the magnitude of thefield on a phasefront of a spherical wave mayvary significantly, but it is the shape ofphasefronts that make it a spherical wave.

A cylindrical wave exhibits phasefronts that formconcentric cylinders, as shown in Figure 9.2. Saiddifferently, the phasefronts of a cylindrical waveare circular in one dimension, and planar in theperpendicular direction. A cylindrical wave isoften a good description of the wave thatemerges from a line-shaped source.

A plane wave exhibits phasefronts that areplanar, with planes that are parallel to eachother as shown in Figure 9.3. There are twoconditions in which waves are well-modeled asplane waves. First, some structures give rise towaves that appear to have planar phasefrontsover a limited area; a good example is the waveradiated by a parabolic reflector, as shown inFigure 9.4. Second, all waves are well-modeled as


Figure 9.2: The phasefronts of a cylindrical waveform concentric cylinders.

plane waves when observed over a small arealocated sufficiently far from the source. Inparticular, spherical waves are “locally planar” inthe sense that they are well-modeled as planarwhen observed over a small portion of thespherical phasefront, as shown in Figure 9.5. (Ananalogy is that the Earth seems “locally flat” toan observer on the ground, even though it isclearly spherical to an observer in orbit.) The“locally planar” approximation is often employedbecause it works well over limited regions andsimplifies analysis.

Most waves are well-modeled as spherical,cylindrical, or plane waves.

Plane waves (having planar phasefronts) areof particular importance due to wide applica-bility of the “locally planar” approximation.

9.4 Uniform Plane Waves:Derivation

[m0038]





Figure 9.3: The phasefronts of a plane wave formparallel planes.

feed

re¬ector

planar ®¯°±²³onts


Figure 9.4: Plane waves formed in the region infront of a parabolic reflector antenna.

"locally-planar" phasefronts

spherical phasefront


Figure 9.5: “Locally planar” approximation of aspherical wave over a limited area.

Section 9.2 showed how Maxwell’s Equationscould be reduced to a pair of phasor-domain“wave equations”, namely:

∇2E+ β2E = 0 (9.41)

∇2H+ β2H = 0 (9.42)

where β = ω√µǫ, assuming unbounded

homogeneous, isotropic, lossless, and source-freemedia. In this section we solve these equationsfor the special case of a uniform plane wave. Auniform plane wave is one for which both E andH have constant magnitude and phase in aspecified plane. Despite being a special case, thesolution turns out to be broadly applicable,appearing as a common building block in manypractical and theoretical problems in unguidedpropagation (as explained in Section 9.3) as wellas in more than a few transmission line andwaveguide problems.

To begin, let us assume that the plane overwhich E and H have constant magnitude andphase is a plane of constant z, and work theproblem using the Cartesian coordinate system.First note that we may make this assumptionwith no loss of generality. For example, we couldalternatively select a plane of constant y, solvethe problem, and then simply exchange variables





to get a solution for planes of constant z (or x).1

Furthermore the solution for any planarorientation not corresponding to a plane ofconstant x, y, or z may be similarly obtained bya rotation of coordinates, since the physics of theproblem does not depend on the orientation ofthis plane: If it does, then the medium is notisotropic!

We may express the constraint that themagnitude and phase of E and H are constantover a plane which is perpendicular to the z axisas follows:

∂

∂xE =

∂

∂yE =

∂

∂xH =

∂

∂yH = 0 (9.43)

Let us identify the Cartesian components of eachof these fields as follows:

E = xEx + yEy + zEz , and (9.44)

H = xHx + yHy + zHz (9.45)

Now Equation 9.43 may be interpreted in detailfor E as follows:

∂

∂xEx =

∂

∂xEy =

∂

∂xEz = 0 (9.46)

∂

∂yEx =

∂

∂yEy =

∂

∂yEz = 0 (9.47)

and for H as follows:

∂

∂xHx =

∂

∂xHy =

∂

∂xHz = 0 (9.48)

∂

∂yHx =

∂

∂yHy =

∂

∂yHz = 0 (9.49)

The wave equation for E (Equation 9.41) writtenexplicitly in Cartesian coordinates is

[∂2

∂x2+

∂2

∂y2+

∂2

∂z2

](xEx + yEy + zEz

)

+β2(xEx + yEy + zEz

)= 0

(9.50)

Decomposing this equation into separateequations for each of the three coordinates, we

1By the way, this is a highly-recommended exercise forthe student.

obtain the following:

[∂2

∂x2+

∂2

∂y2+

∂2

∂z2

]Ex + β2Ex = 0 (9.51)

[∂2

∂x2+

∂2

∂y2+

∂2

∂z2

]Ey + β2Ey = 0 (9.52)

[∂2

∂x2+

∂2

∂y2+

∂2

∂z2

]Ez + β2Ez = 0 (9.53)

Applying the constraints of Equations 9.46 and9.47, we note that many of the terms inEquations 9.51–9.53 are zero. We are left with:

∂2

∂z2Ex + β2Ex = 0 (9.54)

∂2

∂z2Ey + β2Ey = 0 (9.55)

∂2

∂z2Ez + β2Ez = 0 (9.56)

Now we will show that Equation 9.43 also impliesthat Ez must be zero. To show this, we useAmpere’s Law for a source-free region(Section 9.2):

∇× H = +jωǫE (9.57)

and take the dot product with z on both sides:

z ·(∇× H

)= +jωǫEz

∂

∂yHx −

∂

∂xHy = +jωǫEz (9.58)

Again applying the constraints of Equation 9.43,the left side of Equation 9.58 must be zero;therefore Ez = 0. The exact same procedureapplied to H (using the Maxwell-FaradayEquation; also given in Section 9.2) reveals that

Hz is also zero.2 Here is what we have found:

If a wave is uniform over a plane, then theelectric and magnetic field vectors must lie inthis plane.

This conclusion is a direct consequence of thefact that Maxwell’s Equations require the electric

2Showing this is a highly-recommended exercise for thereader.


field to be proportional to the curl of themagnetic field, and vice-versa.

