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 ¬elds

and electric currents, Maxwell worked out an elaborate conceptual model

in which magnetic ¬elds 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

¬elds. That source is the changing electric ¬‚ux in the Ampere“Maxwell law.

Most texts use one of three approaches to demonstrating the need for

the changing-¬‚ux 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 I ¬‚ows as the battery charges the capacitor. This current produces

92 A student™s guide to Maxwell™s Equations

Amperian loop

I

Switch Capacitor

Battery

Figure 4.6 Charging capacitor.

a magnetic ¬eld around the wires, and the circulation of that ¬eld is given

by Ampere™s law

I

˜ d˜ ¼ l0 °Ienc Þ:

Bl

C

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 C is a boundary. However, you™ll get very

different answers for the enclosed current if you choose a ¬‚at 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 I penetrates the ¬‚at 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 ¬eld 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 ¬‚ows 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 ¬‚ows between the capacitor plates,

what else might be going on in that region that would serve as the source

of a magnetic ¬eld?

Since charge is accumulating on the plates as the capacitor charges up,

you know that the electric ¬eld between the plates must be changing with

93

The Ampere“Maxwell law

(a) Amperian loop

Current I penetrates

this surface

I

Capacitor

Switch

Battery

(b) Amperian loop

I

Switch

Capacitor

Battery

Figure 4.7 Alternative surfaces for determining enclosed current.

time. This means that the electric ¬‚ux through the portion of your

˜˜stocking cap™™ surface between the plates must also be changing, and

you can use Gauss™s law for electric ¬elds to determine the change in ¬‚ux.

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 ¬eld and over which the electric ¬eld is either uniform or zero.

Neglecting edge effects, the electric ¬eld between two charged conducting

plates is ˜ = (r/e0) ^, where r is the charge density on the plates (Q/A),

E n

making the electric ¬‚ux through the surface

Z Z Z

r Q Q

UE ¼ ˜ ^ da ¼ da ¼ da ¼ : °4:4Þ

En

e0 e0

Ae0 S

S S

94 A student™s guide to Maxwell™s Equations

Time-varying

electric field

Special Gaussian “

+

surface + “

+ “

+ “

+Q “Q

Figure 4.8 Changing electric ¬‚ux between capacitor plates.

The change in electric ¬‚ux over time is therefore,

Z

1 dQ

d dQ

˜ ^ da ¼ ¼ : °4:5Þ

En

dt e0 e0 dt

dt S

Multiplying by the vacuum permittivity makes this

Z

d ˜ ^ da ¼ dQ :

e0 °4:6Þ

En

dt S dt

Thus, the change in electric ¬‚ux 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 ¬eld around your surface boundary. For historical reasons,

the product of the permittivity and the change of electric ¬‚ux through a

surface is called the ˜˜displacement current™™ even though no charge

actually ¬‚ows across the surface. The displacement current is de¬ned by

the relation

Z

d ˜ ^ da :

Id e0 °4:7Þ

En

dt S

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.

95

The Ampere“Maxwell law

À Á

H R

˜ d˜ ¼ l0 Ienc þ e0 d ˜ ^ da Applying the

Bl En

C dt S

Ampere“Maxwell law

(integral form)

Like the electric ¬eld in Gauss™s law, the magnetic ¬eld 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

¬eld 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 ¬nd an Amperian loop over which

you expect ˜ to be uniform and at a constant angle to the loop. However,

B

how do you know what to expect for ˜ before you solve the problem?

B

In many cases, you™ll already have some idea of the behavior of the

magnetic ¬eld on the basis of your past experience or from experimental

evidence. What if that™s not the case “ how are you supposed to ¬gure 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

¬eld 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 ˜ must get smaller as you move away

B

from the wire; otherwise Oersted™s demonstration in Denmark would

have de¬‚ected 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 ¬eld on one side of the wire is any different from the

¬eld on the other side. So if ˜ decreases with distance from the wire and is

B

the same all around the wire, you can safely conclude that one path of

constant ˜ would be a circle centered on the wire and perpendicular to the

B

direction of current ¬‚ow.

However, to deal with the dot product of ˜ and d˜ in Ampere™s law,

B l

you also need to make sure that your path maintains a constant angle

(preferably 0 ) to the magnetic ¬eld. If ˜ were to have both radial

B

and transverse components that vary with distance, the angle between

your path and the magnetic ¬eld might depend on distance from the wire.

