ELECTROMAGNETIC
INDUCTION
INTRODUCTION
In chapter 13, we learned that electric currents generate magnetic fields, and we will now see how
magnetism can generate electric currents.
299
MOTIONAL EMF
The figure below shows a conducting wire of length f , moving with constant velocity v in the plane
of the page through a uniform magnetic field B that's perpendicular to the page. The magnetic field
exerts a force on the moving conduction electrons in the wire. With B pointing into the page, the
direction of v x B is upward, so the magnetic force, FB' on these electrons (which are negativelycharged) is downward.
x
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
B
v
(;
X
X
X
X
X
X
l
FB
As a result, electrons will be pushed to the lower end of the wire, which will leave an excess of
positive charge at its upper end. This separation of charge creates a uniform electric field, E, within
the wire (pointing downward).
x
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X E
X
X
X
X
B
b
(±J
X
X
X
X
X
X
-Y.
X
8
a
X
A charge q in the wire feels two forces: an electric force, FE
X
=qE, and a magnetic force,
FB = q(v x B)
If q is negative, FE is upward and FB is downward; if q is positive, FE is downward and FB is upward.
So, in both cases, the forces act in opposite directions. Once the magnitude of FE equals the magnitude
of FB, the charges in the wire are in electromagnetic equilibrium. This occurs when qE =qvB; that is,
when E =vB.
The presence of the electric field creates a potential difference between the ends of the rod. Since
negative charge accumulates at the lower end (which we'll call point a) and positive charge accumulates at the upper end (point b), point b is at a higher electric potential.
300 •
CRACKING THE AP PHYSICS EXAM
x
x
x
x b x
x
x
x
x
x
x
x
x
X
X
X
x
x
x
B
@
x
x
x
v
xE x
8
X
a
X
x
X
X
The potential difference Vba is equal to E J! and, since E = vB, the potential difference can be written
as vB J! .
Now imagine that the rod is sliding along a pair of conducting rails connected at the left by a
stationary bar. The sliding rod now completes a rectangular circuit, and the potential difference Vba
causes current to flow.
x
B
x
I
x
x
x b x
,.-
0(
x
x
x
x
x
x
x
x
x
-
X
X
X
X
x
x
v
X
X
X
X
a X
x
~
IX
x
~
The motion of the sliding rod through the magnetic field creates an electromotive force, called
motional emf:
£ = vB J!
The existence of a current in the sliding rod causes the magnetic field to exert a force on it. Using
the formula FB =I( £ x B), the fact that £ points upward (in the direction of the current) and B is into
the page, tells us that the direction of FB on the rod is to the left. An external agent must provide this
same amount of force to the right to maintain the rod's constant velocity and keep the current
flowing. The power that the external agent must supply is P = Fv = I J! Bv, and the electrical power
delivered to the circuit is P = IVba = IE = IvB J! . Notice that these two expressions are identical. The
energy provided by the external agent is transformed first into electrical energy and then thermal
energy as the conductors making up the circuit dissipate heat.
ElECTROMAGNETIC INDUCTION .
301
FARADAY'S LAW OF ELECTROMAGNETIC INDUCTION
Electromotive force can be created by the motion of a conductor through a magnetic field, but there's
another way to create an emf from a magnetic field.
Magnetic Flux
Think back to chapter 10 and electric flux. The electric flux through a surface of area A is equal to the
product of A and the electric field that's perpendicular to it. That is, <PE = E 1- A = E . A = EA cos e.
If E varies over the area A, then we write <P E = E· dA .
The idea of magnetic flux is exactly the same. The magnetic flux, <P B, through an area A is equal
to the product of A and the magnetic field perpendicular to it: CPB = B 1- A = B . A =BA cos e. Again,
,
if B varies over the area, then we must write <P B= B · dA.
f
f
Example 14.1 The figure below shows two views of a circular loop of
radius 3 cm placed within a uniform magnetic field, B (magnitude 0.2 T).
~~l~l
,,
,,
B
A
,,
,,
(a) What's the magnetic flux through the loop?
(b) What would be the magnetic flux through the loop if the loop were
rotated 45°?
B
A
302 •
CRACKING THE AP PHYSICS EXAM
(c) What would be the magnetic flux through the loop if the loop were
rotated 90°?
B
Solution.
(a) Since B is parallel to A, the magnetic flux is equal to BA:
<PB = BA = B . nr2 = (0.2 T) . n(0.03 m)2 = 5.7 x 10-4 T·m2
The SI unit for magnetic flux, the tesla·meter2, is called a weber (abbreviated Wb).
