^ ^

happen if the electric ¬eld had a nonradial component (say in the h or ™

direction); by rotating the sphere about some arbitrary axis, you™d be able

to change the direction of the ¬eld. 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 ¬eld. Faced with this conundrum, you are forced to conclude

that the electric ¬eld of a spherically symmetric charge distribution must

be entirely radial.

To ¬nd the value of this radial ¬eld using Gauss™s law, you™ll have to

imagine a surface that meets the requirements of a special Gaussian

surface; ˜ must be either parallel or perpendicular to the surface normal

E

at all locations, and ˜ must be uniform everywhere on the surface. For a

E

radial electric ¬eld, 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 ¬eld from the surface

integral in Gauss™s law.

Since the radial electric ¬eld is everywhere parallel to the surface

normal, the ˜ ^ term in the integral in Gauss™s law becomes

En

j˜ nj cos°0 Þ, and the electric ¬‚ux over the Gaussian surface S is

Ejj^

I I

˜ ^ da ¼

UE ¼ En E da

S S

Since ˜ has no h or u dependence, it must be constant over S, which

E

means it may be removed from the integral:

I I

UE ¼ E da ¼ E da ¼ E°4pr 2 Þ;

S S

where r is the radius of the special Gaussian surface. You can now use

Gauss™s law to ¬nd the value of the electric ¬eld:

27

Gauss™s law for electric ¬elds

n

Special

Gaussian

surface

Radial

electric

field

n n

Charged

sphere

n

Figure 1.12 A special Gaussian around a charged sphere.

qenc

UE ¼ E°4pr 2 Þ ¼ ;

e0

or qenc

E¼ ;

4pe0 r 2

where qenc is the charge enclosed by your Gaussian surface. You can use

this expression to ¬nd the electric ¬eld both outside and inside the sphere.

To ¬nd the electric ¬eld outside the sphere, construct your Gaussian

surface with radius r > a so that the entire charged sphere is within the

Gaussian surface. This means that the enclosed charge is just the charge

density times the entire volume of the charged sphere: qenc ¼ °4=3Þpa3 q.

Thus,

°4=3Þpa3 q qa3

E¼ ¼ °outside sphereÞ:

4pe0 r 2 3e0 r 2

To ¬nd the electric ¬eld within the charged sphere, construct your

Gaussian surface with r < a. In this case, the enclosed charge is the charge

28 A student™s guide to Maxwell™s Equations

density times the volume of your Gaussian surface: qenc ¼ °4=3Þpr 3 q.

Thus,

°4=3Þpr 3 q qr

E¼ ¼ °inside sphereÞ:

4pe0 r 2 3e0

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 ¬eld.

29

Gauss™s law for electric ¬elds

1.2 The differential form of Gauss™s law

The integral form of Gauss™s law for electric ¬elds relates the electric ¬‚ux

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

˜E q

r˜¼ Gauss™s law for electric fields °differential formÞ:

e0

The left side of this equation is a mathematical description of the

divergence of the electric ¬eld “ the tendency of the ¬eld to ˜˜¬‚ow™™ away

from a speci¬ed 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 ¬eld 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 ¬eld

is not zero are those locations at which charge is present. If positive

charge is present, the divergence is positive, meaning that the electric ¬eld

tends to ˜˜¬‚ow™™ away from that location. If negative charge is present, the

divergence is negative, and the ¬eld lines tend to ˜˜¬‚ow™™ 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 ¬eld and the charge density at individual points

in space, whereas the integral form entails the integral of the normal

component of the electric ¬eld over a surface. Familiarity with both forms

will allow you to use whichever is better suited to the problem you™re

trying to solve.

30 A student™s guide to Maxwell™s Equations

To help you understand the meaning of each symbol in the differential

form of Gauss™s law for electric ¬elds, here™s an expanded view:

Reminder that del is Reminder that the electric

a vector operator field is a vector

The charge density in

coulombs per cubic meter

E=

0

The differential

The electric The electric

operator called

field in N/C permittivity of

“del” or “nabla”

free space

The dot product turns

the del operator into the

divergence

How is the differential form of Gauss™s law useful? In any problem in

which the spatial variation of the vector electric ¬eld is known at a

speci¬ed location, you can ¬nd the volume charge density at that location

using this form. And if the volume charge density is known, the

divergence of the electric ¬eld may be determined.

31

Gauss™s law for electric ¬elds

˜

r Nabla “ the del operator

An inverted uppercase delta appears in the differential form of all four of

Maxwell™s Equations. This symbol represents a vector differential operator

called ˜˜nabla™™ or ˜˜del,™™ and its presence instructs you to take derivatives

of the quantity on which the operator is acting. The exact form of those

˜

derivatives depends on the symbol following the del operator, with ˜˜r™™

˜ ˜

signifying divergence, ˜˜r ·™™ indicating curl, and r signifying 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 H tells you to take the square root of anything that

˜

appears under its roof, r is an instruction to take derivatives in three

directions. Speci¬cally,

˜ i@ j@ k@

r ^ þ^ þ ^ ; °1:18Þ

@x @y @z

where ^ ^ and ^ are the unit vectors in the direction of the Cartesian

i, j, k

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 ¬elds, the del operator is dotted into the electric ¬eld vector,

forming the divergence of ˜ That operation and its results are described

E.

in the next section.

32 A student™s guide to Maxwell™s Equations

˜

r Del dot “ the divergence

The concept of divergence is important in many areas of physics and

engineering, especially those concerned with the behavior of vector ¬elds.