96 A student™s guide to Maxwell™s Equations

If you understand the cross“product between d˜ and ^ in the

l r

Biot“Savart law, you probably suspect that this is not the case. To verify

that, imagine that ˜ has a component pointing directly toward the wire. If

B

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 ¬eld 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 ˜ pointing toward the wire.

B

Now ask yourself, what would happen if you reversed the direction of

the current ¬‚ow. Since the magnetic ¬eld is linearly proportional to the

current (˜ / ˜ according to the Biot“Savart law, reversing the current

B I)

must also reverse the magnetic ¬eld, and ˜ would then point away from

B

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 ¬eld 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 ¬eld

pointing away from the wire.

Comparing notes with your friend, you™d ¬nd a logical inconsistency.

You™d say, ˜˜currents traveling away from me produce a magnetic ¬eld

pointing toward the wire, and currents coming toward me produce a

magnetic ¬eld 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 ¬nd that your original

conclusions were no longer true.

This inconsistency is resolved if the magnetic ¬eld circles around the wire,

having no radial component at all. With ˜ having only a u-component,5 all

B

observers agree that currents traveling away from an observer produce

clockwise magnetic ¬elds as seen by that observer, whereas currents

approaching an observer produce counterclockwise magnetic ¬elds 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 ¬rst place “ to

¬nd the value of the magnetic ¬eld at some location. So make your

5

Remember that there™s a review of cylindrical and spherical coordinates on the book™s

website.

97

The Ampere“Maxwell law

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 ¬nd

the value of the magnetic ¬eld. The following example shows how this

works.

Example 4.1: Given the current in a wire, ¬nd the magnetic ¬eld within

and outside the wire.

Problem: A long, straight wire of radius r0 carries a steady current I

uniformly distributed throughout its cross-sectional area. Find the

magnitude of the magnetic ¬eld as a function of r, where r is the distance

from the center of the wire, for both r > r0 and r < r0.

Solution: Since the current is steady, you can use Ampere™s law in its

original form

I

˜ d˜ ¼ l0 °Ienc Þ:

Bl

C

To ¬nd ˜ at exterior points (r>r0), use the logic described above and

B

draw your loop outside the wire, as shown by Amperian loop #1 in

Figure 4.9. Since both ˜ and d˜ have only u-components and point in the

B l

same direction if you obey the right-hand rule in determining your dir-

ection of integration, the dot product ˜ d˜ becomes |˜ || d˜

B l|cos(0 ).

B l

Furthermore, since |˜ is constant around your loop, it comes out of the

B|

integral:

I I I

˜d˜ ¼ B

B l

˜ d˜ ¼ dl ¼ B°2prÞ,

Bl

C C C

Amperian

loop #1

Amperian

loop #2

I

Wire

Figure 4.9 Amperian loops for current-carrying wire of radius r0.

98 A student™s guide to Maxwell™s Equations

where r is the radius of your Amperian loop.6 Ampere™s law tells you that

the integral of ˜ around your loop is equal to the enclosed current times

B

the permeability of free space, and the enclosed current in this case is all of

I, so

B°2prÞ ¼ l0 Ienc ¼ l0 I

and, since ˜ is in the ™-direction,

B

˜ ¼ l0 I ™,

^

B

2pr

as given in Table 2.1. Note that this means that at points outside the wire

the magnetic ¬eld decreases as 1/r and behaves as if all the current were at

the center of the wire.

To ¬nd the magnetic ¬eld 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 density7 is I/(pr02), and the current passing

through the loop is simply that density times the area of the loop. Thus,

Enclosed current ¼ current density · loop area

or

r2

I

Ienc ¼ pr 2 ¼ I 2 :

pr02 r0

Inserting this into Ampere™s law gives

I 2

˜ d˜ ¼ B°2prÞ ¼ l0 Ienc ¼ l0 I r ,

Bl 2

r0

C

or

l0 Ir

B¼ :

2

2pr0

Thus, inside the wire the magnetic ¬eld increases linearly with distance

from the center of the wire, reaching a maximum at the surface of the

wire.

Another way to understand this is to write ˜ as B™ ™ and d˜ as (rd™) ™, so ˜ d˜ ¼ B™ rd™

^ ^

6

B l B l

R 2p

and 0 B™ rd™ ¼ B™ °2prÞ.

7

If you need a review of current density, you™ll ¬nd a section covering this topic later in this

chapter.

