© Lawrence B. Rees 2007. You may make a single copy of this document for personal use without written permission.
10.0 Introduction
In Lesson 1 we learned how the thread model can explain the origin of the Coulomb force between stationary charged particles. In Lesson 2 we saw that the thread model could be used to find how the motion of a source charge affects its electric field. We also found that if the field charge is moving, we could account for the effects of its motion by including stubs in our model.
In these lessons, however, we always assumed that the charges moved with constant velocity. In this lesson we will consider what the thread model predicts when the source charge accelerates.
10.1 The Fields of a Charge Accelerating from Rest
In Lesson 2 we learned that when a source particle moves at constant velocity, a thread’s head and tail are emitted in the same direction so that the head line and tail line are parallel to each other. This is shown in Fig. 10.1. Because of this, as each thread travels off at the speed of light, its length and direction both remain constant. But what happens when the source particle accelerates? y head r r h
line
P
P tail line r r t
R ray line r r r r l
2
S T r h
r s
U
x
Figure 10.1. A thread emitted by a uniformly moving source particle.
To answer this question, first let’s consider the heads emitted by a uniformly moving source. Equivalently, we can ask how the observers moving at different velocities view a single thread emitted at a particular angle. Consider a source charge at rest emitting a head at an angle of
θ
0
=
45
°
. If an observer moves at a speed in the –x direction, she sees the source moving at a speed in the +x direction. To this observer, the head goes off at an angle
θ <
45
°
. Recall that in Section 2.2, we thought of a thread emitted in a moving train car. Non-relativistically,
1

observers on the ground saw that the y component of the velocity remained unchanged, but the x component increased because of the train’s motion. This meant the angle of the head line is smaller to an observer on the ground than to an observer inside the train car. Relativity modified this conclusion somewhat, but the results were qualitatively the same. The equations that related the angles in the ground and train systems were:
(10.1) tan tan
θ
θ =
0
=
γ
γ sin
(cos
θ
θ
0 sin
(cos
θ
θ
−
0
+ β
β
)
) where
β
is the ratio of the speed of the source charge to the speed of light.
γ =
1 / 1
− β
2
.
Figure 10.2 shows head lines for threads emitted at 45° when the source moves in the +x direction at a range of speeds. y
β =
0
β =
0 .
2
β =
0 .
4
β =
0 .
6
β =
0 .
8
θ
0
θ x
Figure 10.2. Head lines for threads emitted from a source moving at various speeds in the +x direction.
Now let’s return to a thread emitted by an accelerating charge. If an observer travels on the source charge, he sees the head and tail both emitted at the same angle,
θ
0
. In the lab frame, however, the head and tail are emitted at two different angles,
θ h and
θ t
, respectively. One more thing we need to consider is the motion of the source during the time the thread is emitted. We took the source to be initially at rest. If we denote by
∆ t the time it took to emit the thread, the source traveled a distance
∆ x
=
1
2 a )
2 where a is the acceleration of the source. Since the thread is very small, we also know that
∆ t is very small and that ( t )
2
is much, much smaller
2

than that. Hence, we can conclude that
∆ x quadratic in infinitesimal quantities.
≈
0 . In more mathematical words, we ignore terms
This all boils down to the situation illustrated in Fig. 10.3. The head and tail lines both come from the source when it’s located at the origin. The tail leaves a little later than the head, but after that both travel at the speed of light. In the figure, the thread is illustrated at several different times. We can conclude two important things from this figure:
1) The length of the thread increases as it moves away from the source. As long as the thread is not very near to the source, l
∝ r h
.
2) As the thread moves away from the source, its direction becomes perpendicular to the head line.
Note that since the thread is very short, the head and tail lines are nearly parallel, even though it’s hard to draw them that way in the figure. This means that the thread becomes perpendicular to the tail line as well as to the head line. This also suggests that the thread becomes the arc of the very small angle
∆ θ = θ h
− θ t
. The length of the thread is just the arc length, l
= r h
∆ θ
, as shown in Fig. 10.4. Hence we correctly concluded that l
∝ r h
as we noted above. r l y head line tail line
t
θ h x
Figure 10.3. Threads emitted by a source accelerating in the +x direction.
3