SO <PB = 5.7 X 10-4 Wb.
(b) Since the angle between B and A is 45°, the magnetic flux through the loop is
<PB = BA cos 45° = B·nr2 cos 45°= (0.2 T) . n(0.03 m)2 cos 45°= 4.0 x 10-4 Wb
(c) If the angle between B and A is 90°, the magnetic flux through the loop is zero,
since cos 90° = O.
The concept of magnetic flux is crucial, because changes in magnetic flux induce emf. According
to Faraday'S Law of Electromagnetic Induction, the magnitude of the emf induced in a circuit is
equal to the rate of change of the magnetic flux through the circuit. This can be written mathematically in the form
£
1
or, if we let fit
-7
I
avg
=
~<I>B
- ~t
0, we get
This induced emf can produce a current, which will then create its own magnetic field. The direction
of the induced current is determined by the polarity of the induced emf and is given by Lenz's Law:
The induced current will always flow in the direction that opposes the change in magnetic flux that
produced it. If this were not s~), then the magnetic flux created by the induced current would magnify
the change that produced it, and energy would not be conserved. Lenz's Law can be included
mathematically with Faraday's Law by the introduction of a minus sign; this leads to a single
equation that expresses both results:
= _ ~<I>B
£
avg
~t
or
£ = _ d<I>B
dt
ElECTROMAGNETIC INDUCTION .
303
Example 14.2 The circular loop of Example 14.1 rotates at a constant
angular speed through 45° in 0.5 s.
H~~
,
,,
B
,,
,
(a) What's the induced emf in the loop?
(b) In which direction will current be induced to flow?
Solution.
(a) As we found in Example 14.1, the magnetic flux through the loop changes when the
loop rotates. Using the values we determined earlier, Faraday's Law gives
E
= _
avg
~<I>B
~t
= _ (4.0 X 10--4 Wb) - (5.7 X 10--4 Wb) = 3.4 X 10--4 V
0.5 s
(b) The original magnetic flux was 5.7 x 10-4 Wb upward, and was decreased to
4.0 x 10-4 Wb. So the change in magnetic flux is -1.7 X 10-4 Wb upward, or, equivalently, ~ <PB = 1.7 X 10-4 Wb, downward. To oppose this change we would need to
create some magnetic flux upward. The current would be induced in the counterclockwise direction (looking down on the loop), because the right-hand rule tells us
that then the current would produce a magnetic field that would point up.
B
Current will flow only while the loop rotates, because emf is induced only when
magnetic flux is changing. If the loop rotates 45° and then stops, the current will
disappear.
304 •
CRACKING THE AP PHYSICS EXAM
Example 14.3 Again consider the conducting rod that's moving with
constant velocity v along a pair of parallel conducting rails (separated by
a distance f), within a uniform magnetic field, B:
x
x
x
x
x
x
...,
x
X
X
X
X
X
,,
,
B
X
X
X
X
X
X
X
X
~
X
X
X
,,
't
X
X
X
X
X
X
Find the induced emf and the direction of the induced current in the
rectangular circuit.
Solution. The area of the rectangular loop is f x, where x is the distance from the left-hand bar to the
moving rod:
X
X
X
X
X
X
...,
X ,,
X
X
X
X
X
X
X
X
~
X
X
X
X
X
X
X
X
X
X
X
,
X£
,
,,,
X ,,
'I(
X
B
-------- x -- ---- --~
Because the area is changing, the magnetic flux tl\rough the loop is changing, which means ~hat an
emf will be induced in the loop. To calculate the induced emf, we first write <PB= BA = B £ x, then
since I1x/ I1t = v, we get
£
= _ /1«PB = _ /1(B£x) = _ B£ I1x = _ B£v
M
M
/1t
We can figure out the direction of the induced current from Lenz's Law. As the rod slides to the
right, the magnetic flux into the page increases. How do we oppose an increasing into-the-page flux?
By producing out-of-the-page flux. In order for the induced current to generate a magnetic field that
points out of the plane of the page, the current must be directed counterclockwise (according to the
right-hand rule).
avg
ELECTROMAGNETIC INDUCTION .
30S
x
x I x
x
x
x
...,
X ,
,,
,,
X
X
X
X
X
x
x ~
xc,
X
x
--
B
x
,,
,,
,
X
X
X
~ ------- -x -------- ->-
x I
•
x
x
x
X
X
x
x
Notice that the magnitude of the induced emf and the direction of the current agree with the results
we derived earlier, in the section on motional emf.