James Clerk Maxwell coined the term ˜˜convergence™™ to describe the

mathematical operation that measures the rate at which electric ¬eld lines

˜˜¬‚ow™™ 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 ˜˜¬‚ow™™ of electric ¬eld lines away from positive charge.

Both ¬‚ux and divergence deal with the ˜˜¬‚ow™™ of a vector ¬eld, but

with an important difference; ¬‚ux is de¬ned over an area, while diver-

gence applies to individual points. In the case of ¬‚uid ¬‚ow, the divergence

at any point is a measure of the tendency of the ¬‚ow 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 ¬‚uid ¬‚ow, positive electric charge in electro-

statics), while points of negative divergence are sinks (drains in ¬‚uid ¬‚ow,

negative charge in electrostatics).

The mathematical de¬nition of divergence may be understood by

considering the ¬‚ux through an in¬nitesimal surface surrounding the

point of interest. If you were to form the ratio of the ¬‚ux of a vector ¬eld

˜ through a surface S to the volume enclosed by that surface as the

A

volume shrinks toward zero, you would have the divergence of ˜ A:

I

1

AÞ ˜ A

div°˜ ¼ r ˜ lim ˜ ^ da: °1:19Þ

An

DV!0 DV S

While this expression states the relationship between divergence and ¬‚ux,

it is not particularly useful for ¬nding the divergence of a given vector

¬eld. You™ll ¬nd a more user-friendly mathematical expression for

divergence later in this section, but ¬rst you should take a look at the

vector ¬elds shown in Figure 1.13.

To ¬nd the locations of positive divergence in each of these ¬elds, look

for points at which the ¬‚ow 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 ¬‚owing water to

assess the divergence; if the sawdust is dispersed, you have selected a

point of positive divergence, while if it becomes more concentrated,

you™ve picked a location of negative divergence.

33

Gauss™s law for electric ¬elds

(a) (b) (c)

4

6

3 5

2

7

1

Figure 1.13 Vector ¬elds with various values of divergence.

Using such tests, it is clear that locations such as 1 and 2 in Figure 1.13(a)

and location 3 in Figure 1.13(b) are points of positive divergence, while

the divergence is negative at point 4.

The divergence at various points in Figure 1.13(c) is less obvious.

Location 5 is obviously a point of positive divergence, but what about

locations 6 and 7? The ¬‚ow 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 ¬‚ow?

Answering that question requires a useful mathematical form of the

divergence as well as a description of how the vector ¬eld varies from

place to place. The differential form of the mathematical operation of

˜

divergence or ˜˜del dot™™ (r) on a vector ˜ in Cartesian coordinates is

A

À Á

@ ^@ ^@

˜A ^ x þ^ y þ ^ z ;

r˜ ¼ ^ þj þki iA jA kA

@x @y @z

and, since ^ ^ ¼ ^ ^ ¼ ^ ^ ¼ 1; this is

ii jj kk

@Ax @Ay @Az

˜A

r˜ ¼ þ þ : °1:20Þ

@x @y @z

Thus, the divergence of the vector ¬eld ˜ is simply the change in its

A

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 ¬eld is a scalar quantity; it has magnitude but

no direction.

34 A student™s guide to Maxwell™s Equations

You can now apply this to the vector ¬elds in Figure 1.13. In Figure

1.13(a), assume that the magnitude of the vector ¬eld varies sinusoidally

along the x-axis (which is vertical in this case) as ˜ ¼ sin°pxÞ^ while

A i

remaining constant in the y- and z-directions. Thus,

˜ A @Ax ¼ p cos°pxÞ;

r˜ ¼

@x

since Ay and Az are zero. This expression is positive for 0 < x < 1, 0 at

2

x ¼ 1, and negative for 1 < x < 3, just as your visual inspection suggested.

2 2 2

Now consider Figure 1.13(b), which represents a slice through a

spherically symmetric vector ¬eld with amplitude increasing as the square

of the distance from the origin. Thus ˜ ¼ r 2^. Since r2 ¼ (x2 þ y2 þ z2) and

A r

x^ þ y^ þ z^

i j k

^ ¼ p¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬ ;

r

x2 þ y2 þ z2

this means

x^ þ y^ þ z^

i j k

˜ ¼ r 2^ ¼ °x2 þ y2 þ z2 Þ p¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬ ;

A r

x2 þ y2 þ z2

and

@Ax À 2 Á°1=2Þ 1 À2 ÁÀ°1=2Þ

¼ x þ y 2 þ z2 þx x þ y2 þ z2 °2xÞ:

@x 2

Doing likewise for the y- and z-components and adding yields

À Á°1=2Þ x2 þ y2 þ z2 À Á1=2

˜A

r ˜ ¼ 3 x2 þ y2 þ z2 þ p¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬ ¼ 4 x2 þ y2 þ z2 ¼ 4r:

x2 þ y2 þ z2

Thus, the divergence in the vector ¬eld in Figure 1.13(b) is increasing

linearly with distance from the origin.

Finally, consider the vector ¬eld in Figure 1.13(c), which is similar to

the previous case but with the amplitude of the vector ¬eld decreasing as

the square of the distance from the origin. The ¬‚ow 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 ¬eld will affect the value of the

divergence. Since ˜ ¼ °1=r 2 Þ^,

A r

x^ þ y^ þ z^ x^ þ y^ þ z^

1 i j k i j k

˜¼ p¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬ ¼ ;

A

°x2 þ y2 þ z2 Þ x2 þ y2 þ z2 °x2 þ y2 þ z2 Þ°3=2Þ

35

Gauss™s law for electric ¬elds

and

3 À2 ÁÀ°5=2Þ

@Ax 1

¼ x þ y2 þ z2 °2xÞ;

“x

@x °x2 þ y2 þ z2 Þ°3=2Þ 2

Adding in the y- and z-derivatives gives

À Á

3 x2 þ y 2 þ z2

3

˜A

r˜ ¼ À ¼ 0:

°3=2Þ °5=2Þ

°x 2 þ y 2 þ z2 Þ °x2 þ y2 þ z2 Þ

This validates the suspicion that the reduced amplitude of the vector ¬eld

with distance from the origin may compensate for the spreading out of

the ¬‚ow lines. Note that this is true only for the case in which the

amplitude of the vector ¬eld falls off as 1/r2 (this case is especially rele-

vant for the electric ¬eld, which you™ll ¬nd in the next section).