The general solution to Equation 9.54 is:

Ex = E+x0e

−jβz + E−x0e

+jβz (9.59)

where E+x0 and E−

x0 are complex-valuedconstants. The values of these constants aredetermined by boundary conditions – possiblysources – outside the region of interest. Since inthis section we are limiting our scope tosource-free and homogeneous regions, we may forthe moment consider the values of E+

x0 and E−x0

to be arbitrary, since any values will satisfy theassociated wave equation.3

Similarly we have for Ey:

Ey = E+y0e

−jβz + E−y0e

+jβz (9.60)

where E+y0 and E−

y0 are again arbitrary constants.Summarizing, we have found

E = xEx + yEy (9.61)

where Ex and Ey are given by Equations 9.59and 9.60, respectively.

Note that Equations 9.59 and 9.60 are essentiallythe same equations encountered in the study ofwaves in lossless transmission lines; for areminder, see Section 3.6. Specifically, factorscontaining e−jβz describe propagation in the +zdirection, whereas factors containing e+jβz

describe propagation in the −z direction. Weconclude

If a wave is uniform over a plane, then possi-ble directions of propagation are the two di-rections perpendicular to the plane.

Since we previously established that the electricand magnetic field vectors must lie in the plane,we also conclude

The direction of propagation is perpendicularto the electric and magnetic field vectors.

3The reader is encouraged to confirm that these are so-lutions by substitution into the associated wave equation.

This conclusion turns out to be generally true;i.e., it is not limited to uniform plane waves.Although we will not provide a rigorous proof ofthis, one way to see that this is true is to imaginethat any type of wave can be imagined as thesum (formally, a linear combination) of uniformplane waves, so perpendicular orientation of thefield vectors with respect to the direction ofpropagation is inescapable.

The same procedure yields the uniform planewave solution to the wave equation for H, whichis

H = xHx + yHy (9.62)

where

Hx = H+x0e

−jβz +H−x0e

+jβz (9.63)

Hy = H+y0e

−jβz +H−y0e

+jβz (9.64)

and where H+x0, H

−x0, H

+y0 and H−

y0 are arbitraryconstants. Note that the solution is essentiallythe same as that for E, with the sole differencebeing that the arbitrary constants mayapparently have different values.

To this point, we have seen no particularrelationship between the electric and magneticfields, and it may appear that the electric andmagnetic fields are independent of each other.However Maxwell’s Equations – specifically, thetwo curl equations – make it clear that theremust be a more strictly-defined relationshipbetween these fields. Subsequently the arbitraryconstants in the solutions for E and H must alsobe related. In fact, there are two considerationshere:

• The magnitude and phase of E must berelated to the magnitude and phase of H.Since both fields are solutions to the samedifferential (wave) equation, they may differby no more than a multiplicative constant.Since the units of E and H are V/m andA/m respectively, this constant must beexpressible in units of V/m divided by A/m;i.e., in units of Ω, an impedance.

• The direction of E must be related todirection of H.


Let us now address these considerations.Consider an electric field which points in one ofthe cardinal directions – let’s say +x – and makethe definition E0 , E+

x0 for notationalconvenience. Then the electric field intensity maybe written as follows:

E = xE0e−jβz (9.65)

Again, there is no loss of generality here since thecoordinate system could in principle be rotatedin such a way that any uniform plane could bedescribed in this way.

We may now calculate H from E using theMaxwell-Faraday Equation (Section 9.2):

∇× E = −jωµH (9.66)

Solving this equation for H, we find:

H =∇× E

−jωµ =∇×

[xE0e

−jβz]

−jωµ (9.67)

Now let us apply the curl operator. Thecomplete expression for the curl operator inCartesian coordinates is given in Section B.2.Here let us consider one component at a time.First the x component:

x ·(∇× E

)=∂Ez∂y

− ∂Ey∂z

(9.68)

Since Ey = Ez = 0, the above expression is zero

and subsequently Hx = 0. Next the ycomponent:

y ·(∇× E

)=∂Ex∂z

− ∂Ez∂x

(9.69)

Here Ez = 0, so we have simply

y ·(∇× E

)=∂Ex∂z

(9.70)

It is not necessary to repeat this procedure forHz, since we know in advance that H must beperpendicular to to the direction of propagationand subsequently Hz = 0. Returning to

H

E

´ µ ¶

·¸¹ection of prº»¼½¼¾¸º¿

Figure 9.6: Relationship between the electric fielddirection, magnetic field direction, and direction ofpropagation.

Equation 9.67 we obtain:

H = y1

−jωµ∂

∂zEx

= y1

−jωµ∂

∂z

(E0e

−jβz)

= y1

−jωµ(−jβE0e

−jβz)

= yβ

ωµE0e

−jβz (9.71)

Note that H points in the +y direction. So, asexpected, both E and H are perpendicular to thedirection of propagation. However we have nowfound a more specific relationship: E and H areperpendicular to each other. Just as x× y = z,we see that E× H points in the direction ofpropagation. This is illustrated in Figure 9.6.Summarizing:

E, H, and the direction of propagation (E×H) are mutually perpendicular.

Now let us resolve the question of the factorrelating E and H. The factor is now seen to beβ/ωµ in Equation 9.71, which can be simplifiedas follows:

β

ωµ=ω√µǫ

ωµ=

1√µ/ǫ

(9.72)


The factor√µ/ǫ appearing above has units of Ω

is known variously as the wave impedance or theintrinsic impedance of the medium. Assigningthis quantity the symbol “η”, we have:

η ,Ex

Hy

=

√µ

ǫ(9.73)

The ratio of the electric field intensity tothe magnetic field intensity is the waveimpedance η (Equation 9.73; units of Ω). Inlossless media, η is determined by the ratioof permeability of the medium to the permit-tivity of the medium.

The wave impedance in free space, assigned thesymbol η0, is

η0 ,

√µ0

ǫ0∼= 377 Ω. (9.74)

Wrapping up our solution, we find that if E is asgiven by Equation 9.65, then

H = yE0

ηe−jβz (9.75)

9.5 Uniform Plane Waves:Characteristics

[m0039]

In Section 9.4 expressions for the electric andmagnetic fields are determined for a uniformplane wave in lossless media. If this plane isperpendicular to the z axis, then waves maypropagate in either the +z direction of the −zdirection. If we consider only the former, andselect E to point in the +x direction, then wefind

E = +xE0e−jβz (9.76)

H = +yE0

ηe−jβz (9.77)

where β = ω√µǫ is the phase propagation

constant, η =√µ/ǫ is the wave impedance, and

E0 is a complex-valued constant associated withthe magnitude and phase of the source. Thisresult is in fact completely general for uniformplane waves, since any other possibility may beobtained by simply rotating coordinates. In factthis is pretty easy because (as determined in

Section 9.4) E, H, and the direction ofpropagation are mutually perpendicular, with thedirection of propagation pointing in the samedirection as E× H.

