A Student’s Guide To Maxwell’s Equations Daniel Fleisch Student's Maxwell's

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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.
daniel fleisch 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
First published in print format
ISBN-13 978-0-521-87761-9
ISBN-13 978-0-511-39308-2
© D. Fleisch 2008
2008
Information on this title: www.cambridge.org/9780521877619
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
p
ermission of Cambrid
g
e University Press.
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
g
uarantee that any content on such websites is, or will remain, accurate or a
pp
ro
p
riate.
Published in the United States of America by Cambridge University Press, New York
www.cambridge.org
eBook (EBL)
hardback
Contents
Preface page vii
Acknowledgments ix
1 Gauss’s law for electric fields 1
1.1 The integral form of Gauss’s law 1
The electric field 3
The dot product 6
The unit normal vector 7
The component of ~
Enormal to a surface 8
The surface integral 9
The flux of a vector field 10
The electric flux through a closed surface 13
The enclosed charge 16
The permittivity of free space 18
Applying Gauss’s law (integral form) 20
1.2 The differential form of Gauss’s law 29
Nabla – the del operator 31
Del dot – the divergence 32
The divergence of the electric field 36
Applying Gauss’s law (differential form) 38
2 Gauss’s law for magnetic fields 43
2.1 The integral form of Gauss’s law 43
The magnetic field 45
The magnetic flux through a closed surface 48
Applying Gauss’s law (integral form) 50
2.2 The differential form of Gauss’s law 53
The divergence of the magnetic field 54
Applying Gauss’s law (differential form) 55
v
3 Faraday’s law 58
3.1 The integral form of Faraday’s law 58
The induced electric field 62
The line integral 64
The path integral of a vector field 65
The electric field circulation 68
The rate of change of flux 69
Lenz’s law 71
Applying Faraday’s law (integral form) 72
3.2 The differential form of Faraday’s law 75
Del cross – the curl 76
The curl of the electric field 79
Applying Faraday’s law (differential form) 80
4 The Ampere–Maxwell law 83
4.1 The integral form of the Ampere–Maxwell law 83
The magnetic field circulation 85
The permeability of free space 87
The enclosed electric current 89
The rate of change of flux 91
Applying the Ampere–Maxwell law (integral form) 95
4.2 The differential form of the Ampere–Maxwell law 101
The curl of the magnetic field 102
The electric current density 105
The displacement current density 107
Applying the Ampere–Maxwell law (differential form) 108
5 From Maxwell’s Equations to the wave equation 112
The divergence theorem 114
Stokes’ theorem 116
The gradient 119
Some useful identities 120
The wave equation 122
Appendix: Maxwell’s Equations in matter 125
Further reading 131
Index 132
Contentsvi
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 dif-
ference 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
meaning of ~
E^
nin Gauss’s Law or why it is only the enclosed currents
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
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 text-
books, 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 sim-
plified 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.
Prefaceviii
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 sug-
gestions 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 sig-
nificant 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:
IS
~
E^
nda ¼qenc
e0
Gauss’s law for electric fields (integral form).
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
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:
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.
ˆ
0
=
S
qenc
danE
Reminder that this
integral is over a
closed surface
The electric
field in N/C
Reminder that this is a surface
integral (not a volume or a line integral)
Reminder that the
electric field is a
vector
The unit vector normal
to the surface
The amount of
charge in coulombs
Reminder that only
the enclosed charge
contributes
An increment of
surface area in m
2
Tells you to sum up the
contributions from each
p
ortion of the surface
The electric
permittivity
of the free space
Dot product tells you to find
the part of E parallel to n
(perpendicular to the surface)
ˆ
∫
A student’s guide to Maxwell’s Equations2
~
EThe 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 def-
inition 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
~
E~
Fe
q0
;ð1:1Þ
where ~
Feis the electrical force on a small
1
charge q0. This definition
makes clear two important characteristics of the electric field:
(1) ~
Eis 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) ~
Ehas units of newtons per coulomb (N/C), which are the same as
volts per meter (V/m), since volts ¼newtons ·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.
Gauss’s law for electric fields 3
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 charges
2
:
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
Positive point charge Negative point charge Infinite line of
positive charge
Infinite plane of
negative charge
Positively charged
conducting sphere
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.
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.
A student’s guide to Maxwell’s Equations4
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 ~
Ein Gauss’s law represent? It represents the
total electric field at each point on the surface under consideration. The sur-
face 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.
Table 1.1. Electric field equations for simple objects
Point charge (charge ¼q)~
E¼1
4pe0
q
r2
^
r(at distance rfrom q)
Conducting sphere (charge ¼Q)~
E¼1
4pe0
Q
r2
^
r(outside, distance rfrom
center)
~
E¼0(inside)
Uniformly charged insulating
sphere (charge ¼Q, radius ¼r
0
)~
E¼1
4pe0
Q
r2
^
r(outside, distance rfrom
center)
~
E¼1
4pe0
Qr
r3
0
^
r(inside, distance rfrom
center)
Infinite line charge (linear
charge density ¼k)~
E¼1
2pe0
k
r
^r(distance rfrom line)
Infinite flat plane (surface
charge density ¼r)~
E¼r
2e0
^
n
Gauss’s law for electric fields 5
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 ~
Eand ^
nin 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 ~
Eand ^
nrepresents the dot product
(or ‘‘scalar product’’) between the electric field vector ~
Eand 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~
B¼AxBxþAyByþAzBz:ð1:2Þ
Or, if you know the angle hbetween the vectors, you can use
~
A~
B¼j
~
Ajj~
Bjcos h;ð1:3Þ
where j~
Ajand j~
Bjrepresent 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
~
Aand ~
Bthat differ in direction by angle h, as shown in Figure 1.2(a).
For these vectors, the projection of ~
Aonto ~
Bis j~
Ajcos h, as shown
in Figure 1.2(b). Multiplying this projection by the length of ~
Bgives
j~
Ajj~
Bjcos h. Thus, the dot product ~
A~
Brepresents the projection of ~
A
onto the direction of ~
Bmultiplied by the length of ~
B.
3
The usefulness of
this operation in Gauss’s law will become clear once you understand the
meaning of the vector ^
n.
A
(a) (b) A
BB
uu
The projection of A onto B:|A| cos u
multiplied by the length of B:3|B|
gives the dot product AB:
|
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 ~
Bonto the direction
of ~
Aand then multiplying by the length of ~
A.
A student’s guide to Maxwell’s Equations6
^
nThe 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 per-
pendicular 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~
arather
than ^
nda. 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~
aand ^
nda serve the same
purpose.
Figure 1.3 Unit normal vectors for planar and spherical surfaces.
Gauss’s law for electric fields 7
~
E^
nThe component of ~
Enormal to a surface
If you understand the dot product and unit normal vector, the meaning of
~
E^
nshould 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 ~
Eand ^
nis simply the pro-
jection 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^
nj¼1),
so that
~
E^
n¼j
~
Ejj^
njcos h¼j
~
Ejcos h;ð1:4Þ
where his the angle between the unit normal ^
nand ~
E. This is the com-
ponent of the electric field vector perpendicular to the surface, as illus-
trated in Figure 1.4.
Thus, if h¼90,~
Eis perpendicular to ^
n, which means that the electric
field is parallel to the surface, and ~
E^
n¼j
~
Ejcosð90Þ¼0. So in this case
the component of ~
Eperpendicular to the surface is zero.
Conversely, if h¼0,~
Eis parallel to ^
n, meaning the electric field is
perpendicular to the surface, and ~
E^
n¼j
~
Ejcosð0Þ¼j
~
Ej. In this case,
the component of ~
Eperpendicular 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.
n
E
Component of E normal
to surface is E n
Surface
^
^
Figure 1.4 Projection of ~
Eonto direction of ^
n.
A student’s guide to Maxwell’s Equations8
RSðÞda 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 spe-
cified surface (this type of integral is also called the ‘‘surface integral’’).
The time you spend understanding this important mathematical oper-
ation 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 xand 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 r
i
and area dA
i
, the mass of
each segment is r
i
dA
i
, and the mass of the entire surface of Nsegments is
given by PN
i¼1ridAi. As you can imagine, the smaller you make the area
segments, the closer this gets to the true mass, since your approximation
of constant ris more accurate for smaller segments. If you let the seg-
ment area dA approach zero and Napproach infinity, the summation
becomes integration, and you have
Mass ¼ZS
rðx;yÞdA:
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)
s1s2s3
sΝ
Density = s(x,y) Mass = s1dA1+ s2dA2+ . . . + sNdAN.
xy
Figure 1.5 Finding the mass of a variable-density surface.
Gauss’s law for electric fields 9
Rs~
A^
nda 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 ~
Aand a surface Sper-
pendicular to the direction of the field, the flux Uis defined as the product
of the field magnitude and the area of the surface:
U¼j
~
Aj·surface area:ð1:5Þ
This case is shown in Figure 1.6(a). Note that if ~
Ais 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 ~
Aperpendicular 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.
n
n
A
(a) (b)
A
Figure 1.6 Flux of a vector field through a surface.
A student’s guide to Maxwell’s Equations10
Consider the curved surface and vector field ~
Ashown in Figure 1.7(a).
Imagine that ~
Arepresents the flow of a real fluid and Sa porous mem-
brane; 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 ~
Aas 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), ~
Amust 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 ~
Aby 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
u
ni
A
A
A
A
Surface
S
(a) (b)
Component of A perpendicular
to this surface element is A ° ni
Figure 1.7 Component of ~
Aperpendicular to surface.
Gauss’s law for electric fields 11
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 com-
ponent of the flow perpendicular to the surface.
Of course, you know how to find the component of ~
Aperpendicular
to the surface; simply form the dot product of ~
Aand ^
n, the unit normal to
the surface. But since the surface is curved, the direction of ^
ndepends 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 ^
nand ~
Aare constant over each segment.
Let ^
nirepresent the unit normal for the ith segment (of area da
i
); the
flow through segment iis (~
Ai^
ni)da
i
, and the total is
flow through entire surface ¼P
i
~
Ai^
nidai:
It should come as no surprise that if you now let the size of each
segment shrink to zero, the summation becomes integration.
Flow through entire surface ¼ZS
~
A^
nda:ð1:7Þ
For a closed surface, the integral sign includes a circle:
IS
~
A^
nda:ð1:8Þ
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 ~
Awith the electric field ~
Eto make the expressions identical.
A student’s guide to Maxwell’s Equations12
HS~
E^
ndaThe electric flux through a closed surface
On the basis of the results of the previous section, you should understand
that the flux U
E
of vector field ~
Ethrough surface Scan be determined
using the following equations:
UE¼j
~
Ej·ðsurface areaÞ~
Eis uniform and perpendicular to S;ð1:9Þ
UE¼~
E^
n·ðsurface areaÞ~
Eis uniform and at an angle to S;ð1:10Þ
UE¼ZS
~
E^
nda ~
Eis non-uniform and at a variable angle to S:ð1:11Þ
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 specif-
ically, 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. Inte-
grating 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 ðUEÞ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
Gauss’s law for electric fields 13
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 ^
nhas been established, field line com-
ponents 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 penetra-
tion 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^
nmust be negative when the angle between
~
Eand ^
nis greater than 90. This is precisely cancelled by the outward flux
(lines exiting the box), which is positive, since ~
E^
nis positive when the
angle between ~
Eand ^
nis 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
Zero net flux
(a) (b) (c)
Positive flux Negative flux
Figure 1.8 Flux lines penetrating closed surfaces.
A student’s guide to Maxwell’s Equations14
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.
Gauss’s law for electric fields 15
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 out-
side 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 ¼X
i
qi:
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).
Enclosing
sphere
enclosing
cube
Charged
line Charged
plane
Point
charge Multiple
point charges
Enclosing
cylinder
Enclosing
pillbox
Figure 1.9 Surface enclosing known charges.
A student’s guide to Maxwell’s Equations16
If these quantities are constant over the length, area, or volume under
consideration, finding the enclosed charge requires only a single multi-
plication:
1-D: qenc ¼kLðL¼enclosed length of charged lineÞ;ð1:12Þ
2-D: qenc ¼rAðA¼enclosed area of charged surfaceÞ;ð1:13Þ
3-D: qenc ¼qVðV¼enclosed 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,
1-D: qenc ¼ZL
kdl where kvaries along a line;ð1:15Þ
2-D: qenc ¼ZS
rda where rvaries over a surface;ð1:16Þ
3-D: qenc ¼ZV
qdV where qvaries over a volume:ð1:17Þ
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 e
0
, the permittivity of free space. The
physical meaning of that parameter is described in the next section.
Dimensions Terminology Symbol Units
1 Linear charge
density
kC/m
2 Area charge density rC/m
2
3 Volume charge
density
qC/m
3
Gauss’s law for electric fields 17
e0The 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 e
0
, 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 ·10
12
coulombs per volt-meter (C/Vm); you will sometimes see the
units of permittivity given as farads per meter (F/m), or, more funda-
mentally, (C
2
s
2
/kg m
3
). A more precise value for the permittivity of free
space is
e
0
¼8.8541878176 ·10
12
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
A student’s guide to Maxwell’s Equations18
C¼eA
d;
where Ais the plate area, dis the plate separation, and eis 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 per-
mittivity, which is the factor by which the material’s permittivity exceeds
that of free space:
relative permittivity er¼e=e0:
Some texts refer to relative permittivity as ‘‘dielectric constant,’’ although
the variation in permittivity with frequency suggests that the word ‘‘con-
stant’’ 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.
N
o dielectric present
Displaced
charges
Induced
field
Dielectric
External
electric
field
+
+
+
+
–
–
–
–
Figure 1.10 Electric field induced in a dielectric.
Gauss’s law for electric fields 19
Hs~
E^
nda¼qenc=e0Applying Gauss’s law (integral form)
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 situ-
ations. 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 q
1
¼þ3 nC, q
2
¼2 nC, q
3
¼þ2 nC, q
4
¼þ4 nC,
and q
5
¼1 nC, find the total flux through S.
Solution: From Gauss’s law,
UE¼IS
~
E^
nda¼qenc
e0
:
For discrete charges, you know that the total charge is just the sum of the
individual charges. Thus,
qenc ¼Total enclosed charge ¼Xiqi
¼ð32þ2þ41Þ·109C
¼6·109C
and
UE¼qenc
e0
¼6·109C
8:85 ·1012 C=Vm ¼678 Vm:
This is the total flux through any closed surface surrounding this group of
charges.
S
q1
q3
q2
q5
q4
A student’s guide to Maxwell’s Equations20
Example 1.2: Given the flux through a closed surface, find the enclosed
charge.
Problem: A line charge with linear charge density k¼10
12
C/m passes
through the center of a sphere. If the flux through the surface of the
sphere is 1.13 ·10
3
Vm, what is the radius Rof the sphere?
Solution: The charge on a line charge of length Lis given by q¼kL. Thus,
UE¼qenc
e0
¼kL
e0
;
and
L¼UEe0
k:
Since Lis twice the radius of the sphere, this means
2R¼UEe0
kor R¼UEe0
2k:
Inserting the values for U
E
,e
0
and k, you will find that R¼5·10
3
m.
Example 1.3: Find the flux through a section of a closed surface.
Problem: A point source of charge qis placed at the center of curvature of
a spherical section that extends from spherical angle h
1
to h
2
and from u
1
to u
2
. 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.
Charged
line
L
Sphere
encloses
p
ortion of
line
Gauss’s law for electric fields 21
The electric flux U
E
is RS~
E^
nda, where Sis the spherical section of
interest and ~
Eis the electric field on the surface due to the point charge
at the center of curvature, a distance rfrom the section of interest.
From Table 1.1, you know that the electric field at a distance rfrom a
point charge is
~
E¼1
4pe0
q
r2
^
r:
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 ^
nfor a spherical section points in the outward radial direction
(the ^
rdirection), as may be seen in Figure 1.11. This means that ~
Eand ^
n
are parallel, and the flux is given by
UE¼ZS
~
E^
nda¼ZS
j~
Ejj^
njcosð0Þda ¼ZS
j~
Ejda ¼ZS
1
4pe0
q
r2da:
Since you are integrating over a spherical section in this case, the logical
choice for coordinate system is spherical. This makes the area element r
2
sin hdhdU, and the surface integral becomes
rdu
r sin udf
da n
df
du
f
r sin u
u
da = (rdu)(r sin udf)
Figure 1.11 Geomentry of sperical section.
A student’s guide to Maxwell’s Equations22
UE¼ZhZf
1
4pe0
q
r2r2sin hdhdf¼q
4pe0Zh
sin hdhZf
df;
which is easily integrated to give
UE¼q
4pe0
ðcos h1cos h2Þðf2f1Þ:
As a check on this result, take the entire sphere as the section (h
1
¼0,
h
2
¼p,u
1
¼0, and u
2
¼2p). This gives
UE¼q
4pe0
ð1ð1ÞÞ ð2p0Þ¼ q
e0
;
exactly as predicted by Gauss’s law.
Example 1.4: Given ~
Eover a surface, find the flux through the surface
and the charge enclosed by the surface.
Problem: The electric field at distance rfrom an infinite line charge with
linear charge density kis given in Table 1.1 as
~
E¼1
2pe0
k
r
^
r:
Use this expression to find the electric flux through a cylinder of radius r
and height hsurrounding a portion of an infinite line charge, and then use
Gauss’s law to verify that the enclosed charge is kh.
h
r
Gauss’s law for electric fields 23
Solution: Problems like this are best approached by considering the flux
through each of three surfaces that comprise the cylinder: the top, bot-
tom, and curved side surfaces. The most general expression for the
electric flux through any surface is
UE¼ZS
~
E^
nda;
which in this case gives
UE¼ZS
1
2pe0
k
r
^
r^
nda:
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, ~
Eis
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
UE;top ¼ZS
1
2pe0
k
r
^
r^
ntop da ¼0;
UE;bottom ¼ZS
1
2pe0
k
r
^
r^
nbottom da ¼0;
UE;side ¼ZS
1
2pe0
k
r
^
r^
nside da ¼1
2pe0
k
rZS
da;
n
n
n
n
n n
n
A student’s guide to Maxwell’s Equations24
and, since the area of the curved side of the cylinder is 2prh, this gives
UE;side ¼1
2pe0
k
rð2prhÞ¼kh
e0
:
Gauss’s law tells you that this must equal q
enc
/e
0
,which verifies that the
enclosed charge q
enc
¼khin 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 propo-
rtional 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 sym-
metry. 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 distribu-
tions, 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 approxi-
mated 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.
Gauss’s law for electric fields 25
Problem: Use Gauss’s law to find the electric field at a distance rfrom the
center of a sphere with uniform volume charge density qand 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 ^
hor ^
’
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 distrib-
uted 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; ~
Emust be either parallel or perpendicular to the surface normal
at all locations, and ~
Emust 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^
nterm in the integral in Gauss’s law becomes
j~
Ejj^
njcosð0Þ, and the electric flux over the Gaussian surface Sis
UE¼IS
~
E^
nda¼IS
Eda
Since ~
Ehas no hor udependence, it must be constant over S, which
means it may be removed from the integral:
UE¼IS
Eda¼EIS
da ¼Eð4pr2Þ;
where ris the radius of the special Gaussian surface. You can now use
Gauss’s law to find the value of the electric field:
A student’s guide to Maxwell’s Equations26
UE¼Eð4pr2Þ¼qenc
e0
;
or
E¼qenc
4pe0r2;
where q
enc
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>aso 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 ¼ð4=3Þpa3q.
Thus,
E¼ð4=3Þpa3q
4pe0r2¼qa3
3e0r2ðoutside sphereÞ:
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
Radial
electric
field
Charged
sphere
n
Special
Gaussian
surface
n
nn
Figure 1.12 A special Gaussian around a charged sphere.
Gauss’s law for electric fields 27
density times the volume of your Gaussian surface: qenc ¼ð4=3Þpr3q.
Thus,
E¼ð4=3Þpr3q
4pe0r2¼qr
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.
A student’s guide to Maxwell’s Equations28
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
~
r~
E¼q
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.
Don’t be concerned if the del operator (~
r) or the concept of divergence
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.
Gauss’s law for electric fields 29
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:
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.
E
0
=
Reminder that del is
a vector operator
Reminder that the electric
field is a vector
The differential
operator called
“del” or “nabla”
The dot product turns
the del operator into the
divergence
The electric
field in N/C
The electric
permittivity of
free space
The charge density in
coulombs per cubic meter
A student’s guide to Maxwell’s Equations30
~
rNabla – 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’’
signifying divergence, ‘‘~
r·’’ indicating curl, and ~
rsignifying gradient.
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 Htells you to take the square root of anything that
appears under its roof, ~
ris an instruction to take derivatives in three
directions. Specifically,
~
r^
i@
@xþ^
j@
@yþ^
k@
@z;ð1:18Þ
where ^
i,^
j, and ^
kare 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.
Gauss’s law for electric fields 31
~
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 diver-
gence 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 (fau-
cets in situations involving fluid flow, positive electric charge in electro-
statics), 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
~
Athrough a surface Sto the volume enclosed by that surface as the
volume shrinks toward zero, you would have the divergence of ~
A:
divð~
AÞ¼~
r~
Alim
DV!0
1
DVIS
~
A^
nda:ð1:19Þ
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.
A student’s guide to Maxwell’s Equations32
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 loca-
tions, 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
divergence or ‘‘del dot’’ (~
r) on a vector ~
Ain Cartesian coordinates is
~
r~
A¼^
i@
@xþ^
j@
@yþ^
k@
@z

