Main A Student's Guide to Maxwell's Equations
A Student's Guide to Maxwell's EquationsDaniel Fleisch
Gauss's law for electric fields, Gauss's law for magnetic fields, Faraday's law, and the Ampere-Maxwell law are four of the most influential equations in science. In this guide for students, each equation is the subject of an entire chapter, with detailed, plain-language explanations of the physical meaning of each symbol in the equation, for both the integral and differential forms. The final chapter shows how Maxwell's equations may be combined to produce the wave equation, the basis for the electromagnetic theory of light. This book is a wonderful resource for undergraduate and graduate courses in electromagnetism and electromagnetics. A website hosted by the author at www.cambridge.org/9780521701471 contains interactive solutions to every problem in the text as well as audio podcasts to walk students through each chapter.
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This page intentionally left blank A Student’s Guide to Maxwell’s Equations Maxwell’s Equations are four of the most influential equations in science: Gauss’s law for electric fields, Gauss’s law for magnetic fields, Faraday’s law, and the Ampere–Maxwell law. In this guide for students, each equation is the subject of an entire chapter, with detailed, plain-language explanations of the physical meaning of each symbol in the equation, for both the integral and differential forms. The final chapter shows how Maxwell’s Equations may be combined to produce the wave equation, the basis for the electromagnetic theory of light. This book is a wonderful resource for undergraduate and graduate courses in electromagnetism and electromagnetics. A website hosted by the author, and available through www.cambridge.org/9780521877619, contains interactive solutions to every problem in the text. Entire solutions can be viewed immediately, or a series of hints can be given to guide the student to the final answer. The website also contains audio podcasts which walk students through each chapter, pointing out important details and explaining key concepts. da n i e l fl eis ch is Associate Professor in the Department of Physics at Wittenberg University, Ohio. His research interests include radar cross-section measurement, radar system analysis, and ground-penetrating radar. He is a member of the American Physical Society (APS), the American Association of Physics Teachers (AAPT), and the Institute of Electrical and Electronics Engineers (IEEE). A Student’s Guide to Maxwell’s Equations DANIEL FLEISCH Wittenberg University CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521877619 © D. Fleisch 2008 This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2008 ISBN-13 978-0-511-39308-2 eBook (EBL) ISBN-13 hardback 978-0-521-87761-9 Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Contents page vii ix Preface Acknowledgments 1 1.1 1.2 2 2.1 2.2 Gauss’s law for electric fields The integral form of Gauss’s law The electric field The dot product The unit normal vector ~ normal to a surface The component of E The surface integral The flux of a vector field The electric flux through a closed surface The enclosed charge The permittivity of free space Applying Gauss’s law (integral form) The differential form of Gauss’s law Nabla – the del operator Del dot – the divergence The divergence of the electric field Applying Gauss’s law (differential form) Gauss’s law for magnetic fields The integral form of Gauss’s law The magnetic field The magnetic flux through a closed surface Applying Gauss’s law (integral form) The differential form of Gauss’s law The divergence of the magnetic field Applying Gauss’s law (differential form) v 1 1 3 6 7 8 9 10 13 16 18 20 29 31 32 36 38 43 43 45 48 50 53 54 55 vi 3 3.1 3.2 4 4.1 4.2 5 Contents Faraday’s law The integral form of Faraday’s law The induced electric field The line integral The path integral of a vector field The electric field circulation The rate of change of flux Lenz’s law Applying Faraday’s law (integral form) The differential form of Faraday’s law Del cross – the curl The curl of the electric field Applying Faraday’s law (differential form) The Ampere–Maxwell law The integral form of the Ampere–Maxwell law The magnetic field circulation The permeability of free space The enclosed electric current The rate of change of flux Applying the Ampere–Maxwell law (integral form) The differential form of the Ampere–Maxwell law The curl of the magnetic field The electric current density The displacement current density Applying the Ampere–Maxwell law (differential form) From Maxwell’s Equations to the wave equation The divergence theorem Stokes’ theorem The gradient Some useful identities The wave equation 58 58 62 64 65 68 69 71 72 75 76 79 80 83 83 85 87 89 91 95 101 102 105 107 108 112 114 116 119 120 122 Appendix: Maxwell’s Equations in matter Further reading Index 125 131 132 Preface This book has one purpose: to help you understand four of the most influential equations in all of science. If you need a testament to the power of Maxwell’s Equations, look around you – radio, television, radar, wireless Internet access, and Bluetooth technology are a few examples of contemporary technology rooted in electromagnetic field theory. Little wonder that the readers of Physics World selected Maxwell’s Equations as “the most important equations of all time.” How is this book different from the dozens of other texts on electricity and magnetism? Most importantly, the focus is exclusively on Maxwell’s Equations, which means you won’t have to wade through hundreds of pages of related topics to get to the essential concepts. This leaves room for in-depth explanations of the most relevant features, such as the difference between charge-based and induced electric fields, the physical meaning of divergence and curl, and the usefulness of both the integral and differential forms of each equation. You’ll also find the presentation to be very different from that of other books. Each chapter begins with an “expanded view” of one of Maxwell’s Equations, in which the meaning of each term is clearly called out. If you’ve already studied Maxwell’s Equations and you’re just looking for a quick review, these expanded views may be all you need. But if you’re a bit unclear on any aspect of Maxwell’s Equations, you’ll find a detailed explanation of every symbol (including the mathematical operators) in the sections following each expanded view. So if you’re not sure of the ~ n^ in Gauss’s Law or why it is only the enclosed currents meaning of E that contribute to the circulation of the magnetic field, you’ll want to read those sections. As a student’s guide, this book comes with two additional resources designed to help you understand and apply Maxwell’s Equations: an interactive website and a series of audio podcasts. On the website, you’ll find the complete solution to every problem presented in the text in vii viii Preface interactive format – which means that you’ll be able to view the entire solution at once, or ask for a series of helpful hints that will guide you to the final answer. And if you’re the kind of learner who benefits from hearing spoken words rather than just reading text, the audio podcasts are for you. These MP3 files walk you through each chapter of the book, pointing out important details and providing further explanations of key concepts. Is this book right for you? It is if you’re a science or engineering student who has encountered Maxwell’s Equations in one of your textbooks, but you’re unsure of exactly what they mean or how to use them. In that case, you should read the book, listen to the accompanying podcasts, and work through the examples and problems before taking a standardized test such as the Graduate Record Exam. Alternatively, if you’re a graduate student reviewing for your comprehensive exams, this book and the supplemental materials will help you prepare. And if you’re neither an undergraduate nor a graduate science student, but a curious young person or a lifelong learner who wants to know more about electric and magnetic fields, this book will introduce you to the four equations that are the basis for much of the technology you use every day. The explanations in this book are written in an informal style in which mathematical rigor is maintained only insofar as it doesn’t get in the way of understanding the physics behind Maxwell’s Equations. You’ll find plenty of physical analogies – for example, comparison of the flux of electric and magnetic fields to the flow of a physical fluid. James Clerk Maxwell was especially keen on this way of thinking, and he was careful to point out that analogies are useful not because the quantities are alike but because of the corresponding relationships between quantities. So although nothing is actually flowing in a static electric field, you’re likely to find the analogy between a faucet (as a source of fluid flow) and positive electric charge (as the source of electric field lines) very helpful in understanding the nature of the electrostatic field. One final note about the four Maxwell’s Equations presented in this book: it may surprise you to learn that when Maxwell worked out his theory of electromagnetism, he ended up with not four but twenty equations that describe the behavior of electric and magnetic fields. It was Oliver Heaviside in Great Britain and Heinrich Hertz in Germany who combined and simplified Maxwell’s Equations into four equations in the two decades after Maxwell’s death. Today we call these four equations Gauss’s law for electric fields, Gauss’s law for magnetic fields, Faraday’s law, and the Ampere– Maxwell law. Since these four laws are now widely defined as Maxwell’s Equations, they are the ones you’ll find explained in the book. Acknowledgments This book is the result of a conversation with the great Ohio State radio astronomer John Kraus, who taught me the value of plain explanations. Professor Bill Dollhopf of Wittenberg University provided helpful suggestions on the Ampere–Maxwell law, and postdoc Casey Miller of the University of Texas did the same for Gauss’s law. The entire manuscript was reviewed by UC Berkeley graduate student Julia Kregenow and Wittenberg undergraduate Carissa Reynolds, both of whom made significant contributions to the content as well as the style of this work. Daniel Gianola of Johns Hopkins University and Wittenberg graduate Melanie Runkel helped with the artwork. The Maxwell Foundation of Edinburgh gave me a place to work in the early stages of this project, and Cambridge University made available their extensive collection of James Clerk Maxwell’s papers. Throughout the development process, Dr. John Fowler of Cambridge University Press has provided deft guidance and patient support. ix 1 Gauss’s law for electric fields In Maxwell’s Equations, you’ll encounter two kinds of electric field: the electrostatic field produced by electric charge and the induced electric field produced by a changing magnetic field. Gauss’s law for electric fields deals with the electrostatic field, and you’ll find this law to be a powerful tool because it relates the spatial behavior of the electrostatic field to the charge distribution that produces it. 1.1 The integral form of Gauss’s law There are many ways to express Gauss’s law, and although notation differs among textbooks, the integral form is generally written like this: I qenc ~ Gauss’s law for electric fields (integral form). E^ n da ¼ e0 S The left side of this equation is no more than a mathematical description of the electric flux – the number of electric field lines – passing through a closed surface S, whereas the right side is the total amount of charge contained within that surface divided by a constant called the permittivity of free space. If you’re not sure of the exact meaning of ‘‘field line’’ or ‘‘electric flux,’’ don’t worry – you can read about these concepts in detail later in this chapter. You’ll also find several examples showing you how to use Gauss’s law to solve problems involving the electrostatic field. For starters, make sure you grasp the main idea of Gauss’s law: Electric charge produces an electric field, and the flux of that field passing through any closed surface is proportional to the total charge contained within that surface. 