l
= r h
∆
y r r h
t
θ h x
Figure 10.4. A thread of Fig. 10.3 far from the source.
As we learned in Lesson 2, the thread of a stationary source has no stub associated with it. However, the thread of a source accelerating from rest does have a stub. The same relationship we found earlier applies here as well:
(10.2 General stub formula) s r
= r ˆ h
× r l .
The direction of the stub is determined by the cross product. In Fig. 10.4, the stub points out of the screen. For threads that are not too near the source, the thread is perpendicular to the head line as we discussed above. Since the vectors of Eq. (10.2) are perpendicular and rˆ is a unit h vector, this means that length of the stub is:
(10.3 stub of a charge accelerating from rest) s
= l .
Now that we have found the threads and stubs, we can use Eqs. (2.11) and (2.12) to find the electric and magnetic fields. These are: r
E
= e
ε
0
ν r r l and B
=
1 c e
ε
0
ν s r
.
Before we continue, we need to consider the density of states. A particle, either at rest or moving at constant velocity, emits some threads during the time interval t
0
to t
0
+ ∆ t . Later some of the threads emitted in this interval arrive at a field point P. The density of threads near the field point is just the number of threads per unit volume located in the vicinity of P. If the particle is accelerating, but has the same velocity at t as in the previous case, the same number of heads is
0 emitted as before and the heads are emitted in the same directions as before. All the acceleration does is change the position of the tails with respect to the heads. The conclusion is that the
4 r r t
∆ θ

acceleration doesn’t change the density of threads; the density only depends on the velocity of the source and the location of P. r r h
Now, we can put these pieces together. Since the source is initially at rest, we can call let
≡ r r
without confusion. First, for a source that remains at rest Coulomb’s Law tells us:
E
= e
ε
0
ν l
0
=
1
4
πε
0 q r
2
.
Since the thread length is a constant, we see that the density of threads is proportional to conclusion we reached in Lesson 1. This means that for the electric field of the accelerating
2
, a charge:
E
= e
ε
0
ν l
= e
ε
0
ν r
∆ θ ∝
1
.
r
Furthermore:
B
=
1 c
ε e
0
ν s
=
1 c
ε e
0
ν l
=
1
E c
∝
1
.
r
We’ll give equations for the electric and magnetic fields of accelerating point charges later in the lesson; however, these equations give us some very important facts for fields of accelerating charges at field points far from the source:
1) The electric and magnetic fields of accelerating charges fall off as 1/r rather than 1/r
2
.
2) The magnetic field of accelerating charges is related to the electric field as B=E/c in SI units.
3) The electric and magnetic fields are both perpendicular to the radial direction. Note that the directions of the fields are related to the radial direction by
× = r ˆ .
Things to remember:
• The larger the velocity of a source particle, the smaller the angle of a head line or tail line becomes.
• When a source particle accelerates, the head line and the tail line go off at different directions.
• For accelerating charges, the length of threads is proportional to r.
• The electric and magnetic fields of accelerating charges fall of as 1 / r rather than
2
.
• As a thread moves farther away from the source, its direction becomes perpendicular to the head line.
• The stub is perpendicular to both the head line and the thread.
• The directions are related by E
ˆ ×
B
= r ˆ .
5

10.2 The Fields of Accelerating Point Charges
In Section 10.1, we considered the fields of point charges accelerating from rest. When the source particle is not initially at rest, the situation becomes more complicated, as you might expect. However, it turns out that we can separate the electric field (or the magnetic field) into two parts: one that results from constant velocity motion and one that results from acceleration.
For the acceleration part of the field, the conclusions of Section 10.1 still hold.
First, let’s make a drawing similar to Fig. 10.1 for an accelerating point charge. At point
S a thread is emitted. A short time later the tail is emitted from point P. The head line and tail line are no longer parallel, due to the acceleration of the source. Note that the velocity and acceleration of the source at the time the thread is emitted affect the threads that arrive at point P; however, the subsequent motion of the source can in no way affect the threads already emitted. It turns out that it is still convenient to define a point U to be the point where the source would have been when the thread arrived at point P if the source had continued moving at constant
velocity from point S. y
S T tail line head line r r t r h
β r s
2 r r h
U ray
P line
R r r r
P thread r l x
Figure 10.5. A thread emitted by an accelerating soure.
The first question we wish to ask is where point T is located. We know the distance an accelerating particle moves is:
∆ x
= v
0
∆ t
+
1
2 a (
∆ t )
2
.
If we again ignore the term that is quadratic in the infinitesimal quantity
∆ t
∆ x
≈ v
0
∆ t ,
, this becomes:
6