As you consider the divergence of the electric ¬eld, 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

1 @Af @Az

1@

˜A

r˜ ¼ °rAr Þ þ þ °cylindricalÞ; °1:21Þ

r @r r @f @z

and

˜ A 1@ 1@

r ˜ ¼ 2 °r 2 Ar Þ þ °Ah sin hÞ

r @r r sin h @h

°1:22Þ

1 @Af

þ °sphericalÞ:

r sin h @f

If you doubt the ef¬cacy of choosing the proper coordinate system, you

should rework the last two examples in this section using spherical

coordinates.

36 A student™s guide to Maxwell™s Equations

˜E

r ˜ The divergence of the electric ¬eld

This expression is the entire left side of the differential form of Gauss™s

law, and it represents the divergence of the electric ¬eld. In electrostatics,

all electric ¬eld 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 ¬eld of the positive point charge; the electric ¬eld

lines originate on the positive charge, and you know from Table 1.1 that

the electric ¬eld is radial and decreases as 1/r2:

˜ ¼ 1 q ^:

E r

4pe0 r 2

This is analogous to the vector ¬eld shown in Figure 1.13(c), for which

the divergence is zero. Thus, the spreading out of the electric ¬eld lines is

exactly compensated by the 1/r2 reduction in ¬eld amplitude, and the

divergence of the electric ¬eld is zero at all points away from the origin.

The reason the origin (where r ¼ 0) is not included in the previous

analysis is that the expression for the divergence includes terms con-

taining r in the denominator, and those terms become problematic as r

approaches zero. To evaluate the divergence at the origin, use the formal

de¬nition of divergence:

I

1

˜E

r ˜ lim ˜ ^ da:

En

DV!0 DV

S

Considering a special Gaussian surface surrounding the point charge

q, this is

0 1

I

1 1

q q

˜E

r ˜ lim @ daA ¼ lim °4pr Þ

2

DV!0 DV 4pe0 r 2 DV!0 DV 4pe0 r 2

S

1q

¼ lim :

DV!0 DV e0

But q/DV is just the average charge density over the volume DV, and as

DV shrinks to zero, this becomes equal to q, the charge density at the

origin. Thus, at the origin the divergence is

˜E q

r˜ ¼ ;

e0

in accordance with Gauss™s law.

37

Gauss™s law for electric ¬elds

It is worth your time to make sure you understand the signi¬cance of

this last point. A casual glance at the electric ¬eld 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 ¬elds that

decrease in amplitude as 1/r2 actually have zero divergence everywhere

except at the source. The key factor in determining the divergence at any

point is not simply the spacing of the ¬eld lines at that point, but whether

the ¬‚ux out of an in¬nitesimally small volume around the point is greater

than, equal to, or less than the ¬‚ux into that volume. If the outward ¬‚ux

exceeds the inward ¬‚ux, the divergence is positive at that point. If the

outward ¬‚ux is less than the inward ¬‚ux, the divergence is negative, and if

the outward and inward ¬‚uxes are equal the divergence is zero at that

point.

In the case of a point charge at the origin, the ¬‚ux through an in¬ni-

tesimally small surface is nonzero only if that surface contains the point

charge. Everywhere else, the ¬‚ux into and out of that tiny surface must be

the same (since it contains no charge), and the divergence of the electric

¬eld must be zero.

38 A student™s guide to Maxwell™s Equations

˜E

r ˜ ¼ q=e0 Applying Gauss™s law (differential form)

The problems you™re most likely to encounter that can be solved using the

differential form of Gauss™s law involve calculating the divergence of the

electric ¬eld and using the result to determine the charge density at a

speci¬ed 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 ¬eld, ¬nd the

divergence of the ¬eld at a speci¬ed location.

Problem: If the vector ¬eld of Figure 1.13(a) were changed to

˜ ¼ sin p y ^ À sin p x ^

A i j;

2 2

in the region À 0.5 < x < þ 0.5 and À 0.5 < y < þ 0.5, how would the ¬eld

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 ¬eld. A better approach is to think about the ¬eld for a

moment and to attempt to visualize the ¬eld lines “ a task that may be

dif¬cult 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 ¬eld. Using the

˜˜quiver™™ command in MATLAB’ shows that the ¬eld looks as shown in

Figure 1.14.

If you™re surprised by the direction of the ¬eld, consider that the

x-component of the ¬eld depends on y (so the ¬eld points to the right

above the x-axis and to the left below the x-axis), while the y-component

of the ¬eld depends on the negative of x (so the ¬eld points up on the left

of the y-axis and down on the right of the y-axis). Combining these

features leads to the ¬eld depicted in Figure 1.14.