1 2 A student’s guide to Maxwell’s Equations In other words, if you have a real or imaginary closed surface of any size and shape and there is no charge inside the surface, the electric flux through the surface must be zero. If you were to place some positive charge anywhere inside the surface, the electric flux through the surface would be positive. If you then added an equal amount of negative charge inside the surface (making the total enclosed charge zero), the flux would again be zero. Remember that it is the net charge enclosed by the surface that matters in Gauss’s law. To help you understand the meaning of each symbol in the integral form of Gauss’s law for electric fields, here’s an expanded view: Reminder that the electric field is a vector Reminder that this integral is over a closed surface Dot product tells you to find the part of E parallel to nˆ (perpendicular to the surface) The unit vector normal to the surface qenc E nˆ da = 0 S ∫ Tells you to sum up the contributions from each portion of the surface The electric field in N/C An increment of surface area in m2 The amount of charge in coulombs Reminder that only the enclosed charge contributes The electric permittivity of the free space Reminder that this is a surface integral (not a volume or a line integral) How is Gauss’s law useful? There are two basic types of problems that you can solve using this equation: (1) Given information about a distribution of electric charge, you can find the electric flux through a surface enclosing that charge. (2) Given information about the electric flux through a closed surface, you can find the total electric charge enclosed by that surface. The best thing about Gauss’s law is that for certain highly symmetric distributions of charges, you can use it to find the electric field itself, rather than just the electric flux over a surface. Although the integral form of Gauss’s law may look complicated, it is completely understandable if you consider the terms one at a time. That’s exactly what you’ll find in the following sections, starting with ~ E, the electric field. Gauss’s law for electric fields 3 ~ E The electric field To understand Gauss’s law, you first have to understand the concept of the electric field. In some physics and engineering books, no direct definition of the electric field is given; instead you’ll find a statement that an electric field is ‘‘said to exist’’ in any region in which electrical forces act. But what exactly is an electric field? This question has deep philosophical significance, but it is not easy to answer. It was Michael Faraday who first referred to an electric ‘‘field of force,’’ and James Clerk Maxwell identified that field as the space around an electrified object – a space in which electric forces act. The common thread running through most attempts to define the electric field is that fields and forces are closely related. So here’s a very pragmatic definition: an electric field is the electrical force per unit charge exerted on a charged object. Although philosophers debate the true meaning of the electric field, you can solve many practical problems by thinking of the electric field at any location as the number of newtons of electrical force exerted on each coulomb of charge at that location. Thus, the electric field may be defined by the relation ~ Fe ~ E ; q0 ð1:1Þ where ~ Fe is the electrical force on a small1 charge q0 . This definition makes clear two important characteristics of the electric field: (1) ~ E is a vector quantity with magnitude directly proportional to force and with direction given by the direction of the force on a positive test charge. (2) ~ E has units of newtons per coulomb (N/C), which are the same as volts per meter (V/m), since volts ¼ newtons · meters/coulombs. In applying Gauss’s law, it is often helpful to be able to visualize the electric field in the vicinity of a charged object. The most common approaches to constructing a visual representation of an electric field are to use a either arrows or ‘‘field lines’’ that point in the direction of the field at each point in space. In the arrow approach, the strength of the field is indicated by the length of the arrow, whereas in the field line 1 Why do physicists and engineers always talk about small test charges? Because the job of this charge is to test the electric field at a location, not to add another electric field into the mix (although you can’t stop it from doing so). Making the test charge infinitesimally small minimizes the effect of the test charge’s own field. 4 A student’s guide to Maxwell’s Equations + - Positive point charge Negative point charge Infinite plane of negative charge Positively charged conducting sphere Infinite line of positive charge Electric dipole with positive charge on left Figure 1.1 Examples of electric fields. Remember that these fields exist inthree dimensions; full three-dimensional (3-D) visualizations are available on the book’s website. approach, it is the spacing of the lines that tells you the field strength (with closer lines signifying a stronger field). When you look at a drawing of electric field lines or arrows, be sure to remember that the field exists between the lines as well. Examples of several electric fields relevant to the application of Gauss’s law are shown in Figure 1.1. Here are a few rules of thumb that will help you visualize and sketch the electric fields produced by charges2: Electric field lines must originate on positive charge and terminate on negative charge. The net electric field at any point is the vector sum of all electric fields present at that point. Electric field lines can never cross, since that would indicate that the field points in two different directions at the same location (if two or more different sources contribute electric fields pointing in different directions at the same location, the total electric field is the vector sum 2 In Chapter 3, you can read about electric fields produced not by charges but by changing magnetic fields. That type of field circulates back on itself and does not obey the same rules as electric fields produced by charge. Gauss’s law for electric fields 5 Table 1.1. Electric field equations for simple objects Point charge (charge ¼ q) 1 q ~ ^r (at distance r from q) E¼ 4pe0 r 2 Conducting sphere (charge ¼ Q) 1 Q ~ ^r (outside, distance r from E¼ 4pe0 r 2 center) ~ E ¼ 0 (inside) Uniformly charged insulating sphere (charge ¼ Q, radius ¼ r0) 1 Q ~ ^r (outside, distance r from E¼ 4pe0 r 2 center) 1 Qr ~ ^r (inside, distance r from E¼ 4pe0 r03 center) Infinite line charge (linear charge density ¼ k) 1 k ~ ^r (distance r from line) E¼ 2pe0 r Infinite flat plane (surface charge density ¼ r) r ~ ^n E¼ 2e0 of the individual fields, and the electric field lines always point in the single direction of the total field). Electric field lines are always perpendicular to the surface of a conductor in equilibrium. Equations for the electric field in the vicinity of some simple objects may be found in Table 1.1. So exactly what does the ~ E in Gauss’s law represent? It represents the total electric field at each point on the surface under consideration. The surface may be real or imaginary, as you’ll see when you read about the meaning of the surface integral in Gauss’s law. But first you should consider the dot product and unit normal that appear inside the integral. 6 A student’s guide to Maxwell’s Equations The dot product When you’re dealing with an equation that contains a multiplication symbol (a circle or a cross), it is a good idea to examine the terms on both sides of that symbol. If they’re printed in bold font or are wearing vector hats (as are ~ E and ^ n in Gauss’s law), the equation involves vector multiplication, and there are several different ways to multiply vectors (quantities that have both magnitude and direction). In Gauss’s law, the circle between ~ E and ^n represents the dot product (or ‘‘scalar product’’) between the electric field vector ~ E and the unit normal vector ^ n (discussed in the next section). If you know the Cartesian components of each vector, you can compute this as ~ A~ B ¼ A x Bx þ Ay By þ Az Bz : ð1:2Þ Or, if you know the angle h between the vectors, you can use ~ A~ B ¼ j~ Ajj~ Bj cos h; ð1:3Þ where j~ Aj and j~ Bj represent the magnitude (length) of the vectors. Notice that the dot product between two vectors gives a scalar result. To grasp the physical significance of the dot product, consider vectors ~ A and ~ B that differ in direction by angle h, as shown in Figure 1.2(a). For these vectors, the projection of ~ A onto ~ B is j~ Aj cos h, as shown in Figure 1.2(b). Multiplying this projection by the length of ~ B gives j~ Ajj~ Bj cos h. Thus, the dot product ~ A~ B represents the projection of ~ A 3 ~ ~ onto the direction of B multiplied by the length of B. The usefulness of this operation in Gauss’s law will become clear once you understand the meaning of the vector ^ n. (a) (b) A A B u B u The projection of A onto B: |A| cos u 3|B| multiplied by the length of B: gives the dot product A B: |A||B|cos u Figure 1.2 The meaning of the dot product. 3 You could have obtained the same result by finding the projection of ~ B onto the direction of ~ A and then multiplying by the length of ~ A. Gauss’s law for electric fields ^n 7 The unit normal vector The concept of the unit normal vector is straightforward; at any point on a surface, imagine a vector with length of one pointing in the direction perpendicular to the surface. Such a vector, labeled ^ n, is called a ‘‘unit’’ vector because its length is unity and ‘‘normal’’ because it is perpendicular to the surface. The unit normal for a planar surface is shown in Figure 1.3(a). Certainly, you could have chosen the unit vector for the plane in Figure 1.3(a) to point in the opposite direction – there’s no fundamental difference between one side of an open surface and the other (recall that an open surface is any surface for which it is possible to get from one side to the other without going through the surface). For a closed surface (defined as a surface that divides space into an ‘‘inside’’ and an ‘‘outside’’), the ambiguity in the direction of the unit normal has been resolved. By convention, the unit normal vector for a closed surface is taken to point outward – away from the volume enclosed by the surface. Some of the unit vectors for a sphere are shown in Figure 1.3(b); notice that the unit normal vectors at the Earth’s North and South Pole would point in opposite directions if the Earth were a perfect sphere. You should be aware that some authors use the notation d~ a rather than ^ n da. In that notation, the unit normal is incorporated into the vector area element d~ a, which has magnitude equal to the area da and direction along the surface normal ^ n. Thus d~ a and ^n da serve the same purpose. Figure 1.3 Unit normal vectors for planar and spherical surfaces. 8 A student’s guide to Maxwell’s Equations ~ E ^n The component of ~ E normal to a surface If you understand the dot product and unit normal vector, the meaning of ~ E^ n should be clear; this expression represents the component of the electric field vector that is perpendicular to the surface under consideration. If the reasoning behind this statement isn’t apparent to you, recall that the dot product between two vectors such as ~ E and ^n is simply the projection of the first onto the second multiplied by the length of the second. Recall also that by definition the length of the unit normal is one ðj^nj ¼ 1), so that ~ E^ n ¼ j~ Ejj^ nj cos h ¼ j~ Ej cos h; ð1:4Þ where h is the angle between the unit normal n^ and ~ E. This is the component of the electric field vector perpendicular to the surface, as illustrated in Figure 1.4. Thus, if h ¼ 90 , ~ E is perpendicular to ^ n, which means that the electric field is parallel to the surface, and ~ E^ n ¼ j~ Ej cosð90 Þ ¼ 0. So in this case the component of ~ E perpendicular to the surface is zero. E is parallel to ^ n, meaning the electric field is Conversely, if h ¼ 0 , ~ ~ Ej. In this case, perpendicular to the surface, and E ^ n ¼ j~ Ej cosð0 Þ ¼ j~ the component of ~ E perpendicular to the surface is the entire length of ~ E. The importance of the electric field component normal to the surface will become clear when you consider electric flux. To do that, you should make sure you understand the meaning of the surface integral in Gauss’s law. Component of E normal to surface is E n^ n^ Surface Figure 1.4 Projection of ~ E onto direction of ^ n. E 9 Gauss’s law for electric fields R S ðÞda The surface integral Many equations in physics and engineering – Gauss’s law among them – involve the area integral of a scalar function or vector field over a specified surface (this type of integral is also called the ‘‘surface integral’’). The time you spend understanding this important mathematical operation will be repaid many times over when you work problems in mechanics, fluid dynamics, and electricity and magnetism (E&M). The meaning of the surface integral can be understood by considering a thin surface such as that shown in Figure 1.5. Imagine that the area density (the mass per unit area) of this surface varies with x and y, and you want to determine the total mass of the surface. You can do this by dividing the surface into two-dimensional segments over each of which the area density is approximately constant. For individual segments with area density ri and area dAi, the mass of each segment is ri dAi, and the mass of the entire surface of N segments is PN given by i¼1 ri dAi . As you can imagine, the smaller you make the area segments, the closer this gets to the true mass, since your approximation of constant r is more accurate for smaller segments. If you let the segment area dA approach zero and N approach infinity, the summation becomes integration, and you have Z Mass ¼ rðx; yÞ dA: S This is the area integral of the scalar function r(x, y) over the surface S. It is simply a way of adding up the contributions of little pieces of a function (the density in this case) to find a total quantity. To understand the integral form of Gauss’s law, it is necessary to extend the concept of the surface integral to vector fields, and that’s the subject of the next section. Area density (s) varies across surface Density approximately constant over each of these areas (dA1, dA2, . . . , dAN) s1 x y Density = s(x,y) s2 s3 sΝ Mass = s1 dA1+ s2 dA2+ . . . + sN dAN. Figure 1.5 Finding the mass of a variable-density surface. 