the same result that we had without any acceleration.
What this means is that we can think of the thread as the sum of two other threads. The head of the first thread is the head of the actual thread and its tail is the tail that would have been emitted if the source had moved from S to T at constant velocity. The head of the second thread is the same as the tail of the first thread, and its tail is the actual tail of the actual thread. Thus the head and tail of the second thread are emitted simultaneously, but at different angles. The first thread is called the velocity thread, the second head is called the acceleration thread, and the true or total thread is the vector sum of the velocity and acceleration threads. This process is depicted in Fig. 10.6. y
P
P total thread head line r r h original tail line r a
∆ θ velocity thread acceleration thread tail line ray line r r r r r t x
S T U
Figure 10.6. The velocity thread, the acceleration thread, and the total thread.
Qualitatively, we can say that the acceleration “pulls” the tail of a thread to one side or the other, as shown in Fig. 10.7. By making use of this mnemonic device, you can tell if the angle between the head line and the tail line is positive or negative, and in turn deduce the direction of the electric field far from the source.
7

(a) (b) (c) (d)
Figure 10.7. How acceleration affects the direction of threads. The long, black arrows indicate the head and tail directions. The thick red arrow is the acceleration direction. (a) There is no acceleration, so the head and tail directions are the same. (b) The acceleration pulls the tail to the left. (The thread with no acceleration is shown in light green for comparison.) (c) The acceleration pulls the tail to the right. (d) The acceleration is in the direction of the thread and leaves the tail direction unchanged.
The formulas of Lesson 2 still apply for the velocity threads and the associated fields.
From the acceleration threads, we can obtain formulas for the field caused by the acceleration.
The mathematical details are somewhat involved, so I have put them in Appendix C rather than in the text. The bottom line is that the formulas for the velocity and acceleration fields are as follows:
General Equations for Fields
(10.4) r
E r
B
=
= r
E v
1 r ˆ h c
+ r
E a
× r
E
= q
4
πε s
0
γ s
2 r r r
ρ
3
+ q
4
πε s
0 r r h
× c
( r r r
2
ρ
×
3 r a s
) where
c is the speed of light. r r r h
is the head line, the vector from point S to point P.
β s
is the velocity of the source at point S (divided by the speed of light).
γ r a s
=
1 / 1
− β s
2
.
r r r s
is the acceleration of the source. r
The vector from S to U is r h
β s
is the ray line, the line from
.
U to P. r r r
= r r h
−
ψ
is the angle the ray line makes with respect to r h r
β
β s r s
.
. (See Fig. 10.5.)
ρ = r r
1
− β s
2 sin
2
ψ
8

Things to remember:
• We can think of the acceleration as “pulling the tail” of the thread in the direction of the acceleration.
• Electric and magnetic fields can be considered as the sum of “velocity” fields and
“acceleration” fields.
10.3 The Acceleration Fields for Slow Source Charges
Equation (10.4) is the most general equation for the fields of a point charge, but we often
In this case, we call the head line r for simplicity and note that:
β r
γ r r r s
ρ s
≈
≈
0
≈
≈ r r
1 r r
B a
≈
1 r ˆ
× c r
E
=
1 c r ˆ
× r
E a r
E v
= q
4
πε s
0 r ˆ
(10.5) r
E a r
B a
=
= q
4
πε s
0 q s
4
πε
0 r
2 r ˆ
×
( r ˆ
× r a s
) r ˆ
×
[ c r ˆ
2 r
×
( r ˆ
× r a s
)
] c
3 r
Note that the electric velocity field r
E v is just the Coulomb’s law field. When r is sufficiently large, the velocity field becomes much smaller than the acceleration field. We’ll generally use these results qualitatively, so it most important to recognize that:
(10.6) r
E r a
B a
∝
∝ r r ˆ
×
ˆ
×
( r r ˆ
E a
× r a s
)
Things to remember:
• The electric field is in the direction r ˆ
×
( r ˆ
×
)
.
• The magnetic field is in the direction r ˆ
×
E .
• The magnitude of the electric and magnetic fields are related by: E a
= cB a
(in SI units).
9