Examining the ¬eld closely reveals that the ¬‚ow lines neither converge

nor diverge, but simply circulate back on themselves. Calculating the

divergence con¬rms this

h i h i

˜ ˜ ¼ @ sin p y “ @ sin p x ¼ 0:

rA

@x @y

2 2

39

Gauss™s law for electric ¬elds

0.5

0.4

0.3

0.2

0.1

0

y

“0.1

“0.2

“0.3

“0.4

“0.5

“0.5 “0.4 “0.3 “0.2 “0.1 0 0.1 0.2 0.3 0.4 0.5

x

Figure 1.14 Vector ¬eld ˜ ¼ sin°p yÞ^ À sin°p xÞ^

A 2i 2 j:

Electric ¬elds that circulate back on themselves are produced not by

electric charge, but rather by changing magnetic ¬elds. Such ˜˜solenoidal™™

¬elds are discussed in Chapter 3.

Example 1.7: Given the vector electric ¬eld in a speci¬ed region, ¬nd the

density of electric charge at a location within that region.

Problem: Find the charge density at x ¼ 2 m and x ¼ 5 m if the electric ¬eld

in the region is given by

˜ ¼ ax2^ V for x ¼ 0 to 3 m;

E i

m

and

˜ ¼ b^ V for x > 3 m:

E i

m

Solution: By Gauss™s law, in the region x ¼ 0 to 3 m,

q @ ^@ ^ @

˜E

r˜¼ ¼ ^ þj þk °ax2^

i iÞ;

e0 @x @y @k

40 A student™s guide to Maxwell™s Equations

q @°ax2 Þ

¼ ¼ 2xa;

e0 @x

and

q ¼ 2xae0 :

Thus at x ¼ 2 m, q ¼ 4ae0.

In the region x > 3 m,

q @ ^@ ^ @

˜E

r˜¼ ¼ ^ þj þk °b^ ¼ 0;

i iÞ

e0 @x @y @k

so q ¼ 0 at x ¼ 5 m.

Problems

The following problems will test your understanding of Gauss™s law for

electric ¬elds. Full solutions are available on the book™s website.

1.1 Find the electric ¬‚ux through the surface of a sphere containing 15

protons and 10 electrons. Does the size of the sphere matter?

1.2 A cube of side L contains a ¬‚at plate with variable surface charge

density of r ¼ À3xy. If the plate extends from x ¼ 0 to x ¼ L and from

y ¼ 0 to y ¼ L, what is the total electric ¬‚ux through the walls of the

cube?

z

s

L

y

L

L

x

1.3 Find the total electric ¬‚ux through a closed cylinder containing a line

charge along its axis with linear charge density k ¼ k0(1Àx/h) C/m if

the cylinder and the line charge extend from x ¼ 0 to x ¼ h.

41

Gauss™s law for electric ¬elds

1.4 What is the ¬‚ux through any closed surface surrounding a charged

sphere of radius a0 with volume charge density of q ¼ q0(r/a0), where r

is the distance from the center of the sphere?

1.5 A circular disk with surface charge density 2 · 10À10 C/m2 is

surrounded by a sphere with radius of 1 m. If the ¬‚ux through the

sphere is 5.2 · 10À2 Vm, what is the diameter of the disk?

1.6 A 10 cm · 10 cm ¬‚at plate is located 5 cm from a point charge of

10À8 C. What is the electric ¬‚ux through the plate due to the point

charge?

10 cm

5 cm

10 cm

1.7 Find the electric ¬‚ux through a half-cylinder of height h owing to an

in¬nitely long line charge with charge density k running along the

axis of the cylinder.

1.8 A proton rests at the center of the rim of a hemispherical bowl

of radius R. What is the electric ¬‚ux through the surface of the

bowl?

1.9 Use a special Gaussian surface around an in¬nite line charge to ¬nd

the electric ¬eld of the line charge as a function of distance.

42 A student™s guide to Maxwell™s Equations

1.10 Use a special Gaussian surface to prove that the magnitude of the

electric ¬eld of an in¬nite ¬‚at plane with surface charge density r is

j˜ ¼ r=2e0 .

Ej

1.11 Find the divergence of the ¬eld given by ˜ ¼ °1=rÞ^ in spherical

A r

coordinates.

1.12 Find the divergence of the ¬eld given by ˜ ¼ r^ in spherical

A r

coordinates.

1.13 Given the vector ¬eld

p^

˜ ¼ cos py “ i þ sin° pxÞ^

A j;

2

sketch the ¬eld lines and ¬nd the divergence of the ¬eld.

1.14 Find the charge density in a region for which the electric ¬eld in

cylindrical coordinates is given by

˜ ¼ az ^ þ br f þ cr 2 z2^

^

E r z

r

1.15 Find the charge density in a region for which the electric ¬eld in

spherical coordinates is given by

˜ ¼ ar 2^ þ b cos °hÞ h þ cf:

^ ^

E r

r

2

Gauss™s law for magnetic ¬elds

Gauss™s law for magnetic ¬elds is similar in form but different in content

from Gauss™s law for electric ¬elds. For both electric and magnetic ¬elds,

the integral form of Gauss™s law involves the ¬‚ux of the ¬eld over a closed

surface, and the differential form speci¬es the divergence of the ¬eld at a

point.

The key difference in the electric ¬eld and magnetic ¬eld 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 ¬‚ux and on the

divergence of the magnetic ¬eld.

2.1 The integral form of Gauss™s law

Notation differs among textbooks, but the integral form of Gauss™s law is

generally written as follows:

I

˜ ^ da ¼ 0 Gauss™s law for magnetic fields °integral formÞ:

Bn

S

As described in the previous chapter, the left side of this equation is a

mathematical description of the ¬‚ux of a vector ¬eld through a closed

surface. In this case, Gauss™s law refers to magnetic ¬‚ux “ the number of

magnetic ¬eld 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

¬eld case, and you will ¬nd some examples of how to use the magnetic

43

44 A Student™s guide to Maxwell™s Equations

version to solve problems “ but ¬rst you should make sure you understand

the main idea of Gauss™s law for magnetic ¬elds:

The total magnetic ¬‚ux 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 ¬‚ux through that surface must be zero.