99

The Ampere“Maxwell law

Example 4.2: Given the time-dependent charge on a capacitor, ¬nd the

rate of change of the electric ¬‚ux between the plates and the magnitude

of the resulting magnetic ¬eld at a speci¬ed location.

Problem: A battery with potential difference DV charges a circular

parallel-plate capacitor of capacitance C and plate radius r0 through a

wire with resistance R. Find the rate of change of the electric ¬‚ux between

the plates as a function of time and the magnetic ¬eld at a distance r from

the center of the plates.

Solution: From Equation 4.5, the rate of change of electric ¬‚ux between

the plates is

0 1

Z

1 dQ

dUE d

¼ @ ˜ ^ daA ¼ ,

En

e0 dt

dt dt

S

where Q is 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

Q°tÞ ¼ CDV°1 À eÀt=RC Þ,

where DV, R, and C represent the potential difference, the series resist-

ance, and the capacitance, respectively. Thus,

i 1

1 dh DV Àt=RC

1 Àt=RC

dUE

CDV°1 À eÀt=RC Þ ¼

¼ ¼ :

CDV e e

e0 dt e0 e0 R

dt RC

This is the rate of change of the total electric ¬‚ux between the plates. To

¬nd the magnetic ¬eld at a distance r from the center of the plates, you™re

going to have to construct a special Amperian loop to help you extract

the magnetic ¬eld from the integral in the Ampere“Maxwell law:

I Z

d

˜ d˜ ¼ l0 Ienc þ e0 ˜ ^ da :

Bl En

dt S

C

Since no charge ¬‚ows between the capacitor plates, Ienc=0, and

I Z

d

˜ d˜ ¼ l0 ˜ ^ da :

e0

Bl En

dt S

C

100 A student™s guide to Maxwell™s Equations

“

+

+ “

+ “

+ “

Figure 4.10 Amperian loop between capacitor plates.

As in the previous example, you™re faced with the challenge of designing a

special Amperian loop around which the magnetic ¬eld 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 ¬nd the magnetic ¬eld. Of course, not all of the

¬‚ux between the plates passes through this loop, so you will have to

modify your expression for the ¬‚ux change accordingly. The fraction of

the total ¬‚ux that passes through a loop of radius r is simply the ratio of

the loop area to the capacitor plate area, which is pr2/ pr02, so the rate

of change of ¬‚ux through the loop is

DV Àt=RC r 2

dUE

¼ :

e

dt Loop e0 R 2

r0

Inserting this into the Ampere“Maxwell law gives

!

I

DV Àt=RC r 2 l0 DV Àt=RC r 2

˜ d˜ ¼ l0 e0 ¼ :

Bl e e

e0 R 2 2

R

r0 r0

C

Moreover, since you™ve chosen your Amperian loop so as to allow ˜ to

B

come out of the dot product and the integral using the same symmetry

arguments as in Example 4.1,

I

l0 DV Àt=RC r 2

˜ d˜ ¼ B°2prÞ ¼ ,

Bl e 2

R r0

C

which gives

l0 DV Àt=RC r 2 l0 DV Àt=RC r

B¼ ¼ ,

e e

2 2

2prR 2pR

r0 r0

meaning that the magnetic ¬eld increases linearly with distance from the

center of the capacitor plates and decreases exponentially with time,

reaching 1/e of its original value at time t=RC.

101

The Ampere“Maxwell law

4.2 The differential form of the Ampere“Maxwell law

The differential form of the Ampere“Maxwell law is generally written as

!

@˜

E

˜B

r · ˜ ¼ l0 ˜ þ e0 The Ampere“Maxwell law:

J

@t

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

the magnetic ¬eld “ the tendency of the ¬eld 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 ¬eld.

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 ¬eld is produced by an electric current and by

an electric ¬eld 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:

Reminder that the Reminder that the

magnetic field is current density is

Reminder that the del The rate of change of

a vector a vector

operator is a vector the electric field

with time

*E

—B = J+ 0

0

*t

The differential The magnetic The magnetic The electric current The electric

operator called field in teslas permeability density in amperes permittivity

“del” or “nabla” of free space per square meter of free space

The cross-product turns

the del operator

into the curl

102 A student™s guide to Maxwell™s Equations

˜B

r · ˜ The curl of the magnetic ¬eld

The left side of the differential form of the Ampere“Maxwell law

represents the curl of the magnetic ¬eld. All magnetic ¬elds, whether

produced by electrical currents or by changing electric ¬elds, circulate

back upon themselves and form continuous loops. In addition all ¬elds

that circulate back on themselves must include at least one location about

which the path integral of the ¬eld is nonzero. For the magnetic ¬eld,

locations of nonzero curl are locations at which current is ¬‚owing or an

electric ¬eld is changing.