10.4 Visualizing the Fields of Accelerating Charges
The physics of this lesson is all contained in Eq. (10.5), but this is too complicated an expression to really get any intuitive idea about acceleration fields from the equation itself. This becomes is particularly apparent when we remember that to apply Eq. (10.5) at a particular point and time, we have to figure out where the source was located when the threads arriving at P were emitted, and what the velocity and acceleration of the source were at that time.
First, let’s consider the threads emitted from a source moving at constant speed. We saw earlier in the lesson that the angle between the head lines and the direction of motion gets smaller as the velocity of the source gets larger. On the other hand, we saw in Lesson 2 that the angle between the ray lines and the direction of motion approaches 90° as the velocity of the source gets larger. Nonrelativistically this angle, , would be the same as
θ
0
; however, the contraction of distance and the slowing of clocks combine to bunch up the threads in the direction perpendicular to the direction of motion. This can be seen in Fig. 10.8, where threads for three different source velocities are compared. Note that we have aligned the threads into field lines for convenience in this and several of the following figures.
(a)
β =
0 .
3
(b)
β =
0 .
7
10

=
0 .
9 (c)
β
Figure 10.8. Aligned threads emitted from a source moving to the right at three different speeds. of
β =
Let’s now consider a particle that is initially at rest and then accelerates to a final speed
0 .
7 over a short period of time. The threads emitted when the particle is at rest are uniformly distributed in space and continue moving radially outward from the initial position of the source at the speed of light, as seen in Fig. 10.9 (a). After the source moves at
β =
0 .
7 , the threads are like those of Fig. 10.9 (b). These threads still travel at the speed of light from the points where they are emitted, so they are located within a sphere that fits inside the hollow region of Fig. 10.9 (a). The threads that are emitted during the acceleration simply join these two sets of threads, as seen in Fig. 10.10.
(a)
11

(b)
Figure 10.9. Aligned threads of (a) a source at rest and (b) a source moving at =0.7.
(c)
Figure 10.10. Aligned threads of a source accelerating from rest to =0.7.
Since the aligned threads are the field lines, we can easily see that the acceleration threads in Fig. 10.10 have the characteristics we described in the previous section: they are close
12

together, showing that the field is stronger in this region than in regions to either side, and their direction is perpendicular to the radial direction.
One interesting question to ask is what would happen if we treated the thread nonrelativistically. This result is shown in Fig. 10.11. Note that the velocity of the threads is larger than the speed of light in front of the source and smaller than the speed of light behind the source. Also, the field lines don’t tend to bunch in the plane perpendicular to the velocity of the source. The net result is something rather bizarre. In fact, it’s physically meaningless since the field lines cross one another in the forward direction.
Figure 10.11. The field lines for a source moving at =0.7, ignoring relativity.
Hopefully by now you’re beginning to get a feeling of how acceleration disrupts the regular field lines of a source charge, creating bumps and wiggles in the electric field. Now see if you can make sense of the fields shown in Figs. 10.12-10.14.
13

Fig. 10.12. A source is initially at rest, then experience a short force to the right, it moves with a constant speed to the right for a time, then later experiences a short force to the left, bringing it back to rest.
Fig. 10.13. A source is initially at rest, then experience a short force to the right, and immediately afterward experiences a short force to the left, bringing it back to rest.
14

Figure 10.14. A source charge oscillates along the x axis.
Think About It
If you were standing at the red point in Fig. 10.14, describe how the electric field would change as time went on. What would the magnetic field do? (Remember that the magnetic field r is in the r ˆ
×
E direction.)
Notice how the electromagnetic fields of oscillating charges form waves. These waves travel at the speed of light. They are termed “electromagnetic waves” or “electromagnetic radiation.” It probably comes as little surprise to you that radio waves, light, and even x rays and gamma rays are forms of electromagnetic radiation. We will study electromagnetic radiation in more detail in Lesson 13.
Since threads and field lines are not stationary objects, it is easier to understand the figures of this section by viewing animations. The following table lists AVI animations that are available online. The Windows Media Player can play the AVI files, however, you may have to save the file to your computer to get past the ‘Windows Media Player cannot play the file” error message.
15