10 A student’s guide to Maxwell’s Equations R ~ s A ^n da The flux of a vector field In Gauss’s law, the surface integral is applied not to a scalar function (such as the density of a surface) but to a vector field. What’s a vector field? As the name suggests, a vector field is a distribution of quantities in space – a field – and these quantities have both magnitude and direction, meaning that they are vectors. So whereas the distribution of temperature in a room is an example of a scalar field, the speed and direction of the flow of a fluid at each point in a stream is an example of a vector field. The analogy of fluid flow is very helpful in understanding the meaning of the ‘‘flux’’ of a vector field, even when the vector field is static and nothing is actually flowing. You can think of the flux of a vector field over a surface as the ‘‘amount’’ of that field that ‘‘flows’’ through that surface, as illustrated in Figure 1.6. In the simplest case of a uniform vector field ~ A and a surface S perpendicular to the direction of the field, the flux U is defined as the product of the field magnitude and the area of the surface: U ¼ j~ Aj · surface area: ð1:5Þ This case is shown in Figure 1.6(a). Note that if ~ A is perpendicular to the surface, it is parallel to the unit normal ^ n: If the vector field is uniform but is not perpendicular to the surface, as in Figure 1.6(b), the flux may be determined simply by finding the component of ~ A perpendicular to the surface and then multiplying that value by the surface area: U¼~ A^ n · ðsurface areaÞ: ð1:6Þ While uniform fields and flat surfaces are helpful in understanding the concept of flux, many E&M problems involve nonuniform fields and curved surfaces. To work those kinds of problems, you’ll need to understand how to extend the concept of the surface integral to vector fields. (a) A (b) n A n Figure 1.6 Flux of a vector field through a surface. 11 Gauss’s law for electric fields (a) (b) ni Component of A perpendicular to this surface element is A ° ni u A A Surface S A Figure 1.7 Component of ~ A perpendicular to surface. Consider the curved surface and vector field ~ A shown in Figure 1.7(a). Imagine that ~ A represents the flow of a real fluid and S a porous membrane; later you’ll see how this applies to the flux of an electric field through a surface that may be real or purely imaginary. Before proceeding, you should think for a moment about how you might go about finding the rate of flow of material through surface S. You can define ‘‘rate of flow’’ in a few different ways, but it will help to frame the question as ‘‘How many particles pass through the membrane each second?’’ To answer this question, define ~ A as the number density of the fluid (particles per cubic meter) times the velocity of the flow (meters per second). As the product of the number density (a scalar) and the velocity (a vector), ~ A must be a vector in the same direction as the velocity, with units of particles per square meter per second. Since you’re trying to find the number of particles per second passing through the surface, dimensional analysis suggests that you multiply ~ A by the area of the surface. But look again at Figure 1.7(a). The different lengths of the arrows are meant to suggest that the flow of material is not spatially uniform, meaning that the speed may be higher or lower at various locations within the flow. This fact alone would mean that material flows through some portions of the surface at a higher rate than other portions, but you must also consider the angle of the surface to the direction of flow. Any portion of the surface lying precisely along the direction of flow will necessarily have zero particles per second passing through it, since the flow lines must penetrate the surface to carry particles from one side to 12 A student’s guide to Maxwell’s Equations the other. Thus, you must be concerned not only with the speed of flow and the area of each portion of the membrane, but also with the component of the flow perpendicular to the surface. Of course, you know how to find the component of ~ A perpendicular to the surface; simply form the dot product of ~ A and ^n, the unit normal to the surface. But since the surface is curved, the direction of ^n depends on which part of the surface you’re considering. To deal with the different ^n (and ~ A) at each location, divide the surface into small segments, as shown in Figure 1.7(b). If you make these segments sufficiently small, you can assume that both ^ n and ~ A are constant over each segment. Let ^ ni represent the unit normal for the ith segment (of area dai); the flow through segment i is (~ Ai ^ ni ) dai, and the total is flow through entire surface ¼ P ~ Ai ^ni dai : i It should come as no surprise that if you now let the size of each segment shrink to zero, the summation becomes integration. Z Flow through entire surface ¼ ~ A ^n da: ð1:7Þ S For a closed surface, the integral sign includes a circle: I ~ A^ n da: ð1:8Þ S This flow is the particle flux through a closed surface S, and the similarity to the left side of Gauss’s law is striking. You have only to replace the vector field ~ A with the electric field ~ E to make the expressions identical. 13 Gauss’s law for electric fields H ~^ n da SE The electric flux through a closed surface On the basis of the results of the previous section, you should understand that the flux UE of vector field ~ E through surface S can be determined using the following equations: UE ¼ j~ Ej · ðsurface areaÞ ~ E is uniform and perpendicular to S; ð1:9Þ UE ¼ ~ E^ n · ðsurface areaÞ ~ E is uniform and at an angle to S; ð1:10Þ Z UE ¼ ~ E^ n da ~ E is non-uniform and at a variable angle to S: ð1:11Þ S These relations indicate that electric flux is a scalar quantity and has units of electric field times area, or Vm. But does the analogy used in the previous section mean that the electric flux should be thought of as a flow of particles, and that the electric field is the product of a density and a velocity? The answer to this question is ‘‘absolutely not.’’ Remember that when you employ a physical analogy, you’re hoping to learn something about the relationships between quantities, not about the quantities themselves. So, you can find the electric flux by integrating the normal component of the electric field over a surface, but you should not think of the electric flux as the physical movement of particles. How should you think of electric flux? One helpful approach follows directly from the use of field lines to represent the electric field. Recall that in such representations the strength of the electric field at any point is indicated by the spacing of the field lines at that location. More specifically, the electric field strength can be considered to be proportional to the density of field lines (the number of field lines per square meter) in a plane perpendicular to the field at the point under consideration. Integrating that density over the entire surface gives the number of field lines penetrating the surface, and that is exactly what the expression for electric flux gives. Thus, another way to define electric flux is electric flux ðUE Þ number of field lines penetrating surface. There are two caveats you should keep in mind when you think of electric flux as the number of electric field lines penetrating a surface. The first is that field lines are only a convenient representation of the electric field, which is actually continuous in space. The number of field lines you 14 A student’s guide to Maxwell’s Equations (a) Zero net flux (b) Positive flux (c) Negative flux Figure 1.8 Flux lines penetrating closed surfaces. choose to draw for a given field is up to you, so long as you maintain consistency between fields of different strengths – which means that fields that are twice as strong must be represented by twice as many field lines per unit area. The second caveat is that surface penetration is a two-way street; once the direction of a surface normal ^ n has been established, field line components parallel to that direction give a positive flux, while components in the opposite direction (antiparallel to ^ n) give a negative flux. Thus, a surface penetrated by five field lines in one direction (say from the top side to the bottom side) and five field lines in the opposite direction (from bottom to top) has zero flux, because the contributions from the two groups of field lines cancel. So, you should think of electric flux as the net number of field lines penetrating the surface, with direction of penetration taken into account. If you give some thought to this last point, you may come to an important conclusion about closed surfaces. Consider the three boxes shown in Figure 1.8. The box in Figure 1.8(a) is penetrated only by electric field lines that originate and terminate outside the box. Thus, every field line that enters must leave, and the flux through the box must be zero. Remembering that the unit normal for closed surfaces points away from the enclosed volume, you can see that the inward flux (lines entering the box) is negative, since ~ E^ n must be negative when the angle between ~ E and ^ n is greater than 90 . This is precisely cancelled by the outward flux (lines exiting the box), which is positive, since ~ E ^n is positive when the angle between ~ E and ^ n is less than 90 . Now consider the box in Figure 1.8(b). The surfaces of this box are penetrated not only by the field lines originating outside the box, but also by a group of field lines that originate within the box. In this case, the net number of field lines is clearly not zero, since the positive flux of the lines Gauss’s law for electric fields 15 that originate in the box is not compensated by any incoming (negative) flux. Thus, you can say with certainty that if the flux through any closed surface is positive, that surface must contain a source of field lines. Finally, consider the box in Figure 1.8(c). In this case, some of the field lines terminate within the box. These lines provide a negative flux at the surface through which they enter, and since they don’t exit the box, their contribution to the net flux is not compensated by any positive flux. Clearly, if the flux through a closed surface is negative, that surface must contain a sink of field lines (sometimes referred to as a drain). Now recall the first rule of thumb for drawing charge-induced electric field lines; they must originate on positive charge and terminate on negative charge. So, the point from which the field lines diverge in Figure 1.8(b) marks the location of some amount of positive charge, and the point to which the field lines converge in Figure 1.8(c) indicates the existence of negative charge at that location. If the amount of charge at these locations were greater, there would be more field lines beginning or ending on these points, and the flux through the surface would be greater. And if there were equal amounts of positive and negative charge within one of these boxes, the positive (outward) flux produced by the positive charge would exactly cancel the negative (inward) flux produced by the negative charge. So, in this case the flux would be zero, just as the net charge contained within the box would be zero. You should now see the physical reasoning behind Gauss’s law: the electric flux passing through any closed surface – that is, the number of electric field lines penetrating that surface – must be proportional to the total charge contained within that surface. Before putting this concept to use, you should take a look at the right side of Gauss’s law. 16 A student’s guide to Maxwell’s Equations qenc The enclosed charge If you understand the concept of flux as described in the previous section, it should be clear why the right side of Gauss’s law involves only the enclosed charge – that is, the charge within the closed surface over which the flux is determined. Simply put, it is because any charge located outside the surface produces an equal amount of inward (negative) flux and outward (positive) flux, so the net contribution to the flux through the surface must be zero. How can you determine the charge enclosed by a surface? In some problems, you’re free to choose a surface that surrounds a known amount of charge, as in the situations shown in Figure 1.9. In each of these cases, the total charge within the selected surface can be easily determined from geometric considerations. For problems involving groups of discrete charges enclosed by surfaces of any shape, finding the total charge is simply a matter of adding the individual charges. Total enclosed charge ¼ X qi : i While small numbers of discrete charges may appear in physics and engineering problems, in the real world you’re far more likely to encounter charged objects containing billions of charge carriers lined along a wire, slathered over a surface, or arrayed throughout a volume. In such cases, counting the individual charges is not practical – but you can determine the total charge if you know the charge density. Charge density may be specified in one, two, or three dimensions (1-, 2-, or 3-D). Point charge Enclosing sphere Multiple point charges enclosing cube Figure 1.9 Surface enclosing known charges. Charged line Charged plane Enclosing pillbox Enclosing cylinder 17 Gauss’s law for electric fields Dimensions Terminology 1 Linear charge density Area charge density Volume charge density 2 3 Symbol Units k C/m r q C/m2 C/m3 If these quantities are constant over the length, area, or volume under consideration, finding the enclosed charge requires only a single multiplication: 1-D : qenc ¼ k L ðL ¼ enclosed length of charged lineÞ; ð1:12Þ 2-D : qenc ¼ rA ðA ¼ enclosed area of charged surfaceÞ; ð1:13Þ 3-D : qenc ¼ qV ðV ¼ enclosed portion of charged volumeÞ: ð1:14Þ You are also likely to encounter situations in which the charge density is not constant over the line, surface, or volume of interest. In such cases, the integration techniques described in the ‘‘Surface Integral’’ section of this chapter must be used. Thus, Z 1-D : qenc ¼ k dl where k varies along a line; ð1:15Þ L Z 2-D : qenc ¼ r da where r varies over a surface; ð1:16Þ q dV where q varies over a volume: ð1:17Þ S Z 3-D : qenc ¼ V You should note that the enclosed charge in Gauss’s law for electric fields is the total charge, including both free and bound charge. You can read about bound charge in the next section, and you’ll find a version of Gauss’s law that depends only on free charge in the Appendix. Once you’ve determined the charge enclosed by a surface of any size and shape, it is very easy to find the flux through that surface; simply divide the enclosed charge by e0, the permittivity of free space. The physical meaning of that parameter is described in the next section. 18 A student’s guide to Maxwell’s Equations e0 The permittivity of free space The constant of proportionality between the electric flux on the left side of Gauss’s law and the enclosed charge on the right side is e0, the permittivity of free space. The permittivity of a material determines its response to an applied electric field – in nonconducting materials (called ‘‘insulators’’ or ‘‘dielectrics’’), charges do not move freely, but may be slightly displaced from their equilibrium positions. The relevant permittivity in Gauss’s law for electric fields is the permittivity of free space (or ‘‘vacuum permittivity’’), which is why it carries the subscript zero. The value of the vacuum permittivity in SI units is approximately 8.85 · 1012 coulombs per volt-meter (C/Vm); you will sometimes see the units of permittivity given as farads per meter (F/m), or, more fundamentally, (C2s2/kg m3). A more precise value for the permittivity of free space is e0 ¼ 8.8541878176 · 1012 C/Vm. Does the presence of this quantity mean that this form of Gauss’s law is only valid in a vacuum? No, Gauss’s law as written in this chapter is general, and applies to electric fields within dielectrics as well as those in free space, provided that you account for all of the enclosed charge, including charges that are bound to the atoms of the material. The effect of bound charges can be understood by considering what happens when a dielectric is placed in an external electric field. Inside the dielectric material, the amplitude of the total electric field is generally less than the amplitude of the applied field. The reason for this is that dielectrics become ‘‘polarized’’ when placed in an electric field, which means that positive and negative charges are displaced from their original positions. And since positive charges are displaced in one direction (parallel to the applied electric field) and negative charges are displaced in the opposite direction (antiparallel to the applied field), these displaced charges give rise to their own electric field that opposes the external field, as shown in Figure 1.10. This makes the net field within the dielectric less than the external field. It is the ability of dielectric materials to reduce the amplitude of an electric field that leads to their most common application: increasing the capacitance and maximum operating voltage of capacitors. As you may recall, the capacitance (ability to store charge) of a parallel-plate capacitor is 19 Gauss’s law for electric fields Induced field Dielectric No dielectric present External electric field – + – + – – + + Displaced charges Figure 1.10 Electric field induced in a dielectric. C¼ eA ; d where A is the plate area, d is the plate separation, and e is the permittivity of the material between the plates. High-permittivity materials can provide increased capacitance without requiring larger plate area or decreased plate spacing. The permittivity of a dielectric is often expressed as the relative permittivity, which is the factor by which the material’s permittivity exceeds that of free space: relative permittivity er ¼ e=e0 : Some texts refer to relative permittivity as ‘‘dielectric constant,’’ although the variation in permittivity with frequency suggests that the word ‘‘constant’’ is better used elsewhere. The relative permittivity of ice, for example, changes from approximately 81 at frequencies below 1 kHz to less than 5 at frequencies above 1 MHz. Most often, it is the low-frequency value of permittivity that is called the dielectric constant. One more note about permittivity; as you’ll see in Chapter 5, the permittivity of a medium is a fundamental parameter in determining the speed with which an electromagnetic wave propagates through that medium. 20 A student’s guide to Maxwell’s Equations H ~^ n da ¼ qenc =e0 Applying Gauss’s law (integral form) sE A good test of your understanding of an equation like Gauss’s law is whether you’re able to solve problems by applying it to relevant situations. At this point, you should be convinced that Gauss’s law relates the electric flux through a closed surface to the charge enclosed by that surface. Here are some examples of what can you actually do with that information. Example 1.1: Given a charge distribution, find the flux through a closed surface surrounding that charge. Problem: Five point charges are enclosed in a cylindrical surface S. If the values of the charges are q1 ¼ þ3 nC, q2 ¼ 2 nC, q3 ¼ þ2 nC, q4 ¼ þ4 nC, and q5 ¼ 1 nC, find the total flux through S. q1 q2 Solution: From Gauss’s law, I UE ¼ S S q3 q5 q4 qenc ~ E^ n da ¼ : e0 For discrete charges, you know that the total charge is just the sum of the individual charges. Thus, qenc ¼ Total enclosed charge ¼ X i qi ¼ ð3 2 þ 2 þ 4 1Þ · 109 C ¼ 6 · 109 C and UE ¼ qenc 6 · 109 C ¼ 678 Vm: ¼ 8:85 · 1012 C=Vm e0 This is the total flux through any closed surface surrounding this group of charges. Gauss’s law for electric fields 21 Example 1.2: Given the flux through a closed surface, find the enclosed charge. Problem: A line charge with linear charge density k ¼ 1012 C/m passes through the center of a sphere. If the flux through the surface of the sphere is 1.13 · 103 Vm, what is the radius R of the sphere? Charged line L Sphere encloses portion of line Solution: The charge on a line charge of length L is given by q ¼ kL. Thus, UE ¼ qenc kL ¼ ; e0 e0 and L¼ U E e0 : k Since L is twice the radius of the sphere, this means 2R ¼ U E e0 k or R ¼ UE e0 : 2k Inserting the values for UE, e0 and k, you will find that R ¼ 5 · 103 m. Example 1.3: Find the flux through a section of a closed surface. Problem: A point source of charge q is placed at the center of curvature of a spherical section that extends from spherical angle h1 to h2 and from u1 to u2. Find the electric flux through the spherical section. Solution: Since the surface of interest in this problem is open, you’ll have to find the electric flux by integrating the normal component of the electric field over the surface. You can then check your answer using Gauss’s law by allowing the spherical section to form a complete sphere that encloses the point charge. 22 A student’s guide to Maxwell’s Equations R E^ n da, where S is the spherical section of The electric flux UE is S ~ ~ interest and E is the electric field on the surface due to the point charge at the center of curvature, a distance r from the section of interest. From Table 1.1, you know that the electric field at a distance r from a point charge is 1 q ~ ^r : E¼ 4pe0 r 2 Before you can integrate this over the surface of interest, you have to consider ~ E ^n (that is, you must find the component of the electric field perpendicular to the surface). That is trivial in this case, because the unit normal ^ n for a spherical section points in the outward radial direction (the ^r direction), as may be seen in Figure 1.11. This means that ~ E and ^n are parallel, and the flux is given by Z ~ E^ n da ¼ UE ¼ Z S j~ Ejj^ nj cosð0 Þ da ¼ S Z j~ Ej da ¼ S Z S 1 q da: 4pe0 r 2 Since you are integrating over a spherical section in this case, the logical choice for coordinate system is spherical. This makes the area element r2 sin h dh dU, and the surface integral becomes rs da in n r du u u r sin du u df da = (rdu)(r sin udf) df f Figure 1.11 Geomentry of sperical section. 