Note that this does not mean that zero magnetic ¬eld lines penetrate the

surface “ it means that for every magnetic ¬eld line that enters the volume

enclosed by the surface, there must be a magnetic ¬eld line leaving that

volume. Thus the inward (negative) magnetic ¬‚ux must be exactly bal-

anced by the outward (positive) magnetic ¬‚ux.

Since many of the symbols in Gauss™s law for magnetic ¬elds are the

same as those covered in the previous chapter, in this chapter you™ll ¬nd

only those symbols peculiar to this law. Here™s an expanded view:

Reminder that the Dot product tells you to find the part of

ˆ

magnetic field is a vector B parallel to n (perpendicular to the surface)

«

The unit vector normal to the surface

B n da = 0

ˆ

S

Reminder that this The magnetic An increment of

surface area in m2

integral is over a field in Teslas

closed surface Reminder that this is a surface integral

(not a volume or a line integral)

Tells you to sum up the contributions

from each portion of the surface

Gauss™s law for magnetic ¬elds 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 ¬eld

lines, just as electric charge does for electric ¬eld lines. In that case,

enclosing a single magnetic pole within a closed surface would produce

nonzero ¬‚ux through the surface (exactly as you can produce nonzero

electric ¬‚ux 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 ¬elds is identically zero.

Knowing that the total magnetic ¬‚ux through a closed surface must be

zero may allow you to solve problems involving complex surfaces, par-

ticularly if the ¬‚ux through one portion of the surface can be found by

integration.

45

Gauss™s law for magnetic ¬elds

˜ The magnetic ¬eld

B

Just as the electric ¬eld may be de¬ned by considering the electric force

on a small test charge, the magnetic ¬eld may be de¬ned 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 ¬eld, as shown by the Lorentz

equation for magnetic force:

˜B ¼ q˜ · ˜ °2:1Þ

F vB

where ˜B is the magnetic force, q is the particle™s charge, ˜ is the particle™s

F v

velocity (with respect to ˜ and ˜ is the magnetic ¬eld.

B), B

Using the de¬nition of the vector cross-product which says that

˜ · ˜ ¼ j˜ ˜ sin°hÞ, where h is the angle between ˜ and ˜ the magnitude

ab ajjbj a b,

of the magnetic ¬eld may be written as

j˜B j

F

j˜ ¼ °2:2Þ

Bj

qj˜ sin°hÞ

vj

where h is the angle between the velocity vector ˜ and the magnetic ¬eld

v

˜ The terminology for magnetic quantities is not as standardized as that

B.

of electric quantities, so you are likely to ¬nd texts that refer to ˜ as the

B

˜˜magnetic induction™™ or the ˜˜magnetic ¬‚ux density.™™ Whatever it is

called, ˜ has units equivalent to N/(C m/s), which include Vs/m2,

B

N/(Am), kg/(Cs), or most simply, Tesla (T).

Comparing Equation 2.2 to the relevant equation for the electric ¬eld,

Equation (1.1), several important distinctions between magnetic and

electric ¬elds become clear:

Like the electric ¬eld, the magnetic ¬eld is directly proportional to the

magnetic force. But unlike ˜ which is parallel or antiparallel to the

E,

electric force, the direction of ˜ is perpendicular to the magnetic force.

B

˜ the magnetic ¬eld may be de¬ned through the force

Like E,

experienced by a small test charge, but unlike ˜ the speed and

E,

direction of the test charge must be taken into consideration when

relating magnetic forces and ¬elds.

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 ¬eld is therefore always zero.

Whereas electrostatic ¬elds are produced by electric charges, magneto-

static ¬elds are produced by electric currents.

46 A Student™s guide to Maxwell™s Equations

I

B

B

N

B I

S

Current loop

Current-carrying Bar magnet

straight wire

B

I

I

Horseshoe

Torus

Solenoid

magnet

Figure 2.1 Examples of magnetic ¬elds.

Magnetic ¬elds may be represented using ¬eld lines whose density in a

plane perpendicular to the line direction is proportional to the strength of

the ¬eld. Examples of several magnetic ¬elds 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 ¬elds produced by currents:

Magnetic ¬eld lines do not originate and terminate on charges; they

form closed loops.

The magnetic ¬eld 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 ¬eld lines run between the poles).

The net magnetic ¬eld at any point is the vector sum of all magnetic

¬elds present at that point.

Magnetic ¬eld lines can never cross, since that would indicate that the

¬eld points in two different directions at the same location “ if the ¬elds

from two or more sources overlap at the same location, they add (as

vectors) to produce a single, total ¬eld at that point.

47

Gauss™s law for magnetic ¬elds

Table 2.1. Magnetic ¬eld equations for simple objects

˜ ¼ l0 I ™

In¬nite straight wire carrying

^

B

current I (at distance r) 2pr

l Id˜· ^

Segment of straight wire carrying current I lr

d˜ ¼ 0

B

(at distance r) 4p r 2

l0 IR2

Circular loop of radius R carrying current I ˜¼ ^

B x

(loop in yz plane, at distance x along x-axis) 2°x2 þ R2 Þ3=2

˜ ¼ l0 NI ^ °insideÞ

Solenoid with N turns and length l

B x

carrying current I l

˜ ¼ l0 NI ™ (inside)

Torus with N turns and radius r

^

B

carrying current I 2pr

P

Point at which

magnetic field

is determined

r

^

r

Current

element

dl

Figure 2.2 Geometry for Biot“Savart law.