Content Link
Asource initially at rest accelerates to the right briefly until = 0.3.
A source initially at rest accelerates to the right briefly until = 0.7. Fig. 10.10
A source initially at rest accelerates to the right briefly until = 0.9.
A source initially at rest accelerates to the right to = 0.5, remains at that speed for a time, then accelerates to the left until its speed is zero. Fig. 10.12.
A source initially at rest accelerates to the right to = 0.7, remains at that speed for a time, then accelerates to the left until its speed is zero.
A source initially at rest accelerates to the right to = 0.5, then quickly accelerates to the left until its speed is http://www.physics.byu.edu/faculty/rees/AVIfiles/right30.avi
http://www.physics.byu.edu/faculty/rees/AVIfiles/right70.avi
http://www.physics.byu.edu/faculty/rees/AVIfiles/right90.avi
http://www.physics.byu.edu/faculty/rees/AVIfiles/square50.avi
http://www.physics.byu.edu/faculty/rees/AVIfiles/square70.avi
http://www.physics.byu.edu/faculty/rees/AVIfiles/pulse50.avi
zero. Fig. 10.13.
A source initially at rest accelerates to the right to = 0.7, then quickly accelerates to the left until its speed is zero. http://www.physics.byu.edu/faculty/rees/AVIfiles/pulse70.avi
A source oscillates along the x axis with a maximum speed of = 0.5. Fig. 10.14. http://www.physics.byu.edu/faculty/rees/AVIfiles/oscillate50.avi
A source oscillates along the x axis with a maximum speed of = 0.7. Fig. 10.14. http://www.physics.byu.edu/faculty/rees/AVIfiles/oscillate70.avi
Things to remember:
• Review the online animations and qualitatively understand these animations.
• Remember that the faster a particle goes, the smaller the angle of the head line becomes, but the larger the angle of the ray line becomes.
• Oscillating charges lead to oscillating fields and electromagnetic waves.
• The wave travels in the rˆ direction.
• The electric and magnetic fields are still perpendicular to rˆ , even though they oscillate.
10.5 Accelerating Charges in a Wire
To this point, we have mostly thought about the acceleration of individual charges. In
Sections 10.5 and 10.6, we will consider some things that happen when charges accelerate in electrical circuits. It’s relatively easy to accelerate charges in circuits because whenever the current in a circuit changes, charges accelerate. We know that accelerating charges produce radiation, so one consequence of changing currents is radiation. It turns out that when current changes rapidly, radiation becomes very important. However, even when current is changing
16

fairly slowly, a phenomenon called “retardation” affects the fields. Since we’ve been studying radiation, let’s consider it first.
We think of a circuit as a collection of positive charge carriers all moving through a wire at the drift velocity. We know, of course, that electrons carry charge and they dart around quite rapidly. Fortunately, the electric and magnetic fields that are caused by the random motion of electrons in circuits all cancel out because there are billions of randomly moving electrons everywhere in the circuit. So let’s take a wire with a current traveling to the right as shown in
Fig. 10.15:
L r a i
Figure 10.15. Current increasing in a wire.
We will assume that the current is increasing in time. For convenience, we draw a box of length
L around the wire. Let’s let be the linear charge density of the current-carrying charges in the wire. This means that the total charge on charge carriers in the box is:
∆ q
= λ
L .
Let’s say that the charges on the left side of the box take a time
∆ t box. The drift velocity of the charge carriers in the wire is to get to the right side of the v
=
L
∆ t
.
We also know that all the charge carriers initially in the box pass through the right face of the box during
∆ t . Since the current is the charge that passes a point in the circuit per unit time, we can find an expression for the current in terms of the drift velocity:
(10.7) i
=
∆ q
∆ t
=
λ
L
L / v
= λ v .
Differentiating both sides of Eq. (10.7) with respect to time gives:
17