23 Gauss’s law for electric fields Z Z UE ¼ h f 1 q 2 q r sin h dh df ¼ 4pe0 r 2 4pe0 Z h Z sin h dh df; f which is easily integrated to give UE ¼ q ðcos h1 cos h2 Þðf2 f1 Þ: 4pe0 As a check on this result, take the entire sphere as the section (h1 ¼ 0, h2 ¼ p, u1 ¼ 0, and u2 ¼ 2p). This gives UE ¼ q q ð1 ð1ÞÞ ð2p 0Þ ¼ ; 4pe0 e0 exactly as predicted by Gauss’s law. Example 1.4: Given ~ E over a surface, find the flux through the surface and the charge enclosed by the surface. Problem: The electric field at distance r from an infinite line charge with linear charge density k is given in Table 1.1 as 1 k ~ ^r : E¼ 2pe0 r r h Use this expression to find the electric flux through a cylinder of radius r and height h surrounding a portion of an infinite line charge, and then use Gauss’s law to verify that the enclosed charge is kh. 24 A student’s guide to Maxwell’s Equations Solution: Problems like this are best approached by considering the flux through each of three surfaces that comprise the cylinder: the top, bottom, and curved side surfaces. The most general expression for the electric flux through any surface is Z UE ¼ ~ E^ n da; S which in this case gives Z UE ¼ S 1 k ^r ^n da: 2pe0 r n n n n n n n Consider now the unit normal vectors of each of the three surfaces: since the electric field points radially outward from the axis of the cylinder, ~ E is perpendicular to the normal vectors of the top and bottom surfaces and parallel to the normal vectors for the curved side of the cylinder. You may therefore write Z 1 k ^r ^ UE; top ¼ ntop da ¼ 0; 2pe 0r S Z UE; bottom ¼ S Z UE; side ¼ S 1 k ^r ^ nbottom da ¼ 0; 2pe0 r 1 k 1 k ^r ^ nside da ¼ 2pe0 r 2pe0 r Z da; S Gauss’s law for electric fields 25 and, since the area of the curved side of the cylinder is 2prh, this gives UE;side ¼ 1 k kh ð2prhÞ ¼ : 2pe0 r e0 Gauss’s law tells you that this must equal qenc /e0, which verifies that the enclosed charge qenc ¼ kh in this case. Example 1.5: Given a symmetric charge distribution, find ~ E: Finding the electric field using Gauss’s law may seem to be a hopeless task. After all, while the electric field does appear in the equation, it is only the normal component that emerges from the dot product, and it is only the integral of that normal component over the entire surface that is proportional to the enclosed charge. Do realistic situations exist in which it is possible to dig the electric field out of its interior position in Gauss’s law? Happily, the answer is yes; you may indeed find the electric field using Gauss’s law, albeit only in situations characterized by high symmetry. Specifically, you can determine the electric field whenever you’re able to design a real or imaginary ‘‘special Gaussian surface’’ that encloses a known amount of charge. A special Gaussian surface is one on which (1) the electric field is either parallel or perpendicular to the surface normal (which allows you to convert the dot product into an algebraic multiplication), and (2) the electric field is constant or zero over sections of the surface (which allows you to remove the electric field from the integral). Of course, the electric field on any surface that you can imagine around arbitrarily shaped charge distributions will not satisfy either of these requirements. But there are situations in which the distribution of charge is sufficiently symmetric that a special Gaussian surface may be imagined. Specifically, the electric field in the vicinity of spherical charge distributions, infinite lines of charge, and infinite planes of charge may be determined by direct application of the integral form of Gauss’s law. Geometries that approximate these ideal conditions, or can be approximated by combinations of them, may also be attacked using Gauss’s law. The following problem shows how to use Gauss’s law to find the electric field around a spherical distribution of charge; the other cases are covered in the problem set, for which solutions are available on the website. 26 A student’s guide to Maxwell’s Equations Problem: Use Gauss’s law to find the electric field at a distance r from the center of a sphere with uniform volume charge density q and radius a. Solution: Consider first the electric field outside the sphere. Since the distribution of charge is spherically symmetric, it is reasonable to expect the electric field to be entirely radial (that is, pointed toward or away from the sphere). If that’s not obvious to you, imagine what would ^ happen if the electric field had a nonradial component (say in the h^ or ’ direction); by rotating the sphere about some arbitrary axis, you’d be able to change the direction of the field. But the charge is uniformly distributed throughout the sphere, so there can be no preferred direction or orientation – rotating the sphere simply replaces one chunk of charge with another, identical chunk – so this can have no effect whatsoever on the electric field. Faced with this conundrum, you are forced to conclude that the electric field of a spherically symmetric charge distribution must be entirely radial. To find the value of this radial field using Gauss’s law, you’ll have to imagine a surface that meets the requirements of a special Gaussian surface; ~ E must be either parallel or perpendicular to the surface normal at all locations, and ~ E must be uniform everywhere on the surface. For a radial electric field, there can be only one choice; your Gaussian surface must be a sphere centered on the charged sphere, as shown in Figure 1.12. Notice that no actual surface need be present, and the special Gaussian surface may be purely imaginary – it is simply a construct that allows you to evaluate the dot product and remove the electric field from the surface integral in Gauss’s law. Since the radial electric field is everywhere parallel to the surface normal, the ~ E^ n term in the integral in Gauss’s law becomes ~ jEjj^ nj cosð0 Þ, and the electric flux over the Gaussian surface S is I UE ¼ ~ E^ n da ¼ S I E da S Since ~ E has no h or u dependence, it must be constant over S, which means it may be removed from the integral: I UE ¼ I E da ¼ E S da ¼ Eð4pr 2 Þ; S where r is the radius of the special Gaussian surface. You can now use Gauss’s law to find the value of the electric field: 27 Gauss’s law for electric fields n Special Gaussian surface Radial electric field n n Charged sphere n Figure 1.12 A special Gaussian around a charged sphere. UE ¼ Eð4pr 2 Þ ¼ or E¼ qenc ; e0 qenc ; 4pe0 r 2 where qenc is the charge enclosed by your Gaussian surface. You can use this expression to find the electric field both outside and inside the sphere. To find the electric field outside the sphere, construct your Gaussian surface with radius r > a so that the entire charged sphere is within the Gaussian surface. This means that the enclosed charge is just the charge density times the entire volume of the charged sphere: qenc ¼ ð4=3Þpa3 q. Thus, E¼ ð4=3Þpa3 q qa3 ¼ ðoutside sphereÞ: 2 4pe0 r 3e0 r 2 To find the electric field within the charged sphere, construct your Gaussian surface with r < a. In this case, the enclosed charge is the charge 28 A student’s guide to Maxwell’s Equations density times the volume of your Gaussian surface: qenc ¼ ð4=3Þpr 3 q. Thus, E¼ ð4=3Þpr 3 q qr ¼ 4pe0 r 2 3e0 ðinside sphereÞ: The keys to successfully employing special Gaussian surfaces are to recognize the appropriate shape for the surface and then to adjust its size to ensure that it runs through the point at which you wish to determine the electric field. Gauss’s law for electric fields 29 1.2 The differential form of Gauss’s law The integral form of Gauss’s law for electric fields relates the electric flux over a surface to the charge enclosed by that surface – but like all of Maxwell’s Equations, Gauss’s law may also be cast in differential form. The differential form is generally written as q ~ ~ r E¼ e0 Gauss’s law for electric fields ðdifferential formÞ: The left side of this equation is a mathematical description of the divergence of the electric field – the tendency of the field to ‘‘flow’’ away from a specified location – and the right side is the electric charge density divided by the permittivity of free space. ~ or the concept of divergence Don’t be concerned if the del operator (r) isn’t perfectly clear to you – these are discussed in the following sections. For now, make sure you grasp the main idea of Gauss’s law in differential form: The electric field produced by electric charge diverges from positive charge and converges upon negative charge. In other words, the only places at which the divergence of the electric field is not zero are those locations at which charge is present. If positive charge is present, the divergence is positive, meaning that the electric field tends to ‘‘flow’’ away from that location. If negative charge is present, the divergence is negative, and the field lines tend to ‘‘flow’’ toward that point. Note that there’s a fundamental difference between the differential and the integral form of Gauss’s law; the differential form deals with the divergence of the electric field and the charge density at individual points in space, whereas the integral form entails the integral of the normal component of the electric field over a surface. Familiarity with both forms will allow you to use whichever is better suited to the problem you’re trying to solve. 30 A student’s guide to Maxwell’s Equations To help you understand the meaning of each symbol in the differential form of Gauss’s law for electric fields, here’s an expanded view: Reminder that del is a vector operator Reminder that the electric field is a vector The charge density in coulombs per cubic meter E= 0 The differential operator called “del” or “nabla” The electric field in N/C The electric permittivity of free space The dot product turns the del operator into the divergence How is the differential form of Gauss’s law useful? In any problem in which the spatial variation of the vector electric field is known at a specified location, you can find the volume charge density at that location using this form. And if the volume charge density is known, the divergence of the electric field may be determined. Gauss’s law for electric fields ~ r 31 Nabla – the del operator An inverted uppercase delta appears in the differential form of all four of Maxwell’s Equations. This symbol represents a vector differential operator called ‘‘nabla’’ or ‘‘del,’’ and its presence instructs you to take derivatives of the quantity on which the operator is acting. The exact form of those ~ derivatives depends on the symbol following the del operator, with ‘‘r’’ ~ ·’’ indicating curl, and r ~ signifying gradient. signifying divergence, ‘‘r Each of these operations is discussed in later sections; for now we’ll just consider what an operator is and how the del operator can be written in Cartesian coordinates. Like all good mathematical operators, del is an action waiting to happen. Just as H tells you to take the square root of anything that ~ is an instruction to take derivatives in three appears under its roof, r directions. Specifically, ~ ^i @ þ ^j @ þ ^k @ ; r @x @y @z ð1:18Þ where ^i, ^j, and ^k are the unit vectors in the direction of the Cartesian coordinates x, y, and z. This expression may appear strange, since in this form it is lacking anything on which it can operate. In Gauss’s law for electric fields, the del operator is dotted into the electric field vector, forming the divergence of ~ E. That operation and its results are described in the next section. 32 A student’s guide to Maxwell’s Equations ~ r Del dot – the divergence The concept of divergence is important in many areas of physics and engineering, especially those concerned with the behavior of vector fields. James Clerk Maxwell coined the term ‘‘convergence’’ to describe the mathematical operation that measures the rate at which electric field lines ‘‘flow’’ toward points of negative electric charge (meaning that positive convergence was associated with negative charge). A few years later, Oliver Heaviside suggested the use of the term ‘‘divergence’’ for the same quantity with the opposite sign. Thus, positive divergence is associated with the ‘‘flow’’ of electric field lines away from positive charge. Both flux and divergence deal with the ‘‘flow’’ of a vector field, but with an important difference; flux is defined over an area, while divergence applies to individual points. In the case of fluid flow, the divergence at any point is a measure of the tendency of the flow vectors to diverge from that point (that is, to carry more material away from it than is brought toward it). Thus points of positive divergence are sources (faucets in situations involving fluid flow, positive electric charge in electrostatics), while points of negative divergence are sinks (drains in fluid flow, negative charge in electrostatics). The mathematical definition of divergence may be understood by considering the flux through an infinitesimal surface surrounding the point of interest. If you were to form the ratio of the flux of a vector field ~ A through a surface S to the volume enclosed by that surface as the volume shrinks toward zero, you would have the divergence of ~ A: I 1 ~ ~ ~ divð~ AÞ ¼ r A lim A ^n da: ð1:19Þ DV!0 DV S While this expression states the relationship between divergence and flux, it is not particularly useful for finding the divergence of a given vector field. You’ll find a more user-friendly mathematical expression for divergence later in this section, but first you should take a look at the vector fields shown in Figure 1.13. To find the locations of positive divergence in each of these fields, look for points at which the flow vectors either spread out or are larger pointing away from the location and shorter pointing toward it. Some authors suggest that you imagine sprinkling sawdust on flowing water to assess the divergence; if the sawdust is dispersed, you have selected a point of positive divergence, while if it becomes more concentrated, you’ve picked a location of negative divergence. 