^
iAxþ^
jAyþ^
kAz

;
and, since ^
i^
i¼^
j^
j¼^
k^
k¼1;this is
~
r~
A¼@Ax
@xþ@Ay
@yþ@Az
@z

:ð1:20Þ
Thus, the divergence of the vector field ~
Ais 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.
1
4
35
2
6
7
(a) (b) (c)
Figure 1.13 Vector fields with various values of divergence.
Gauss’s law for electric fields 33
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 ~
A¼sinðpxÞ^
iwhile
remaining constant in the y- and z-directions. Thus,
~
r~
A¼@Ax
@x¼pcosðpxÞ;
since A
y
and A
z
are zero. This expression is positive for 0 <x<1
2,0at
x¼1
2, and negative for 1
2<x<3
2, 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 ~
A¼r2^
r. Since r
2
¼(x
2
þy
2
þz
2
) and
^
r¼x^
iþy^
jþz^
k
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2þy2þz2
p;
this means
~
A¼r2^
r¼ðx2þy2þz2Þx^
iþy^
jþz^
k
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2þy2þz2
p;
and
@Ax
@x¼x2þy2þz2

ð1=2Þþx1
2
x2þy2þz2

ð1=2Þ2xðÞ:
Doing likewise for the y- and z-components and adding yields
~
r~
A¼3x2þy2þz2

ð1=2Þþx2þy2þz2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2þy2þz2
p¼4x2þy2þz2

1=2¼4r:
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 ~
A¼ð1=r2Þ^
r,
~
A¼1
x2þy2þz2
ðÞ
x^
iþy^
jþz^
k
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2þy2þz2
p¼x^
iþy^
jþz^
k
x2þy2þz2
ðÞ
ð3=2Þ;
A student’s guide to Maxwell’s Equations34
and
@Ax
@x¼1
x2þy2þz2
ðÞ
ð3=2Þ–x3
2
x2þy2þz2

ð5=2Þ2xðÞ;
Adding in the y- and z-derivatives gives
~
r~
A¼3
x2þy2þz2
ðÞ
ð3=2Þ3x2þy2þz2

x2þy2þz2
ðÞ
ð5=2Þ¼0:
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/r
2
(this case is especially rele-
vant 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 non-
Cartesian coordinate systems. The divergence may be calculated in
cylindrical and spherical coordinate systems using
~
r~
A¼1
r
@
@rðrArÞþ1
r
@Af
@fþ@Az
@zðcylindricalÞ;ð1:21Þ
and
~
r~
A¼1
r2
@
@rðr2ArÞþ 1
rsin h
@
@hðAhsin hÞ
þ1
rsin h
@Af
@fðsphericalÞ:
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.
Gauss’s law for electric fields 35
~
r~
EThe 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/r
2
:
~
E¼1
4pe0
q
r2
^
r:
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/r
2
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 r¼0) is not included in the previous
analysis is that the expression for the divergence includes terms con-
taining rin the denominator, and those terms become problematic as r
approaches zero. To evaluate the divergence at the origin, use the formal
definition of divergence:
~
r~
Elim
DV!0
1
DVI
S
~
E^
nda:
Considering a special Gaussian surface surrounding the point charge
q, this is
~
r~
Elim
DV!0
1
DV
q
4pe0r2I
S
da
0
@1
A¼lim
DV!0
1
DV
q
4pe0r2ð4pr2Þ

¼lim
DV!0
1
DV
q
e0

:
But q/DVis just the average charge density over the volume DV, and as
DVshrinks to zero, this becomes equal to q, the charge density at the
origin. Thus, at the origin the divergence is
~
r~
E¼q
e0
;
in accordance with Gauss’s law.
A student’s guide to Maxwell’s Equations36
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/r
2
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 infini-
tesimally 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.
Gauss’s law for electric fields 37
~
r~
E¼q=e0Applying 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
~
A¼sin p
2y

^
isin p
2x

^
j;
in the region 0.5 <x<þ0.5 and 0.5 <y<þ0.5, 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
~
r~
A¼@
@xsin p
2y
hi
–@
@ysin p
2x
hi
¼0:
A student’s guide to Maxwell’s Equations38
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 x¼2 m and x¼5 m if the electric field
in the region is given by
~
E¼ax2^
iV
mfor x¼0to3m;
and
~
E¼b
^
iV
mfor x>3m:
Solution: By Gauss’s law, in the region x¼0to3m,
~
r~
E¼q
e0
¼^
i@
@xþ^
j@
@yþ^
k@
@k

ðax2^
iÞ;
0.5
0.4
0.3
0.2
0.1
0
–0.1
–0.1
–0.2
–0.2
–0.3
–0.3
–0.4
–0.4
–0.5
–0.5 0 0.1 0.2 0.3 0.4 0.5
y
x
Figure 1.14 Vector field ~
A¼sinðp
2yÞ^
isinðp
2xÞ^
j:
Gauss’s law for electric fields 39
q
e0
¼@ðax2Þ
@x¼2xa;
and
q¼2xae0:
Thus at x¼2m, q¼4ae
0
.
In the region x>3m,
~
r~
E¼q
e0
¼^
i@
@xþ^
j@
@yþ^
k@
@k

ðb^
iÞ¼0;
so q¼0atx¼5m.
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 Lcontains a flat plate with variable surface charge
density of r¼3xy. If the plate extends from x¼0tox¼Land from
y¼0toy¼L, what is the total electric flux through the walls of the
cube?
1.3 Find the total electric flux through a closed cylinder containing a line
charge along its axis with linear charge density k¼k
0
(1x/h) C/m if
the cylinder and the line charge extend from x¼0tox¼h.
s
x
y
z
L
L
L
A student’s guide to Maxwell’s Equations40
1.4 What is the flux through any closed surface surrounding a charged
sphere of radius a
0
with volume charge density of q¼q
0
(r/a
0
), where r
is the distance from the center of the sphere?
1.5 A circular disk with surface charge density 2 ·10
10
C/m
2
is
surrounded by a sphere with radius of 1 m. If the flux through the
sphere is 5.2 ·10
2
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
10
8
C. What is the electric flux through the plate due to the point
charge?
1.7 Find the electric flux through a half-cylinder of height howing to an
infinitely long line charge with charge density krunning 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.
10 cm
10 cm
5 cm
Gauss’s law for electric fields 41
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 ris
j~
Ej¼r=2e0.
1.11 Find the divergence of the field given by ~
A¼ð1=rÞ^
rin spherical
coordinates.
1.12 Find the divergence of the field given by ~
A¼r^
rin spherical
coordinates.
1.13 Given the vector field
~
A¼cos py–p
2

^
iþsin pxðÞ
^
j;
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
~
E¼az
r
^
rþbr ^
fþcr2z2^z
1.15 Find the charge density in a region for which the electric field in
spherical coordinates is given by
~
E¼ar2^
rþbcos ðhÞ
r
^
hþc^
f:
A student’s guide to Maxwell’s Equations42
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:
IS
~
B^
nda ¼0Gauss’s law for magnetic fields ðintegral formÞ:
As de scribed in the prev ious chapter , the left side of this equati on 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
ˆ=
S
da 0
n
B
Reminder that this
integral is over a
closed surface
The magnetic
field in Teslas
Reminder that the
magnetic field is a vector ˆ
Dot product tells you to find the part of
B parallel to n (perpendicular to the surface)
The unit vector normal to the surface
Reminder that this is a surface integral
(not a volume or a line integral)
An increment of
surface area in m
2
Tells you to sum up the contributions
from each portion of the surface
∫
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 bal-
anced 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 pre vious chap ter, in this ch apter you ’ll find
only those symbols peculiar to this law. Here’s an expanded view:
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, par-
ticularly if the flux through one portion of the surface can be found by
integration.
A Student’s guide to Maxwell’s Equations44
~
BThe 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:
~
FB¼q~
v·~
Bð2:1Þ
where ~
FBis the magnetic force, qis the particle’s charge, ~
vis the particle’s
velocity (with respect to ~
B), and ~
Bis the magnetic field.
Using the definition of the vector cross-product which says that
~
a·~
b¼j
~
ajj~
bjsinðhÞ, where his the angle between ~
aand ~
b, the magnitude
of the magnetic field may be written as
j~
Bj¼ j~
FBj
qj
~
vjsinðhÞð2:2Þ
where his the angle between the velocity vector ~
vand 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 ~
Bas the
‘‘magnetic induction’’ or the ‘‘magnetic flux density.’’ Whatever it is
called, ~
Bhas units equivalent to N/(C m/s), which include Vs/m
2
,
N/(Am), kg/(Cs), or most simply, Tesla (T).
Comparing Equation 2.2 to the relevant equation for the electric field,
Equation (1.1 ), severa l important dist inction s between magnet ic 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 ~
Bis 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, magneto-
static fields are produced by electric currents.
Gauss’s law for magnetic fields 45
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.
I
I
B
N
S
I
I
Current-carrying
straight wire Current loop Bar magnet
Solenoid Torus Horseshoe
magnet
B
B
B
Figure 2.1 Examples of magnetic fields.
A Student’s guide to Maxwell’s Equations46
All static magnetic fields are produced by moving electric charge. The
contribution d~
Bto the magnetic field at a specified point P from a small
element of electric current is given by the Biot–Savart law:
d~
B¼l0
4p
Id
~
l·^
r
r2
In this equation, l
0
is the permeability of free space, Iis the current
through the small element, d
~
lis a vector with the length of the current
element and pointing in the direction of the current, ^
ris a unit vector
pointing from the current element to the point P at which the field is
being calculated, and ris 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.
Table 2.1. Magnetic field equations for simple objects
Infinite straight wire carrying
current I(at distance r)~
B¼l0I
2pr
^
’
Segment of straight wire carrying current I
(at distance r)d~
B¼l0
4p
Id
~
l·^
r
r2
Circular loop of radius Rcarrying current I
(loop in yz plane, at distance xalong x-axis) ~
B¼l0IR2
2ðx2þR2Þ3=2
^
x
Solenoid with Nturns and length l
carrying current I~
B¼l0NI
l
^
xðinsideÞ
Torus with Nturns and radius r
carrying current I~
B¼l0NI
2pr
^
’(inside)
r
P
dl
r
Current
element
Point at whic
h
magnetic field
is determined
^
Figure 2.2 Geometry for Biot–Savart law.
Gauss’s law for magnetic fields 47
HS~
B^
nda The magnetic flux through a closed surface
Like the electric flux U
E
, the magnetic flux U
B
through a surface may be
thought of as the ‘‘amount’’ of magnetic field ‘‘flowing’’ through the sur-
face. How this quantity is calculated depends on the situation:
UB¼j
~
Bj·(surface area) ~
Buniform and perpendicular to S;ð2:3Þ
UB¼~
B^
n·(surface area) ~
Buniform and at an angle to S;ð2:4Þ
UB¼ZS
~
B^
nda ~
Bnonuniform and at variable angle to S:ð2:5Þ
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’’ (abbre-
viated Wb and which, by any of the relations shown above, must be
equivalent to T m
2
).
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 espe-
cially important to remember the caveat that surface penetration is a two-
way 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
A Student’s guide to Maxwell’s Equations48
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.
B
Figure 2.3 Magnetic flux lines penetrating closed surfaces.
Gauss’s law for magnetic fields 49
HS~
B^
nda ¼0Applying 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 hand radius Ris placed in a mag-
netic field given by ~
B¼B0ð^
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 por-
tions. 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 þUB;Bottom þUB;Sides ¼0:
The magnetic flux through any surface is
UB¼Z
S
~
B^
nda:
For the top surface, ^
n¼^
k,so
~
B^
n¼ðB0
^
jB0
^
kÞ^
k¼B0:
Thus
UB;Top ¼Z
S
~
B^
nda ¼B0Z
S
da ¼B0ðpR2Þ:
A similar analysis for the bottom surface (for which ^
n¼
^
k) gives
UB;Bottom ¼Z
S
~
B^
nda ¼þB0Z
S
da ¼þB0ðpR2Þ:
Since U
B,Top
¼U
B,Bottom
, you can conclude that U
B,Sides
¼0.
A Student’s guide to Maxwell’s Equations50
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.
Solution: At distance rfrom a current-carrying wire, the magnetic field is
given by
~
B¼l0I
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
UB¼Z
S
~
B^
nda:
In this case, ^
n¼
^
’,so
~
B^
n¼l0I
2pr
^
’ð^
’Þ¼l0I
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 ¼dy dz.Notice
also that on the flat face the distance increment dr ¼dy,soda ¼dr dz and the
flux integral is
UB;Flat ¼Z
S
~
B^
nda¼ZS
l0I
2prdr dz ¼l0I
2pZh
z¼0Zy1þ2R
r¼y1
dr
rdz :
y1
x
y
z
h
I
Gauss’s law for magnetic fields 51
Thus
UB;Flat ¼l0I
2pln y1þ2R
y1

ðhÞ¼l0Ih
2pln 1þ2R
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
UB;Curved side ¼l0Ih
2pln 1þ2R
y1

:
A Student’s guide to Maxwell’s Equations52
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~
B¼0Gauss’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 fol-
lowing 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:
B0
=
Reminder that the
del operator is a vector
Reminder that the magnetic
field is a vector
The differential
operator called
“del” or “nabla”
The dot product turns
the del operator into the
divergence
The magnetic
field in Teslas
Gauss’s law for magnetic fields 53
~
r~
BThe 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 diver-
gence 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 ~
Bis always
zero.
To verify this for the case of the magnetic field around a long, current-
carrying wire, take the divergence of the expression for the wire’s mag-
netic field as given in Table 2.1:
divð~
BÞ¼~
r~
B¼~
r l0I
2pr
^
’