di a
=
= λ dv
= λ a
(10.8) dt
1
λ dt di dt
We won’t do anything quantitative with this expression, but will make use of the following result:
Direction of Acceleration in Circuits
If the current is increasing, the acceleration of positive charge carriers is in the direction of the current.
If the current is decreasing, the acceleration of positive charge carriers is opposite the direction of the current.
To find the fields at a point P near the wire of Fig. 10.15, we really need to consider how every point along the wire contributes to the fields at P at a given time. This requires integration along the wire, much we did in Lessons 1 and 8. However, since we are only interested in qualitative results, we’ll take a shortcut that works in many applications. We will simply take a point of the wire nearest the field point P and calculate the direction of the fields produced by this point. We know the electric acceleration field is in the direction of
×
( )
where Rˆ is a unit vector in the direction of the slice to point Pm as shown in Fig.10.16. Note that this gives us r an electric field pointing opposite the acceleration direction. The magnetic field is in the
×
E , and so it points out of the page. r
E P r
R i r a
Figure 10.16. The electric acceleration field of a wire.
Integration of the fields produced by each slice along the wire indeed gives a total electric field that points to the left.
18

Think About It
How does the strength of the acceleration field of Fig. 10.16 vary with distance from the wire?
Although the definitive answer is given by direct integration, we can deduce the correct behavior with a little thinking. For stationary charges, the field of a point charge falls of as 1/r
2
and the field of an infinitely long rod falls off as 1/r
1
. (This is easily obtained from Gauss’s law, just in case you forgot the result.) The radiation field of a point charge falls of as 1/r
1
, so it seems likely that the radiation field of an infinitely ling wire falls off as 1/r
0
. This is the correct result.
The second effect of acceleration comes from retardation. Consider again the case of current increasing in a wire. Electrons at every point along the wire continuously emit threads.
The threads that arrive at a field point P at a given time come from all along the wire, so these threads were emitted at many different times. That means that the electric field at point P is dependent on the current in the wire over a range of times in the past. The effects of past currents giving rise to present fields are called “retardation effects.” We usually word this in a more sophisticated, if less understandable, way:
Retardation Effects
Retardation effects are the effects on electric and magnetic fields caused by the finite propagation speed of electromagnetic fields.
If the current in a wire is constant, then all the threads are emitted by source charges moving with the same drift speed. In this case, the vector sum of all the threads from moving and stationary charges at any point P is exactly zero. This means that, as we discussed in Lesson 2, a current-carrying wire produces no electric field. However, if charges accelerate in a wire, the threads arriving at point P at any given time come from sources moving at many different speeds. The threads no longer sum to zero. When we integrate the velocity field along the wire, we find that, like radiation field from the wire, the electric field points opposite the direction of the acceleration. The electric velocity field of an infinitely long wire is proportional to the acceleration and drops off as 1/r (like the static electric field of a charged rod).
The total electric field of an infinitely long wire then looks something like the following:
19

i r
E r a
Figure 10.17. The electric field of a wire with accelerating current.
Let’s put a rectangular loop of wire in the electric field of the wire, as shown in Fig. 10.18. The electric field in the wire pushes charges to the left on the bottom and to the right on the top, but since the field on the bottom of the loop is larger, it wins out and current goes around the loop in a clockwise direction. An electric field or current produced in this way by moving charges is called an “induced field” or an “induced current.” i r
E i r a
Figure 10.18. Induced current flowing in a loop near a wire with increasing current.
Let’s now consider the loop of Fig. 10.18. to be an Amperian loop. We wish to evaluate the electric line integral around this loop,
Λ
E
= ∫ r
E
⋅ d r l .
Let’s go around the loop in a counterclockwise sense, as usual. The right and left sides contribute nothing to the line integral, as the electric field is perpendicular to the path. The top and bottom segments of the loop give us:
20