In this section we identify some importantcharacteristics of uniform plane waves, includingwavelength and phase velocity. Chances are thatmuch of what appears here will be familiar to thereader; if not, a quick review of Sections 1.3(“Fundamentals of Waves”) and 3.8 (“WavePropagation on a Transmission Line”) arerecommended.

First, recall that E and H are phasorsrepresenting physical (real-valued) fields, and arenot the field values themselves. The actual,physical electric field intensity is

E = ReEejωt

(9.78)

= RexE0e

−jβzejωt

= x |E0| cos (ωt− βz + ψ)

where ψ is the phase of E0. Similarly:

H = y|E0|η

cos (ωt− βz + ψ) (9.79)

This result is illustrated in Figure 9.7. Note thatboth E and H (as well as their phasorrepresentations) have the same phase and havethe same frequency and position dependencecos (ωt− βz + ψ). Since β is a real-valuedconstant for lossless media, we also see thatfrequency affects only the phase, and not themagnitude. If ω → 0, then β → 0 and the fieldsno longer depend on z; in this case the field isnot propagating. For ω > 0, cos (ωt− βz + ψ) isperiodic in z; specifically, it has the same valueeach time z increases or decreases by 2π/β. Thisis, by definition, the wavelength λ:

λ ,2π

β(9.80)

9.5. UNIFORM PLANE WAVES: CHARACTERISTICS 195

À

Á

Â

E

H

c© E. Boutet (Modified) CC BY SA 3.0

Figure 9.7: Relationship between the electric fielddirection, magnetic field direction, and direction ofpropagation (here, +z).

If we observe this wave at some fixed point (i.e.,hold z constant), we find that the electric andmagnetic fields are also periodic in time;specifically, they have the same value each time tincreases by 2π/ω. We may characterize thespeed at which the wave travels by comparingthe distance required to experience 2π of phaserotation at a fixed time, which is 1/β; to thetime it takes to experience 2π of phase rotationat a fixed location, which is 1/ω. This is knownas the phase velocity4 vp:

vp ,1/β

1/ω=ω

β(9.81)

Note that vp has the units expected from itsdefinition, namely (rad/s)/(rad/m) = m/s. If wemake the substitution β = ω

√µǫ, we find

vp =ω

ω√µǫ

=1√µǫ

(9.82)

Note that vp, like the wave impedance η, dependsonly on material properties. For example, thephase velocity of an electromagnetic wave in freespace, given the special symbol c, is

c , vp|µ=µ0,ǫ=ǫ0=

1√µ0ǫ0

∼= 3.00× 108 m/s

(9.83)This constant is commonly referred to as thespeed of light, but in fact it is the phase velocity

4We acknowledge that this is a misnomer, since velocityis properly defined as speed in a specified direction, andvp by itself does not specify direction. In fact, the phasevelocity in this case should properly be +zvp. Nevertheless,we adopt the prevailing terminology.

of an electromagnetic field at any frequency (notjust optical frequencies) in free space. Since thepermittivity ǫ and permeability µ of any materialis greater than that of a vacuum, vp in anymaterial is less than the phase velocity in freespace. Summarizing:

Phase velocity is the speed at which any pointof constant phase appears to travel along thedirection of propagation.

Phase velocity is maximum (= c) in freespace, and slower by a factor of 1/

√µrǫr in

any other lossless medium.

Finally, we note the following relationshipbetween frequency f , wavelength, and phasevelocity:

λ =2π

β=

2π

ω/vp=

2π

2πf/vp=vpf

(9.84)

Thus given any two of the parameters f , λ, andvp, we may quickly solve for the remainingquantity. Also note that as a consequence of theinverse-proportional relationship between λ andvp, we find:

At a given frequency, the wavelength in anymaterial is shorter than the wavelength in freespace.

Furthermore, note that it is wavelength andphase velocity that is changing in response tomaterial properties, not frequency.

Example 9.1. Wave propagation in alossless dielectric.

Polyethylene is a low-loss dielectric havingǫr ∼= 2.3. What is the phase velocity inpolyethylene? What is wavelength inpolyethylene? The frequency of interest is1 GHz.

Solution. Low-loss dielectrics exhibitµr ∼= 1 and σ ≈ 0. Therefore the phase



velocity is

vp =1√

µ0ǫrǫ0=

c√ǫr

∼= 1.98× 108 m/s

(9.85)i.e., very nearly two-thirds of the speed oflight in free space. The wavelength at 1 GHzis

λ =vpf

∼= 19.8 cm (9.86)

Again, about two-thirds of a wavelength atthe same frequency in free space.

Returning to polarization and magnitude, it isuseful to note that Equation 9.79 could bewritten in terms of E as follows:

H =1

ηz×E (9.87)

i.e., H is perpendicular to both the direction ofpropagation (here, +z) and E, with magnitudethat is different from E by the wave impedance η.This simple expression is very useful for quicklydetermining H given E when the direction ofpropagation is known. In fact it is so useful thatit commonly explicitly generalized as follows:

H =1

ηk×E (9.88)

where k (here, +z) is the direction of

propagation. Similarly, given H, k, and η, wemay find E using

E = −ηk×H (9.89)

These spatial relationships can be readily verifiedusing Figure 9.7. We have a name for this specialrelationship between the directions andmagnitudes of the electric and magnetic fields:

Equations 9.88 and 9.89 are known as theplane wave relationships.

The plane wave relationships apply equally wellto the phasor representations of E and H; i.e.,

E = −ηk× H (9.90)

H =1

ηk× E (9.91)

These equations can be readily verified byapplying the definition of the associated phasors(e.g., Equation 9.78). It also turns out that theserelationships usually apply even if the waves arenot planar, uniform, or in lossless media.

Example 9.2. Analysis of aradially-directed plane wave.