33 Gauss’s law for electric fields (a) (b) (c) 4 6 3 5 2 7 1 Figure 1.13 Vector fields with various values of divergence. Using such tests, it is clear that locations such as 1 and 2 in Figure 1.13(a) and location 3 in Figure 1.13(b) are points of positive divergence, while the divergence is negative at point 4. The divergence at various points in Figure 1.13(c) is less obvious. Location 5 is obviously a point of positive divergence, but what about locations 6 and 7? The flow lines are clearly spreading out at those locations, but they’re also getting shorter at greater distance from the center. Does the spreading out compensate for the slowing down of the flow? Answering that question requires a useful mathematical form of the divergence as well as a description of how the vector field varies from place to place. The differential form of the mathematical operation of ~ on a vector ~ divergence or ‘‘del dot’’ (r) A in Cartesian coordinates is @ @ @ ~ ~ r A ¼ ^i þ ^j þ ^k ^iAx þ ^jAy þ ^kAz ; @x @y @z and, since ^i ^i ¼ ^j ^j ¼ ^k ^k ¼ 1; this is @Ax @Ay @Az ~ ~ r A¼ þ þ : @x @y @z ð1:20Þ Thus, the divergence of the vector field ~ A is simply the change in its x-component along the x-axis plus the change in its y-component along the y-axis plus the change in its z-component along the z-axis. Note that the divergence of a vector field is a scalar quantity; it has magnitude but no direction. 34 A student’s guide to Maxwell’s Equations You can now apply this to the vector fields in Figure 1.13. In Figure 1.13(a), assume that the magnitude of the vector field varies sinusoidally along the x-axis (which is vertical in this case) as ~ A ¼ sinðpxÞ^i while remaining constant in the y- and z-directions. Thus, @Ax ~ ~ r A¼ ¼ p cosðpxÞ; @x since Ay and Az are zero. This expression is positive for 0 < x < 12, 0 at x ¼ 12, and negative for 12 < x < 32, just as your visual inspection suggested. Now consider Figure 1.13(b), which represents a slice through a spherically symmetric vector field with amplitude increasing as the square of the distance from the origin. Thus ~ A ¼ r 2^r . Since r2 ¼ (x2 þ y2 þ z2) and x^i þ y^j þ z^k ^r ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ; x2 þ y2 þ z2 this means x^i þ y^j þ z^k ~ A ¼ r 2^r ¼ ðx2 þ y2 þ z2 Þ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ; x2 þ y2 þ z2 and ð1=2Þ @Ax 2 1 2 2 2 ð1=2Þ ¼ x þy þz þx ð2xÞ: x þ y2 þ z2 2 @x Doing likewise for the y- and z-components and adding yields ð1=2Þ x2 þ y2 þ z2 1=2 ~ ~ þ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ 4 x2 þ y2 þ z2 ¼ 4r: r A ¼ 3 x2 þ y2 þ z2 2 2 2 x þy þz Thus, the divergence in the vector field in Figure 1.13(b) is increasing linearly with distance from the origin. Finally, consider the vector field in Figure 1.13(c), which is similar to the previous case but with the amplitude of the vector field decreasing as the square of the distance from the origin. The flow lines are spreading out as they were in Figure 1.13(b), but in this case you might suspect that the decreasing amplitude of the vector field will affect the value of the divergence. Since ~ A ¼ ð1=r 2 Þ^r , ~ A¼ ðx2 1 x^i þ y^j þ z^k x^i þ y^j þ z^k pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ ; 2 2 þ y þ z Þ x2 þ y2 þ z2 ðx2 þ y2 þ z2 Þð3=2Þ 35 Gauss’s law for electric fields and ð5=2Þ @Ax 1 3 2 ¼ –x ð2xÞ; x þ y2 þ z2 ð3=2Þ 2 2 2 2 @x ðx þ y þ z Þ Adding in the y- and z-derivatives gives ~ ~ r A¼ 3 ðx 2 þ y 2 þ z2 Þ ð3=2Þ 3 x2 þ y 2 þ z2 ðx2 þ y2 þ z2 Þ ð5=2Þ ¼ 0: This validates the suspicion that the reduced amplitude of the vector field with distance from the origin may compensate for the spreading out of the flow lines. Note that this is true only for the case in which the amplitude of the vector field falls off as 1/r2 (this case is especially relevant for the electric field, which you’ll find in the next section). As you consider the divergence of the electric field, you should remember that some problems may be solved more easily using nonCartesian coordinate systems. The divergence may be calculated in cylindrical and spherical coordinate systems using 1@ 1 @Af @Az ~ ~ r A ¼ ðrAr Þ þ þ r @r r @f @z ðcylindricalÞ; ð1:21Þ and 1 @ 1 @ ~ ~ ðAh sin hÞ r A ¼ 2 ðr 2 Ar Þ þ r @r r sin h @h 1 @Af þ ðsphericalÞ: r sin h @f ð1:22Þ If you doubt the efficacy of choosing the proper coordinate system, you should rework the last two examples in this section using spherical coordinates. 36 A student’s guide to Maxwell’s Equations ~ ~ r E The divergence of the electric field This expression is the entire left side of the differential form of Gauss’s law, and it represents the divergence of the electric field. In electrostatics, all electric field lines begin on points of positive charge and terminate on points of negative charge, so it is understandable that this expression is proportional to the electric charge density at the location under consideration. Consider the electric field of the positive point charge; the electric field lines originate on the positive charge, and you know from Table 1.1 that the electric field is radial and decreases as 1/r2: 1 q ~ ^r : E¼ 4pe0 r 2 This is analogous to the vector field shown in Figure 1.13(c), for which the divergence is zero. Thus, the spreading out of the electric field lines is exactly compensated by the 1/r2 reduction in field amplitude, and the divergence of the electric field is zero at all points away from the origin. The reason the origin (where r ¼ 0) is not included in the previous analysis is that the expression for the divergence includes terms containing r in the denominator, and those terms become problematic as r approaches zero. To evaluate the divergence at the origin, use the formal definition of divergence: I 1 ~ ~ ~ r E lim E ^n da: DV!0 DV S Considering a special Gaussian surface surrounding the point charge q, this is 0 1 I 1 q q 2 ~ ~ A ¼ lim 1 r E lim @ da ð4pr Þ DV!0 DV 4pe0 r 2 DV!0 DV 4pe0 r 2 S 1 q ¼ lim : DV!0 DV e0 But q/DV is just the average charge density over the volume DV, and as DV shrinks to zero, this becomes equal to q, the charge density at the origin. Thus, at the origin the divergence is q ~ ~ r E¼ ; e0 in accordance with Gauss’s law. Gauss’s law for electric fields 37 It is worth your time to make sure you understand the significance of this last point. A casual glance at the electric field lines in the vicinity of a point charge suggests that they ‘‘diverge’’ everywhere (in the sense of getting farther apart). But as you’ve seen, radial vector fields that decrease in amplitude as 1/r2 actually have zero divergence everywhere except at the source. The key factor in determining the divergence at any point is not simply the spacing of the field lines at that point, but whether the flux out of an infinitesimally small volume around the point is greater than, equal to, or less than the flux into that volume. If the outward flux exceeds the inward flux, the divergence is positive at that point. If the outward flux is less than the inward flux, the divergence is negative, and if the outward and inward fluxes are equal the divergence is zero at that point. In the case of a point charge at the origin, the flux through an infinitesimally small surface is nonzero only if that surface contains the point charge. Everywhere else, the flux into and out of that tiny surface must be the same (since it contains no charge), and the divergence of the electric field must be zero. 38 A student’s guide to Maxwell’s Equations ~ ~ r E ¼ q=e0 Applying Gauss’s law (differential form) The problems you’re most likely to encounter that can be solved using the differential form of Gauss’s law involve calculating the divergence of the electric field and using the result to determine the charge density at a specified location. The following examples should help you understand how to solve problems of this type. Example 1.6: Given an expression for the vector electric field, find the divergence of the field at a specified location. Problem: If the vector field of Figure 1.13(a) were changed to p p ~ A ¼ sin y ^i sin x ^j; 2 2 in the region 0.5 < x < þ 0.5 and 0.5 < y < þ 0.5, how would the field lines be different from those of Figure 1.13(a), and what is the divergence in this case? Solution: When confronted with a problem like this, you may be tempted to dive in and immediately begin taking derivatives to determine the divergence of the field. A better approach is to think about the field for a moment and to attempt to visualize the field lines – a task that may be difficult in some cases. Fortunately, there exist a variety of computational tools such as MATLAB and its freeware cousin Octave that are immensely helpful in revealing the details of a vector field. Using the ‘‘quiver’’ command in MATLAB shows that the field looks as shown in Figure 1.14. If you’re surprised by the direction of the field, consider that the x-component of the field depends on y (so the field points to the right above the x-axis and to the left below the x-axis), while the y-component of the field depends on the negative of x (so the field points up on the left of the y-axis and down on the right of the y-axis). Combining these features leads to the field depicted in Figure 1.14. Examining the field closely reveals that the flow lines neither converge nor diverge, but simply circulate back on themselves. Calculating the divergence confirms this @ h p i @ h p i ~ ~ r A¼ sin y – sin x ¼ 0: @x 2 @y 2 39 Gauss’s law for electric fields 0.5 0.4 0.3 0.2 y 0.1 0 –0.1 –0.2 –0.3 –0.4 –0.5 –0.5 –0.4 –0.3 –0.2 –0.1 0 x 0.1 0.2 0.3 0.4 0.5 Figure 1.14 Vector field ~ A ¼ sinðp2 yÞ^i sinðp2 xÞ^j: Electric fields that circulate back on themselves are produced not by electric charge, but rather by changing magnetic fields. Such ‘‘solenoidal’’ fields are discussed in Chapter 3. Example 1.7: Given the vector electric field in a specified region, find the density of electric charge at a location within that region. Problem: Find the charge density at x ¼ 2 m and x ¼ 5 m if the electric field in the region is given by V ~ E ¼ ax2^i m for x ¼ 0 to 3 m; and V ~ E ¼ b^i m for x > 3 m: Solution: By Gauss’s law, in the region x ¼ 0 to 3 m, q @ @ @ ~ ~ r E ¼ ¼ ^i þ ^j þ ^k ðax2^iÞ; e0 @x @y @k 40 A student’s guide to Maxwell’s Equations q @ðax2 Þ ¼ 2xa; ¼ e0 @x and q ¼ 2xae0 : Thus at x ¼ 2 m, q ¼ 4ae0. In the region x > 3 m, q @ @ @ ~ ~ r E ¼ ¼ ^i þ ^j þ ^k ðb^iÞ ¼ 0; e0 @x @y @k so q ¼ 0 at x ¼ 5 m. Problems The following problems will test your understanding of Gauss’s law for electric fields. Full solutions are available on the book’s website. 1.1 Find the electric flux through the surface of a sphere containing 15 protons and 10 electrons. Does the size of the sphere matter? 1.2 A cube of side L contains a flat plate with variable surface charge density of r ¼ 3xy. If the plate extends from x ¼ 0 to x ¼ L and from y ¼ 0 to y ¼ L, what is the total electric flux through the walls of the cube? z L s L y L x 1.3 Find the total electric flux through a closed cylinder containing a line charge along its axis with linear charge density k ¼ k0(1x/h) C/m if the cylinder and the line charge extend from x ¼ 0 to x ¼ h. Gauss’s law for electric fields 41 1.4 What is the flux through any closed surface surrounding a charged sphere of radius a0 with volume charge density of q ¼ q0(r/a0), where r is the distance from the center of the sphere? 1.5 A circular disk with surface charge density 2 · 1010 C/m2 is surrounded by a sphere with radius of 1 m. If the flux through the sphere is 5.2 · 102 Vm, what is the diameter of the disk? 1.6 A 10 cm · 10 cm flat plate is located 5 cm from a point charge of 108 C. What is the electric flux through the plate due to the point charge? 10 cm 5 cm 10 cm 1.7 Find the electric flux through a half-cylinder of height h owing to an infinitely long line charge with charge density k running along the axis of the cylinder. 1.8 A proton rests at the center of the rim of a hemispherical bowl of radius R. What is the electric flux through the surface of the bowl? 1.9 Use a special Gaussian surface around an infinite line charge to find the electric field of the line charge as a function of distance. 42 A student’s guide to Maxwell’s Equations 1.10 Use a special Gaussian surface to prove that the magnitude of the electric field of an infinite flat plane with surface charge density r is j~ Ej ¼ r=2e0 . 1.11 Find the divergence of the field given by ~ A ¼ ð1=rÞ^r in spherical coordinates. 1.12 Find the divergence of the field given by ~ A ¼ r^r in spherical coordinates. 1.13 Given the vector field p^ ~ A ¼ cos py – i þ sinð pxÞ^j; 2 sketch the field lines and find the divergence of the field. 