Λ
E
= +
E top l
−
E bottom l where is the length of the top and bottom wire segments. The important thing to note here is that the line integral is non-zero. Furthermore, as we let the size of the Amperian loop shrink to zero, this becomes: lim
∆ a
→
0
Λ
E
∆ a
= ∇ × r
E
≠
0 .
That is, the curl of the electric field at a point P is also non-zero. But in the terminology of
Maxwell’s equations, when a field has non-zero curl at a point, it means that there must be a source of looping electric fields at point P. There are no charges and no currents – in fact there is no matter of any kind – at point P. The only “thing” that is at P other than the electric field is the magnetic field. But we know that neither permanent magnets nor wires carrying a steady current produce looping electric fields. Therefore, magnetic fields themselves cannot be the source of looping electric fields. As it turns out, the “source” of looping electric fields is the change in the magnetic fields. The relation between these fields is the third of Maxwell’s Equations, Faraday’s law of induction:
Faraday’s Law of Induction, Differential Form
(10.9)
∇ × r
E
= − r
∂
B
∂ t
Of course the actual source of a looping electric field is the sum of all threads emitted by accelerating electrons along the wire. Faraday’s law simply expresses a relationship between threads and stubs that is valid either for individual point charges or for collections of charges. It’s meaning is simply that the curl of the electric field is equal in magnitude and opposite the direction of the rate the magnetic field changes at that point. Or less accurately, we often say,
“The source of looping electric fields is changing magnetic fields.”
There is a very useful theorem called “Stokes’s theorem” from vector calculus that we can apply to the differential form of Faraday’s law to give us the law in integral form:
Stokes’s Theorem
(10.10) ∫
(
∇ × r
F
⋅
) r d A
=
∫ r
F
⋅ r d l where r
F is any vector field
the line integral is taken over an Amperian loop and the surface integral is taken over the area bounded by the loop
As an example, let’s apply Stoke’s theorem to the differential form of Ampère’s law:
21

r
∫ B
⋅ r d l
Λ
B
=
∫
=
µ
0
∫ r j
⋅
× r d A
= r
⋅
µ d A
0 r
∫ j
⋅ r d A
= µ
0 i enc
Note how Stokes’s theorem gives us the integral form of Ampère’s law easily.
Now, let’s apply Stokes’s theorem to Faraday’s law. r
∫ E
⋅ r d l
Λ
E
= ∫
=
−
∫
∂
( ) r
∂
B
⋅
⋅ r d A t d A r
= −
∂
∂ t r
∫ B
⋅ r d A
= −
∂ Φ
∂ t
B
In words, this tells us that the electric line integral around an Amperian loop is proportional to the change in the number magnetic field lines passing through the loop. In fact, Faraday originally had the notion that when a wire cuts a magnetic field line, it “bleeds” electricity into a wire.
Faraday’s Law of Induction, Integral Form
(10.11)
Λ
E
= r
∫ E
⋅ r d l
= −
∂ Φ
∂ t
B
We’ll study Faraday’s law in much more detail in Lesson 11. For now, I just want you to understand that accelerating charges give rise to new kinds of fields, and that Faraday’s law of induction is a useful way to describe some of these new fields.
Things to remember:
• When current is increasing, charges accelerate in the direction of the current. When current is decreasing, charges accelerate opposite the direction of a current.
• Know the example of accelerating charges in a wire.
• If current increases or decreases in a wire, an induced electric field points opposite the acceleration.
• If a loop of wire is placed near a wire with changing current, induced current flows around the loop.
• This leads to Faraday’s Law of Induction:
∇ r
×
E
= −
∂
∂ r
B
, t
Λ
E
= r
E
⋅ d l r
= − d
Φ d t
B
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10.6 Displacement Current and the Magnetic Field in a Capacitor
A second example of the fields of accelerating charges is given by a charging capacitor.
Let’s consider a parallel-plate capacitor with circular plates. As the capacitor charges, the current flowing into the capacitor decreases until the capacitor is fully charged. This means that acceleration fields will be produced between the capacitor plates. i r a r
E r
R
P r
E a r a i
Figure 10.19. The acceleration fields of a charging capacitor.
Let’s look at Fig. 10.19 in detail. Consider the point P between the capacitor plates and two small regions of the capacitor plates just above and just below P. Current is flowing outward in the plate above point P, but the current is decreasing in time, so the charges in the plate are accelerating toward the center of the plate. In the lower plate, current is flowing to the left, but since it is also decreasing, the charges accelerate to the right. Since the electric field is in the direction
×
( )
, the electric field from the region above P is to the right. The field from the region below P is, however, to the left and both fields cancel out. The magnetic fields are given r by
×
E , so the magnetic fields from both regions are out of the page. By symmetry, this gives magnetic field lines that form closed loops much as the magnetic field lines around a wire.
Of course, we would have to do a direct integration over all charges moving in the wires and the capacitor plates to reliable generalize this result, but the answer turns out that there are circular field lines within the capacitor.
If we draw an Amperian loop along one of these filed lines in the middle of the capacitor, however, there is no current passing through the Amperian loop. That means that Ampère’s law must be modified. James Clerk Maxwell first noticed this about 1860 and added what we now know as “Maxwell’s term” to Ampère’s law. Using the same logic as before, we conclude that the electric field of the capacitor plates is the only “thing” between the plates that could “cause” the magnetic field. Again, electric field by itself does not produce magnetic fields; however, the change in the electric field could do this.
23