This example also shows how a violation of Lenz's Law would lead directly to a violation of the
Law of Conservation of Energy. The current in the sliding rod is directed upward, as given by Lenz's
Law, so the conduction electrons are drifting downward. The force on these drifting electrons- and
thus the rod itself-is directed to the left, opposing the force that's pulling the rod to the right. If the
current were directed downward, in violation of Lenz's Law, then the magnetic force on the rod
would be to the right, causing the rod to accelerate to the right with ever-increasing speed and kinetic
energy, without the input of an equal amount of energy from an external agent.
Example 14.4 A permanent magnet creates a magnetic field in the
surrounding space. The end of the magnet at which the field lines emerge
south pole (5):
is designated the north pole (N), and the other end is
the
B
306 • CRACKING THE AP PHYSICS EXAM
(a) The figure below shows a bar magnet moving down, through a
circular loop of wire. What will be the direction of the induced
current in the wire?
(b) What will be the direction of the induced current in the wire if the
magnet is moved as shown in the following diagram?
Solution.
/
(a) The magnetic flux down, through the loop, increases as the magnet is moved. By
Lenz's Law, the induced emf will generate a current that opposes this change. How
do we oppose a change of more flux downward? By creating flux upward. So,
according to the right-hand rule, the induced current must flow counterclockwise
(because this current will generate an upward-pointing magnetic field):
ElECTROMAGNETIC INDUCTION .
307
------------------------------------ - - - - -- - - - - - - - - - - - - - - - -
(b) In this case, the magnetic flux through the loop is upward and, as the south pole
moves closer to the loop, the magnetic field strength increases so the magnetic flux
through the loop increases upward. How do we oppose a change of more flux
upward? By creating flux downward. Therefore, in accordance with the right-hand
rule, the induced current will flow clockwise (because this current will generate a
downward-pointing magnetic field):
I
Example 14.5 A square loop of wire 2 em on each side contains 5 tight
turns and has a total resistance of 0.0002 Q. It is placed 20 em from a
long, straight, current-carrying wire. If the current in the straight wire is
increased at a steady rate from 20 A to 50 A in 2 8, determine the direction
of the current induced in the square loop. (Because the square loop is at
such a great distance from the straight wire, assume that the magnetic
field through the loop is uniform and equal to the magnetic field at its
center.)
I
D<Of--------------------------------------r= 20cm
308 •
CRACKING THE AP PHYSICS EXAM
Solution. At the position of the square loop, the magnetic field due to the straight wire is directed
out of the plane of the page and its strength is given by the equation B =(1l/21t )(I! r). As the current
in the straight wire increases, the magnetic flux through the turns of the square loop changes,
inducing an emf and current. There are N =5 turns; Faraday's Law becomes E avg =-N( Ll <pBI Llt),
and
E
avg
Ll<l>B
Ll(BA)
LlB
110 M
= - N - = - N - - = -NA- = -NA-M
M
M
21tr M
Substituting the given numerical values, we get
110 M
Eavg =-NA-21tr M
=-(5)(0.02 m)2 41t x 10-7 T · m I A (50 A - 20 A)
2s
21t(0.20 m)
=-6xlO-9 V
The magnetic flux through the loop is out of the page and increases as the current in the straight wire
increases. To oppose an increasing out-of-the-page flux, the direction of the induced current should
be clockwise, thereby generating an into-the-page magnetic field (and flux) .
I
I
B (due to--
--8
• ______________________________________ _
straight wire)
/
r = 20 cm
The value of the current in the loop will be
I
E 6xlO-9 V
R
0.0002 Q
5
3xlO- A
ELECTROMAGNETIC INDUCTION .
309
Example 14.6 [C] A rectangular loop of wire 10 cm by 4 cm has a total
resistance of 0.005 n. It is placed 2 cm from a long, straight, currentcarrying wire. If the current in the straight wire is increased at a steady
rate from 20 A to 50 A in 2 s, determine the direction of the current
induced in the rectangular loop.
1
~--
. ----- e=10 cm -------->
...