It is important that you understand that just because magnetic ¬elds

circulate, you should not conclude that the curl is nonzero everywhere in

the ¬eld. A common misconception is that the curl of a vector ¬eld is

nonzero wherever the ¬eld appears to curve.

To understand why that is not correct, consider the magnetic ¬eld of

the in¬nite line current shown in Figure 2.1. The magnetic ¬eld lines

circulate around the current, and you know from Table 2.1 that the

^

magnetic ¬eld points in the u direction and decreases as 1/r

˜ ¼ l0 I u :

^

B

2pr

Finding the curl of this ¬eld is particularly straightforward in cylindrical

coordinates

1 @Bz @B™ 1 @°rB™ Þ @Br

@Br @Bz

˜B

r ·˜ ¼ ^þ ^

À À ™þ À ^:

r z

r @™ @z @z @r @r @™

r

Since Br and Bz are both zero, this is

@Bu 1 @°rBu Þ @ °l0 I=2pr Þ 1 @ °r l0 I=2pr Þ

˜B

r ·˜ ¼ À ^þ ^þ

^¼À ^ ¼ 0:

r z r z

@z @r @z @r

r r

However, doesn™t the differential form of the Ampere“Maxwell law tell

us that the curl of the magnetic ¬eld is nonzero in the vicinity of electric

currents and changing electric ¬elds?

No, it doesn™t. It tells us that the curl of ˜ is nonzero exactly at the

B

location through which an electric current is ¬‚owing, or at which an

electric ¬eld is changing. Away from that location, the ¬eld de¬nitely

does curve, but the curl at any given point is precisely zero, as you just

found from the equation for the magnetic ¬eld of an in¬nite line current.

How can a curving ¬eld have zero curl? The answer lies in the amplitude

as well as the direction of the magnetic ¬eld, as you can see in Figure 4.11.

103

The Ampere“Maxwell law

(a) (b) Weaker push

to the right

Weaker field

Upward push Downward push

Upward- Downward-

pointing pointing

field field

Stronger push

Stronger field

to the right

Figure 4.11 Offsetting components of the curl of ˜

B:

Using the ¬‚uid-¬‚ow and small paddlewheel analogy, imagine the forces

on the paddlewheel placed in the ¬eld shown in Figure 4.11(a). The

center of curvature is well below the bottom of the ¬gure, and the

spacing of the arrows indicates that the ¬eld is getting weaker with

distance from the center. At ¬rst glance, it may seem that this pad-

dlewheel would rotate clockwise owing to the curvature of the ¬eld,

since the ¬‚ow lines are pointing slightly upward at the left paddle and

slightly downward at the right. However, consider the effect of the

weakening of the ¬eld above the axis of the paddlewheel: the top paddle

receives a weaker push from the ¬eld 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 ¬eld is offset by the weakening of the ¬eld with

distance from the center of curvature. And if the ¬eld 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 ¬eld lines are curved.

The key concept in this explanation is that the magnetic ¬eld may be

curved at many different locations, but only at points at which current is

¬‚owing (or the electric ¬‚ux is changing) is the curl of ˜ nonzero. This is

B

2

analogous to the 1/r reduction in electric ¬eld amplitude with distance

from a point charge, which keeps the divergence of the electric ¬eld as

zero at all points away from the location of the charge.

As in the electric ¬eld 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 r in the denominator, and those terms become

104 A student™s guide to Maxwell™s Equations

in¬nite at the origin. To evaluate the curl at the origin, use the formal

de¬nition of curl as described in Chapter 3:

I

˜ · ˜ lim 1 ˜ d˜

rB Bl

DS!0 DS

C

Considering a special Amperian loop surrounding the current, this is

!

I

1 l0˜

1 1˜

I

˜B ˜ d˜ ¼ lim

r · ˜ lim °2prÞ ¼ lim l0 I :

Bl

DS!0 DS DS!0 DS 2pr DS!0 DS

C

However, ˜I=DS is just the average current density over the surface DS, and

as DS shrinks to zero, this becomes equal to ˜ the current density at the

J,

origin. Thus, at the origin

˜B

r · ˜ ¼ l0˜

J

in accordance with Ampere™s law.