1.14 Find the charge density in a region for which the electric field in cylindrical coordinates is given by az ^ þ cr 2 z2^z ~ E ¼ ^r þ br f r 1.15 Find the charge density in a region for which the electric field in spherical coordinates is given by b cos ðhÞ ^ ^ ~ E ¼ ar 2^r þ h þ cf: r 2 Gauss’s law for magnetic fields Gauss’s law for magnetic fields is similar in form but different in content from Gauss’s law for electric fields. For both electric and magnetic fields, the integral form of Gauss’s law involves the flux of the field over a closed surface, and the differential form specifies the divergence of the field at a point. The key difference in the electric field and magnetic field versions of Gauss’s law arises because opposite electric charges (called ‘‘positive’’ and ‘‘negative’’) may be isolated from one another, while opposite magnetic poles (called ‘‘north’’ and ‘‘south’’) always occur in pairs. As you might expect, the apparent lack of isolated magnetic poles in nature has a profound impact on the behavior of magnetic flux and on the divergence of the magnetic field. 2.1 The integral form of Gauss’s law Notation differs among textbooks, but the integral form of Gauss’s law is generally written as follows: I ~ B^ n da ¼ 0 Gauss’s law for magnetic fields ðintegral formÞ: S As described in the previous chapter, the left side of this equation is a mathematical description of the flux of a vector field through a closed surface. In this case, Gauss’s law refers to magnetic flux – the number of magnetic field lines – passing through a closed surface S. The right side is identically zero. In this chapter, you will see why this law is different from the electric field case, and you will find some examples of how to use the magnetic 43 44 A Student’s guide to Maxwell’s Equations version to solve problems – but first you should make sure you understand the main idea of Gauss’s law for magnetic fields: The total magnetic flux passing through any closed surface is zero. In other words, if you have a real or imaginary closed surface of any size or shape, the total magnetic flux through that surface must be zero. Note that this does not mean that zero magnetic field lines penetrate the surface – it means that for every magnetic field line that enters the volume enclosed by the surface, there must be a magnetic field line leaving that volume. Thus the inward (negative) magnetic flux must be exactly balanced by the outward (positive) magnetic flux. Since many of the symbols in Gauss’s law for magnetic fields are the same as those covered in the previous chapter, in this chapter you’ll find only those symbols peculiar to this law. Here’s an expanded view: Reminder that the magnetic field is a vector Reminder that this integral is over a closed surface Dot product tells you to find the part of B parallel to nˆ (perpendicular to the surface) ∫B S The unit vector normal to the surface nˆ da = 0 The magnetic field in Teslas An increment of surface area in m2 Reminder that this is a surface integral (not a volume or a line integral) Tells you to sum up the contributions from each portion of the surface Gauss’s law for magnetic fields arises directly from the lack of isolated magnetic poles (‘‘magnetic monopoles’’) in nature. Were such individual poles to exist, they would serve as the sources and sinks of magnetic field lines, just as electric charge does for electric field lines. In that case, enclosing a single magnetic pole within a closed surface would produce nonzero flux through the surface (exactly as you can produce nonzero electric flux by enclosing an electric charge). To date, all efforts to detect magnetic monopoles have failed, and every magnetic north pole is accompanied by a magnetic south pole. Thus the right side of Gauss’s law for magnetic fields is identically zero. Knowing that the total magnetic flux through a closed surface must be zero may allow you to solve problems involving complex surfaces, particularly if the flux through one portion of the surface can be found by integration. Gauss’s law for magnetic fields 45 ~ B The magnetic field Just as the electric field may be defined by considering the electric force on a small test charge, the magnetic field may be defined using the magnetic force experienced by a moving charged particle. As you may recall, charged particles experience magnetic force only if they are in motion with respect to the magnetic field, as shown by the Lorentz equation for magnetic force: ~ FB ¼ q~ v ·~ B ð2:1Þ v is the particle’s where ~ FB is the magnetic force, q is the particle’s charge, ~ velocity (with respect to ~ B), and ~ B is the magnetic field. Using the definition of the vector cross-product which says that ~ a ·~ b ¼ j~ ajj~ bj sinðhÞ, where h is the angle between ~ a and ~ b, the magnitude of the magnetic field may be written as j~ FB j j~ Bj ¼ qj~ vj sinðhÞ ð2:2Þ where h is the angle between the velocity vector ~ v and the magnetic field ~ B. The terminology for magnetic quantities is not as standardized as that of electric quantities, so you are likely to find texts that refer to ~ B as the ‘‘magnetic induction’’ or the ‘‘magnetic flux density.’’ Whatever it is called, ~ B has units equivalent to N/(C m/s), which include Vs/m2, N/(Am), kg/(Cs), or most simply, Tesla (T). Comparing Equation 2.2 to the relevant equation for the electric field, Equation (1.1), several important distinctions between magnetic and electric fields become clear: Like the electric field, the magnetic field is directly proportional to the magnetic force. But unlike ~ E, which is parallel or antiparallel to the electric force, the direction of ~ B is perpendicular to the magnetic force. ~ Like E, the magnetic field may be defined through the force experienced by a small test charge, but unlike ~ E, the speed and direction of the test charge must be taken into consideration when relating magnetic forces and fields. Because the magnetic force is perpendicular to the velocity at every instant, the component of the force in the direction of the displacement is zero, and the work done by the magnetic field is therefore always zero. Whereas electrostatic fields are produced by electric charges, magnetostatic fields are produced by electric currents. 46 A Student’s guide to Maxwell’s Equations I B N B B I S Current loop Current-carrying straight wire Bar magnet B I I Solenoid Torus Horseshoe magnet Figure 2.1 Examples of magnetic fields. Magnetic fields may be represented using field lines whose density in a plane perpendicular to the line direction is proportional to the strength of the field. Examples of several magnetic fields relevant to the application of Gauss’s law are shown in Figure 2.1. Here are a few rules of thumb that will help you visualize and sketch the magnetic fields produced by currents: Magnetic field lines do not originate and terminate on charges; they form closed loops. The magnetic field lines that appear to originate on the north pole and terminate on the south pole of a magnet are actually continuous loops (within the magnet, the field lines run between the poles). The net magnetic field at any point is the vector sum of all magnetic fields present at that point. Magnetic field lines can never cross, since that would indicate that the field points in two different directions at the same location – if the fields from two or more sources overlap at the same location, they add (as vectors) to produce a single, total field at that point. 47 Gauss’s law for magnetic fields Table 2.1. Magnetic field equations for simple objects l I ~ ^ B¼ 0 ’ 2pr l Id~l · ^r d~ B¼ 0 4p r 2 l0 IR2 ~ ^x B¼ 2 2ðx þ R2 Þ3=2 l NI ~ B ¼ 0 ^x ðinsideÞ l l NI ~ ^ (inside) B¼ 0 ’ 2pr Infinite straight wire carrying current I (at distance r) Segment of straight wire carrying current I (at distance r) Circular loop of radius R carrying current I (loop in yz plane, at distance x along x-axis) Solenoid with N turns and length l carrying current I Torus with N turns and radius r carrying current I P r r^ dl Point at which magnetic field is determined Current element Figure 2.2 Geometry for Biot–Savart law. All static magnetic fields are produced by moving electric charge. The contribution d~ B to the magnetic field at a specified point P from a small element of electric current is given by the Biot–Savart law: l Id~l · ^r d~ B¼ 0 4p r 2 In this equation, l0 is the permeability of free space, I is the current through the small element, d~l is a vector with the length of the current element and pointing in the direction of the current, ^r is a unit vector pointing from the current element to the point P at which the field is being calculated, and r is the distance between the current element and P, as shown in Figure 2.2. Equations for the magnetic field in the vicinity of some simple objects may be found in Table 2.1. 48 A Student’s guide to Maxwell’s Equations H S ~ B ^n da The magnetic flux through a closed surface Like the electric flux UE, the magnetic flux UB through a surface may be thought of as the ‘‘amount’’ of magnetic field ‘‘flowing’’ through the surface. How this quantity is calculated depends on the situation: UB ¼ j~ Bj · (surface area) ~ B uniform and perpendicular to S; ð2:3Þ ~ B uniform and at an angle to S; ð2:4Þ ~ B nonuniform and at variable angle to S: ð2:5Þ UB ¼ ~ B^ n · (surface area) Z ~ B^ n da UB ¼ S Magnetic flux, like electric flux, is a scalar quantity, and in the magnetic case, the units of flux have been given the special name ‘‘webers’’ (abbreviated Wb and which, by any of the relations shown above, must be equivalent to T m2). As in the case of electric flux, the magnetic flux through a surface may be considered to be the number of magnetic field lines penetrating that surface. When you think about the number of magnetic field lines through a surface, don’t forget that magnetic fields, like electric fields, are actually continuous in space, and that ‘‘number of field lines’’ only has meaning once you’ve established a relationship between the number of lines you draw and the strength of the field. When considering magnetic flux through a closed surface, it is especially important to remember the caveat that surface penetration is a twoway street, and that outward flux and inward flux have opposite signs. Thus equal amounts of outward (positive) flux and inward (negative) flux will cancel, producing zero net flux. The reason that the sign of outward and inward flux is so important in the magnetic case may be understood by considering a small closed surface placed in any of the fields shown in Figure 2.1. No matter what shape of surface you choose, and no matter where in the magnetic field you place that surface, you’ll find that the number of field lines entering the volume enclosed by the surface is exactly equal to the number of field lines leaving that volume. If this holds true for all magnetic fields, it can only mean that the net magnetic flux through any closed surface must always be zero. Of course, it does hold true, because the only way to have field lines enter a volume without leaving it is to have them terminate within the Gauss’s law for magnetic fields 49 B Figure 2.3 Magnetic flux lines penetrating closed surfaces. volume, and the only way to have field lines leave a volume without entering it is to have them originate within the volume. But unlike electric field lines, magnetic field lines do not originate and terminate on charges – instead, they circulate back on themselves, forming continuous loops. If one portion of a loop passes through a closed surface, another portion of that same loop must pass through the surface in the opposite direction. Thus the outward and inward magnetic flux must be equal and opposite through any closed surface. Consider the closer view of the field produced by a bar magnet shown in Figure 2.3. Irrespective of the shape and location of the closed surfaces placed in the field, all field lines entering the enclosed volume are offset by an equal number of field lines leaving that volume. The physical reasoning behind Gauss’s law should now be clear: the net magnetic flux passing through any closed surface must be zero because magnetic field lines always form complete loops. The next section shows you how to use this principle to solve problems involving closed surfaces and the magnetic field. 