Consider the scenario illustrated inFigure 9.8. Here a uniform plane wavehaving frequency f = 3 GHz is propagatingalong a ray of constant φ, where is φ isknown but not specified. The phase of theelectric field is π/2 radians at ρ = 0 andt = 0. The material is an effectivelyunbounded region of free space exhibitingnegligible loss. The electric field is polarizedin the +z direction and has peak magnitudeof 1 mV/m. Find (a) the electric fieldintensity in phasor representation, (b) themagnetic field intensity in phasorrepresentation, and (c) the actual, physicalelectric field along the radial.

Solution: First, realize that a“radially-directed” plane wave is a planewave, and not a cylindrical wave. It maywell we be that if we “zoom out” farenough, we are able to perceive a cylindricalwave (for more on this idea, see Section 9.3);or it might simply be that wave is exactlyplanar, and cylindrical coordinates justhappen to be a convenient coordinatesystem for this application. In either case,the direction of propagation k = +ρ and thesolution to this example will be the same.

Here’s the phasor representation for theelectric field intensity of a uniform planewave in a lossless medium, analogous toEquation 9.76:

E = +zE0e−jβρ (9.92)

From the problem statement,|E0| = 1 mV/m. Also from the problemstatement, the phase of E0 is π/2 radians; infact, we could just write E0 = +j |E0|. Thus

9.6. WAVE POLARIZATION 197

the answer to (a) is

E = +zj |E0| e−jβρ (9.93)

where the propagation constantβ = ω

√µǫ = 2πf

√µ0ǫ0 ∼= 62.9 rad/m.

The answer to (b) is easiest to obtain fromthe plane wave relationship:

H =1

ηk×E

=1

ηρ×

(+zj |E0| e−jβρ

)

= −φ j |E0|η

e−jβρ (9.94)

where η here is√µ0/ǫ0 ∼= 377 Ω. Thus the

answer to part (b) is

H = −φjH0e−jβρ (9.95)

where H0∼= 2.65 µA/m. At this point you

should check vector directions: E× H shouldpoint in the direction of propagation +ρ.

Here we find z×(−φ)= +ρ, as expected.

The answer to (c) is obtained by applyingthe defining relationship for phasors to theanswer to part (a):

E = ReEejωt

= Re(

+zj |E0| e−jβρ)ejωt

= +z |E0|Reejπ/2e−jβρejωt

= +z |E0| cos(ωt− βρ+

π

2

)(9.96)

Additional Reading:

• “Electromagnetic radiation” on Wikipedia.

9.6 Wave Polarization

[m0131]

y

x

E

ϕ

Figure 9.8: A radially-directed plane wave.

Polarization refers to the orientation of theelectric field vector. For waves, the term“polarization” refers specifically to theorientation of this vector with increasing distancealong the direction of propagation; or,equivalently, the orientation of this vector withincreasing time at a fixed point in space. Therelevant concepts are readily demonstrated foruniform plane waves, as shown in this section.Therefore a review of Section 9.5 (“UniformPlane Waves: Characteristics”) is recommendedbefore reading further.

To begin, consider the following uniform planewave, described in terms of the phasorrepresentation of its electric field intensity:

Ex = xExe−jβz (9.97)

Here Ex is a complex-valued constant and β isthe positive real-valued propagation constant.Therefore this wave is propagating in the +zdirection in lossless media. This wave is said toexhibit linear polarization (and “linearlypolarized”) because the electric field alwayspoints in the same direction, namely +x. Nowconsider the wave

Ey = yEye−jβz (9.98)

Note that Ey is identical to Ex except thatelectric field vector now points in the +ydirection, and has magnitude and phase that isdifferent by the factor Ey/Ex. This wave too issaid to exhibit linear polarization, again because

https://en.wikipedia.org/wiki/Electromagnetic_radiation


by RJB1 (Modified)

Figure 9.9: Linear polarization. Here Ex is shownin blue, Ey is shown in green, and E is shown inred with black vector symbols. In this examplethe phases of Ex and Ey are zero and φ = −π/4.

the direction of the electric field is constant withboth time and position. In fact, alllinearly-polarized uniform plane wavespropagating in the +z direction in lossless mediacan be described as follows:

E = ρEρe−jβz (9.99)

This is so because ρ could be x, y, or any otherdirection which is perpendicular to z. If one isdetermined to use Cartesian coordinates, theabove expression may be rewritten using(Section 4.3)

ρ = x cosφ+ y cosφ (9.100)

yielding

E = (x cosφ+ y cosφ)Eρe−jβz (9.101)

When written in this form, φ = 0 corresponds toE = Ex, φ = π/2 corresponds to E = Ey, andany other value of φ corresponds to some otherconstant orientation of the electric field vector;see Figure 9.9 for an example.

A wave is said to exhibit linear polarization ifthe direction of the electric field vector doesnot vary with either time or position.

Linear polarization arises when the source of thewave is linearly polarized. A common example isthe wave radiated by a straight wire antenna,such as a dipole or a monopole. Linearpolarization may also be created by passing aplane wave through a polarizer ; this isparticularly common at optical frequencies (see“Additional Reading” at the end of this section).

A commonly-encountered alternative to linearpolarization is circular polarization. For anexplanation, let us return to thelinearly-polarized plane waves Ex and Ey definedearlier. If both of these waves existsimultaneously, then the total electric fieldintensity is simply the sum:

E = Ex + Ey

= (xEx + yEy) e−jβz (9.102)

If the phase of Ex and Ey is the same, thenEx = Eρ cosφ, Ey = Eρ sinφ, and the aboveexpression is essentially the same asEquation 9.101. In this case, E is linearlypolarized. But what if the phases of Ex and Eyare different? In particular, let’s consider thefollowing case: Let Ex = E0, somecomplex-valued constant; and let Ey = +jE0,which is E0 phase-shifted by +π/2 radians. With

no further math, it is apparent that Ex and Eyare different only in that one is phase-shifted byπ/2 radians relative to the other. For thephysical (real-valued) fields, this means that Exhas maximum magnitude when Ey is zero, andvice versa. As a result the direction ofE = Ex +Ey will rotate in the x− y plane, asshown in Figure 9.10

The rotation of the electric field vector can alsobe identified mathematically. When Ex = E0

and Ey = +jE0, Equation 9.102 can be written:

E = (x+ jy)E0e−jβz (9.103)

Now reverting from phasor notation to the

9.6. WAVE POLARIZATION 199

by Dave3457

Figure 9.10: A circularly-polarized wave (inred, with black vector symbols) resulting from theaddition of orthogonal linearly-polarized waves(shown in green and blue) which are are phase-shifted by π/2 radians.

physical field:

E = ReEejωt

= Re(x+ jy)E0e

−jβzejωt

= x |E0| cos (ωt− βz) + y |E0| cos(ωt− βz +

π

2

)

(9.104)

As anticipated, we see that both Ex and Ey varysinusoidally, but are π/2 radians out of phaseresulting in rotation in the plane perpendicularto the direction of propagation.