All static magnetic ¬elds are produced by moving electric charge. The

contribution d˜ to the magnetic ¬eld at a speci¬ed point P from a small

B

element of electric current is given by the Biot“Savart law:

˜r

˜ ¼ l0 Idl · ^

dB

4p r 2

In this equation, l0 is the permeability of free space, I is the current

through the small element, d˜ is a vector with the length of the current

l

element and pointing in the direction of the current, ^ is a unit vector

r

pointing from the current element to the point P at which the ¬eld is

being calculated, and r is the distance between the current element and P,

as shown in Figure 2.2.

Equations for the magnetic ¬eld in the vicinity of some simple objects

may be found in Table 2.1.

48 A Student™s guide to Maxwell™s Equations

H

˜ ^ da

Bn The magnetic ¬‚ux through a closed surface

S

Like the electric ¬‚ux UE, the magnetic ¬‚ux UB through a surface may be

thought of as the ˜˜amount™™ of magnetic ¬eld ˜˜¬‚owing™™ through the sur-

face. How this quantity is calculated depends on the situation:

UB ¼ j˜ · (surface area) ˜ uniform and perpendicular to S; °2:3Þ

Bj B

UB ¼ ˜ ^ · (surface area) ˜ uniform and at an angle to S; °2:4Þ

Bn B

Z

˜ ^ da ˜ nonuniform and at variable angle to S:

UB ¼ °2:5Þ

Bn B

S

Magnetic ¬‚ux, like electric ¬‚ux, is a scalar quantity, and in the magnetic

case, the units of ¬‚ux have been given the special name ˜˜webers™™ (abbre-

viated Wb and which, by any of the relations shown above, must be

equivalent to T m2).

As in the case of electric ¬‚ux, the magnetic ¬‚ux through a surface may

be considered to be the number of magnetic ¬eld lines penetrating that

surface. When you think about the number of magnetic ¬eld lines

through a surface, don™t forget that magnetic ¬elds, like electric ¬elds, are

actually continuous in space, and that ˜˜number of ¬eld lines™™ only has

meaning once you™ve established a relationship between the number of

lines you draw and the strength of the ¬eld.

When considering magnetic ¬‚ux through a closed surface, it is espe-

cially important to remember the caveat that surface penetration is a two-

way street, and that outward ¬‚ux and inward ¬‚ux have opposite signs.

Thus equal amounts of outward (positive) ¬‚ux and inward (negative) ¬‚ux

will cancel, producing zero net ¬‚ux.

The reason that the sign of outward and inward ¬‚ux is so important in

the magnetic case may be understood by considering a small closed

surface placed in any of the ¬elds shown in Figure 2.1. No matter what

shape of surface you choose, and no matter where in the magnetic ¬eld

you place that surface, you™ll ¬nd that the number of ¬eld lines entering

the volume enclosed by the surface is exactly equal to the number of ¬eld

lines leaving that volume. If this holds true for all magnetic ¬elds, it can

only mean that the net magnetic ¬‚ux through any closed surface must

always be zero.

Of course, it does hold true, because the only way to have ¬eld lines

enter a volume without leaving it is to have them terminate within the

49

Gauss™s law for magnetic ¬elds

B

Figure 2.3 Magnetic ¬‚ux lines penetrating closed surfaces.

volume, and the only way to have ¬eld lines leave a volume without

entering it is to have them originate within the volume. But unlike

electric ¬eld lines, magnetic ¬eld 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 ¬‚ux must be equal and

opposite through any closed surface.

Consider the closer view of the ¬eld produced by a bar magnet shown

in Figure 2.3. Irrespective of the shape and location of the closed surfaces

placed in the ¬eld, all ¬eld lines entering the enclosed volume are offset by

an equal number of ¬eld lines leaving that volume.

The physical reasoning behind Gauss™s law should now be clear: the

net magnetic ¬‚ux passing through any closed surface must be zero

because magnetic ¬eld lines always form complete loops. The next section

shows you how to use this principle to solve problems involving closed

surfaces and the magnetic ¬eld.

50 A Student™s guide to Maxwell™s Equations

H

˜ ^ da ¼ 0

Bn Applying Gauss™s law (integral form)

S

In situations involving complex surfaces and ¬elds, ¬nding the ¬‚ux by

integrating the normal component of the magnetic ¬eld over a speci¬ed

surface can be quite dif¬cult. In such cases, knowing that the total

magnetic ¬‚ux 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 ¬eld and a surface

geometry, ¬nd the ¬‚ux through a speci¬ed portion of that surface.

Problem: A closed cylinder of height h and radius R is placed in a mag-

netic ¬eld given by ˜ ¼ B0 ° ^ À ^ If the axis of the cylinder is aligned

B j kÞ.

along the z-axis, ¬nd the ¬‚ux 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 ¬‚ux through the entire

surface must be zero, so if you™re able to ¬gure out the ¬‚ux through some

portions of the surface, you can deduce the ¬‚ux through the other por-

tions. In this case, the ¬‚ux through the top and bottom of the cylinder are

relatively easy to ¬nd; whatever additional amount it takes to make the

total ¬‚ux equal to zero must come from the curved sides of the cylinder.