So just as you might be fooled into thinking that charge-based electric

¬eld vectors ˜˜diverge™™ everywhere because they get farther apart, you

might also think that magnetic ¬eld 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 ¬eld lines at that

point, but how the change in the ¬eld from one side of the point to the

other (say from left to right) compares to the change in the ¬eld 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 ¬eld away from the wire exactly compensates for the

curvature of the ¬eld lines. Thus, the curl of the magnetic ¬eld is zero

everywhere except at the wire itself, where electric current is ¬‚owing.

105

The Ampere“Maxwell law

˜ The electric current density

J

The right side of the differential form of the Ampere“Maxwell law

contains two source terms for the circulating magnetic ¬eld; the ¬rst

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/m3 for mass density or C/m3

for charge density.

This is not the case for current density, which is de¬ned as the vector

current ¬‚owing 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/m2).

To understand the concept of current density, recall that in the dis-

cussion of ¬‚ux in Chapter 1, the quantity ˜ is de¬ned as the number

A

density of the ¬‚uid (particles per cubic meter) times the velocity of the

¬‚ow (meters per second). As the product of the number density (a scalar)

and the velocity (a vector), ˜ is a vector in the same direction as the

A

velocity, with units of particles per square meter per second. To ¬nd the

number of particles per second passing through a surface in the simplest

case (˜ uniform and perpendicular to the surface), you simply multiply ˜

A 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 n and the

charge per carrier is q, then the amount of charge passing through a unit

area perpendicular to the ¬‚ow per second is

˜ ¼ nq˜d °C=m2 s or A=m2 Þ; °4:8Þ

J v

where ˜d is the average drift velocity of charge carriers. Thus, the direc-

v

tion of the current density is the direction of current ¬‚ow, and the

magnitude is the current per unit area, as shown in Figure 4.12.

The complexity of the relationship between the total current I through

a surface and the current density ˜ depends on the geometry of the

J

situation. If the current density ˜ is uniform over a surface S and is

J

everywhere perpendicular to the surface, the relationship is

I ¼ ˜ · °surface areaÞ ˜ uniform and perpendicular to S: °4:9Þ

J J

106 A student™s guide to Maxwell™s Equations

J

Charge Imaginary surface

carriers within wire

Figure 4.12 Charge ¬‚ow and current density.

If ˜ is uniform over a surface S but is not necessarily perpendicular to

J

the surface, to ¬nd the total current I through S you™ll have to determine

the component of the current density perpendicular to the surface. This

makes the relationship between I and ˜ J:

I ¼ ˜ ^ · °surface areaÞ ˜ uniform and at an angle to S: °4:10Þ

Jn J

And, if ˜ is nonuniform and not perpendicular to the surface, then

J

Z

I ¼ ˜ ^ da ˜ nonuniform and at a variable angle to S: °4:11Þ

Jn J

S

This expression explains why some texts refer to electric current as ˜˜the

¬‚ux 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.

107

The Ampere“Maxwell law

˜

e0 @ E The displacement current density

@t

The second source term for the magnetic ¬eld in the Ampere“Maxwell

law involves the rate of change of the electric ¬eld 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 de¬ned 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 ¬eld has retained that name over the

years. Furthermore, the displacement current density is a vector quantity

that has the same relationship to the magnetic ¬eld as does ˜ the con-

J,

duction current density.

The key concept here is that a changing electric ¬eld produces a

changing magnetic ¬eld even when no charges are present and no phys-

ical current ¬‚ows. Through this mechanism, electromagnetic waves may

propagate through even a perfect vacuum, as changing magnetic ¬elds

induce electric ¬elds, and changing electric ¬elds induce magnetic ¬elds.

The importance of the displacement current term, which arose initially

from Maxwell™s mechanical model, is dif¬cult to overstate. Adding a

changing electric ¬eld as a source of the magnetic ¬eld certainly extended

the scope of Ampere™s law to time-dependent ¬elds 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 ¬rst true ¬eld theory and the

foundation for much of twentieth century physics.

108 A student™s guide to Maxwell™s Equations

˜B ˜

r · ˜ ¼ l0 ˜ þ e0 @ E

J Applying the Ampere“Maxwell

@t

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 ¬eld 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 ¬eld, ¬nd the current density at a

speci¬ed location.

Problem: Use the expressions for the magnetic ¬eld in Table 2.1 to ¬nd

the current density both inside and outside a long, straight wire of radius

r0 carrying current I uniformly throughout its volume in the positive

z-direction.