In the example above, the electric field vectorrotates either clockwise or counter-clockwiserelative to the direction of propagation. Thedirection of this rotation can be identified bypointing the thumb of the left hand in thedirection of propagation; in this case the fingersof the left hand curl in the direction of rotation.For this reason, this particular form of circularpolarization is known as left circular (or”left-hand” circular) polarization (LCP). If weinstead had chosen Ey = −jE0 = −jEx, then thedirection of E rotates in the opposite direction,giving rise to right circular (or “right-hand”circular) polarization (RCP). These twoconditions are illustrated in Figure 9.11.

by Dave3457

Figure 9.11: Left-circular polarization (LCP;top) and right-circular polarization (RCP; bot-tom).


A wave is said to exhibit circular polariza-tion if the electric field vector rotates withconstant magnitude. Left- and right-circularpolarizations may be identified by the direc-tion of rotation with respect to the directionof propagation.

In engineering applications, circular polarizationis useful when the relative orientations oftransmit and receive equipment is variableand/or when the medium is able rotate theelectric field vector. For example, radiocommunications involving satellites innon-geosynchronous orbits typically employcircular polarization. In particular, satellites ofthe U.S. Global Positioning System (GPS)transmit circular polarization because of thevariable geometry of the space-to-earth radiolink, and the tendency of the Earth’s ionosphereto rotate the electric field vector through amechanism known Faraday rotation (sometimescalled the “Faraday effect”). If GPS were insteadto transmit using a linear polarization, then areceiver would need to continuously adjust theorientation of its antenna in order to optimallyreceive the signal. Circularly-polarized radiowaves can be generated (or received) using pairsof perpendicularly-oriented dipoles which are fedthe same signal but with a 90 phase shift, oralternatively by using an antenna which isintrinsically circularly-polarized, such as a helicalantenna (see “Additional Reading” at the end ofthis section).

Linear and circular polarization are certainly notthe only possibilities. Elliptical polarizationresults when Ex and Ey are not either in-phase(leading to linear polarization) or ±π/2 radiansout of phase. Elliptical polarization is typicallynot an intended condition, but rather is mostcommonly observed as a degradation in a systemwhich is nominally linearly- orcircularly-polarized. For example, most antennaswhich are said to be “circularly polarized” in factproduce circular polarization only in onedirection, and various degrees of ellipticalpolarization in all other directions.

Additional Reading:

• “Polarization (waves)” on Wikipedia.

• “Dipole antenna” on Wikipedia.

• “Polarizer” on Wikipedia.

• “Faraday effect” on Wikipedia.

• “Helical antenna” on Wikipedia.

9.7 Wave Power in a LosslessMedium

[m0041]

In many applications involving electromagneticwaves, one is less concerned with theinstantaneous values of the electric and magneticfields than the power associated with the wave.In this section we address this question: Howmuch power is conveyed by an electromagneticwave in a lossless medium? The relevantconcepts are readily demonstrated in the contextof uniform plane waves, as shown in this section.A review of Section 9.5 (“Uniform Plane Waves:Characteristics”) is recommended before readingfurther.

Consider the following uniform plane wave,described in terms of the phasor representationof its electric field intensity:

E = xE0e−jβz (9.105)

Here E0 is a complex-valued constant associatedwith the source of the wave and β is the positivereal-valued propagation constant. Therefore thewave is propagating in the +z direction inlossless media.

The first thing that should be apparent is thatthe amount of power conveyed by this wave isinfinite. The reason is as follows: If the powerpassing through any finite area is greater thanzero, then the total power must be infinitebecause, for a uniform plane wave, the electricand magnetic field intensities are constant over aplane of infinite area. In practice, we neverencounter this situation because all practical

https://en.wikipedia.org/wiki/Polarization_(waves)

https://en.wikipedia.org/wiki/Dipole_antenna

https://en.wikipedia.org/wiki/Polarizer

https://en.wikipedia.org/wiki/Faraday_effect

https://en.wikipedia.org/wiki/Helical_antenna

9.7. WAVE POWER IN A LOSSLESS MEDIUM 201

plane waves are only “locally planar” (seeSection 9.3 for a refresher on this idea).Nevertheless, we seek some way to express thepower associated with such waves.

The solution is not to seek total power, butrather power per unit area. This quantity isknown as the spatial power density, or simply“power density,” and has units of W/m2.5 Thenif we are interested in total power passingthrough some finite area, then we may simplyintegrate the power density over this area. Let’sskip to the answer, and then consider where thisanswer comes from. It turns out that theinstantaneous power density of a uniform planewave is the magnitude of the Poynting vector

S , E×H (9.106)

Note that this equation is dimensionally correct;i.e. the units of E (V/m) times the units of H(A/m) yield the units of spatial power density(V·A/m2, which is W/m2). Also the direction ofE×H is in the direction of propagation(reminder: Section 9.5), which is the direction inwhich the power is expected to flow. Thus wehave some compelling evidence that |S| is thepower density we seek. However this is not proof– for that, we require the Poynting Theorem,which is a bit outside the scope of the presentsection.

A bit later we’re going to need to know S for auniform plane wave, so let’s work that out now.From the plane wave relationships (Section 9.5)we find that the magnetic field intensityassociated with the electric field inEquation 9.105 is

H = yE0

ηe−jβz (9.107)

where η =√µ/ǫ is the real-valued impedance of

the medium. Let ψ be the phase of E0; i.e.,E0 = |E0|ejψ. Then

E = ReEejωt

= x |E0| cos (ωt− βz + ψ) (9.108)

5Be careful: The quantities spectral power density

(W/Hz) and power flux density (W/(m2·Hz)) are alsosometimes referred to as “power density”. In this sectionwe will limit the scope to spatial power density (W/m2).

and

H = ReHejωt

= y|E0|η

cos (ωt− βz + ψ) (9.109)

Now applying Equation 9.106,

S = z|E0|2η

cos2 (ωt− βz + ψ) (9.110)

As noted earlier |S| is only the instantaneouspower density, which is still not quite what weare looking for. What we are actually looking foris the time-average power density Save; that is,the average value of |S| over one period T of thewave. This may be calculated as follows:

Save =1

T

∫ T

t=0

|S|dt

=|E0|2η

1

T

∫ T

t=0

cos2 (ωt− ks+ ψ) dt

(9.111)

Since ω = 2πf = 2π/T , the definite integralequals T/2. We obtain

Save =|E0|22η

(9.112)

It is useful to check units again at this point:Note (V/m)2 divided by Ω is W/m2, as expected.