Thus

UB;Top þ UB;Bottom þ UB;Sides ¼ 0:

The magnetic ¬‚ux through any surface is

Z

UB ¼ ˜ ^ da:

Bn

S

For the top surface, ^ ¼ ^ so

n k,

˜ ^ ¼ °B0^ À B0 ^ ^ ¼ ÀB0 :

Bn j kÞ k

Thus

Z Z

˜ ^ da ¼ ÀB0

UB;Top ¼ da ¼ ÀB0 °pR2 Þ:

Bn

S S

A similar analysis for the bottom surface (for which n ¼ À^ gives

^ k)

Z Z

UB;Bottom ¼ ˜ ^ da ¼ þB0 da ¼ þB0 °pR2 Þ:

Bn

S S

Since UB,Top ¼ ÀUB,Bottom, you can conclude that UB,Sides ¼ 0.

51

Gauss™s law for magnetic ¬elds

Example 2.2: Given the current in a long wire, ¬nd the magnetic

¬‚ux through nearby surfaces

Problem: Find the magnetic ¬‚ux through the curved surface of a

half-cylinder near a long, straight wire carrying current I.

z

I

h

y1

y

x

Solution: At distance r from a current-carrying wire, the magnetic ¬eld is

given by

˜ ¼ l0 I ™;

^

B

2pr

which means that the magnetic ¬eld lines make circles around the wire,

entering the half-cylinder through the ¬‚at surface and leaving through the

curved surface. Gauss™s law tells you that the total magnetic ¬‚ux through

all faces of the half-cylinder must be zero, so the amount of (negative)

¬‚ux through the ¬‚at surface must equal the amount of (positive) ¬‚ux

leaving the curved surface. To ¬nd the ¬‚ux through the ¬‚at surface, use

the expression for ¬‚ux

Z

UB ¼ ˜ ^ da:

Bn

S

In this case, ^ ¼ À™, so

^

n

l I lI

˜^ ¼ ^ ^

™ °À™Þ ¼ À 0 :

0

Bn

2pr 2pr

To integrate over the ¬‚at face of the half-cylinder, notice that the face lies

in the yz plane, and an element of surface area is therefore da ¼ dy dz. Notice

also that on the ¬‚at face the distance increment dr ¼ dy, so da ¼ dr dz and the

¬‚ux integral is

Z Z Z Z y1 þ2R

l0 I l0 I h dr

UB;Flat ¼ ˜ ^ da ¼ À dr dz ¼ À dz :

Bn

S 2pr 2p z¼0 r¼y1 r

S

52 A Student™s guide to Maxwell™s Equations

Thus

l0 I y1 þ 2R l0 Ih 2R

UB;Flat ¼À °hÞ ¼ À ln 1 þ :

ln

2p 2p

y1 y1

Since the total magnetic ¬‚ux through this closed surface must be zero, this

means that the ¬‚ux through the curved side of the half-cylinder is

l0 Ih 2R

UB;Curved side ¼ ln 1 þ :

2p y1

53

Gauss™s law for magnetic ¬elds

2.2 The differential form of Gauss™s law

The continuous nature of magnetic ¬eld lines makes the differential form

of Gauss™s law for magnetic ¬elds quite simple. The differential form is

written as

˜B

r ˜ ¼ 0 Gauss™s law for magnetic fields °differential formÞ:

The left side of this equation is a mathematical description of the

divergence of the magnetic ¬eld “ the tendency of the magnetic ¬eld to

˜˜¬‚ow™™ more strongly away from a point than toward it “ while the right

side is simply zero.

The divergence of the magnetic ¬eld 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 ¬eld at any point is zero.

One way to understand why this is true is by analogy with the electric

¬eld, 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 ¬eld must also be zero.

To help you understand the meaning of each symbol in Gauss™s law for

magnetic ¬elds, here is an expanded view:

Reminder that the Reminder that the magnetic

del operator is a vector field is a vector

B=0

The magnetic

field in Teslas

The differential The dot product turns

operator called the del operator into the

“del” or “nabla” divergence

54 A Student™s guide to Maxwell™s Equations

˜B

r ˜ The divergence of the magnetic ¬eld

This expression is the entire left side of the differential form of Gauss™s

law, and it represents the divergence of the magnetic ¬eld. Since diver-

gence is by de¬nition the tendency of a ¬eld to ˜˜¬‚ow™™ away from a point

more strongly than toward that point, and since no point sources or sinks

of the magnetic ¬eld have ever been found, the amount of ˜˜incoming™™

¬eld is exactly the same as the amount of ˜˜outgoing™™ ¬eld at every point.

So it should not surprise you to ¬nd that the divergence of ˜ is always

B

zero.

To verify this for the case of the magnetic ¬eld around a long, current-

carrying wire, take the divergence of the expression for the wire™s mag-

netic ¬eld as given in Table 2.1:

l0 I

BÞ ˜ B ˜

div°˜ ¼ r ˜ ¼ r ^

™: °2:6Þ

2pr

This is most easily determined using cylindrical coordinates:

˜ B 1 @ °rBr Þ þ 1 @Bf þ @Bz :

r˜ ¼ °2:7Þ

r @r r @f @z

which, since ˜ has only a u-component, is

B

˜ B 1 @ °l0 I=2pr Þ ¼ 0:

r ˜ ¼ °2:8Þ

@™

r

You can understand this result using the following reasoning: since

the magnetic ¬eld 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 ¬eld has constant amplitude around any circular path

centered on the wire), the ¬‚ux away from any point must be the same as

the ¬‚ux toward that point. This means that the divergence of the mag-

netic ¬eld is zero everywhere.

Vector ¬elds with zero divergence are called ˜˜solenoidal™™ ¬elds, and all

magnetic ¬elds are solenoidal.

55

Gauss™s law for magnetic ¬elds

˜B

r˜ ¼ 0 Applying Gauss™s law (differential form)

Knowing that the divergence of the magnetic ¬eld must be zero allows

you to attack problems involving the spatial change in the components of

a magnetic ¬eld and to determine whether a speci¬ed vector ¬eld could be

a magnetic ¬eld. This section has examples of such problems.