We find experimentally that if there are no real currents:
Faraday’s Term of Ampère’s Law
(10.12)
∇ × r
B
= + µ
0
ε
0 r
∂
E
∂ t
.
Note that this is much like the differential form of Ampère’s law except that
ε
0 r
∂
E
∂ t
replaces the current density j. For this reason this quantity is called the “displacement current density” (The term “displacement” comes from Maxwell’s mechanical model of the ether, and no longer has any physical significance.) Integrating over the entire area of the wire, we get the total
“displacement current”
(10.13 – Displacement current) i dis
= ε
0 ∂
∂ t r
∫ E
⋅ r d A
= ε
0 ∂
∂ t
EA
= ε
0
∂ Φ
∂ t
E .
By applying Stokes’s Equation, we can obtain Ampère’s law in integral form. Let’s write
Ampère’s complete law in both differential and integral forms:
Ampère’s law, complete
(10.14)
∇ r
×
B
Λ
B
=
=
µ
0
µ
0 r j i enc
+
+
µ
µ
0
ε
0
0
ε
0 r
∂
E
∂ t
∂ Φ
E
∂ t
= µ
0
( i enc
+ i dis
)
The easiest way to see how displacement current works is to apply it to the example of the capacitor we have above.
Example 10.1. The magnetic field of a charging capacitor.
A current i(t) flows into a parallel plate capacitor that has circular plates of radius R separated by a distance d. Find the magnetic field inside and outside the capacitor.
Ampère’s law gives us:
Λ
B
=
2
π rB i dis
=
µ
0
=
( i enc
µ
0 i dis
+
ε
0
∂ Φ
∂ t
E
= i dis
)
= µ
0 i dis
ε
0
A
∂
E
∂ t
24

In turn, we know that
E
=
ε
σ
0
=
π q ( t
R
2
ε
)
0
First, let’s find the field outside the capacitor, r<R. In this case the area that appears in the electric flux is the area of the entire capacitor plate, as the electric flux does not extend (much) beyond the edge of the capacitors.
This gives:
2
π rB
= µ
0
ε
0
A
∂
E
∂ t
= µ
0
ε
0
π
R
2
∂
∂ t
 q ( t )
π
R
2
ε
0

= µ
0
∂ q ( t )
∂ t
= µ
0 i ( t )
B
=
µ
0 i ( t )
2
π r
Note that this is just the same result we get for the magnetic field of the wire without the capacitor!
Inside the wire, r>R, we have
2
π rB
= µ
0
ε
0
A
∂
∂
E t
A
= π r
2
:
= µ
0
ε
0
π r
2
∂
∂ t
 q ( t )
π
R
2
ε
0

= µ
0 r
2
R
2
∂ q ( t )
∂ t
= µ
0 r
2
R
2 i ( t )
B
=
µ i ( t
2
0
π r
) r
2
R
2
Although you may not remember this result, this is the magnetic field inside a wire of uniform current density. – In other words, as far as the magnetic field is concerned, we can treat the capacitor just as we would a wire of radius R.
Things to remember:
• As current flows in and out of capacitor plates, charges accelerate.
• The electric acceleration fields from charges on the top and bottom plates cancel.
• The magnetic fields add to give looping magnetic fields.
• These magnetic fields are the same as if they were caused by an actual current passing between the capacitor plates. This is called “displacement current” and is given by the equation i dis
= µ
0
ε
0 d
Φ d t
E
. Equivalently, the displacement current density is given by r j dis
= µ
0
ε
0
∂
∂ t r
E
.
• This leads to Maxwell’s Term in Ampère’s Law. Maxwell’s Term is the second term in
∇ × r
B
= µ
0
( r j
+ r j dis
)
,
Λ
B
=
r
B
⋅ d r l
= µ
0
( i
+ i dis
)
25