Equation 9.112 is the time-average powerdensity (units of W/m2) associated with asinusoidally-varying uniform plane wave.

Note that Equation 9.112 is analogous to awell-known result from electric circuit theory:Recall the time-average power Pave (units of W)

associated with a voltage phasor V across aresistance R is

Pave =|V |22R

(9.113)

which closely resembles Equation 9.112.

Here is a good point at which to identify acommon pitfall: Above, |E0| and |V | are the peak


magnitudes of the associated real-valued physicalquantities. However, these quantities are alsoroutinely given as root mean square (“rms”)quantities. When this is the case, the peakmagnitudes are greater by a factor of

√2, and

then Equations 9.112 and 9.113 lack the factor of1/2. Be careful!

Example 9.3. Power density of a typicalradio wave.

A radio wave transmitted from a distantlocation may be perceived locally as auniform plane wave if there is no nearbystructure to scatter the wave; a goodexample is the wave arriving at the user of acellular telephone in a rural area with nosignificant terrain scattering. The range ofpossible signal strengths varies widely, but atypical value of the electric field intensityarriving at the user’s location is 10 µV/mrms. What is the corresponding powerdensity?

Solution: From the problem statement,|E0| = 10 µV/m rms. We assumepropagation occurs in air, which isindistinguishable from free space at cellularfrequencies. If we use Equation 9.112, thenwe must first convert |E0| from rms to peakmagnitude, which is done by multiplying by√2. Thus:

Save =|E0|22η

∼=(√

2 · 10× 10−6 V/m)2

2 · 377 Ω∼= 2.65× 10−13 W/m2

Alternatively, we can just use a version ofEquation 9.112 which is appropriate for rmsunits:

Save =|E0,rms|2

η∼=(10× 10−6 V/m

)2

377 Ω∼= 2.65× 10−13 W/m2

Either way, we obtain the correct answer,0.265 pW/m2 (that’s picowatts per squaremeter).

Considering the prevalence of phasorrepresentation, it is useful to have an alternativeform of the Poynting vector which yieldstime-average power by operating directly on fieldquantities in phasor form. This is Save, definedas:

Save =1

2ReE× H∗

(9.114)

(Note that the magnetic field intensity phasor isconjugated.) The above expression gives theexpected result for a uniform plane wave: UsingEquations 9.105 and 9.107 we find

Save =1

2Re

(xE0e

−jkz)×(yE0

ηe−jkz

)∗

(9.115)which yields

Save = z|E0|22η

(9.116)

as expected.

Additional Reading:

• “Poynting vector” on Wikipedia.

• “Poynting’s Theorem” on Wikipedia.

[m0035]

https://en.wikipedia.org/wiki/Poynting_vector

https://en.wikipedia.org/wiki/Poynting's_theorem

9.7. WAVE POWER IN A LOSSLESS MEDIUM 203

Image Credits

Fig. 9.1: c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0142 fSphericalPhasefront.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/).

Fig. 9.2 c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0142 fCylindricalPhasefront.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/).

Fig. 9.3 c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0142 fPlanarPhasefront.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/).

Fig. 9.4 c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0142 fPlaneWavesReflector.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/).

Fig. 9.5 c© Y. Qin, https://commons.wikimedia.org/wiki/File:M0142 fLocallyPlanar.svg,CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/).

Fig. 9.7: c© E. Boutet, https://en.wikipedia.org/wiki/File:Onde electromagnetique.svg,CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Heavily modified by author.

Fig. 9.9: RJB1, https://commons.wikimedia.org/wiki/File:Linearly Polarized Wave.svg,public domain. Modified by author.

Fig. 9.10: Dave3457, https://en.wikipedia.org/wiki/File:Circular.Polarization.Circularly.Polarized.Light With.Components Left.Handed.svg,public domain.

Fig. 9.11: Dave3457, https://en.wikipedia.org/wiki/File:Circular.Polarization.Circularly.Polarized.Light Without.Components Left.Handed.svg

and https://en.wikipedia.org/wiki/File:Circular.Polarization.

Circularly.Polarized.Light Without.Components Right.Handed.svg,public domain.

https://commons.wikimedia.org/wiki/File:M0142_fSphericalPhasefront.svg


https://commons.wikimedia.org/wiki/File:M0142_fCylindricalPhasefront.svg


https://commons.wikimedia.org/wiki/File:M0142_fPlanarPhasefront.svg


https://commons.wikimedia.org/wiki/File:M0142_fPlaneWavesReflector.svg


https://commons.wikimedia.org/wiki/File:M0142_fLocallyPlanar.svg


https://en.wikipedia.org/wiki/File:Onde_electromagnetique.svg


https://commons.wikimedia.org/wiki/File:Linearly_Polarized_Wave.svg

Appendix A

Constitutive Parameters of SomeCommon Materials

A.1 Permittivity of SomeCommon Materials

[m0135]

The values below are relative permittivityǫr , ǫ/ǫ0 for a few materials that are commonlyencountered in electrical engineering applications,and for which permittivity emerges as aconsideration. Note that “relative permittivity”is sometimes referred to as dielectric constant.

Here we consider only the physical (real-valued)permittivity, which is the real part of thecomplex permittivity (typically indicated as ǫ′ orǫ′r) for materials exhibiting significant loss.

Permittivity varies significantly as a function offrequency. The values below are representative offrequencies from a few kHz to about 1 GHz. Thevalues given are also representative of opticalfrequencies for materials such as silica which areused in optical applications. Permittivity alsovaries as a function of temperature. Inapplications where precision better than about10% is required, primary references accountingfor frequency and temperature should beconsulted. The values presented here aregathered from a variety of references, includingthose indicated in “Additional References”.