Example 2.3: Given incomplete information about the components of

a magnetic ¬eld, use Gauss™s law to establish relationships between those

components

Problem: A magnetic ¬eld is given by the expression

˜ ¼ axz ^ þ byz ^ þ c^

B i j k

What is the relationship between a and b?

Solution: You know from Gauss™s law for magnetic ¬elds that the

divergence of the magnetic ¬eld must be zero. Thus

˜ B @Bx þ @By þ @Bz ¼ 0:

r˜ ¼

@x @y @z

Thus

@°axzÞ @°byzÞ @c

þ þ ¼0

@x @y @z

and

az þ bz þ 0 ¼ 0;

which means that a ¼ Àb.

Example 2.4: Given an expression for a vector ¬eld, determine whether

that ¬eld could be a magnetic ¬eld.

Problem: A vector ¬eld is given by the expression

˜ yÞ ¼ a cos°bxÞ^ þ aby sin°bxÞ^

A°x; i j:

Could this ¬eld be a magnetic ¬eld?

Solution: Gauss™s law tells you that the divergence of all magnetic ¬elds

must be zero, and checking the divergence of this vector ¬eld gives

56 A Student™s guide to Maxwell™s Equations

@ @

˜A

r ˜ ¼ ½a cos°bxÞ þ ½aby sin°bxÞ

@x @y

¼ Àab sin°bxÞ þ ab sin°bxÞ ¼ 0

which indicates that ˜ could represent a magnetic ¬eld.

A

Problems

The following problems will check your understanding of Gauss™s law

for magnetic ¬elds. Full solutions are available on the book™s website.

2.1 Find the magnetic ¬‚ux produced by the magnetic ¬eld ˜ ¼ 5^ À 3^ þ

B i j

4^ through the top, bottom, and side surfaces of the ¬‚ared cylinder

knT

shown in the ¬gure.

z

B

Rtop

y

Rbottom

x

2.2 What is the change in magnetic ¬‚ux 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 ¬‚ux through all ¬ve surfaces of the wedge shown

in the ¬gure if the magnetic ¬eld in the region is given by

˜ ¼ 0:002^ þ 0:003^ T;

B i j

and show that the total ¬‚ux through the wedge is zero.

z

130 cm

y

70 cm

50 cm

x

57

Gauss™s law for magnetic ¬elds

2.4 Find the ¬‚ux of the Earth™s magnetic ¬eld through each face of a

cube with 1-m sides, and show that the total ¬‚ux through the cube

is zero. Assume that at the location of the cube the Earth™s

magnetic ¬eld has amplitude of 4 · 10À5 T and points upward at

an angle of 30 with respect to the horizontal. You may orient the

cube in any way you choose.

2.5 A cylinder of radius r0 and height h is placed inside an ideal solenoid

with the cylinder™s axis parallel to the axis of the solenoid. Find the

¬‚ux through the top, bottom, and curved surfaces of the cylinder and

show that the total ¬‚ux through the cylinder is zero.

2.6 Determine whether the vector ¬elds given by the following

expressions in cylindrical coordinates could be magnetic ¬elds:

˜ ™; zÞ ¼ a cos2 °™Þ^;

A°r; r

(a)

r

˜ ™; zÞ ¼ a cos2 °™Þ^:

A°r; r

(b)

r2

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 ¬‚ux enclosed by the circuit. That discovery is

made even more useful when extended to the general statement that a

changing magnetic ¬eld produces an electric ¬eld. Such ˜˜induced™™ elec-

tric ¬elds are very different from the ¬elds produced by electric charge,

and Faraday™s law of induction is the key to understanding their

behavior.

3.1 The integral form of Faraday™s law

In many texts, the integral form of Faraday™s law is written as

I Z

˜ d˜ ¼ À d ˜ ^ da Faraday™s law °integral formÞ:

El Bn

dt

C S

Some authors feel that this form is misleading because it confounds two

distinct phenomena: magnetic induction (involving a changing magnetic

¬eld) and motional electromotive force (emf) (involving movement of a

charged particle through a magnetic ¬eld). In both cases, an emf is

produced, but only magnetic induction leads to a circulating electric ¬eld

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 ˜ rep-E

resents the electric ¬eld in the rest frame of each segment d˜ of the path of

l

integration.

58

59

Faraday™s law

A version of Faraday™s law that separates the two effects and makes

clear the connection between electric ¬eld circulation and a changing

magnetic ¬eld is

Z

d ˜ ^ da

emf ¼ À Flux rule,

Bn

dt S

I Z

@˜

B

˜ d˜ ¼ À ^ da Faraday™s law °alternate formÞ:

El n

@t

C S

Note that in this version of Faraday™s law the time derivative operates

only on the magnetic ¬eld rather than on the magnetic ¬‚ux, and both ˜ E

and ˜ are measured in the laboratory reference frame.

B

Don™t worry if you™re uncertain of exactly what emf is or how it is

related to the electric ¬eld; that™s all explained in this chapter. There are

also examples of how to use the ¬‚ux rule and Faraday™s law to solve

problems involving induction “ but ¬rst you should make sure you

understand the main idea of Faraday™s law:

Changing magnetic ¬‚ux through a surface induces an emf in any

boundary path of that surface, and a changing magnetic ¬eld induces

a circulating electric ¬eld.

In other words, if the magnetic ¬‚ux through a surface changes, an electric

¬eld is induced along the boundary of that surface. If a conducting

material is present along that boundary, the induced electric ¬eld pro-

vides an emf that drives a current through the material. Thus quickly