:ð2:6Þ
This is most easily determined using cylindrical coordinates:
~
r~
B¼1
r
@
@rrBr
ðÞþ
1
r
@Bf
@fþ@Bz
@z:ð2:7Þ
which, since ~
Bhas only a u-component, is
~
r~
B¼1
r
@l0I=2prðÞ
@’ ¼0: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 mag-
netic field is zero everywhere.
Vector fields with zero divergence are called ‘‘solenoidal’’ fields, and all
magnetic fields are solenoidal.
A Student’s guide to Maxwell’s Equations54
~
r~
B¼0Applying 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
~
B¼axz^
iþbyz^
jþc^
k
What is the relationship between aand b?
Solution: You know from Gauss’s law for magnetic fields that the
divergence of the magnetic field must be zero. Thus
~
r~
B¼@Bx
@xþ@By
@yþ@Bz
@z¼0:
Thus
@ðaxzÞ
@xþ@ðbyzÞ
@yþ@c
@z¼0
and
az þbz þ0¼0;
which means that a¼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;yÞ¼acosðbxÞ^
iþaby 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
Gauss’s law for magnetic fields 55
~
r~
A¼@
@x½acosðbxÞ þ @
@y½aby sinðbxÞ
¼ab sinðbxÞþab sinðbxÞ¼0
which indicates that ~
Acould 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 ~
B¼5^
i3^
jþ
4^
knTthrough the top, bottom, and side surfaces of the flared cylinder
shown in the figure.
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
~
B¼0:002^
iþ0:003^
jT;
and show that the total flux through the wedge is zero.
Rtop
Rbottom
x
y
z
B
x
y
z
50 cm
70 cm
130 cm
A Student’s guide to Maxwell’s Equations56
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 ·10
5
T and points upward at
an angle of 30with respect to the horizontal. You may orient the
cube in any way you choose.
2.5 A cylinder of radius r
0
and height his 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) ~
Aðr;’;zÞ¼a
rcos2ð’Þ^
r;
(b) ~
Aðr;’;zÞ¼ a
r2cos2ð’Þ^
r:
Gauss’s law for magnetic fields 57
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’’ elec-
tric 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
IC
~
Ed
~
l¼d
dt ZS
~
B^
nda Faraday’s law ðintegral formÞ:
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 ~
Erep-
resents the electric field in the rest frame of each segment d
~
lof the path of
integration.
58
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d
dt ZS
~
B^
nda Flux rule,
IC
~
Ed
~
l¼ZS
@~
B
@t^
nda 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 ~
Bare 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 pro-
vides 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.
Faraday’s law 59
Here’s an expanded view of the standard form of Faraday’s law:
Note that ~
Ein this expression is the induced electric field at each segment
d
~
lof the path Cmeasured in the reference frame in which that segment is
stationary.
And here is an expanded view of the alternative form of Faraday’s law:
=
C
d
–
dl dt
E
The electric
field in V/m
Reminder that the
electric field is a
vector
Reminder that this is a line
integral (not a surface or a
volume integral)
An incremental segment
of path C
Tells you to sum up the
contributions from each
p
ortion of the closed path C
in a direction given by the
right-hand rule
ˆ
S
danB
Dot product tells you to find
the part of E parallel to dl
(along path C)
The rate of change
with time
The magnetic flux
through any surface
bounded by C
∫∫
=
C
–
dl
E
The electric
field in V/m
Reminder that the
electric field is a
vector
Dot product tells you to find
the part of E parallel to dl
(along path C)
Reminder that this is a line
integral (not a surface or a
volume integral)
Tells you to sum up the
contributions from each
p
ortion of the closed path C
ˆ
S
dan
The rate of change of the
magnetic field with time
An incremental segment
of path C
The flux of the time
rate of change of
the magnetic field
∫∫
A student’s guide to Maxwell’s Equations60
In this case, ~
Erepresents the electric field in the laboratory frame of
reference (the same frame in which ~
Bis measured).
Faraday’s law and the flux rule can be used to solve a variety of
problems involving changing magnetic flux and induced electric fields, in
particular problems of two types:
(1) Given information about the changing magnetic flux, find the
induced emf.
(2) Given the induced emf on a specified path, determine the rate of
change of the magnetic field magnitude or direction or the area
bounded by the path.
In situations of high symmetry, in addition to finding the induced emf, it
is also possible to find the induced electric field when the rate of change of
the magnetic field is known.
Faraday’s law 61
~
EThe induced electric field
The electric field in Faraday’s law is similar to the electrostatic field in its
effect on electric charges, but quite different in its structure. Both types of
electric field accelerate electric charges, both have units of N/C or V/m,
and both can be represented by field lines. But charge-based electric fields
have field lines that originate on positive charge and terminate on
negative charge (and thus have non-zero divergence at those points),
while induced electric fields produced by changing magnetic fields have
field lines that loop back on themselves, with no points of origination or
termination (and thus have zero divergence).
It is important to understand that the electric field that appears in the
common form of Faraday’s law (the one with the full derivative of the
magnetic flux on the right side) is the electric field measured in the ref-
erence frame of each segment d
~
lof the path over which the circulation is
calculated. The reason for making this distinction is that it is only in this
frame that the electric field lines actually circulate back on themselves.
E
+
–
Electric field lines
orginate on positive
charges and terminate
on negative charges
(a)
N
Magnet motion
B
E
Electric field lines
form complete loops
around boundary
As magnet moves to
right, magnetic flux
through surface
decreases
Surface may be real
or purely imaginary
S
(b)
Figure 3.1 Charge-based and induced electric fields. As always, you should
remember that these fields exist in three dimensions, and you can see full 3-D
visualizations on the book’s website.
A student’s guide to Maxwell’s Equations62
Examples of a charge-based and an induced electric field are shown in
Figure 3.1.
Note that the induced electric field in Figure 3.1(b) is directed so as to
drive an electric current that produces magnetic flux that opposes the
change in flux due to the changing magnetic field. In this case, the motion of
the magnet to the right means that the leftward magnetic flux is decreasing,
so the induced current produces additional leftward magnetic flux.
Here are a few rules of thumb that will help you visualize and sketch
the electric fields produced by changing magnetic fields:
Induced electric field lines produced by changing magnetic fields must
form complete loops.
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.
In summary, the ~
Ein Faraday’s law represents the induced electric field at
each point along path C, a boundary of the surface through which the
magnetic flux is changing over time. The path may be through empty
space or through a physical material – the induced electric field exists in
either case.
Faraday’s law 63
HCðÞdl The line integral
To understand Faraday’s law, it is essential that you comprehend the
meaning of the line integral. This type of integral is common in physics
and engineering, and you have probably come across it before, perhaps
when confronted with a problem such as this: find the total mass of a wire
for which the density varies along its length. This problem serves as a
good review of line integrals.
Consider the variable-density wire shown in Figure 3.2(a). To deter-
mine the total mass of the wire, imagine dividing the wire into a series of
short segments over each of which the linear density k(mass per unit
length) is approximately constant, as shown in Figure 3.2(b). The mass of
each segment is the product of the linear density of that segment times the
segment length dx
i
, and the mass of the entire wire is the sum of the
segment masses.
For Nsegments, this is
Mass ¼X
N
i¼1
kidxi:ð3:1Þ
Allowing the segment length to approach zero turns the summation of
the segment masses into a line integral:
Mass ¼Z
L
0
kðxÞdx:ð3:2Þ
This is the line integral of the scalar function k(x). To fully comprehend
the left side of Faraday’s law, you’ll have to understand how to extend
this concept to the path integral of a vector field, which you can read
about in the next section.
x
0L
Density varies with x:
λ
=
λ
(x)
123
dx1dx2dx3dxN
(a)
(b)
λλλ
N
λ
Figure 3.2 Line integral for a scalar function.
A student’s guide to Maxwell’s Equations64
HC~
Ad
~
lThe path integral of a vector field
The line integral of a vector field around a closed path is called the
‘‘circulation’’ of the field. A good way to understand the meaning of this
operation is to consider the work done by a force as it moves an object
along a path.
As you may recall, work is done when an object is displaced under the
influence of a force. If the force ð~
FÞis constant and in the same direction
as the displacement ðd
~
lÞ, the amount of work (W) done by the force is
simply the product of the magnitudes of the force and the displacement:
W¼j
~
Fjjd
~
lj:ð3:3Þ
This situation is illustrated in Figure 3.3(a). In many cases, the dis-
placement is not in the same direction as the force, and it then becomes
necessary to determine the component of the force in the direction of the
displacement, as shown in Figure 3.3(b).
In this case, the amount of work done by the force is equal to the
component of the force in the direction of the displacement multiplied by
the amount of displacement. This is most easily signified using the dot
product notation described in Chapter 1:
W¼~
Fd
~
l¼j
~
Fjjd
~
ljcosðhÞ;ð3:4Þ
where his the angle between the force and the displacement.
In the most general case, the force ~
Fand the angle between the force
and the displacement may not be constant, which means that the pro-
jection of the force on each segment may be different (it is also possible
that the magnitude of the force may change along the path). The general
case is illustrated in Figure 3.4. Note that as the path meanders from the
starting point to the end, the component of the force in the direction of
the displacement varies considerably.
F
F
dl dl
u
Work = F ° dl = |F| |dl| cos u
Work = |F| |dl|
(a) (b)
Figure 3.3 Object moving under the influence of a force.
Faraday’s law 65
To find the work in this case, the path may be thought of as a series of
short segments over each of which the component of the force is constant.
The incremental work (dW
i
) done over each segment is simply the com-
ponent of the force along the path at that segment times the segment
length (dl
i
) – and that’s exactly what the dot product does. Thus,
dWi¼~
Fd
~
li;ð3:5Þ
and the work done along the entire path is then just the summation of the
incremental work done at each segment, which is
W¼X
N
i¼1
dWi¼X
N
i¼1
~
Fd
~
li:ð3:6Þ
As you’ve probably guessed, you can now allow the segment length to
shrink toward zero, converting the sum to an integral over the path:
W¼Z
P
~
Fd
~
l:ð3:7Þ
Thus, the work in this case is the path integral of the vector ~
Fover path
P. This integral is similar to the line integral you used to find the mass of
a variable-density wire, but in this case the integrand is the dot product
between two vectors rather than the scalar function k.
Start End
Path of object
F
Path divided into
N segments
dl1
dl8
123
8
N
F
F
u8
u1
Component of F
in direction of dl8
dl8
F
u8
Force
Figure 3.4 Component of force along object path.
A student’s guide to Maxwell’s Equations66
Although the force in this example is uniform, the same analysis
pertains to a vector field of force that varies in magnitude and direction
along the path. The integral on the right side of Equation 3.7 may be
defined for any vector field ~
Aand any path C. If the path is closed, this
integral represents the circulation of the vector field around that path:
Circulation I
C
~
Ad
~
l:ð3:8Þ
The circulation of the electric field is an important part of Faraday’s law,
as described in the next section.
Faraday’s law 67
HC~
Ed
~
lThe electric field circulation
Since the field lines of induced electric fields form closed loops, these
fields are capable of driving charged particles around continuous circuits.
Charge moving through a circuit is the very definition of electric current,
so the induced electric field may act as a generator of electric current. It is
therefore understandable that the circulation of the electric field around a
circuit has come to be known as an ‘‘electromotive force’’:
electromotive force ðemf Þ¼IC
~
Ed
~
l:ð3:9Þ
Of course, the path integral of an electric field is not a force (which must
have SI units of newtons), but rather a force per unit charge integrated
over a distance (with units of newtons per coulomb times meters, which
are the same as volts). Nonetheless, the terminology is now standard, and
‘‘source of emf’’ is often applied to induced electric fields as well as to
batteries and other sources of electrical energy.
So, exactly what is the circulation of the induced electric field around a
path? It is just the work done by the electric field in moving a unit charge
around that path, as you can see by substituting ~
F=qfor ~
Ein the circu-
lation integral:
IC
~
Ed
~
l¼IC
~
F
qd
~
l¼HC~
Fd
~
l
q¼W
q:ð3:10Þ
Thus, the circulation of the induced electric field is the energy given to
each coulomb of charge as it moves around the circuit.
A student’s guide to Maxwell’s Equations68
d
dt RS~
B^
ndaThe rate of change of flux
The right side of the common form of Faraday’s law may look intimi-
dating at first glance, but a careful inspection of the terms reveals that the
largest portion of this expression is simply the magnetic flux ðUBÞ:
UB¼Z
S
~
B^
nda:
If you’re tempted to think that this quantity must be zero according to
Gauss’s law for magnetic fields, look more carefully. The integral in this
expression is over any surface S, whereas the integral in Gauss’s law is
specifically over a closed surface. The magnetic flux (proportional to the
number of magnetic field lines) through an open surface may indeed be
nonzero – it is only when the surface is closed that the number of mag-
netic field lines passing through the surface in one direction must equal
the number passing through in the other direction.
So the right side of this form of Faraday’s law involves the magnetic
flux through any surface S– more specifically, the rate of change with
time (d/dt) of that flux. If you’re wondering how the magnetic flux
through a surface might change, just look at the equation and ask
yourself what might vary with time in this expression. Here are three
possibilities, each of which is illustrated in Figure 3.5:
The magnitude of ~
Bmight change: the strength of the magnetic field
may be increasing or decreasing, causing the number of field lines
penetrating the surface to change.
The angle between ~
Band the surface normal might change: varying the
direction of either ~
Bor the surface normal causes ~
B^
nto change.
The area of the surface might change: even if the magnitude of ~
Band
the direction of both ~
Band ^
nremain the same, varying the area of
surface Swill change the value of the flux through the surface.
Each of these changes, or a combination of them, causes the right side of
Faraday’s law to become nonzero. And since the left side of Faraday’s
law is the induced emf, you should now understand the relationship
between induced emf and changing magnetic flux.
To connect the mathematical statement of Faraday’s law to physical
effects, consider the magnetic fields and conducting loops shown in
Figure 3.5. As Faraday discovered, the mere presence of magnetic flux
Faraday’s law 69
through a circuit does not produce an electric current within that circuit.
Thus, holding a stationary magnet near a stationary conducting loop
induces no current (in this case, the magnetic flux is not a function of
time, so its time derivative is zero and the induced emf must also be zero).
Of course, Faraday’s law tells you that changing the magnetic flux
through a surface does induce an emf in any circuit that is a boundary to
that surface. So, moving a magnet toward or away from the loop, as in
Figure 3.5(a), causes the magnetic flux through the surface bounded by
the loop to change, resulting in an induced emf around the circuit.
4
In Figure 3.5(b), the change in magnetic flux is produced not by
moving the magnet, but by rotating the loop. This changes the angle
between the magnetic field and the surface normal, which changes ~
B^
n.In
Figure 3.5(c), the area enclosed by the loop is changing ove r tim e, whi ch
changes the flux through the surface. In each of these cases, you should
note that the magnitude of the induced emf does not depend on the total
amount of magnetic flux through the loop – it depends only on how fast
the flux changes.
Before looking at some examples of how to use Faraday’s law to solve
problems, you should consider the direction of the induced electric field,
which is provided by Lenz’s law.
B
Rotating loop
B
Loop of decreasing radius
N
Magnet motion
B
Induced current Induced current Induced current
(a) (b) (c)
Figure 3.5 Magnetic flux and induced current.
4
For simplicity, you can imagine a planar surface stretched across the loop, but Faraday’s
law holds for any surface bounded by the loop.
A student’s guide to Maxwell’s Equations70
Lenz’s law
There’s a great deal of physics wrapped up in the minus sign on the right
side of Faraday’s law, so it is fitting that it has a name: Lenz’s law. The
name comes from Heinrich Lenz, a German physicist who had an
important insight concerning the direction of the current induced by
changing magnetic flux.
Lenz’s insight was this: currents induced by changing magnetic flux
always flow in the direction so as to oppose the change in flux. That is, if
the magnetic flux through the circuit is increasing, the induced current
produces its own magnetic flux in the opposite direction to offset the
increase. This situation is shown in Figure 3.6(a), in which the magnet
is moving toward the loop. As the leftward flux due to the magnet
increases, the induced current flows in the direction shown, which
produces rightward magnetic flux that opposes the increased flux from
the magnet.
The alternative situation is shown in Figure 3.6(b), in which the magnet
is moving away from the loop and the leftward flux through the circuit is
decreasing. In this case, the induced current flows in the opposite direc-
tion, contributing leftward flux to make up for the decreasing flux from
the magnet.
It is important for you to understand that changing magnetic flux
induces an electric field whether or not a conducting path exists in which
a current may flow. Thus, Lenz’s law tells you the direction of the cir-
culation of the induced electric field around a specified path even if no
conduction current actually flows along that path.
N S
Magnet motion
B
(a) (b)
Leftward flux
increases as
magnet approaches
Current produces
rightward flux
N S
Magnet motion
Leftward flux
decreases as
magnet recedes
Current produces
more leftward flux
B
Figure 3.6 Direction of induced current.
Faraday’s law 71
HC~
Ed
~
l¼d
dt RS~
B^
ndaApplying Faraday’s
law (integral form)
The following examples show you how to use Faraday’s law to solve
problems involving changing magnetic flux and induced emf.
Example 3.1: Given an expression for the magnetic field as a function of
time, determine the emf induced in a loop of specified size.
Problem: For a magnetic field given by
~
Bðy;tÞ¼B0
t
t0
y
y0
^z:
Find the emf induced in a square loop of side Llying in the xy-plane
with one corner at the origin. Also, find the direction of current flow in
the loop.
Solution: Using Faraday’s flux rule,
emf ¼d
dt ZS
~
B^
nda:
For a loop in the xy-plane, ^
n¼^zand da =dx dy,so
emf ¼d
dt ZL
y¼0ZL
x¼0
B0
t
t0
y
y0
^z^zdxdy,
and
emf ¼d
dt LZL
y¼0
B0
t
t0
y
y0
dy