Free Space (vacuum): ǫr , 1

Solid Dielectrics:

Material ǫr Common usesStyrofoam1 1.1Teflon2 2.1Polyethylene 2.3 coaxial cablePolypropylene 2.3Silica 2.4 optical fiber3

Polystyrene 2.6Polycarbonate 2.8Rogers RO3003 3.0 PCB substrateFR4 (glass epoxy laminate) 4.5 PCB substrate

1 Properly known as extruded polystyrene foam(XPS).2 Properly known as polytetrafluoroethylene(PTFE).3 Typically doped with small amounts of othermaterials to slightly raise or lower the index ofrefraction (=

√ǫr).

Non-conducting spacing materials used indiscrete capacitors exhibit ǫr ranging from about5 to 50.

Semiconductors commonly appearing inelectronics – including carbon, silicon, geranium,indium phosphide, and so on – typically exhibitǫr in the range 5–15.

Glass exhibits ǫr in the range 4–10, dependingon composition.

Gasses, including air, typically exhibit ǫr ∼= 1 towithin a tiny fraction of a percent.

Liquid water typically exhibits ǫr in the range72–81. Distilled water exhibits ǫr ≈ 81 at room



A.2. PERMEABILITY OF SOME COMMON MATERIALS 205

temperature, whereas sea water tends to be atthe lower end of the range.

Other liquids typically exhibit ǫr in the range10–90, with considerable variation as a functionof temperature and frequency. Animal flesh andblood consists primarily of liquid matter and soalso exhibits permittivity in this range.

Soil typically exhibits ǫr in the range 2.5–3.5when dry, and higher when wet. Thepermittivity of soil varies considerably dependingon composition.

Additional Reading:

• CRC Handbook of Chemistry and Physics.

• “Relative permittivity” on Wikipedia.

• “Miscellaneous Dielectric Constants” onmicrowaves101.com.

A.2 Permeability of SomeCommon Materials

[m0136]

The values below are relative permeabilityµr , µ/µ0 for a few materials that are commonlyencountered in electrical engineeringapplications, and for which µr is significantlydifferent from 1. These materials arepredominantly ferromagnetic metals and (in thecase of ferrites) materials containing significantferromagnetic metal content. Nearly all othermaterials exhibit µr which is not significantlydifferent from that of free space.

The values presented here are gathered from avariety of references, including those indicated in“Additional References” at the end of thissection. Be aware that permeability may varysignificantly with frequency; values given here areapplicable to the frequency ranges forapplications in which these materials aretypically used. Also be aware that materialsexhibiting high permeability are also typically

non-linear; that is, permeability depends on themagnitude of the magnetic field. Again, valuesreported here are those applicable to applicationsin which these materials are typically used.

Free Space (vacuum): µr , 1.

Iron (also referred to by the element notation“Fe”) appears as a principal ingredient in manymaterials and alloys employed in electricalstructures and devices. Iron exhibits µr which isvery high, but which decreases with decreasingpurity. 99.95% pure iron exhibits µr ∼ 200, 000.This decreases to ∼ 5000 at 99.8% purity, and istypically below 100 for purity less than 99%.

Steel is an iron alloy that comes in many forms,with a correspondingly broad range ofpermeabilites. Electrical steel, commonly used inelectrical machinery and transformers when highpermeability is desired, exhibits µr ∼ 4000.Stainless steel, encompassing a broad range ofalloys used in mechanical applications, exhibitsµr in the range 750–1800. Carbon steel, includinga broad class of alloys commonly used instructural applications, exhibits µr on the orderof 100.

Ferrites include a broad range of ceramicmaterials which are chemically combined withiron and various combinations of other metals,and are used as magnets and magnetic devices invarious electrical systems. Common ferritesexhibit µr in the range 16–640.

Additional Reading:

• Section 7.16 (“Magnetic Materials”)

• CRC Handbook of Chemistry and Physics .

• “Magnetic Materials” onmicrowaves101.com.

• “Permeability (electromagnetism)” onWikipedia.

• “Iron” on Wikipedia.

• “Electrical steel” on Wikipedia.

• “Ferrite (magnet)” on Wikipedia.

http://hbcponline.com

https://en.wikipedia.org/wiki/Relative_permittivity

http://www.microwaves101.com/encyclopedias/miscellaneous-dielectric-constants

http://www.microwaves101.com


https://www.microwaves101.com/encyclopedias/magnetic-materials

http://www.microwaves101.com

https://en.wikipedia.org/wiki/Permeability_(electromagnetism)

https://en.wikipedia.org/wiki/Iron

https://en.wikipedia.org/wiki/Electrical_steel

https://en.wikipedia.org/wiki/Ferrite_(magnet)

206 APPENDIX A. CONSTITUTIVE PARAMETERS OF SOME COMMON MATERIALS

A.3 Conductivity of SomeCommon Materials

[m0137]

The values below are conductivity σ for a fewmaterials that are commonly encountered inelectrical engineering applications, and for whichconductivity emerges as a consideration.

Note that materials in some applications aredescribed instead in terms of resistivity, which issimply the reciprocal of conductivity.

Conductivity may vary significantly as a functionof frequency. The values below are representativeof frequencies from a few kHz to a few GHz.Conductivity also varies as a function oftemperature. In applications where precise valuesare required, primary references accounting forfrequency and temperature should be consulted.The values presented here are gathered from avariety of references, including those indicated in“Additional References” at the end of thissection.

Free Space (vacuum): σ , 0.

Commonly encountered elements:Material σ (S/m)Copper 5.8× 107

Gold 4.4× 107

Aluminum 3.7× 107

Iron 1.0× 107

Platinum 0.9× 107

Carbon 1.3× 105

Silicon 4.4× 10−4

Water exhibits σ ranging from about 6 µS/mfor highly distilled water (thus, a very poorconductor) to about 5 S/m for seawater (thus, arelatively good conductor), varying also withtemperature and pressure. Tap water is typicallyin the range 5–50 mS/m, depending on the levelof impurities present.

Soil typically exhibits σ in the range 10−4 S/mfor dry soil to about 10−1 S/m for wet soil,varying also due to chemical composition.