Solution: From Table 2.1 and Example 4.1, the magnetic ¬eld inside a

long, straight wire is

˜ ¼ l0 Ir ™,

^

B 2

2pr0

where I is the current in the wire and r0 is the wire™s radius. In cylindrical

coordinates, the curl of ˜ is

B

1 @Bz @Bu 1 @°rBu Þ @Br

@Br @Bz

˜B

r ·˜ ^

^þ

À À uþ À ^:

r z

r @u @z @z @r @r @u

r

And, since ˜ has only a ™-component in this case,

^

B

0 1

@ r° l0 Ir=2pr0 Þ

2

˜ · ˜ À @B™ ^ þ 1 @°rB™ Þ ^ ¼ 1 @ A^

rB r z z

@z @r @r

r r

l0 I l0 I

1

¼ ^¼ ^:

2r z z

pr0

2 2

2pr0

r

Using the static version of the Ampere“Maxwell law (since the current is

steady), you can ¬nd ˜ from the curl of ˜

J B:

ÀÁ

˜B

r · ˜ ¼ l0 ˜ :

J

109

The Ampere“Maxwell law

Thus,

1 l0 I I

˜¼ ^ ¼ 2 ^;

J z z

l0 pr0 pr0

2

which is the current density within the wire. Taking the curl of the

expression for ˜ outside the wire, you™ll ¬nd that ˜ ¼ 0, as expected.

B J

Example 4.4: Given the magnetic ¬eld, ¬nd the displacement current

density.

Problem: The expression for the magnetic ¬eld of a circular parallel-plate

capacitor found in Example 4.2 is

l0 DV Àt=RC r

˜¼ ^

™:

B e 2

2pR r0

Use this result to ¬nd the displacement current density between the plates.

Solution: Once again you can use the curl of ˜ in cylindrical coordinates:

B

1 @Bz @B™ 1 @°rB™ Þ @Br

@Br @Bz

˜B

r ·˜ ^

^þ

À À uþ À ^:

r z

r @™ @z @z @r @r @™

r

And, once again ˜ has only a ™-component:

^

B

2 3

Àt=RC r

1 @ °rl0 DV=2pRÞe

@B™ 1 @°rB™ Þ 2

˜B ^¼ 4 5^

r ·˜ ¼ À

r0

^þ

r z z

@z @r @r

r r

! !

l0 DV Àt=RC 1 l0 DV Àt=RC 1

1

¼ 2r ^¼ ^:

e z e z

pR

2 2

2pR

r r0 r0

Since there is no conduction current between the plates, ˜ ¼ 0 in this case

J

and the Ampere“Maxwell law is

!

@˜

E

˜B

r · ˜ ¼ l0 e0 ;

@t

from which you can ¬nd the displacement current density,

! !

E ˜B

@˜ r · ˜ 1 l0 DV Àt=RC 1 DV Àt=RC 1

e0 ¼ ¼ ^¼ ^:

e z e z

@t l0 l0 pR pr0

2 2

R

r0

Problems

The following problems will test your understanding of the Ampere“

Maxwell law. Full solutions are available on the book™s website.

110 A student™s guide to Maxwell™s Equations

4.1 Two parallel wires carry currents I1 and 2I1 in opposite directions.

Use Ampere™s law to ¬nd the magnetic ¬eld at a point midway

between the wires.

4.2 Find the magnetic ¬eld inside a solenoid (Hint: use the Amperian

loop shown in the ¬gure, and use the fact that the ¬eld is parallel to

the axis of the solenoid and negligible outside).

4.3 Use the Amperian loop shown in the ¬gure to ¬nd the magnetic

¬eld within a torus.

4.4 The coaxial cable shown in the ¬gure carries current I1 in the

direction shown on the inner conductor and current I2 in the

opposite direction on the outer conductor. Find the magnetic ¬eld

in the space between the conductors as well as outside the cable if

the magnitudes of I1 and I2 are equal.

111

The Ampere“Maxwell law

4.5 Find the displacement current produced between the plates of a

discharging capacitor for which the charge varies as

Q°tÞ ¼ Q0 eÀt=RC ,

where Q0 is the initial charge, C is the capacitance of the capacitor,

and R is the resistance of the circuit through which the capacitor is

discharging.

A magnetic ¬eld of ˜ ¼ a sin°byÞebx^ is produced by an electric

4.6 B z

current. What is the density of that current?