¼d
dt B0
t
t0
L3
2y0

:
Taking the time derivative gives
emf ¼B0
L3
2t0y0
:
Since upward magnetic flux is increasing with time, the current will
flow in a direction that produces flux in the downward ð^zÞdirection.
This means the current will flow in the clockwise direction as seen
from above.
A student’s guide to Maxwell’s Equations72
Example 3.2: Given an expression for the change in orientation of a
conducting loop in a fixed magnetic field, find the emf induced in the loop.
Problem: A circular loop of radius r
0
rotates with angular speed xin a
fixed magnetic field as shown in the figure.
(a) Find an expression for the emf induced in the loop.
(b) If the magnitude of the magnetic field is 25 lT, the radius of the loop
is 1 cm, the resistance of the loop is 25 W, and the rotation rate xis 3
rad/s, what is the maximum current in the loop?
Solution: (a) By Faraday’s flux rule, the emf is
emf ¼d
dt ZS
~
B^
nda
Since the magnetic field and the area of the loop are constant, this becomes
emf ¼ZS
d
dt ð~
B^
nÞda ¼ZS
~
B
d
dt ðcos hÞda:
Using h=xt,thisis
emf ¼ZS
~
B
d
dt ðcos xtÞda ¼~
B
d
dt ðcos xtÞZS
da:
Taking the time derivative and performing the integration gives
emf ¼~
B
xðsin xtÞðpr2
0Þ:
(b) By Ohm’s law, the current is the emf divided by the resistance of the
circuit, which is
I¼emf
R¼~
B
xðsin xtÞðpr2
0Þ
R:
For maximum current, sin(xt) = 1, so the current is
I¼ð25 ·106Þð3Þ½pð0:012Þ
25 ¼9:4·1010 A:
B
v
Faraday’s law 73
Example 3.3: Given an expression for the change in size of a conducting
loop in a fixed magnetic field, find the emf induced in the loop.
Problem: A circular loop lying perpendicular to a fixed magnetic field
decreases in size over time. If the radius of the loop is given by r(t)=
r
0
(1t/t
0
), find the emf induced in the loop.
Solution: Since the loop is perpendicular to the magnetic field, the loop
normal is parallel to ~
B, and Faraday’s flux rule is
emf ¼d
dt ZS
~
B^
nda¼~
B
d
dt ZS
da ¼~
B
d
dt ðpr2Þ:
Inserting r(t) and taking the time derivative gives
emf ¼~
B
d
dt pr2
01t
t0

2¼~
B
pr2
0ð2Þ1t
t0

1
t0

;
or
emf ¼2~
B
pr2
0
t0
1t
t0

:
A student’s guide to Maxwell’s Equations74
3.2 The differential form of Faraday’s law
The differential form of Faraday’s law is generally written as
~
r·~
E¼@~
B
@tFaraday’s law:
The left side of this equation is a mathematical description of the curl of
the electric field – the tendency of the field lines to circulate around a
point. The right side represents the rate of change of the magnetic field
over time.
The curl of the electric field is discussed in detail in the following
section. For now, make sure you grasp the main idea of Faraday’s law in
differential form:
A circulating electric field is produced by a magnetic field that changes
with time.
To help you understand the meaning of each symbol in the differential
form of Faraday’s law, here’s an expanded view:
E
=
Reminder that the
del operator is a vector
Reminder that the electric
field is a vector
The differential
operator called
“del” or “nabla”
The cross-product turns
the del operator into the
curl
The electric
field in V/m
×B
–t
The rate of change
of the magnetic fiel
d
with time
Faraday’s law 75
~
r·Del cross – the curl
The curl of a vector field is a measure of the field’s tendency to circulate
about a point – much like the divergence is a measure of the tendency of
the field to flow away from a point. Once again we have Maxwell to
thank for the terminology; he settled on ‘‘curl’’ after considering several
alternatives, including ‘‘turn’’ and ‘‘twirl’’ (which he thought was some-
what racy).
Just as the divergence is found by considering the flux through an
infinitesimal surface surrounding the point of interest, the curl at a spe-
cified point may be found by considering the circulation per unit area
over an infinitesimal path around that point. The mathematical definition
of the curl of a vector field ~
Ais
curlð~
AÞ¼ ~
r·~
Alim
DS!0
1
DSI
C
~
Ad
~
l;ð3:11Þ
where Cis a path around the point of interest and DSis the surface area
enclosed by that path. In this definition, the direction of the curl is the
normal direction of the surface for which the circulation is a maximum.
This expression is useful in defining the curl, but it doesn’t offer much
help in actually calculating the curl of a specified field. You’ll find an
alternative expression for curl later in this section, but first you should
consider the vector fields shown in Figure 3.7.
To find the locations of large curl in each of these fields, imagine that
the field lines represent the flow lines of a fluid. Then look for points at
1
3
5
4
7
2
6
(a) (b) (c)
Figure 3.7 Vector fields with various values of curl.
A student’s guide to Maxwell’s Equations76
which the flow vectors on one side of the point are significantly different
(in magnitude, direction, or both) from the flow vectors on the opposite
side of the point.
To aid this thought experiment, imagine holding a tiny paddlewheel at
each point in the flow. If the flow would cause the paddlewheel to rotate,
the center of the wheel marks a point of nonzero curl. The direction of the
curl is along the axis of the paddlewheel (as a vector, curl must have both
magnitude and direction). By convention, the positive-curl direction is
determined by the right-hand rule: if you curl the fingers of your right hand
along the circulation, your thumb points in the direction of positive curl.
Using the paddlewheel test, you can see that points 1, 2, and 3 in
Figure 3.7(a) and points 4 and 5 in Figure 3.7(b) are high-curl locations.
The uniform flow around point 6 in Figure 3.7(b) and the diverging flow
lines around point 7 in Figure 3.7(b) would not cause a tiny paddlewheel
to rotate, meaning that these are points of low or zero curl.
To make this quantitative, you can use the differential form of the curl
or ‘‘del cross’’ ð~
r·Þoperator in Cartesian coordinates:
~
r·~
A¼^
i@
@xþ^
j@
@yþ^
k@
@z

·ð^
iAxþ^
jAyþ^
kAzÞ:ð3:12Þ
The vector cross-product may be written as a determinant:
~
r·~
A¼
^
i^
j^
k
@
@x
@
@y
@
@z
AxAyAz
;ð3:13Þ
which expands to
~
r·~
A¼@Az
@y@Ay
@z

^
iþ@Ax
@z@Az
@x

^
jþ@Ay
@x@Ax
@y

^
k:ð3:14Þ
Note that each component of the curl of ~
Aindicates the tendency of the
field to rotate in one of the coordinate planes. If the curl of the field at a
point has a large x-component, it means that the field has significant
circulation about that point in the y–zplane. The overall direction of the
curl represents the axis about which the rotation is greatest, with the
sense of the rotation given by the right-hand rule.
If you’re wondering how the terms in this equation measure rotation,
consider the vector fields shown in Figure 3.8. Look first at the field in
Faraday’s law 77
Figure 3.8(a) and the x-component of the curl in the equation: this term
involves the change in A
z
with yand the change in A
y
with z. Proceeding
along the y-axis from the left side of the point of interest to the right, A
z
is
clearly increasing (it is negative on the left side of the point of interest and
positive on the right side), so the term @A
z
/@ymust be positive. Looking
now at A
y
, you can see that it is positive below the point of interest and
negative above, so it is decreasing along the zaxis. Thus, @A
y
/@zis
negative, which means that it increases the value of the curl when it is
subtracted from @A
z
/@y. Thus the curl has a large value at the point of
interest, as expected in light of the circulation of ~
Aabout this point.
The situation in Figure 3.8(b) is quite different. In this case, both @A
y
/@z
and @A
z
/@yare positive, and subtracting @A
y
/@zfrom @A
z
/@ygives a small
result. The value of the x-component of the curl is therefore small in this
case. Vector fields with zero curl at all points are called ‘‘irrotational.’’
Here are expressions for the curl in cylindrical and spherical coordinates:
r·~
A1
r
@Az
@’ @A’
@z

^
rþ@Ar
@z@Az
@r

^
’
þ1
r
@ðrA’Þ
@r@Ar
@’

^zðcylindricalÞ;
ð3:15Þ
r·~
A1
rsin h
@ðA’sin hÞ
@h@Ah
@’

^
rþ1
r
1
sin h
@Ar
@’ @ðrA’Þ
@r

^
h
þ1
r
@ðrAhÞ
@r@Ar
@h

^
’ðsphericalÞ:
ð3:16Þ
x
y
z
x
y
z
(a) (b)
Figure 3.8 Effect of @A
y
/@zand @A
z
/@Yon value of curl.
A student’s guide to Maxwell’s Equations78
~
r·~
EThe curl of the electric field
Since charge-based electric fields diverge away from points of positive
charge and converge toward points of negative charge, such fields cannot
circulate back on themselves. You can understand that by looking at the
field lines for the electric dipole shown in Figure 3.9(a). Imagine moving
along a closed path that follows one of the electric field lines diverging
from the positive charge, such as the dashed line shown in the figure. To
close the loop and return to the positive charge, you’ll have to move
‘‘upstream’’ against the electric field for a portion of the path. For that
segment, ~
Ed
~
lis negative, and the contribution from this part of the
path subtracts from the positive value of ~
Ed
~
lfor the portion of the path
in which ~
Eand d
~
lare in the same direction. Once you’ve gone all the way
around the loop, the integration of ~
Ed
~
lyields exactly zero.
Thus, the field of the electric dipole, like all electrostatic fields, has
no curl.
Electric fields induced by changing magnetic fields are very different, as
you can see in Figure 3.9(b). Wherever a changing magnetic field exists, a
circulating electric field is induced. Unlike charge-based electric fields,
induced fields have no origination or termination points – they are
continuous and circulate back on themselves. Integrating ~
Ed
~
laround
any boundary path for the surface through which ~
Bis changing produces
a nonzero result, which means that induced electric fields have curl. The
faster ~
Bchanges, the larger the magnitude of the curl of the induced
electric field.
12
∂B/∂t
(a) (b)
E
E
E E
Figure 3.9 Closed paths in charge-based and induced electric fields.
Faraday’s law 79
~
r·~
E¼@~
B
@tApplying Faraday’s law (differential form)
The differential form of Faraday’s law is very useful in deriving the
electromagnetic wave equation, which you can read about in Chapter 5.
You may also encounter two types of problems that can be solved using
this equation. In one type, you’re provided with an expression for the
magnetic field as a function of time and asked to find the curl of the
induced electric field. In the other type, you’re given an expression for the
induced vector electric field and asked to determine the time rate of
change of the magnetic field. Here are two examples of such problems.
Example 3.4: Given an expression for the magnetic field as a function of
time, find the curl of the electric field.
Problem: The magnetic field in a certain region is given by the expression
~
BðtÞ¼B0cosðkz xtÞ^
j.
(a) Find the curl of the induced electric field at that location.
(b) If the E
z
is known to be zero, find E
x
.
Solution: (a) By Faraday’s law, the curl of the electric field is the negative
of the derivative of the vector magnetic field with respect to time. Thus,
~
r·~
E¼@~
B
@t¼@B0cosðkz xtÞ½
^
j
@t;
or
~
r·~
E¼xB0sinðkz xtÞ^
j:
(b) Writing out the components of the curl gives
@Ez
@y@Ey
@z

^
iþ@Ex
@z@Ez
@x

^
jþ@Ey
@x@Ex
@y

^
k¼xB0sinðkz xtÞ^
j:
Equating the ^
jcomponents and setting E
z
to zero gives
@Ex
@z

¼xB0sinðkz xtÞ:
Integrating over zgives
Ex¼ZxB0sinðkz xtÞdz ¼x
kB0cosðkz xtÞ,
to within a constant of integration.
A student’s guide to Maxwell’s Equations80
Example 3.5: Given an expression for the induced electric field, find the
time rate of change of the magnetic field.
Problem: Find the rate of change with time of the magnetic field at a
location at which the induced electric field is given by
~
Eðx;y;zÞ¼E0
z
z0

2
^
iþx
x0

2
^
jþy
y0

2
^
k
"#
:
Solution: Faraday’s law tells you that the curl of the induced electric field
is equal to the negative of the time rate of change of the magnetic field.
Thus
@~
B
@t¼~
r·~
E;
which in this case gives
@~
B
@t¼ @Ez
@y@Ey
@z

^
i@Ex
@z@Ez
@x

^
j@Ey
@x@Ex
@y

^
k,
@~
B
@t¼E0
2y
y2
0

^
iþ2z
z2
0

^
jþ2x
x2
0

^
k

:
Problems
You can exercise your understanding of Faraday’s law on the following
problems. Full solutions are available on the book’s website.
3.1 Find the emf induced in a square loop with sides of length alying in
the yz-plane in a region in which the magnetic field changes over time
as ~
BðtÞ¼B0e5t=t0^
i.
3.2 A square conducting loop with sides of length Lrotates so that the
angle between the normal to the plane of the loop and a fixed
magnetic field ~
Bvaries as h(t)¼h
0
(t/t
0
); find the emf induced in the
loop.
3.3 A conducting bar descends with speed vdown conducting rails in the
presence of a constant, uniform magnetic field pointing into the page,
as shown in the figure.
(a) Write an expression for the emf induced in the loop.
(b) Determine the direction of current flow in the loop.
Faraday’s law 81
3.4 A square loop of side amoves with speed vinto a region in which a
magnetic field of magnitude B
0
exists perpendicular to the plane of the
loop, as shown in the figure. Make a plot of the emf induced in the loop
as it enters, moves through, and exits the region of the magnetic field.
3.5 A circular loop of wire of radius 20 cm and resistance of 12 Wsurrounds
a 5-turn solenoid of length 38 cm and radius 10 cm, as shown in the
figure. If the current in the solenoid increases linearly from 80 to
300 mA in 2 s, what is the maximum current induced in the loop?
3.6 A 125-turn rectangular coil of wire with sides of 25 and 40 cm rotates
about a horizontal axis in a vertical magnetic field of 3.5 mT. How
fast must this coil rotate for the induced emf to reach 5V?
3.7 The current in a long solenoid varies as I(t)=I
0
sin(xt). Use
Faraday’s law to find the induced electric field as a function of r
both inside and outside the solenoid, where risthedistancefrom
the axis of the solenoid.
3.8 The current in a long, straight wire decreases as I(t)¼I
0
e
t/s
. Find the
induced emf in a square loop of wire of side slying in the plane of the
current-carrying wire at a distance d, as shown in the figure.
02a
B
a
a
v
I
(t)
d
s
s
A student’s guide to Maxwell’s Equations82
4
The Ampere–Maxwell law
For thousands of years, the only known sources of magnetic fields were
certain iron ores and other materials that had been accidentally or
deliberately magnetized. Then in 1820, French physicist Andre-Marie
Ampere heard that in Denmark Hans Christian Oersted had deflected a
compass needle by passing an electric current nearby, and within one
week Ampere had begun quantifying the relationship between electric
currents and magnetic fields.
‘‘Ampere’s law’’ relating a steady electric current to a circulating
magnetic field was well known by the time James Clerk Maxwell began
his work in the field in the 1850s. However, Ampere’s law was known to
apply only to static situations involving steady currents. It was Maxwell’s
addition of another source term – a changing electric flux – that extended
the applicability of Ampere’s law to time-dependent conditions. More
importantly, it was the presence of this term in the equation now called
the Ampere–Maxwell law that allowed Maxwell to discern the electro-
magnetic nature of light and to develop a comprehensive theory of
electromagnetism.
4.1 The integral form of the Ampere–Maxwell law
The integral form of the Ampere–Maxwell law is generally written as
IC
~
Bd
~
l¼l0Ienc þe0
d
dt ZS
~
E^
nda

