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electric charges, but apparently devoid of the magnetic equivalent.
Maxwell and his contemporaries did realize that Ampere™s law as ori-
ginally conceived applies only to steady electric currents, since it is con-
sistent with the principle of conservation of charge only under static
conditions. To better understand the relationship between magnetic ¬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 ·˜  ^

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 ·˜  ^

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

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

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

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