50 A Student’s guide to Maxwell’s Equations H S ~ B ^n da ¼ 0 Applying Gauss’s law (integral form) In situations involving complex surfaces and fields, finding the flux by integrating the normal component of the magnetic field over a specified surface can be quite difficult. In such cases, knowing that the total magnetic flux through a closed surface must be zero may allow you to simplify the problem, as demonstrated by the following examples. Example 2.1: Given an expression for the magnetic field and a surface geometry, find the flux through a specified portion of that surface. Problem: A closed cylinder of height h and radius R is placed in a magnetic field given by ~ B ¼ B0 ð ^j ^kÞ. If the axis of the cylinder is aligned along the z-axis, find the flux through (a) the top and bottom surfaces of the cylinder and (b) the curved surface of the cylinder. Solution: Gauss’s law tells you that the magnetic flux through the entire surface must be zero, so if you’re able to figure out the flux through some portions of the surface, you can deduce the flux through the other portions. In this case, the flux through the top and bottom of the cylinder are relatively easy to find; whatever additional amount it takes to make the total flux equal to zero must come from the curved sides of the cylinder. Thus UB;Top þ UB;Bottom þ UB;Sides ¼ 0: The magnetic flux through any surface is Z B^ n da: UB ¼ ~ For the top surface, ^ n ¼ ^k, so S ~ B^ n ¼ ðB0^j B0 ^kÞ ^k ¼ B0 : Thus Z ~ B^ n da ¼ B0 UB;Top ¼ S Z da ¼ B0 ðpR2 Þ: S A similar analysis for the bottom surface (for which n^ ¼ ^k) gives Z Z ~ n da ¼ þB0 da ¼ þB0 ðpR2 Þ: UB;Bottom ¼ B ^ S S Since UB,Top ¼ UB,Bottom, you can conclude that UB,Sides ¼ 0. Gauss’s law for magnetic fields 51 Example 2.2: Given the current in a long wire, find the magnetic flux through nearby surfaces Problem: Find the magnetic flux through the curved surface of a half-cylinder near a long, straight wire carrying current I. z I h y1 y x Solution: At distance r from a current-carrying wire, the magnetic field is given by lI ~ ^ B ¼ 0 ’; 2pr which means that the magnetic field lines make circles around the wire, entering the half-cylinder through the flat surface and leaving through the curved surface. Gauss’s law tells you that the total magnetic flux through all faces of the half-cylinder must be zero, so the amount of (negative) flux through the flat surface must equal the amount of (positive) flux leaving the curved surface. To find the flux through the flat surface, use the expression for flux Z UB ¼ ~ B^ n da: S ^ so In this case, ^ n ¼ ’, l I l I 0 ^ ð’Þ ^ ¼ 0 : ’ 2pr 2pr To integrate over the flat face of the half-cylinder, notice that the face lies in the yz plane, and an element of surface area is therefore da ¼ dy dz. Notice also that on the flat face the distance increment dr ¼ dy, so da ¼ dr dz and the flux integral is Z Z y1 þ2R Z Z l0 I l I h dr UB;Flat ¼ ~ dr dz ¼ 0 dz : B^ n da ¼ 2p z¼0 r¼y1 r S 2pr ~ B^ n¼ S 52 A Student’s guide to Maxwell’s Equations Thus UB;Flat l0 I y1 þ 2R l0 Ih 2R ln ln 1 þ ¼ ðhÞ ¼ : 2p y1 2p y1 Since the total magnetic flux through this closed surface must be zero, this means that the flux through the curved side of the half-cylinder is l0 Ih 2R ln 1 þ UB;Curved side ¼ : 2p y1 Gauss’s law for magnetic fields 53 2.2 The differential form of Gauss’s law The continuous nature of magnetic field lines makes the differential form of Gauss’s law for magnetic fields quite simple. The differential form is written as ~ ~ r B ¼ 0 Gauss’s law for magnetic fields ðdifferential formÞ: The left side of this equation is a mathematical description of the divergence of the magnetic field – the tendency of the magnetic field to ‘‘flow’’ more strongly away from a point than toward it – while the right side is simply zero. The divergence of the magnetic field is discussed in detail in the following section. For now, make sure you grasp the main idea of Gauss’s law in differential form: The divergence of the magnetic field at any point is zero. One way to understand why this is true is by analogy with the electric field, for which the divergence at any location is proportional to the electric charge density at that location. Since it is not possible to isolate magnetic poles, you can’t have a north pole without a south pole, and the ‘‘magnetic charge density’’ must be zero everywhere. This means that the divergence of the magnetic field must also be zero. To help you understand the meaning of each symbol in Gauss’s law for magnetic fields, here is an expanded view: Reminder that the del operator is a vector Reminder that the magnetic field is a vector B=0 The magnetic field in Teslas The differential operator called “del” or “nabla” The dot product turns the del operator into the divergence 54 A Student’s guide to Maxwell’s Equations ~ ~ r B The divergence of the magnetic field This expression is the entire left side of the differential form of Gauss’s law, and it represents the divergence of the magnetic field. Since divergence is by definition the tendency of a field to ‘‘flow’’ away from a point more strongly than toward that point, and since no point sources or sinks of the magnetic field have ever been found, the amount of ‘‘incoming’’ field is exactly the same as the amount of ‘‘outgoing’’ field at every point. So it should not surprise you to find that the divergence of ~ B is always zero. To verify this for the case of the magnetic field around a long, currentcarrying wire, take the divergence of the expression for the wire’s magnetic field as given in Table 2.1: ~ ~ ~ l0 I ’ ^ : divð~ BÞ ¼ r B ¼r ð2:6Þ 2pr This is most easily determined using cylindrical coordinates: 1@ 1 @Bf @Bz ~ ~ r B¼ þ : ðrBr Þ þ r @r r @f @z ð2:7Þ which, since ~ B has only a u-component, is 1 @ ðl0 I=2pr Þ ~ ~ ¼ 0: r B¼ r @’ ð2:8Þ You can understand this result using the following reasoning: since the magnetic field makes circular loops around the wire, it has no radial or z-dependence. And since the u-component has no u-dependence (that is, the magnetic field has constant amplitude around any circular path centered on the wire), the flux away from any point must be the same as the flux toward that point. This means that the divergence of the magnetic field is zero everywhere. Vector fields with zero divergence are called ‘‘solenoidal’’ fields, and all magnetic fields are solenoidal. Gauss’s law for magnetic fields ~ ~ r B¼0 55 Applying Gauss’s law (differential form) Knowing that the divergence of the magnetic field must be zero allows you to attack problems involving the spatial change in the components of a magnetic field and to determine whether a specified vector field could be a magnetic field. This section has examples of such problems. Example 2.3: Given incomplete information about the components of a magnetic field, use Gauss’s law to establish relationships between those components Problem: A magnetic field is given by the expression ~ B ¼ axz ^i þ byz ^j þ c^k What is the relationship between a and b? Solution: You know from Gauss’s law for magnetic fields that the divergence of the magnetic field must be zero. Thus @Bx @By @Bz ~ ~ r B¼ þ þ ¼ 0: @x @y @z Thus @ðaxzÞ @ðbyzÞ @c þ þ ¼0 @x @y @z and az þ bz þ 0 ¼ 0; which means that a ¼ b. Example 2.4: Given an expression for a vector field, determine whether that field could be a magnetic field. Problem: A vector field is given by the expression ~ Aðx; yÞ ¼ a cosðbxÞ^i þ aby sinðbxÞ^j: Could this field be a magnetic field? Solution: Gauss’s law tells you that the divergence of all magnetic fields must be zero, and checking the divergence of this vector field gives 56 A Student’s guide to Maxwell’s Equations @ @ ~ ~ r A ¼ ½a cosðbxÞ þ ½aby sinðbxÞ @x @y ¼ ab sinðbxÞ þ ab sinðbxÞ ¼ 0 which indicates that ~ A could represent a magnetic field. Problems The following problems will check your understanding of Gauss’s law for magnetic fields. Full solutions are available on the book’s website. 2.1 Find the magnetic flux produced by the magnetic field ~ B ¼ 5^i 3^j þ 4^knT through the top, bottom, and side surfaces of the flared cylinder shown in the figure. z B Rtop Rbottom y x 2.2 What is the change in magnetic flux through a 10 cm by 10 cm square lying 20 cm from a long wire carrying a current that increases from 5 to 15 mA? Assume that the wire is in the plane of the square and parallel to the closest side of the square. 2.3 Find the magnetic flux through all five surfaces of the wedge shown in the figure if the magnetic field in the region is given by ~ B ¼ 0:002^i þ 0:003^j T; and show that the total flux through the wedge is zero. z 130 cm y 70 cm 50 cm x Gauss’s law for magnetic fields 57 2.4 Find the flux of the Earth’s magnetic field through each face of a cube with 1-m sides, and show that the total flux through the cube is zero. Assume that at the location of the cube the Earth’s magnetic field has amplitude of 4 · 105 T and points upward at an angle of 30 with respect to the horizontal. You may orient the cube in any way you choose. 2.5 A cylinder of radius r0 and height h is placed inside an ideal solenoid with the cylinder’s axis parallel to the axis of the solenoid. Find the flux through the top, bottom, and curved surfaces of the cylinder and show that the total flux through the cylinder is zero. 2.6 Determine whether the vector fields given by the following expressions in cylindrical coordinates could be magnetic fields: (a) a ~ Aðr; ’; zÞ ¼ cos2 ð’Þ^r ; r (b) a ~ Aðr; ’; zÞ ¼ 2 cos2 ð’Þ^r : r 3 Faraday’s law In a series of epoch-making experiments in 1831, Michael Faraday demonstrated that an electric current may be induced in a circuit by changing the magnetic flux enclosed by the circuit. That discovery is made even more useful when extended to the general statement that a changing magnetic field produces an electric field. Such ‘‘induced’’ electric fields are very different from the fields produced by electric charge, and Faraday’s law of induction is the key to understanding their behavior. 3.1 The integral form of Faraday’s law In many texts, the integral form of Faraday’s law is written as I d ~ E d~l ¼ dt C Z ~ B^ n da Faraday’s law ðintegral formÞ: S Some authors feel that this form is misleading because it confounds two distinct phenomena: magnetic induction (involving a changing magnetic field) and motional electromotive force (emf) (involving movement of a charged particle through a magnetic field). In both cases, an emf is produced, but only magnetic induction leads to a circulating electric field in the rest frame of the laboratory. This means that this common version of Faraday’s law is rigorously correct only with the caveat that ~ E rep~ resents the electric field in the rest frame of each segment d l of the path of integration. 58 59 Faraday’s law A version of Faraday’s law that separates the two effects and makes clear the connection between electric field circulation and a changing magnetic field is emf ¼ I C ~ E d~l ¼ Z S d dt Z ~ B^ n da Flux rule, S @~ B ^ n da @t Faraday’s law ðalternate formÞ: Note that in this version of Faraday’s law the time derivative operates only on the magnetic field rather than on the magnetic flux, and both ~ E and ~ B are measured in the laboratory reference frame. Don’t worry if you’re uncertain of exactly what emf is or how it is related to the electric field; that’s all explained in this chapter. There are also examples of how to use the flux rule and Faraday’s law to solve problems involving induction – but first you should make sure you understand the main idea of Faraday’s law: Changing magnetic flux through a surface induces an emf in any boundary path of that surface, and a changing magnetic field induces a circulating electric field. In other words, if the magnetic flux through a surface changes, an electric field is induced along the boundary of that surface. If a conducting material is present along that boundary, the induced electric field provides an emf that drives a current through the material. Thus quickly poking a bar magnet through a loop of wire generates an electric field within that wire, but holding the magnet in a fixed position with respect to the loop induces no electric field. And what does the negative sign in Faraday’s law tell you? Simply that the induced emf opposes the change in flux – that is, it tends to maintain the existing flux. This is called Lenz’s law and is discussed later in this chapter. 60 A student’s guide to Maxwell’s Equations Here’s an expanded view of the standard form of Faraday’s law: Reminder that the electric field is a vector The magnetic flux through any surface bounded by C Dot product tells you to find the part of E parallel to dl (along path C) An incremental segment of path C ∫ d dt E dl = – C ∫B S The electric field in V/m Tells you to sum up the contributions from each portion of the closed path C in a direction given by the right-hand rule nˆ da The rate of change with time Reminder that this is a line integral (not a surface or a volume integr