The Ampere–Maxwell law:
The left side of this equation is a mathematical description of the
circulation of the magnetic field around a closed path C. The right side
83
includes two sources for the magnetic field; a steady conduction current
and a changing electric flux through any surface Sbounded by path C.
In this chapter, you’ll find a discussion of the circulation of the mag-
netic field, a description of how to determine which current to include in
calculating ~
B, and an explanation of why the changing electric flux is
called the ‘‘displacement current.’’ There are also examples of how to use
the Ampere–Maxwell law to solve problems involving currents and
magnetic fields. As always, you should begin by reviewing the main idea
of the Ampere–Maxwell law:
An electric current or a changing electric flux through a surface
produces a circulating magnetic field around any path that bounds
that surface.
In other words, a magnetic field is produced along a path if any current is
enclosed by the path or if the electric flux through any surface bounded
by the path changes over time.
It is important that you understand that the path may be real or purely
imaginary – the magnetic field is produced whether the path exists or not.
Here’s an expanded view of the Ampere–Maxwell law:
Of what use is the Ampere–Maxwell law? You can use it to determine the
circulation of the magnetic field if you’re given information about the
enclosed current or the change in electric flux. Furthermore, in highly
symmetric situations, you may be able to extract ~
Bfrom the dot product
and the integral and determine the magnitude of the magnetic field.
C
dlB =
∫∫
0Ienc +0d
dt
S
E
ˆdan
Tells you to sum up the contributions
from each portion of closed path C in
a direction given by the right-hand rule
The magnetic
field in teslas
Dot product tells you to find the
part of B parallel to dl (along path C)
Reminder that the
magnetic field
is a vector
An incremental
segment of path C
The magnetic
permeability
of free space
The electric
current in amperes
Reminder that
only the enclosed
current contributes
The electric
permittivity
of free space
The electric
flux through
a surface
bounded by C
The rate of change
with time
A student’s guide to Maxwell’s Equations84
HC~
Bd
~
lThe magnetic field circulation
Spend a few minutes moving a magnetic compass around a long, straight
wire carrying a steady current, and here’s what you’re likely to find: the
current in the wire produces a magnetic field that circles around the wire
and gets weaker as you get farther from the wire.
With slightly more sophisticated equipment and an infinitely long wire,
you’d find that the magnetic field strength decreases precisely as 1/r,
where ris the distance from the wire. So if you moved your measuring
device in a way that kept the distance to the wire constant, say by circling
around the wire as shown in Figure 4.1, the strength of the magnetic field
wouldn’t change. If you kept track of the direction of the magnetic field
as you circled around the wire, you’d find that it always pointed along
your path, perpendicular to an imaginary line from the wire to your
location.
If you followed a random path around the wire getting closer and
farther from the wire as you went around, you’d find the magnetic field
getting stronger and weaker, and no longer pointing along your path.
Now imagine keeping track of the magnitude and direction of the
magnetic field as you move around the wire in tiny increments. If, at each
incremental step, you found the component of the magnetic field ~
Balong
that portion of your path d
~
l, you’d be able to find ~
Bd
~
l. Keeping track of
each value of ~
Bd
~
land then summing the results over your entire path,
you’d have a discrete version of the left side of the Ampere–Maxwell law.
Making this process continuous by letting the path increment shrink
NS
Long straight wire carrying steady current
Field strength
meter and magnetic
compass
Magnetic field direction
is along circular path
I
Magnetic field
strength depends
only on distance
from wire
Stronger fiel
d
Weaker field
Magnetic field
strength is constant
along this path
Figure 4.1 Exploring the magnetic field around a current-carrying wire.
The Ampere–Maxwell law 85
toward zero would then give you the circulation of the magnetic field:
Magnetic field circulation ¼IC
~
Bd
~
l:ð4:1Þ
The Ampere–Maxwell law tells you that this quantity is proportional to
the enclosed current and rate of change of electric flux through any
surface bounded by your path of integration (C). But if you hope to use
this law to determine the value of the magnetic field, you’ll need to dig ~
B
out of the dot product and out of the integral. That means you’ll have to
choose your path around the wire very carefully – just as you had to
choose a ‘‘special Gaussian surface’’ to extract the electric field from
Gauss’s law, you’ll need a ‘‘special Amperian loop’’ to determine the
magnetic field.
You’ll find examples of how to do that after the next three sections,
which discuss the terms on the right side of the Ampere–Maxwell law.
A student’s guide to Maxwell’s Equations86
l0The permeability of free space
The constant of proportionality between the magnetic circulation on the
left side of the Ampere–Maxwell law and the enclosed current and rate of
flux change on the right side is l
0
, the permeability of free space. Just as
the electric permittivity characterizes the response of a dielectric to an
applied electric field, the magnetic permeability determines a material’s
response to an applied magnetic field. The permeability in the Ampere–
Maxwell law is that of free space (or ‘‘vacuum permeability’’), which is
why it carries the subscript zero.
The value of the vacuum permeability in SI units is exactly 4p·10
7
volt-seconds per ampere-meter (Vs/Am); the units are sometimes given as
newtons per square ampere (N/A
2
) or the fundamental units of (m kg/C
2
).
Therefore, when you use the Ampere–Maxwell law, remember to multiply
both terms on the right side by
l0¼4p·107Vs=Am:
As in the case of electric permittivity in Gauss’s law for electric fields, the
presence of this quantity does not mean that the Ampere–Maxwell law
applies only to sources and fields in a vacuum. This form of the Ampere–
Maxwell law is general, so long as you consider all currents (bound as well
as free). In the Appendix, you’ll find a version of this law that’s more useful
when dealing with currents and fields in magnetic materials.
One interesting difference between the effect of dielectrics on electric fields
and the effect of magnetic substances on magnetic fields is that the magnetic
field is actually stronger than the applied field within many magnetic
materials. The reason for this is that these materials become magnetized
when exposed to an external magnetic field, and the induced magnetic field is
in the same direction as the applied field, as shown in Figure 4.2.
The permeability of a magnetic material is often expressed as the
relative permeability, which is the factor by which the material’s per-
meability exceeds that of free space:
Relative permeability lr¼l=l0:ð4:2Þ
Materials are classified as diamagnetic, paramagnetic, or ferromagnetic
on the basis of relative permeability. Diamagnetic materials have l
r
slightly less than 1.0 because the induced field weakly opposes the applied
field. Examples of diamagnetic materials include gold and silver, which
have l
r
of approximately 0.99997. The induced field within paramagnetic
The Ampere–Maxwell law 87
materials weakly reinforces the applied field, so these materials have l
r
slightly greater than 1.0. One example of a paramagnetic material is
aluminum with l
r
of 1.00002.
The situation is more complex for ferromagnetic materials, for which
the permeability depends on the applied magnetic field. Typical max-
imum values of permeability range from several hundred for nickel and
cobalt to over 5000 for reasonably pure iron.
As you may recall, the inductance of a long solenoid is given by the
expression
L¼lN2A
‘;ð4:3Þ
where lis the magnetic permeability of the material within the solenoid,
Nis the number of turns, Ais the cross-sectional area, and ‘is the length
of the coil. As this expression makes clear, adding an iron core to a
solenoid may increase the inductance by a factor of 5000 or more.
Like electrical permittivity, the magnetic permeability of any medium
is a fundamental parameter in the determination of the speed with which
an electromagnetic wave propagates through that medium. This makes it
possible to determine the speed of light in a vacuum simply by measuring
l
0
and e
0
using an inductor and a capacitor; an experiment for which, to
paraphrase Maxwell, the only use of light is to see the instruments.
Magnetic dipole
moments align with
applied field
Applied magnetic
field produced by
current I
II
Figure 4.2 Effect of magnetic core on field inside solenoid.
A student’s guide to Maxwell’s Equations88
Ienc The enclosed electric current
Although the concept of ‘‘enclosed current’’ sounds simple, the question
of exactly which current to include on the right side of the Ampere–
Maxwell law requires careful consideration.
It should be clear from the first section of this chapter that the
‘‘enclosing’’ is done by the path Caround which the magnetic field is
integrated (if you’re having trouble imagining a path enclosing anything,
perhaps ‘‘encircling’’ is a better word). However, consider for a moment
the paths and currents shown in Figure 4.3; which of the currents are
enclosed by paths C
1
,C
2
, and C
3
, and which are not?
The easiest way to answer that question is to imagine a membrane
stretched across the path, as shown in Figure 4.4. The enclosed current is
then just the net current that penetrates the membrane.
The reason for saying ‘‘net’’ current is that the direction of the current
relative to the direction of integration must be considered. By convention,
the right-hand rule determines whether a current is counted as positive or
negative: if you wrap the fingers of your right hand around the path in the
(a) (b) (c)
C
1
I
1
I
2
C
2
I
3
C
3
Figure 4.3 Currents enclosed (and not enclosed) by paths.
I
1
C
1
I
2
C
2
C
3
I
3
(a) (b) (c)
Figure 4.4 Membranes stretched across paths.
The Ampere–Maxwell law 89
direction of integration, your thumb points in the direction of positive
current. Thus, the enclosed current in Figure 4.4(a) is þI
1
if the inte-
gration around path C
1
is performed in the direction indicated; it would
be I
1
if the integration were performed in the opposite direction.
Using the membrane approach and right-hand rule, you should be able
to see that the enclosed current is zero in both Figure 4.4(b) and 4.4(c).
No net current is enclosed in Figure 4.4(b) , since the sum of the currents is
I
2
þI
2
= 0, and no current penetrates the membrane in either direction
in Figure 4.4(c).
An important concept for you to understand is that the enclosed
current is exactly the same irrespective of the shape of the surface you
choose, provided that the path of integration is a boundary (edge) of that
surface. The surfaces shown in Figure 4.4 are the simplest, but you could
equally well have chosen the surfaces shown in Figure 4.5, and the
enclosed currents would be exactly the same.
Notice that in Figure 4.5(a) current I
1
penetrates the surface at only
one point, so the enclosed current is þI
1
, just as it was for the flat
membrane of Figure 4.4(a). In Figure 4.5(b), current I
2
does not penetrate
the ‘‘stocking cap’’ surface anywhere, so the enclosed current is zero, as it
was for the flat membrane of Figure 4.4(b). The surface in Figure 4.5(c) is
penetrated twice by current I
3
, once in the positive direction and once in
the negative direction, so the net current penetrating the surface remains
zero, as it was in Figure 4.4(c) (for which the current missed the mem-
brane entirely).
Selection of alternate surfaces and determining the enclosed current is
more than just an intellectual diversion. The need for the changing-flux
term that Maxwell added to Ampere’s law can be made clear through just
such an exercise, as you can see in the next section.
I
1C1 I2C2 C3
I3
(a) (b) (c)
Figure 4.5 Alternative surfaces with boundaries C
1
,C
2
, and C
3
.
A student’s guide to Maxwell’s Equations90
d
dt RS~
E^
ndaThe rate of change of flux
This term is the electric flux analog of the changing magnetic flux term in
Faraday’s law, which you can read about in Chapter 3. In that case, a
changing magnetic flux through any surface was found to induce a cir-
culating electric field along a boundary path for that surface.
Purely by symmetry, you might suspect that a changing electric flux
through a surface will induce a circulating magnetic field around a
boundary of that surface. After all, magnetic fields are known to circulate –
Ampere’s law says that any electric current produces just such a circulating
magnetic field. So how is it that several decades went by before anyone saw
fit to write an ‘‘electric induction’’ law to go along with Faraday’s law of
magnetic induction?
For one thing, the magnetic fields induced by changing electric flux are
extremely weak and are therefore very difficult to measure, so in the
nineteenth century there was no experimental evidence on which to base
such a law. In addition, symmetry is not always a reliable predictor
between electricity and magnetism; the universe is rife with individual
electric charges, but apparently devoid of the magnetic equivalent.
Maxwell and his contemporaries did realize that Ampere’s law as ori-
ginally conceived applies only to steady electric currents, since it is con-
sistent with the principle of conservation of charge only under static
conditions. To better understand the relationship between magnetic fields
and electric currents, Maxwell worked out an elaborate conceptual model
in which magnetic fields were represented by mechanical vortices and
electric currents by the motion of small particles pushed along by the
whirling vortices. When he added elasticity to his model and allowed
the magnetic vortices to deform under stress, Maxwell came to understand
the need for an additional term in his mechanical version of Ampere’s law.
With that understanding, Maxwell was able to discard his mechanical
model and rewrite Ampere’s law with an additional source for magnetic
fields. That source is the changing electric flux in the Ampere–Maxwell law.
Most texts use one of three approaches to demonstrating the need for
the changing-flux term in the Ampere–Maxwell law: conservation of
charge, special relativity, or an inconsistency in Ampere’s law when
applied to a charging capacitor. This last approach is the most common,
and is the one explained in this section.
Consider the circuit shown in Figure 4.6. When the switch is closed, a
current Iflows as the battery charges the capacitor. This current produces
The Ampere–Maxwell law 91
a magnetic field around the wires, and the circulation of that field is given
by Ampere’s law
IC
~
Bd
~
l¼l0Ienc
ðÞ:
A serious problem arises in determining the enclosed current. According
to Ampere’s law, the enclosed current includes all currents that penetrate
any surface for which path Cis a boundary. However, you’ll get very
different answers for the enclosed current if you choose a flat membrane
as your surface, as shown in Figure 4.7(a), or a ‘‘stocking cap’’ surface as
shown in Figure 4.7(b).
Although current Ipenetrates the flat membrane as the capacitor
charges, no current penetrates the ‘‘stocking cap’’ surface (since the
charge accumulates at the capacitor plate). Yet the Amperian loop is a
boundary to both surfaces, and the integral of the magnetic field around
that loop must be same no matter which surface you choose.
You should note that this inconsistency occurs only while the capacitor
is charging. No current flows before the switch is thrown, and after the
capacitor is fully charged the current returns to zero. In both of these
circumstances, the enclosed current is zero through any surface you can
imagine. Therefore, any revision to Ampere’s law must retain its correct
behavior in static situations while extending its utility to charging cap-
acitors and other time-dependent situations.
With more than a little hindsight, we might phrase our question this
way: since no conduction current flows between the capacitor plates,
what else might be going on in that region that would serve as the source
of a magnetic field?
Since charge is accumulating on the plates as the capacitor charges up,
you know that the electric field between the plates must be changing with
I
Battery
Capacitor
Amperian loop
Switch
Figure 4.6 Charging capacitor.
A student’s guide to Maxwell’s Equations92
time. This means that the electric flux through the portion of your
‘‘stocking cap’’ surface between the plates must also be changing, and
you can use Gauss’s law for electric fields to determine the change in flux.
By shaping your surface carefully, as in Figure 4.8, you can make it
into a ‘‘special Gaussian surface’’, which is everywhere perpendicular to
the electric field and over which the electric field is either uniform or zero.
Neglecting edge effects, the electric field between two charged conducting
plates is ~
E=(r/e
0
)^
n, where ris the charge density on the plates (Q/A),
making the electric flux through the surface
UE¼ZS
~
E^
nda¼ZS
r
e0
da ¼Q
Ae0ZS
da ¼Q
e0
:ð4:4Þ
Current I penetrates
this surface I
Battery
Capacitor
Amperian loop(a)
Switch
Amperian loop(b)
Battery
Capacitor
Switch
I
Figure 4.7 Alternative surfaces for determining enclosed current.
The Ampere–Maxwell law 93
The change in electric flux over time is therefore,
d
dt ZS
~
E^
nda

¼d
dt
Q
e0

¼1
e0
dQ
dt :ð4:5Þ
Multiplying by the vacuum permittivity makes this
e0
d
dt ZS
~
E^
nda

¼dQ
dt :ð4:6Þ
Thus, the change in electric flux with time multiplied by permittivity has
units of charge divided by time (coulombs per second or amperes in SI
units), which are of course the units of current. Moreover a current-like
quantity is exactly what you might expect to be the additional source of
the magnetic field around your surface boundary. For historical reasons,
the product of the permittivity and the change of electric flux through a
surface is called the ‘‘displacement current’’ even though no charge
actually flows across the surface. The displacement current is defined by
the relation
Ide0
d
dt ZS
~
E^
nda

:ð4:7Þ
Whatever you choose to call it, Maxwell’s addition of this term to
Ampere’s law demonstrated his deep physical insight and set the stage for
his subsequent discovery of the electromagnetic nature of light.
+Q–Q
Special Gaussian
surface
Time-varying
electric field
+
+
+
+
–
–
–
–
Figure 4.8 Changing electric flux between capacitor plates.
A student’s guide to Maxwell’s Equations94
HC~
Bd
~
l¼l0Ienc þe0d
dt RS~
E^
nda

Applying the
Ampere–Maxwell law
(integral form)
Like the electric field in Gauss’s law, the magnetic field in the
Ampere–Maxwell law is buried within an integral and coupled to another
vector quantity by the dot product. As you might expect, it is only in
highly symmetric situations that you’ll be able to determine the magnetic
field using this law. Fortunately, several interesting and realistic geom-
etries possess the requisite symmetry, including long current-carrying
wires and parallel-plate capacitors.
For such problems, the challenge is to find an Amperian loop over which
you expect ~
Bto be uniform and at a constant angle to the loop. However,
how do you know what to expect for ~
Bbefore you solve the problem?
In many cases, you’ll already have some idea of the behavior of the
magnetic field on the basis of your past experience or from experimental
evidence. What if that’s not the case – how are you supposed to figure out
how to draw your Amperian loop?
There’s no single answer to that question, but the best approach is to
use logic to try to reason your way to a useful result. Even for complex
geometries, you may be able to use the Biot–Savart law to discern the
field direction by eliminating some of the components through symmetry
considerations. Alternatively, you can imagine various behaviors for ~
B
and then see if they lead to sensible consequences.
For example, in a problem involving a long, straight wire, you might
reason as follows: the magnitude of ~
Bmust get smaller as you move away
from the wire; otherwise Oersted’s demonstration in Denmark would
have deflected compass needles everywhere in the world, which it clearly
did not. Furthermore, since the wire is round, there’s no reason to expect
that the magnetic field on one side of the wire is any different from the
field on the other side. So if ~
Bdecreases with distance from the wire and is
the same all around the wire, you can safely conclude that one path of
constant ~
Bwould be a circle centered on the wire and perpendicular to the
direction of current flow.
However, to deal with the dot product of ~
Band d
~
lin Ampere’s law,
you also need to make sure that your path maintains a constant angle
(preferably 0) to the magnetic field. If ~
Bwere to have both radial
and transverse components that vary with distance, the angle between
your path and the magnetic field might depend on distance from the wire.
The Ampere–Maxwell law 95
If you understand the cross–product between d
~
land ^
rin the
Biot–Savart law, you probably suspect that this is not the case. To verify
that, imagine that ~
Bhas a component pointing directly toward the wire. If
you were to look along the wire in the direction of the current, you’d see
the current running away from you and the magnetic field pointing at the
wire. Moreover, if you had a friend looking in the opposite direction at
the same time, she’d see the current coming toward her, and of course she
would also see ~
Bpointing toward the wire.
Now ask yourself, what would happen if you reversed the direction of
the current flow. Since the magnetic field is linearly proportional to the
current (~
B/~
I) according to the Biot–Savart law, reversing the current
must also reverse the magnetic field, and ~
Bwould then point away from
the wire. Now looking in your original direction, you’d see a current
coming toward you (since it was going away from you before it was
reversed), but now you’d see the magnetic field pointing away from the
wire. Moreover, your friend, still looking in her original direction, would
see the current running away from her, but with the magnetic field
pointing away from the wire.
Comparing notes with your friend, you’d find a logical inconsistency.
You’d say, ‘‘currents traveling away from me produce a magnetic field
pointing toward the wire, and currents coming toward me produce a
magnetic field pointing away from the wire.’’ Your friend, of course,
would report exactly the opposite behavior. In addittion, if you switched
positions and repeated the experiment, you’d each find that your original
conclusions were no longer true.
This inconsistency is resolved if the magnetic field circles around the wire,
having no radial component at all. With ~
Bhaving only a u-component,
5
all
observers agree that currents traveling away from an observer produce
clockwise magnetic fields as seen by that observer, whereas currents
approaching an observer produce counterclockwise magnetic fields for that
observer.
In the absence of external evidence, this kind of logical reasoning is
your best guide to designing useful Amperian loops. Therefore, for
problems involving a straight wire, the logical choice for your loop is a
circle centered on the wire. How big should you make your loop?
Remember why you’re making an Amperian loop in the first place – to
find the value of the magnetic field at some location. So make your
5
Remember that there’s a review of cylindrical and spherical coordinates on the book’s
website.
A student’s guide to Maxwell’s Equations96
Amperian loop go through that location. In other words, the loop radius
should be equal to the distance from the wire at which you intend to find
the value of the magnetic field. The following example shows how this
works.
Example 4.1: Given the current in a wire, find the magnetic field within
and outside the wire.
Problem: A long, straight wire of radius r
0
carries a steady current I
uniformly distributed throughout its cross-sectional area. Find the
magnitude of the magnetic field as a function of r, where ris the distance
from the center of the wire, for both r>r
0
and r<r
0
.
Solution: Since the current is steady, you can use Ampere’s law in its
original form
IC
~
Bd
~
l¼l0Ienc
ðÞ:
To find ~
Bat exterior points (r>r
0
), use the logic described above and
draw your loop outside the wire, as shown by Amperian loop #1 in
Figure 4.9. Since both ~
Band d
~
lhave only u-components and point in the
same direction if you obey the right-hand rule in determining your dir-
ection of integration, the dot product ~
Bd
~
lbecomes |~
B|| d
~
l|cos(0).
Furthermore, since |~
B| is constant around your loop, it comes out of the
integral:
IC
~
Bd
~
l¼IC
~
B
d
~
l¼BIC
dl ¼Bð2prÞ,
Amperian
loop #1
Wire
I
Amperia
n
loop #2
Figure 4.9 Amperian loops for current-carrying wire of radius r
0
.
The Ampere–Maxwell law 97
where ris the radius of your Amperian loop.
6
Ampere’s law tells you that
the integral of ~
Baround your loop is equal to the enclosed current times
the permeability of free space, and the enclosed current in this case is all of
I,so
Bð2prÞ¼l0Ienc ¼l0I
and, since ~
Bis in the ’-direction,
~
B¼l0I
2pr
^
’,
as given in Table 2.1. Note that this means that at points outside the wire
the magnetic field decreases as 1/rand behaves as if all the current were at
the center of the wire.
To find the magnetic field within the wire, you can apply the same logic
and use a smaller loop, as shown by Amperian loop #2 in Figure 4.9. The
only difference in this case is that not all the current is enclosed by the
loop; since the current is distributed uniformly throughout the wire’s
cross section, the current density
7
is I/(pr
02
), and the current passing
through the loop is simply that density times the area of the loop. Thus,
Enclosed current ¼current density ·loop area
or
Ienc ¼I
pr2
0
pr2¼Ir2
r2
0
:
Inserting this into Ampere’s law gives
IC
~
Bd
~
l¼Bð2prÞ¼l0Ienc ¼l0Ir2
r2
0
,
or
B¼l0Ir
2pr2
0
:
Thus, inside the wire the magnetic field increases linearly with distance
from the center of the wire, reaching a maximum at the surface of the
wire.
6
Another way to understand this is to write ~
Bas B’^’and d
~
las (rd’)^’,so~
Bd
~
l¼B’rd’
and R2p
0B’rd’¼B’ð2prÞ.
7
If you need a review of current density, you’ll find a section covering this topic later in this
chapter.
A student’s guide to Maxwell’s Equations98
Example 4.2: Given the time-dependent charge on a capacitor, find the
rate of change of the electric flux between the plates and the magnitude
of the resulting magnetic field at a specified location.
Problem: A battery with potential difference DVcharges a circular
parallel-plate capacitor of capacitance Cand plate radius r
0
through a
wire with resistance R. Find the rate of change of the electric flux between
the plates as a function of time and the magnetic field at a distance rfrom
the center of the plates.
Solution: From Equation 4.5, the rate of change of electric flux between
the plates is
dUE
dt ¼d
dt Z
S
~
E^
nda
0
@1
A¼1
e0
dQ
dt ,
where Qis the total charge on each plate. So you should begin by
determining how the charge on a capacitor plate changes with time as the
capacitor is charged. If you’ve studied series RC circuits, you may recall
that the relevant expression is
QðtÞ¼CDVð1et=RC Þ,
where DV,R, and Crepresent the potential difference, the series resist-
ance, and the capacitance, respectively. Thus,
dUE
dt ¼1
e0
d
dt CDVð1et=RCÞ
hi
¼1
e0
CDV1
RC et=RC