Non-conductors. Most other materials whichare not well-described as conductors orsemiconductors, and are dry, exhibitσ < 10−12 S/m. Most materials which areconsidered to be insulators, including air andcommon dielectrics, exhibit σ < 10−15 S/m,often by several orders of magnitude.

Additional Reading:

• CRC Handbook of Chemistry and Physics .

• “Conductivity (electrolytic)” on Wikipedia.

• “Electrical resistivity and conductivity” onWikipedia.

• “Soil resistivity” on Wikipedia.


https://en.wikipedia.org/wiki/Conductivity_(electrolytic)

https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity

https://en.wikipedia.org/wiki/Soil_resistivity

Appendix B

Mathematical Formulas

B.1 Trigonometry

[m0138]

ejθ = cos θ + j sin θ (B.1)

cos θ =1

2

(ejθ + e−jθ

)(B.2)

sin θ =1

j2

(ejθ − e−jθ

)(B.3)

cos2 θ =1

2+

1

2cos 2θ (B.4)

sin2 θ =1

2− 1

2cos 2θ (B.5)

B.2 Vector Operators

[m0139]

This section contains a summary of vectoroperators expressed in each of the three majorcoordinate systems:

• Cartesian (x,y,z)

• cylindrical (ρ,φ,z)

• spherical (r,θ,φ)

Associated basis vectors are identified using acaret ( ) over the symbol. The vector operand Ais expressed in terms of components in the basisdirections as follows:

• Cartesian: A = xAx + yAy + zAz

• cylindrical: A = ρAρ + φAφ + zAz

• spherical: A = rAr + θAθ + φAφ

GradientGradient in Cartesian coordinates:

∇f = x∂f

∂x+ y

∂f

∂y+ z

∂f

∂z(B.6)

Gradient in cylindrical coordinates:

∇f = ρ∂f

∂ρ+ φ

1

ρ

∂f

∂φ+ z

∂f

∂z(B.7)

Gradient in spherical coordinates:

∇f = r∂f

∂r+ θ

1

r

∂f

∂θ+ φ

1

r sin θ

∂f

∂φ(B.8)

DivergenceDivergence in Cartesian coordinates:

∇ ·A =∂Ax∂x

+∂Ay∂y

+∂Az∂z

(B.9)

Divergence in cylindrical coordinates:

∇ ·A =1

ρ

∂

∂ρ(ρAρ) +

1

ρ

∂Aφ∂φ

+∂Az∂z

(B.10)



208 APPENDIX B. MATHEMATICAL FORMULAS

Divergence in spherical coordinates:

∇ ·A =1

r2∂

∂r

(r2Ar

)

+1

r sin θ

∂

∂θ(Aθ sin θ)

+1

r sin θ

∂Aφ∂φ

(B.11)

CurlCurl in Cartesian coordinates:

∇×A = x

(∂Az∂y

− ∂Ay∂z

)

+ y

(∂Ax∂z

− ∂Az∂x

)

+ z

(∂Ay∂x

− ∂Ax∂y

)(B.12)

Curl in cylindrical coordinates:

∇×A = ρ

(1

ρ

∂Az∂φ

− ∂Aφ∂z

)

+ φ

(∂Aρ∂z

− ∂Az∂ρ

)

+ z1

ρ

[∂

∂ρ(ρAφ)−

∂Aρ∂φ

](B.13)

Curl in spherical coordinates:

∇×A = r1

r sin θ

[∂

∂θ(Aφ sin θ)−

∂Aθ∂φ

]

+ θ1

r

[1

sin θ

∂Ar∂φ

− ∂

∂r(rAφ)

]

+ φ1

r

[∂

∂r(rAθ)−

∂Ar∂θ

](B.14)

LaplacianLaplacian in Cartesian coordinates:

∇2f =∂2f

∂x2+∂2f

∂y2+∂2f

∂z2(B.15)

Laplacian in cylindrical coordinates:

∇2f =1

ρ

∂

∂ρ

(ρ∂f

∂ρ

)+

1

ρ2∂2f

∂φ2+∂2f

∂z2(B.16)

Laplacian in spherical coordinates:

∇2f =1

r2∂

∂r

(r2∂f

∂r

)

+1

r2 sin θ

∂

∂θ

(∂f

∂θsin θ

)

+1

r2 sin2 θ

∂2f

∂φ2(B.17)

B.3 Vector Identities

[m0140]

Algebraic Identities

A · (B×C) = B · (C×A) = C · (A×B)(B.18)

A× (B×C) = B (A ·C)−C (A ·B) (B.19)

Identities Involving Differential Operators

∇ · (∇×A) = 0 (B.20)

∇× (∇f) = 0 (B.21)

∇× (fA) = f (∇×A) + (∇f)×A (B.22)

∇ · (A×B) = B · (∇×A)−A · (∇×B)(B.23)

∇ · (∇f) = ∇2f (B.24)

∇×∇×A = ∇ (∇ ·A)−∇2A (B.25)

∇2A = ∇ (∇ ·A)−∇× (∇×A) (B.26)

Divergence Theorem: Given a closed surfaceS enclosing a contiguous volume V,

∫

V

(∇ ·A) dv =

∮

S

A · ds (B.27)

where the surface normal ds is pointing out ofthe volume.

Stokes’ Theorem: Given a closed curve Cbounding a contiguous surface S,

∫

S

(∇×A) · ds =∮

C

A · dl (B.28)

where the direction of the surface normal ds isrelated to the direction of integration along C bythe right hand rule.

Appendix C

Physical Constants

[m0141]

The speed of light in free space (c), which isthe phase velocity of any electromagneticradiation in free space, is ∼= 2.9979× 108 m/s.This is commonly rounded up to 3× 108 m/s.This rounding incurs error of ∼= 0.07%, which isusually much less than other errors present inelectrical engineering calculations.

The charge of an electron (often assigned thevariable e and sometimes referred to as the“elementary charge”) is ∼= 1.602× 10−19 C.

The permittivity of free space (ǫ0) is∼= 8.854× 10−12 F/m.

The permeability of free space (µ0) is4π × 10−7 H/m.

The wave impedance of free space (η0) isthe ratio of the magnitude of the electric fieldintensity to that of the magnetic field intensity infree space, and is

√µ0/ǫ0 ∼= 376.7 Ω. This is also

sometimes referred to as the intrinsic impedanceof free space.



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