4.7 Find the electric current density that produces a magnetic ¬eld

given by ˜ ¼ B0 °eÀ2r sin ™Þ^ in cylindrical coordinates.

B z

4.8 What density of current would produce a magnetic ¬eld given by

˜ ¼ °a=r þ b=reÀr þ ceÀr Þ™ in cylindrical coordinates?

^

B

4.9 In this chapter, you learned that the magnetic ¬eld of a long,

straight wire, given by

˜ ¼ l0 I ™,

^

B

2pr

has zero curl everywhere except at the wire itself. Prove that this

would not be true for a ¬eld that decreases as 1/r2 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

¬eld as a function of time that would produce a magnetic ¬eld

given by

˜ ¼ rx DV cos°xtÞ ™,

^

B

2d°c2 Þ

where r is the distance from the center of the capacitor, x is the

angular frequency of the applied voltage DV, d is the plate spacing,

and c is the speed of light.

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 ¬eld 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 ¬rst

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 ¬elds:

I Divergence

˜E

˜ ^ da ¼ qin =e0 r ˜ ¼ q=e0 :

En

theorem

S

Gauss™s law for magnetic ¬elds:

I Divergence

˜B

˜ ^ da ¼ 0 r ˜ ¼ 0:

Bn

theorem

S

Faraday™s law:

I Z

q˜

Stokes™

d B

˜ d˜ ¼ À ˜E

˜ ^ da r·˜¼À :

El Bn

qt

dt S theorem

C

112

113

From Maxwell™s Equations to the wave equation

Ampere“Maxwell law:

!

I Z ˜

˜ þ e 0 qE :

Stokes™

d

˜ d˜ ¼ l0 Ienc þ e0 ˜B

˜ ^ da r · ˜ ¼ l0

Bl En J

qt

dt S theorem

C

Along with the divergence theorem and Stokes™ theorem, you™ll also ¬nd

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 ¬elds:

The electric permittivity

of free space

The vector

The Laplacian The vector electric

electric field

operator field

2

*E

E= 2

00

*t 2

The second derivative The magnetic The second derivative

of the vector electric permeability of the vector electric

field over space of free space field with time

The electric

permittivity

The Laplacian The vector The vector

of free space

operator magnetic field magnetic field

2

*B

B= 2

00

*t 2

The second derivative The magnetic The second derivative

of the vector magnetic permeability of the vector magnetic

field over space of free space field with time

114 A student™s guide to Maxwell™s Equations

H R

˜ ˜ dV

˜n

S A ^ da ¼ V °r AÞ The divergence theorem

The divergence theorem is a vector“calculus relation that equates the ¬‚ux

of a vector ¬eld to the volume integral of the ¬eld™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 ¬nd 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 ¬‚ux of a vector ¬eld through a closed surface S is equal to the

integral of the divergence of that ¬eld over a volume V for which S is

a boundary.

This theorem applies to vector ¬elds 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 de¬ned as the ¬‚ux through a small

surface surrounding that point divided by the volume enclosed by that

surface as it shrinks to zero. Now consider the ¬‚ux through the cubical

cells within the volume V shown 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) ¬‚ux from one

Boundary

surface S

Flux through faces at

boundary

V

Positive flux from an interior

cell is negative flux for an

adjacent cell with shared face

Figure 5.1 Cubical cells within volume V bounded by surface S.

115

From Maxwell™s Equations to the wave equation

cell is identical in amplitude and opposite in sign to the negative (inward)

¬‚ux 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 S of volume V contribute to the ¬‚ux through S.

This means that adding the ¬‚ux through all the faces of all the cells

throughout volume V leaves only the ¬‚ux through the bounding surface S.

Moreover, in the limit of in¬nitesimally small cells, the de¬nition of

divergence tells you that the divergence of the vector ¬eld at any point is

the outward ¬‚ux from that point. So, adding the ¬‚ux of each cell is the

same as integrating the divergence over the entire volume. Thus,

I Z

˜ AÞ

˜ ^ da ¼ °r ˜ dV: °5:1Þ

An

S V

This is the divergence theorem “ the integral of the divergence of a vector

¬eld over V is identical to the ¬‚ux 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 ¬elds, the integral form of Gauss™s

law is

I

˜ ^ da ¼ qenc =e0 :

En

S

Or, since the enclosed charge is the volume integral of the charge

density q,

I Z

1

˜ ^ da ¼ q dV:

En

e0 V

S

Now, apply the divergence theorem to the left side of Gauss™s law,

I Z Z Z

q

˜ ˜ dV ¼ 1

˜ ^ da ¼ rE q dV ¼

En dV:

e0 V V e0

S V

Since this equality must hold for all volumes, the integrands must be

equal. Thus,

˜E

r ˜ ¼ q=e0 ;

which is the differential form of Gauss™s law for electric ¬elds. The same

approach applied to the integral form of Gauss™s law for magnetic ¬elds

leads to

˜B

r˜ ¼ 0

as you might expect.