¼DV
e0Ret=RC :
This is the rate of change of the total electric flux between the plates. To
find the magnetic field at a distance rfrom the center of the plates, you’re
going to have to construct a special Amperian loop to help you extract
the magnetic field from the integral in the Ampere–Maxwell law:
IC
~
Bd
~
l¼l0Ienc þe0
d
dt ZS
~
E^
nda

:
Since no charge flows between the capacitor plates, I
enc
=0, and
IC
~
Bd
~
l¼l0e0
d
dt ZS
~
E^
nda

:
The Ampere–Maxwell law 99
As in the previous example, you’re faced with the challenge of designing a
special Amperian loop around which the magnetic field is constant in
amplitude and parallel to the path of integration around the loop. If you
use similar logic to that for the straight wire, you’ll see that the best
choice is to make a loop parallel to the plates, as shown in Figure 4.10.
The radius of this loop is r, the distance from the center of the plates at
which you are trying to find the magnetic field. Of course, not all of the
flux between the plates passes through this loop, so you will have to
modify your expression for the flux change accordingly. The fraction of
the total flux that passes through a loop of radius ris simply the ratio of
the loop area to the capacitor plate area, which is pr
2
/pr
02
, so the rate
of change of flux through the loop is
dUE
dt

Loop
¼DV
e0Ret=RC r2
r2
0

:
Inserting this into the Ampere–Maxwell law gives
IC
~
Bd
~
l¼l0e0
DV
e0Ret=RC r2
r2
0

¼l0DV
Ret=RC r2
r2
0

:
Moreover, since you’ve chosen your Amperian loop so as to allow ~
Bto
come out of the dot product and the integral using the same symmetry
arguments as in Example 4.1,
IC
~
Bd
~
l¼Bð2prÞ¼l0DV
Ret=RC r2
r2
0

,
which gives
B¼l0DV
2prR et=RC r2
r2
0

¼l0DV
2pRet=RC r
r2
0

,
meaning that the magnetic field increases linearly with distance from the
center of the capacitor plates and decreases exponentially with time,
reaching 1/eof its original value at time t=RC.
+
+
+
+
–
–
–
–
Figure 4.10 Amperian loop between capacitor plates.
A student’s guide to Maxwell’s Equations100
4.2 The differential form of the Ampere–Maxwell law
The differential form of the Ampere–Maxwell law is generally written as
~
r·~
B¼l0~
Jþe0
@~
E
@t
!
The Ampere–Maxwell law:
The left side of this equation is a mathematical description of the curl of
the magnetic field – the tendency of the field lines to circulate around a
point. The two terms on the right side represent the electric current
density and the time rate of change of the electric field.
These terms are discussed in detail in the following sections. For now,
make sure you grasp the main idea of the differential form of the
Ampere–Maxwell law:
A circulating magnetic field is produced by an electric current and by
an electric field that changes with time.
To help you understand the meaning of each symbol in the differential
form of the Ampere–Maxwell law, here’s an expanded view:
BJ
×=+E
t
00
Reminder that the del
operator is a vector
Reminder that the
magnetic field is
a vector
Reminder that the
current density is
a vector The rate of change of
the electric field
with time
The electric
permittivity
of free space
The electric current
density in amperes
per square meter
The magnetic
permeability
of free space
The magnetic
field in teslas
The differential
operator called
“del” or “nabla”
The cross-product turns
the del operator
into the curl
The Ampere–Maxwell law 101
~
r·~
BThe curl of the magnetic field
The left side of the differential form of the Ampere–Maxwell law
represents the curl of the magnetic field. All magnetic fields, whether
produced by electrical currents or by changing electric fields, circulate
back upon themselves and form continuous loops. In addition all fields
that circulate back on themselves must include at least one location about
which the path integral of the field is nonzero. For the magnetic field,
locations of nonzero curl are locations at which current is flowing or an
electric field is changing.
It is important that you understand that just because magnetic fields
circulate, you should not conclude that the curl is nonzero everywhere in
the field. A common misconception is that the curl of a vector field is
nonzero wherever the field appears to curve.
To understand why that is not correct, consider the magnetic field of
the infinit e line current shown in Figure 2. 1. The magnet ic fie ld lines
circulate around the current, and you know from Table 2.1 that the
magnetic field points in the ^
udirection and decreases as 1/r
~
B¼l0I
2pr
^
u:
Finding the curl of this field is particularly straightforward in cylindrical
coordinates
~
r·~
B¼1
r
@Bz
@’ @B’
@z

^
rþ@Br
@z@Bz
@r

^
’þ1
r
@ðrB’Þ
@r@Br
@’

^z:
Since B
r
and B
z
are both zero, this is
~
r·~
B¼
@Bu
@z

^
rþ1
r
@ðrBuÞ
@r

^z¼@l0I=2prðÞ
@z
^
rþ1
r
@rl0I=2prðÞ
@r
^z¼0:
However, doesn’t the differential form of the Ampere–Maxwell law tell
us that the curl of the magnetic field is nonzero in the vicinity of electric
currents and changing electric fields?
No, it doesn’t. It tells us that the curl of ~
Bis nonzero exactly at the
location through which an electric current is flowing, or at which an
electric field is changing. Away from that location, the field definitely
does curve, but the curl at any given point is precisely zero, as you just
found from the equation for the magnetic field of an infinite line current.
How can a curving field have zero curl? The answer lies in the amplitude
as well as the direction of the magnetic field, as you can see in Figure 4.11.
A student’s guide to Maxwell’s Equations102
Using the fluid-flow and small paddlewheel analogy, imagine the forces
on the paddlewheel placed in the field shown in Figure 4.11(a). The
center of curvature is well below the bottom of the figure, and the
spacing of the arrows indicates that the field is getting weaker with
distance from the center. At first glance, it may seem that this pad-
dlewheel would rotate clockwise owing to the curvature of the field,
since the flow lines are pointing slightly upward at the left paddle and
slightly downward at the right. However, consider the effect of the
weakening of the field above the axis of the paddlewheel: the top paddle
receives a weaker push from the field than the bottom paddle, as shown
in Figure 4.11(b). The stronger force on the bottom paddle will attempt
to cause the paddlewheel to rotate counterclockwise. Thus, the down-
ward curvature of the field is offset by the weakening of the field with
distance from the center of curvature. And if the field diminishes as 1/r,
the upward–downward push on the left and right paddles is exactly
compensated by the weaker–stronger push on the top and bottom
paddles. The clockwise and counter-clockwise forces balance, and the
paddlewheel does not turn – the curl at this location is zero, even
though the field lines are curved.
The key concept in this explanation is that the magnetic field may be
curved at many different locations, but only at points at which current is
flowing (or the electric flux is changing) is the curl of ~
Bnonzero. This is
analogous to the 1/r
2
reduction in electric field amplitude with distance
from a point charge, which keeps the divergence of the electric field as
zero at all points away from the location of the charge.
As in the electric field case, the reason the origin (where r=0) is not
included in our previous analysis is that our expression for the curl
includes terms containing rin the denominator, and those terms become
Weaker field
Stronger field
Upward-
pointing
field
Downward-
pointing
field
Weaker push
to the right
Stronger push
to the right
Upward push Downward push
(a) (b)
Figure 4.11 Offsetting components of the curl of ~
B:
The Ampere–Maxwell law 103
infinite at the origin. To evaluate the curl at the origin, use the formal
definition of curl as described in Chapter 3:
~
r·~
Blim
DS!0
1
DSI
C
~
Bd
~
l
Considering a special Amperian loop surrounding the current, this is
~
r·~
Blim
DS!0
1
DSI
C
~
Bd
~
l¼lim
DS!0
1
DS
l0
~
I
2prð2prÞ
!
¼lim
DS!0
1
DSl0
~
I

:
However,~
I=DSis just the average current density over the surface DS,and
as DSshrinks to zero, this becomes equal to ~
J, the current density at the
origin. Thus, at the origin
~
r·~
B¼l0~
J
in accordance with Ampere’s law.
So just as you might be fooled into thinking that charge-based electric
field vectors ‘‘diverge’’ everywhere because they get farther apart, you
might also think that magnetic field vectors have curl everywhere because
they curve around a central point. But the key factor in determining the
curl at any point is not simply the curvature of the field lines at that
point, but how the change in the field from one side of the point to the
other (say from left to right) compares to the change in the field in the
orthogonal direction (below to above). If those spatial derivatives
are precisely equal, then the curl is zero at that point.
In the case of a current-carrying wire, the reduction in the amplitude of
the magnetic field away from the wire exactly compensates for the
curvature of the field lines. Thus, the curl of the magnetic field is zero
everywhere except at the wire itself, where electric current is flowing.
A student’s guide to Maxwell’s Equations104
~
JThe electric current density
The right side of the differential form of the Ampere–Maxwell law
contains two source terms for the circulating magnetic field; the first
involves the vector electric current density. This is sometimes called
the ‘‘volume current density,’’ which can be a source of confusion if
you’re accustomed to ‘‘volume density’’ meaning the amount of
something per unit volume, such as kg/m
3
for mass density or C/m
3
for charge density.
This is not the case for current density, which is defined as the vector
current flowing through a unit cross-sectional area perpendicular to the
direction of the current. Thus, the units of current density are not
amperes per cubic meter, but rather amperes per square meter (A/m
2
).
To understand the concept of current density, recall that in the dis-
cussion of flux in Chapter 1, the quantity ~
Ais defined 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), ~
Ais a vector in the same direction as the
velocity, with units of particles per square meter per second. To find the
number of particles per second passing through a surface in the simplest
case (~
Auniform and perpendicular to the surface), you simply multiply ~
A
by the area of the surface.
These same concepts are relevant for current density, provided we
consider the amount of charge passing through the surface rather than the
number of atoms. If the number density of charge carriers is nand the
charge per carrier is q, then the amount of charge passing through a unit
area perpendicular to the flow per second is
~
J¼nq
~
vdðC=m2sorA=m2Þ;ð4:8Þ
where ~
vdis the average drift velocity of charge carriers. Thus, the direc-
tion of the current density is the direction of current flow, and the
magnitude is the current per unit area, as shown in Figure 4.12.
The complexity of the relationship between the total current Ithrough
a surface and the current density ~
Jdepends on the geometry of the
situation. If the current density ~
Jis uniform over a surface Sand is
everywhere perpendicular to the surface, the relationship is
I¼~
J
·ðsurface areaÞ~
Juniform and perpendicular to S:ð4:9Þ
The Ampere–Maxwell law 105
If ~
Jis uniform over a surface Sbut is not necessarily perpendicular to
the surface, to find the total current Ithrough Syou’ll have to determine
the component of the current density perpendicular to the surface. This
makes the relationship between Iand ~
J:
I¼~
J^
n·ðsurface areaÞ~
Juniform and at an angle to S:ð4:10Þ
And, if ~
Jis nonuniform and not perpendicular to the surface, then
I¼Z
S
~
J^
nda ~
Jnonuniform and at a variable angle to S:ð4:11Þ
This expression explains why some texts refer to electric current as ‘‘the
flux of the current density.’’
The electric current density in the Ampere–Maxwell law includes all
currents, including the bound current density in magnetic materials. You
can read more about Maxwell’s Equations inside matter in the Appendix.
Imaginary surface
within wire
J
Charge
carriers
Figure 4.12 Charge flow and current density.
A student’s guide to Maxwell’s Equations106
e0
@~
E
@tThe displacement current density
The second source term for the magnetic field in the Ampere–Maxwell
law involves the rate of change of the electric field with time. When
multiplied by the electrical permittivity of free space, this term has SI
units of amperes per square meter. These units are identical to those of ~
J,
the conduction current density that also appears on the right side of the
differential form of the Ampere–Maxwell law. Maxwell originally
attributed this term to the physical displacement of electrical particles
caused by elastic deformation of magnetic vortices, and others coined the
term ‘‘displacement current’’ to describe the effect.
However, does the displacement current density represent an actual
current? Certainly not in the conventional sense of the word, since electric
current is defined as the physical movement of charge. But it is easy to
understand why a term that has units of amperes per square meter and
acts as a source of the magnetic field has retained that name over the
years. Furthermore, the displacement current density is a vector quantity
that has the same relationship to the magnetic field as does ~
J, the con-
duction current density.
The key concept here is that a changing electric field produces a
changing magnetic field even when no charges are present and no phys-
ical current flows. Through this mechanism, electromagnetic waves may
propagate through even a perfect vacuum, as changing magnetic fields
induce electric fields, and changing electric fields induce magnetic fields.
The importance of the displacement current term, which arose initially
from Maxwell’s mechanical model, is difficult to overstate. Adding a
changing electric field as a source of the magnetic field certainly extended
the scope of Ampere’s law to time-dependent fields by eliminating the
inconsistency with the principle of conservation of charge. Far more
importantly, it allowed James Clerk Maxwell to establish a comprehen-
sive theory of electromagnetism, the first true field theory and the
foundation for much of twentieth century physics.
The Ampere–Maxwell law 107
~
r·~
B¼l0~
Jþe0@~
E
@t

Applying the Ampere–Maxwell
law (differential form)
The most common applications of the differential form of the Ampere–
Maxwell law are problems in which you’re provided with an expression
for the vector magnetic field and you’re asked to determine the electric
current density or the displacement current. Here are two examples of
this kind of problem.
Example 4.3: Given the magnetic field, find the current density at a
specified location.
Problem: Use the expressions for the magnetic field in Table 2.1 to find
the current density both inside and outside a long, straight wire of radius
r
0
carrying current Iuniformly throughout its volume in the positive
z-direction.
Solution: From Table 2.1 and Example 4.1, the magnetic field inside a
long, straight wire is
~
B¼l0Ir
2pr2
0
^
’,
where Iis the current in the wire and r
0
is the wire’s radius. In cylindrical
coordinates, the curl of ~
Bis
~
r·~
B1
r
@Bz
@u@Bu
@z

^
rþ@Br
@z@Bz
@r

^
uþ1
r
@ðrBuÞ
@r@Br
@u

^z:
And, since ~
Bhas only a ^
’-component in this case,
~
r·~
B
@B’
@z

^
rþ1
r
@ðrB’Þ
@r

^z¼1
r
@rðl0Ir=2pr2
0Þ
@r
0
@1
A^z
¼1
r2rl0I
2pr2
0

^z¼l0I
pr2
0

^z:
Using the static version of the Ampere–Maxwell law (since the current is
steady), you can find ~
Jfrom the curl of ~
B:
~
r·~
B¼l0~
J

:
A student’s guide to Maxwell’s Equations108
Thus,
~
J¼1
l0
l0I
pr2
0

^z¼I
pr2
0
^z;
which is the current density within the wire. Taking the curl of the
expression for ~
Boutside the wire, you’ll find that ~
J¼0, as expected.
Example 4.4: Given the magnetic field, find the displacement current
density.
Problem: The expression for the magnetic field of a circular parallel-plate
capacitor found in Example 4.2 is
~
B¼l0DV
2pRet=RC r
r2
0

^
’:
Use this result to find the displacement current density between the plates.
Solution: Once again you can use the curl of ~
Bin cylindrical coordinates:
~
r·~
B1
r
@Bz
@’ @B’
@z

^
rþ@Br
@z@Bz
@r

^
uþ1
r
@ðrB’Þ
@r@Br
@’

^z:
And, once again ~
Bhas only a ^
’-component:
~
r·~
B¼
@B’
@z

^
rþ1
r
@ðrB’Þ
@r

^z¼1
r
@ðrl0DV=2pRÞet=RC r
r2
0

@r
2
43
5^z
¼1
r2rl0DV
2pRet=RC 1
r2
0

^z¼l0DV
pRet=RC 1
r2
0

^z:
Since there is no conduction current between the plates, ~
J¼0 in this case
and the Ampere–Maxwell law is
~
r·~
B¼l0e0
@~
E
@t
!
;
from which you can find the displacement current density,
e0
@~
E
@t¼~
r·~
B
l0
¼1
l0
l0DV
pRet=RC 1
r2
0

^z¼DV
Ret=RC 1
pr2
0

^z:
Problems
The following problems will test your understanding of the Ampere–
Maxwell law. Full solutions are available on the book’s website.
The Ampere–Maxwell law 109
4.1 Two parallel wires carry currents I
1
and 2I
1
in opposite directions.
Use Ampere’s law to find the magnetic field at a point midway
between the wires.
4.2 Find the magnetic field inside a solenoid (Hint: use the Amperian
loop shown in the figure, and use the fact that the field is parallel to
the axis of the solenoid and negligible outside).
4.3 Use the Amperian loop shown in the figure to find the magnetic
field within a torus.
4.4 The coaxial cable shown in the figure carries current I
1
in the
direction shown on the inner conductor and current I
2
in the
opposite direction on the outer conductor. Find the magnetic field
in the space between the conductors as well as outside the cable if
the magnitudes of I
1
and I
2
are equal.
A student’s guide to Maxwell’s Equations110
4.5 Find the displacement current produced between the plates of a
discharging capacitor for which the charge varies as
QðtÞ¼Q0et=RC ,
where Q
0
is the initial charge, Cis the capacitance of the capacitor,
and Ris the resistance of the circuit through which the capacitor is
discharging.
4.6 A magnetic field of ~
B¼asinðbyÞebx^zis produced by an electric
current. What is the density of that current?
4.7 Find the electric current density that produces a magnetic field
given by ~
B¼B0ðe2rsin ’Þ^zin cylindrical coordinates.
4.8 What density of current would produce a magnetic field given by
~
B¼ða=rþb=rerþcerÞ^
’in cylindrical coordinates?
4.9 In this chapter, you learned that the magnetic field of a long,
straight wire, given by
~
B¼l0I
2pr
^
’,
has zero curl everywhere except at the wire itself. Prove that this
would not be true for a field that decreases as 1/r
2
with distance.
4.10 To directly measure the displacement current, researchers use a
time-varying voltage to charge and discharge a circular parallel-
plate capacitor. Find the displacement current density and electric
field as a function of time that would produce a magnetic field
given by
~
B¼rxDVcosðxtÞ
2dðc2Þ
^
’,
where ris the distance from the center of the capacitor, xis the
angular frequency of the applied voltage DV,dis the plate spacing,
and cis the speed of light.
The Ampere–Maxwell law 111
5
From Maxwell’s Equations to the wave
equation
Each of the four equations that have come to be known as Maxwell’s
Equations is powerful in its own right, for each embodies an important
aspect of electromagnetic field theory. However, Maxwell’s achievement
went beyond the synthesis of these laws or the addition of the displace-
ment current term to Ampere’s law – it was by considering these equa-
tions in combination that he reached his goal of developing a
comprehensive theory of electromagnetism. That theory elucidated the
true nature of light itself and opened the eyes of the world to the full
spectrum of electromagnetic radiation.
In this chapter, you’ll learn how Maxwell’s Equations lead directly to
the wave equation in just a few steps. To make those steps, you’ll first
have to understand two important theorems of vector calculus: the
divergence theorem and Stokes’ theorem. These two theorems make the
transition from the integral form to the differential form of Maxwell’s
Equations quite straightforward:
Gauss’s law for electric fields:
IS
~
E^
nda ¼qin=e0
Divergence
theorem
~
r~
E¼q=e0:
Gauss’s law for magnetic fields:
IS
~
B^
nda ¼0
Divergence
theorem
~
r~
B¼0:
Faraday’s law:
IC
~
Ed
~
l¼d
dt ZS
~
B^
nda
Stokes’
theorem
~
r·~
E¼q~
B
qt:
112
Ampere–Maxwell law:
IC
~
Bd
~
l¼l0Ienc þe0
d
dt ZS
~
E^
nda

Stokes’
theorem
~
r·~
B¼l0~
Jþe0
q~
E
qt
!
:
Along with the divergence theorem and Stokes’ theorem, you’ll also find
a discussion of the gradient operator and some useful vector identities in
this chapter. In addition, since the goal is to arrive at the wave equation,
here are the expanded views of the wave equation for electric and mag-
netic fields:
E
=
00
t
E
2
2
2
The Laplacian
operator The vector electric
field
The electric permittivity
of free space The vector
electric field
The magnetic
permeability
of free space
The second derivative
of the vector electric
field over space
The second derivative
of the vector electric
field with time
B
=
00
t
B
2
2
2
The Laplacian
operator The vector
magnetic field
The electric
permittivity
of free space The vector
magnetic field
The magnetic
permeability
of free space
The second derivative
of the vector magnetic
field with time
The second derivative
of the vector magnetic
field over space
From Maxwell’s Equations to the wave equation 113
HS~
A^
nda¼RVð~
r~
AÞdV The divergence theorem
The divergence theorem is a vector–calculus relation that equates the flux
of a vector field to the volume integral of the field’s divergence. The
relationships between line, surface, and volume integrals were explored by
several of the leading mathematical thinkers of the eighteenth and nine-
teenth centuries, including J. L. LaGrange in Italy, M. V. Ostrogradsky
in Russia, G. Green in England, and C. F. Gauss in Germany. In some
texts, you’ll find the divergence theorem referred to as ‘‘Gauss’s theorem’’
(which you should not confuse with Gauss’s law).
The divergence theorem may be stated as follows:
The flux of a vector field through a closed surface Sis equal to the
integral of the divergence of that field over a volume Vfor which Sis
a boundary.
This theorem applies to vector fields that are ‘‘smooth’’ in the sense of
being continuous and having continuous derivatives.
To understand the physical basis for the divergence theorem, recall
that the divergence at any point is defined as the flux through a small
surface surrounding that point divided by the volume enclosed by that
surface as it shrinks to zero. Now consider the flux through the cubical
cells within the volume Vshown in Figure 5.1.
For interior cells (those not touching the surface of V), the faces are
shared with six adjacent cells (only some of which are shown in Figure 5.1
for clarity). For each shared face, the positive (outward) flux from one
V
Boundary
surface S
Flux through faces a
t
boundary
Positive flux from an interior
cell is negative flux for an
adjacent cell with shared face
Figure 5.1 Cubical cells within volume Vbounded by surface S.
A student’s guide to Maxwell’s Equations114
cell is identical in amplitude and opposite in sign to the negative (inward)
flux of the adjacent cell over that same face. Since all interior cells share
faces with adjacent cells, only those faces that lie along the boundary
surface Sof volume Vcontribute to the flux through S.
This means that adding the flux through all the faces of all the cells
throughout volume Vleaves only the flux through the bounding surface S.
Moreover, in the limit of infinitesimally small cells, the definition of
divergence tells you that the divergence of the vector field at any point is
the outward flux from that point. So, adding the flux of each cell is the
same as integrating the divergence over the entire volume. Thus,
IS
~
A^
nda¼ZV
ð~
r~
AÞdV:ð5:1Þ
This is the divergence theorem – the integral of the divergence of a vector
field over Vis identical to the flux through S. And how is this useful? For
one thing, it can get you from the integral form to the differential form of
Gauss’s laws. In the case of electric fields, the integral form of Gauss’s
law is
IS
~
E^
nda ¼qenc=e0:
Or, since the enclosed charge is the volume integral of the charge
density q,
IS
~
E^
nda ¼1
e0ZV
qdV:
Now, apply the divergence theorem to the left side of Gauss’s law,
IS
~
E^
nda ¼ZV
~
r~
EdV ¼1
e0ZV
qdV ¼ZV
q
e0
dV:
Since this equality must hold for all volumes, the integrands must be
equal. Thus,
~
r~
E¼q=e0;
which is the differential form of Gauss’s law for electric fields. The same
approach applied to the integral form of Gauss’s law for magnetic fields
leads to
~
r~
B¼0
as you might expect.
From Maxwell’s Equations to the wave equation 115
HC~
Ad
~
l¼RSð~
r·~
AÞ^
ndaStokes’ theorem
Whereas the divergence theorem relates a surface integral to a volume
integral, Stokes’ theorem relates a line integral to a surface integral.
William Thompson (later Lord Kelvin) referred to this relation in a letter
in 1850, and it was G. G. Stokes who made it famous by setting its proof
as an exam question for students at Cambridge. You may encounter
generalized statements of Stokes’ theorem, but the form relevant to
Maxwell’s Equations (sometimes called the ‘‘Kelvin–Stokes theorem’’)
may be stated as follows:
The circulation of a vector field over a closed path Cis equal to the
integral of the normal component of the curl of that field over a
surface Sfor which Cis a boundary.
This theorem applies to vector fields that are ‘‘smooth’’ in the sense of
being continuous and having continuous derivatives.
The physical basis for Stokes’ theorem may be understood by recalling
that the curl at any point is defined as the circulation around a small path
surrounding that point divided by the surface area enclosed by that path
as it shrinks to zero. Consider the circulation around the small squares on
the surface Sshown in Figure 5.2.
For interior squares (those not touching the edge of the surface), each
edge is shared with an adjacent square. For each shared edge, the circulation
from one square is identical in amplitude and opposite in sign to the
Boundary
path C
Circulation at shared edges
of adjacent interior squares
is in opposite directions
Circulation along
edges at boundary
Figure 5.2 Squares on surface Sbounded by path C.
A student’s guide to Maxwell’s Equations116
circulation of the adjacent square over that same edge. It is only those edges
that lie along the boundary path Cof surface Sthat are not shared with an
adjacent square, and which contribute to the circulation around C.
Thus, adding the circulation around all the edges of all the squares over
surface Sleaves only the circulation around the bounding path C.In
addition, in the limit of infinitesimally small squares, the definition of curl
tells you that adding the circulation of each square is the same as inte-
grating the normal component of the curl of the vector field over the
surface. So,
IC
~
Ad
~
l¼ZS
ð~
r·~
AÞ^
nda:ð5:2Þ
Stokes’ theorem does for line integrals and the curl what the divergence
theorem does for surface integrals and the divergence. In this case, the
integral of the normal component of the curl over Sis identical to the
circulation around C. Moreover, just as the divergence theorem leads
from the integral to the differential form of Gauss’s laws, Stokes’ the-
orem can be applied to the integral form of Faraday’s law and the
Ampere–Maxwell law.
Consider the integral form of Faraday’s law, which relates the circu-
lation of the electric field around a path Cto the change in magnetic flux
through a surface Sfor which Cis a boundary,
IC
~
Ed
~
l¼d
dt ZS
~
B^
nda:
Applying Stokes’ theorem to the circulation on the left side gives
IC
~
Ed
~
l¼ZS
ð~
r·~
EÞ^
nda
Thus, Faraday’s law becomes
ZS
ð~
r·~
EÞ^
nda¼d
dt ZS
~
B^
nda:
For stationary geometries, the time derivative can be moved inside the
integral, so this is
ZS
ð~
r·~
EÞ^
nda ¼ZS
@~
B
@t^
n
!
da;
where the partial derivative indicates that the magnetic field may change
over space as well as time. Since this equality must hold over all surfaces,
the integrands must be equal, giving
From Maxwell’s Equations to the wave equation 117
~
r·~
E¼@~
B
@t;
which is the differential form of Faraday’s law, relating the curl of the
electric field at a point to the time rate of change of the magnetic field at
that point.
Stokes’ theorem may also be used to find the differential form of the
Ampere–Maxwell law. Recall that the integral form relates the circula-
tion of the magnetic field around a path Cto the current enclosed by that
path and the time rate of change of electric flux through a surface S
bound by path C:
IC
~
Bd
~
l¼l0Ienc þe0
d
dt ZS
~
E^
nda

:
Applying Stokes’ theorem to the circulation gives
IC
~
Bd
~
l¼ZS
ð~
r·~
BÞ^
nda;
which makes the Ampere–Maxwell law
ZS
ð~
r·~
BÞ^
nda ¼l0Ienc þe0
d
dt ZS
~
E^
nda

:
The enclosed current may be written as the integral of the normal com-
ponent of the current density
Ienc ¼ZS
~
J^
nda;
and the Ampere–Maxwell law becomes
ZS
ð~
r·~
BÞ^
nda ¼l0ZS
~
J^
ndaþZS
e0
@~
E
@t^
nda
!
:
Once again, for this equality to hold over all surfaces, the integrands must
be equal, meaning
~
r·~
B¼l0~
Jþe0
@~
E
@t
!
:
This is the differential form of the Ampere–Maxwell law, relating the curl
of the magnetic field at a point to the current density and time rate of
change of the electric field at that point.
A student’s guide to Maxwell’s Equations118
~
rðÞ The gradient
To understand how Maxwell’s Equations lead to the wave equation, it is
necessary to comprehend a third differential operation used in vector
calculus – the gradient. Similar to the divergence and the curl, the gradient
involves partial derivatives taken in three orthogonal directions. However,
whereas the divergence measures the tendency of a vector field to flow
away from a point and the curl indicates the circulation of a vector field
around a point, the gradient applies to scalar fields. Unlike a vector field, a
scalar field is specified entirely by its magnitude at various locations: one
example of a scalar field is the height of terrain above sea level.
What does the gradient tell you about a scalar field? Two important
things: the magnitude of the gradient indicates how quickly the field is
changing over space, and the direction of the gradient indicates the dir-
ection in that the field is changing most quickly with distance.
Therefore, although the gradient operates on a scalar field, the result of
the gradient operation is a vector, with both magnitude and direction. Thus,
if the scalar field represents terrain height, the magnitude of the gradient at
any location tells you how steeply the ground is sloped at that location, and
the direction of the gradient points uphill along the steepest slope.
The definition of the gradient of the scalar field wis
gradðwÞ¼~
rw^
i@w
@xþ^
j@w
@yþ^
k@w
@zðCartesianÞ:ð5:3Þ
Thus, the x-component of the gradient of windicates the slope of the
scalar field in the x-direction, the y-component indicates the slope in the
y-direction, and the z-component indicates the slope in the z-direction.
The square root of the sum of the squares of these components provides
the total steepness of the slope at the location at which the gradient is
taken.
In cylindrical and spherical coordinates, the gradient is
~
rw^
r@w
@rþ^
’1
r
@w
@uþ^z@w
@zðcylindricalÞð5:4Þ
and
~
rw^
r@w
@rþ^
h1
r
@w
@hþ^
’1
rsin h
@w
@uðsphericalÞ:ð5:5Þ
From Maxwell’s Equations to the wave equation 119
~
r;~
r;~
r·Some useful identities
Here is a quick review of the del differential operator and its three uses
relevant to Maxwell’s Equations:
Del:
~
r^
i@
@xþ^
j@
@yþ^
k@
@z
Del (nabla) represents a multipurpose differential operator that can
operate on scalar or vector fields and produce scalar or vector results.
Gradient:
~
rw^
i@w
@xþ^
j@w
@yþ^
k@w
@z
The gradient operates on a scalar field and produces a vector result
that indicates the rate of spatial change of the field at a point and the
direction of steepest increase from that point.
Divergence:
~
r~
A@Ax
@xþ@Ay
@yþ@Az
@z
The divergence operates on a vector field and produces a scalar result
that indicates the tendency of the field to flow away from a point.
Curl:
~
r·~
A@Az
@y@Ay
@z

^
iþ@Ax
@z@Az
@x

^
jþ@Ay
@x@Ax
@y

^
k
The curl operates on a vector field and produces a vector result that
indicates the tendency of the field to circulate around a point and the
direction of the axis of greatest circulation.
A student’s guide to Maxwell’s Equations120
Once you’re comfortable with the meaning of each of these operators,
you should be aware of several useful relations between them (note that
the following relations apply to fields that are continuous and that have
continuous derivatives).
The curl of the gradient of any scalar field is zero.
~
r·~
rw¼0;ð5:6Þ
which you may readily verify by taking the appropriate derivatives.
Another useful relation involves the divergence of the gradient of a
scalar field; this is called the Laplacian of the field:
~
r~
rw¼r
2w¼@2w
@x2þ@2w
@y2þ@2w
@z2ðCartesianÞ:ð5:7Þ:
The usefulness of these relations can be illustrated by applying them to
the electric field as described by Maxwell’s Equations. Consider, for
example, the fact that the curl of the electrostatic field is zero (since
electric field lines diverge from positive charge and converge upon
negative charge, but do not circulate back upon themselves). Equation 5.6
indicates that as a curl-free (irrotational) field, the electrostatic field ~
E
may be treated as the gradient of another quantity called the scalar
potential V:
~
E¼~
rV;ð5:8Þ
where the minus sign is needed because the gradient points toward the
greatest increase in the scalar field, and by convention the electric force
on a positive charge is toward lower potential. Now apply the differential
form of Gauss’s law for electric fields:
~
r~
E¼q
e0
;
which, combined with Equation 5.8, gives
r2V¼q
e0
:ð5:9Þ
This is called Laplace’s equation, and it is often the best way to find the
electrostatic field when you are not able to construct a special Gaussian
surface. In such cases, it may be possible to solve Laplace’s Equation for
the electric potential Vand then determine ~
Eby taking the gradient of the
potential.
From Maxwell’s Equations to the wave equation 121
r2~
A¼1
v2@2~
A
@t2The wave equation
With the differential form of Maxwell’s Equations and several vector
operator identities in hand, the trip to the wave equation is a short
one. Begin by taking the curl of both sides of the differential form of
Faraday’s law
~
r·ð~
r·~
EÞ¼~
r·@~
B
@t
!
¼@ð~
r·~
BÞ
@t:ð5:10Þ
Notice that the curl and time derivatives have been interchanged in the
final term; as in previous sections, the fields are assumed to be sufficiently
smooth to permit this.
Another useful vector operator identity says that the curl of the curl of
any vector field equals the gradient of the divergence of the field minus
the Laplacian of the field:
~
r·ð~
r·~
AÞ¼~
rð~
r~
AÞr
2~
A:ð5:11Þ
This relation uses a vector version of the Laplacian operator that is
constructed by applying the Laplacian to the components of a vector
field:
r2~
A¼@2Ax
@x2þ@2Ay
@y2þ@2Az
@z2ðCartesianÞ:ð5:12Þ
Thus,
~
r·ð~
r·~
EÞ¼~
rð~
r~
EÞr
2~
E¼@ð~
r·~
BÞ
@t:ð5:13Þ
However, you know the curl of the magnetic field from the differential
form of the Ampere–Maxwell law:
~
r·~
B¼l0~
Jþe0
@~
E
@t
!
:
So,
~
r·ð~
r·~
EÞ¼~
rð~
r~
EÞr
2~
E¼@l0~
Jþe0ð@~
E=@tÞ