116 A student™s guide to Maxwell™s Equations

H R

˜˜ ˜ AÞ n

°r · ˜ ^ da

C A dl ¼ Stokes™ theorem

S

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 ¬eld over a closed path C is equal to the

integral of the normal component of the curl of that ¬eld over a

surface S for which C is a boundary.

This theorem applies to vector ¬elds 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 de¬ned 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 S shown 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

Circulation at shared edges

of adjacent interior squares

Circulation along

is in opposite directions

edges at boundary

Boundary

path C

Figure 5.2 Squares on surface S bounded by path C.

117

From Maxwell™s Equations to the wave equation

circulation of the adjacent square over that same edge. It is only those edges

that lie along the boundary path C of surface S that 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 S leaves only the circulation around the bounding path C. In

addition, in the limit of in¬nitesimally small squares, the de¬nition 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 ¬eld over the

surface. So,

I Z

˜ d˜ ¼ ˜ AÞ n

°r · ˜ ^ da: °5:2Þ

Al

C S

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 S is 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 ¬eld around a path C to the change in magnetic ¬‚ux

through a surface S for which C is a boundary,

I Z

d

˜ d˜ ¼ À ˜ ^ da:

E l Bn

dt S

C

Applying Stokes™ theorem to the circulation on the left side gives

I Z

˜ d˜ ¼ ˜ EÞ n

°r · ˜ ^ da

El

C S

Thus, Faraday™s law becomes

Z Z

d

˜ EÞ n

°r · ˜ ^ da ¼ À ˜ ^ da:

Bn

dt S

S

For stationary geometries, the time derivative can be moved inside the

integral, so this is

!

Z Z ˜

@B

˜ EÞ n

°r · ˜ ^ da ¼ ^ da;

À n

@t

S S

where the partial derivative indicates that the magnetic ¬eld may change

over space as well as time. Since this equality must hold over all surfaces,

the integrands must be equal, giving

118 A student™s guide to Maxwell™s Equations

@˜

B

˜E

r·˜¼À ;

@t

which is the differential form of Faraday™s law, relating the curl of the

electric ¬eld at a point to the time rate of change of the magnetic ¬eld at

that point.

Stokes™ theorem may also be used to ¬nd the differential form of the

Ampere“Maxwell law. Recall that the integral form relates the circula-

tion of the magnetic ¬eld around a path C to the current enclosed by that

path and the time rate of change of electric ¬‚ux through a surface S

bound by path C:

I Z

d

˜ d˜ ¼ l0 Ienc þ e0 ˜ ^ da :

Bl En

dt S

C

Applying Stokes™ theorem to the circulation gives

I Z

˜ d˜ ¼ ˜ BÞ n

°r · ˜ ^ da;

Bl

C S

which makes the Ampere“Maxwell law

Z Z

d

˜ BÞ n

°r · ˜ ^ da ¼ l0 Ienc þ e0 ˜ ^ da :

En

dt S

S

The enclosed current may be written as the integral of the normal com-

ponent of the current density

Z

Ienc ¼ ˜ ^ da;

Jn

S

and the Ampere“Maxwell law becomes

!

Z Z Z ˜

@E

˜ BÞ n

°r · ˜ ^ da ¼ l0 ˜ ^ da þ ^ da :

e0

Jn n

@t

S S S

Once again, for this equality to hold over all surfaces, the integrands must

be equal, meaning

!

˜

@E

˜B

r · ˜ ¼ l0 ˜ þ e0 :

J

@t

This is the differential form of the Ampere“Maxwell law, relating the curl

of the magnetic ¬eld at a point to the current density and time rate of

change of the electric ¬eld at that point.

119

From Maxwell™s Equations to the wave equation

˜

r°Þ 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 ¬eld to ¬‚ow

away from a point and the curl indicates the circulation of a vector ¬eld

around a point, the gradient applies to scalar ¬elds. Unlike a vector ¬eld, a

scalar ¬eld is speci¬ed entirely by its magnitude at various locations: one