@t:
A student’s guide to Maxwell’s Equations122
This looks difficult, but one simplification can be achieved using Gauss’s
law for electric fields:
~
r~
E¼q
e0
;
which means
~
r·ð~
r·~
EÞ¼~
rq
e0

r
2~
E¼@l0~
Jþe0ð@~
E=@tÞ

@t
¼l0
@~
J
@tl0e0
@2~
E
@t2:
Gathering terms containing the electric field on the left side of this
equation gives
r2~
El0e0
@2~
E
@t2¼~
rq
e0

þl0
@~
J
@t:
In a charge- and current-free region, q= 0 and ~
J¼0, so
r2~
E¼l0e0
@2~
E
@t2:ð5:14Þ
This is a linear, second-order, homogeneous partial differential equation
that describes an electric field that travels from one location to another –
in short, a propagating wave. Here is a quick reminder of the meaning of
each of the characteristics of the wave equation:
Linear: The time and space derivatives of the wave function (~
Ein this
case) appear to the first power and without cross terms.
Second-order: The highest derivative present is the second derivative.
Homogeneous: All terms involve the wave function or its derivatives, so
no forcing or source terms are present.
Partial: The wave function is a function of multiple variables (space and
time in this case).
A similar analysis beginning with the curl of both sides of the Ampere–
Maxwell law leads to
r2~
B¼l0e0
@2~
B
@t2;ð5:15Þ
which is identical in form to the wave equation for the electric field.
This form of the wave equation doesn’t just tell you that you have a
wave – it provides the velocity of propagation as well. It is right there in
From Maxwell’s Equations to the wave equation 123
the constants multiplying the time derivative, because the general form of
the wave equation is this
r2~
A¼1
v2
@2~
A
@t2;ð5:16Þ
where vis the speed of propagation of the wave. Thus, for the electric and
magnetic fields
1
v2¼l0e0;
or
v¼ffiffiffiffiffiffiffiffiffi
1
l0e0
s:ð5:17Þ
Inserting values for the magnetic permeability and electric permittivity of
free space,
v¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
½4p·107mkg=C2½8:8541878 ·1012 C2s2=kg m3;
s
or
v¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
8:987552·1016 m2=s2
q¼2:9979 ·108m=s:
It was the agreement of the calculated velocity of propagation with the
measured speed of light that caused Maxwell to write, ‘‘light is an elec-
tromagnetic disturbance propagated through the field according to
electromagnetic laws.’’
A student’s guide to Maxwell’s Equations124
Appendix: Maxwell’s Equations
in matter
Maxwell’s Equations as presented in Chapters 1–4apply to electric and
magnetic fields in matter as well as in free space. However, when you’re
dealing with fields inside matter, remember the following points:
The enclosed charge in the integral form of Gauss’s law for electric
fields (and current density in the differential form) includes ALL
charge – bound as well as free.
The enclosed current in the integral form of the Ampere–Maxwell law
(and volume current density in the differential form) includes ALL
currents – bound and polarization as well as free.
Since the bound charge may be difficult to determine, in this Appendix
you’ll find versions of the differential and integral forms of Gauss’s law
for electric fields that depend only on the free charge. Likewise, you’ll find
versions of the differential and integral form of the Ampere–Maxwell law
that depend only on the free current.
What about Gauss’s law for magnetic fields and Faraday’s law? Since
those laws don’t directly involve electric charge or current, there’s no
need to derive more “matter friendly” versions of them.
Gauss’s law for electric fields: Within a dielectric material, positive and
negative charges may become slightly displaced when an electric field is
applied. When a positive charge Qis separated by distance sfrom an
equal negative charge −Q, the electric “dipole moment” is given by
~
p¼Q
~
s;ðA:1Þ
where ~
sis a vector directed from the negative to the positive charge with
magnitude equal to the distance between the charges. For a dielectric
125
material with Nmolecules per unit volume, the dipole moment per unit
volume is
~
P¼N~
p;ðA:2Þ
a quantity which is also called the “electric polarization” of the material.
If the polarization is uniform, bound charge appears only on the surface
of the material. But if the polarization varies from point to point within
the dielectric, there are accumulations of charge within the material, with
volume charge density given by
qb¼
~
r~
P;ðA:3Þ
where q
b
represents the volume density of bound charge (charge that’s
displaced by the electric field but does not move freely through the
material).
What is the relevance of this to Gauss’s law for electric fields? Recall
that in the differential form of Gauss’s law, the divergence of the electric
field is
~
r~
E¼q
e0
;
where qis the total charge density. Within matter, the total charge density
consists of both free and bound charge densities:
q¼qfþqb;ðA:4Þ
where qis the total charge density, q
f
is the free charge density, and q
b
is
the bound charge density. Thus, Gauss’s law may be written as
~
r~
E¼q
e0
¼qfþqb
e0
:ðA:5Þ
Substituting the negative divergence of the polarization for the bound
charge and multiplying through by the permittivity of free space gives
~
re0~
E¼qfþqb¼qf~
r~
P;ðA:6Þ
or
~
re0~
Eþ~
r~
P¼qf:ðA:7Þ
Collecting terms within the divergence operator gives
~
rðe0~
Eþ~
PÞ¼qf:ðA:8Þ
A student’s guide to Maxwell’s Equations126
In this form of Gauss’s law, the term in parentheses is often written as a
vector called the “displacement,” which is defined as
~
D¼e0~
Eþ~
P:ðA:9Þ
Substituting this expression into equation (A.8) gives
~
r~
D¼qf;ðA:10Þ
which is a version of the differential form of Gauss’s law that depends
only on the density of free charge.
Using the divergence theorem gives the integral form of Gauss’s law
for electric fields in terms of the flux of the displacement and enclosed free
charge:
IS
~
D^
nda¼qfree;enc:ðA:11Þ
What is the physical significance of the displacement ~
D?Infreespace,the
displacement is a vector field proportional to the electric field – pointing in
the same direction as ~
Eand with magnitude scaled by the vacuum permit-
tivity. But in polarizable matter, the displacement field may differ signifi-
cantly from the electric field. You should note, for example, that the
displacement is not necessarily irrotational – it will have curl if the polar-
ization does, as can be seen by taking the curl of both sides of Equation A.9.
The usefulness of ~
Dcomes about in situations for which the free charge
is known and for which symmetry considerations allow you to extract the
displacement from the integral of Equation A.11. In those cases, you may
be able to determine the electric field within a linear dielectric material
by finding ~
Don the basis of the free charge and then dividing by the
permittivity of the medium to find the electric field.
The Ampere–Maxwell law: Just as applied electric fields induce polar-
ization (electric dipole moment per unit volume) within dielectrics, applied
magnetic fields induce “magnetization” (magnetic dipole moment per unit
volume) within magnetic materials. And just as bound electric charges act
as the source of additional electric fields within the material, bound cur-
rents may act as the source of additional magnetic fields. In that case, the
bound current density is given by the curl of the magnetization:
~
Jb¼~
r·~
M:ðA:12Þ
where ~
Jbis the bound current density and ~
Mrepresents the magnetization
of the material.
Appendix 127
Another contribution to the current density within matter comes
from the time rate of change of the polarization, since any movement of
charge constitutes an electric current. The polarization current density is
given by
~
JP¼@~
P
@t:ðA:13Þ
Thus, the total current density includes not only the free current density,
but the bound and polarization current densities as well:
~
J¼~
Jfþ~
Jbþ~
JP:ðA:14Þ
Thus, the Ampere–Maxwell law in differential form may be written as
~
r·~
B¼l0~
Jfþ~
Jbþ~
JPþe0
@~
E
@t
!
:ðA:15Þ
Inserting the expressions for the bound and polarization current and
dividing by the permeability of free space
1
l0
~
r·~
B¼~
Jfþ~
r·~
Mþ@~
P
@tþe0
@~
E
@t:ðA:16Þ
Gathering curl terms and time-derivative terms gives
~
r·~
B
l0
~
r·~
M¼~
Jfþ@~
P
@tþ@ðe0~
EÞ
@t:ðA:17Þ
Moving the terms inside the curl and derivative operators makes this
~
r·~
B
l0
~
M
!
¼~
Jfþ@ðe0~
Eþ~
PÞ
@t:ðA:18Þ
In this form of the Ampere–Maxwell law, the term in parentheses on the
left side is written as a vector sometimes called the “magnetic field
intensity” or “magnetic field strength” and defined as
~
H¼~
B
l0
~
M:ðA:19Þ
Thus, the differential form of the Ampere–Maxwell law in terms of ~
H,~
D,
and the free current density is
A student’s guide to Maxwell’s Equations128
~
r·~
H¼~
Jfree þ@~
D
@t:ðA:20Þ
Using Stokes’ theorem gives the integral form of the Ampere–Maxwell
law:
IC
~
H~
dl ¼Ifree;enc þd
dt ZS
~
D^
nda ðA:21Þ
What is the physical significance of the magnetic intensity ~
H? In free
space, the intensity is a vector field proportional to the magnetic field –
pointing in the same direction as ~
Band with magnitude scaled by the
vacuum permeability. But just as ~
Dmay differ from ~
Einside dielectric
materials, ~
Hmay differ significantly from ~
Bin magnetic matter. For
example, the magnetic intensity is not necessarily solenoidal – it will have
divergence if the magnetization does, as can be seen by taking the
divergence of both sides of Equation A.19.
As is the case for electric displacement, the usefulness of ~
Hcomes
about in situations for which you know the free current and for which
symmetry considerations allow you to extract the magnetic intensity from
the integral of Equation A.21. In such cases, you may be able to deter-
mine the magnetic field within a linear magnetic material by finding ~
Hon
the basis of free current and then multiplying by the permeability of the
medium to find the magnetic field.
Appendix 129
Here is a summary of the integral and differential forms of all of
Maxwell’s Equations in matter:
Gauss’s law for electric fields:
IS
~
D^
nda¼qfree;enc ðintegral formÞ;
~
r~
D¼qfree ðdifferential formÞ:
Gauss’s law for magnetic fields:
IS
~
B^
nda¼0ðintegral formÞ;
~
r~
B¼0ðdifferential formÞ:
Faraday’s law:
IC
~
Ed~
l¼d
dt ZS
~
B^
nda ðintegral formÞ;
~
r·~
E¼@~
B
@tðdifferential formÞ:
Ampere–Maxwell law:
IC
~
Hd~
l¼Ifree;enc þd
dt ZS
~
D^
nda ðintegral formÞ;
~
r·~
H¼~
Jfree þ@~
D
@tðdifferential formÞ:
A student’s guide to Maxwell’s Equations130
Further reading
If you’re looking for a comprehensive treatment of electricity and magnetism, you
have several excellent texts from which to choose. Here are some that you may
find useful:
Cottingham W. N. and Greenwood D. A., Electricity and Magnetism. Cambridge
University Press, 1991; A concise survey of a wide range of topics in
electricity and magnetism.
Griffiths, D. J., Introduction to Electrodynamics. Prentice-Hall, New Jersey, 1989;
The standard undergraduate text at the intermediate level, with clear
explanations and informal style.
Jackson, J. D., Classical Electrodynamics. Wiley & Sons, New York, 1998; The
standard graduate text, but you must be solidly prepared before embarking.
Lorrain, P., Corson, D., and Lorrain, F., Electromagnetic Fields and Waves.
Freeman, New York, 1988; Another excellent intermediate-level text, with
detailed explanations supported by helpful diagrams.
Purcell, E. M., Electricity and Magnetism Berkeley Physics Course, Vol. 2.
McGraw-Hill, New York, 1965; Probably the best of the introductory-level
texts; elegantly written and carefully illustrated.
Wangsness, R. K., Electromagnetic Fields. Wiley, New York, 1986; Also a great
intermediate-level text, especially useful as preparation for Jackson.
And for a comprehensible introduction to vector operators, with many examples
drawn from electrostatics, check out:
Schey, H. M., Div, Grad, Curl, and All That. Norton, New York, 1997.
131
Index
Ampere, Andre-Marie 83
Ampere-- Maxwell law 83 -- 111
differential form 101-- 9
applying 108-- 9
expanded view 101
main idea 101
version involving only free
current 128
integral form 83-- 100
applying 95-- 100
expanded view 84
main idea 84
usefulness of 84
version involving only free
current 129
Ampere’s law 83
Biot-- Savart law 47
capacitance 18-- 19
of a parallel-plate capacitor 18
charge, electric
bound 18,126
density 16
enclosed 16-- 17
relationship to flux 17
circulation 65
of electric field 68
of magnetic field 85-- 6
closed surface 7
curl 76-- 8
in Cartesian coordinates 77
of electric field 79
locating regions of 76-- 7
of magnetic field 102-- 4
main idea 120
in non-Cartesian coordinates 78
relationship to circulation 76
current density 105-- 6
bound 127
polarization 128
units of 105
del cross (see curl)
del dot (see divergence)
del operator (see nabla)
dielectric constant 19
dielectrics 18-- 19
dipole moment, electric 125
displacement 127
physical significance of 127
usefulness of 127
displacement current 94
units of 107
divergence 32-- 5
in Cartesian coordinates 33
locating regions of 32
main idea 120
in non-Cartesian coordinates 35
relationship to flux 32
divergence theorem 114-- 15
main idea 114
dot product 6
how to compute 6
physical significance 6
electric field
definition of 3
electrostatic vs. induced 1,62-- 3
equations for simple objects 5
induced 62-- 3
direction of 63
units of 3,62
electromotive force (emf) 68
units of 68
enclosed current 89-- 90
Faraday, Michael
demonstration of induction 58
refers to ‘‘field of force’’ 3
132
Faraday’s law 58-- 82
differential form 75-- 81
applying 79-- 81
expanded view 75
main idea 75
integral form 58-- 74
applying 72-- 4
expanded view 60
main idea 59
usefulness of 61
field lines 3,13-- 14
flux
electric 13-- 15
rate of change of 91-- 4
units of 13
magnetic 48-- 9
rate of change of 69-- 70
units of (see webers)
as number of field lines 13-- 14
of a vector field 10-- 12
Gauss, C. F. 114
Gauss’s law for electric fields 1-- 41
differential form 29-- 40
applying 38-- 40
expanded view 30
main idea 29
usefulness of 30
version involving only free
charge 127
integral form 1-- 28
applying 20-- 8
expanded view 2
main idea 1
usefulness of 2
version involving only free charge 127
Gauss’s law for magnetic fields 43-- 57
differential form 53-- 6
applying 55-- 6
expanded view 53
main idea 53
integral form 43-- 52
applying 50-- 2
expanded view 44
main idea 44
usefulness of 44
Gauss’s theorem 114
gradient 119-- 20
in Cartesian coordinates 119
in non-Cartesian coordinates 119
main idea 120
Green, G. 114
Heaviside, Oliver 32
induced current 68
direction of 71
inductance 88
insulators (see dielectrics)
irrotational fields 78
Kelvin-- Stokes theorem 116
LaGrange, J. L. 114
Laplacian operator 121
vector version 122
Laplace’s equation 121
Lenz, Heinrich 71
Lenz’s law 71
line integral 64
Lorentz equation 45
magnetic field
definition of 45
distinctions from electric field 45
equations for simple objects 47
intensity 128
physical significance of 129
usefulness of 129
units of 45
magnetic flux density 45
magnetic induction 45
magnetic poles 43
always in pairs 44
magnetization 127
Maxwell, James Clerk
coining of ‘‘convergence’’ 32
coining of ‘‘curl’’ 76
definition of electric field 3
electromagnetic theory 112
use of magnetic vortex model 91
nabla 31
main idea 120
Oersted, Hans Christian 83
Ohm’s law 73
open surface 7
Ostrogradsky, M. V. 114
path integral 65-- 7
permeability
of free space 87-- 8
relative 87-- 8
permittivity
of free space 18-- 19
relative 19
polarization, electric 126
relationship to bound charge 126
scalar field 119
scalar potential 121
scalar product (see dot product)
solenoidal fields 54
special Amperian loop 86,95-- 100
Index 133
special Gaussian surface 25
speed of light 124
Stokes, G. G. 116
Stokes’ theorem 116-- 19
main idea 116
surface integral 9
Thompson, William 116
unit normal vector 7
direction of 7
vacuum permittivity (see permittivity
of free space)
vector cross product 45
vector field 10
wave equation 122-- 4
for electric fields 113
for magnetic fields 113
webers 48
work done along a path 65-- 6
Index134

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