,,
,
y=4cm
,
,,
~--------------~y
--- ->
r=2cm
Solution. Unlike in the previous example, in this case the magnetic field varies greatly over the
interior of the rectangular loop, so we have to use integration to calculate the magnetic flux. Take a
narrow strip of width dx and height y, whose area is dA =Ydx:
I
~--- - ----
e= 10 em -- ------>
...,
,
----- ---- ----->
X
--
~----~--------~
----->
r = 2em
310 •
CRACKING THE AP PHYSICS EXAM
dx
y=4em
,,
,,
y
The magnetic field strength everywhere in this strip is B = ()lol2 7t )(Ij x), so the magnetic flux
through the strip is
dcI>B
=B·dA =B·(ydx) = !lo i. ydx
21t x
Integrating from x = r to x = r + £ gives the total magnetic flux through the loop:
cI>B
f
re
= dcI>B =J x=r+t!lo i .y dx = !loIy J +dx = !loIy [lnx ]~+e = !loIy In r + £
x=r 21t X
21t r x
21t
21t
r
Faraday's Law then gives us:
f
= _ dcI>B = _~(!loIY In r+ £) = _(!loY In r+ £)dI
dt
dt
21t
r
21t
r
dt
Ignoring the minus sign (which only reminds us that Lenz's Law has to be obeyed), the value of the
current induced in the loop is
I=~=(!lOY lnr+ £)dI
r
21tR
R
dt
7
= (41tx10- T ·m/ A)(0.04 m).ln 2 cm+10 cm. 50 A-20 A
21t(0.005 Q)
=
2 cm
2s
5
4.3 x 10- A
At the position of the rectangular loop, the magnetic field due to the straight wire is directed into
the plane of the page. So the magnetic flux through the loop is into the page and increases as the
current in the straight wire increases. To oppose an increasing into-the-page flux, the direction of the
induced current will be counterclockwise, generating an out-of-the-page magnetic field (and flux).
[(] Induced Electric Fields
In Examples 14.4, 14.5, and 14.6, electric current was induced in a stationary conducting loop by a
changing magnetic flux through the loop, but what was the source of the force that pushed these
charges around the loop? It was not the magnetic force, because the loop was stationary. It must
therefore have been an electric force that was produced by an electric field. The changing magnetic
field produced the electric field, which acted on the free ~harges in the conducting loop and caused
the current.
Let's look at a specific situation. The figure below is a view down the central axis of an ideal
solenoid (ideal means that the magnetic field is uniform and parallel to the axis and that no magnetic
field exists outside). Surrounding the solenoid is a loop of wire.
ElECTROMAGNETIC INDUCTION .
311
Now suppose that the current in the solenoid is increased, which increases the strength of its
magnetic field. This changes the magnetic flux through the wire and induces a current (which is
directed counterclockwise). The force that pushes the charges around this wire is FE = qE; E is the
induced electric field.
E
E
----,o-='--E
+q
The work done on a charge q as it makes one revolution is FE times the distance around the wire,
FE' 21tr =qE·(21tr). So the work per unit charge, which is the definition of emf, is equal to E(21tr). But
Faraday's Law says that the emf is equal to -d <I> BI dt:
E(21tr)
= -d<I>Bldt
This equation can be generalized by saying that the work done per unit charge by the induced electric
field around a closed path is equal to E· dl. Therefore,
f
,( E. dt = _ dCl>B
j
dt
nonconservative E-field
This is a restatement of Faraday's Law that includes the electric field, E, induced by the changing
magnetic flux. Notice that this electric field is different from the ones we studied in the previous
chapters. Electric fields created by stationary source charges, called electrostatic fields, are conservative, meaning that the work done by them on charges moved along closed paths is always zero.
However, as we've just seen in the example of th~ solenoid above, the electric field induced by a
changing magnetic flux does not share this property. The work done by Einduced on a charge as it moves
around a closed path is equal to E· dl, which is not zero because d <I> sf dt is not zero. Because of this,
the electric field induced by a changing magnetic flux is nonconservative.
f
Example 14.7 For the situation just described, assume that the solenoid
has 15,000 turns per meter and a radius of r = 2 cm. The radius of the
circular loop is R =4 cm. If the current in the solenoid is increased at a
rate of 10 AI S, what is the magnitude of the induced electric field at each
position along the circular wire?
312 •
CRACKING THE AP PHYSICS EXAM
Solution. From Faraday's Law, we have
J. E.dt = _ deb B
'J
dt
E(21tR) =_dcI>B
dt
lEI = _1 IdcI>B I
21tR dt
Assuming that the magnetic field exists only inside the solenoid, the magnetic flux through the
circular wire is BA = B( 1tr2), where B = !lonI:
<l>B = (!lon!) . (1tr2)
Therefore,
IEI=1 IdcI>B I
21tR dt
1
d
2
= 21tR . dt (llonI ·1tr )
ll nr 2 dI
2R dt
=o- - 7
= (41t X 10- T· m I A)(15,000 I m)(0.02 m)2 (10 A I s)
2(0.04 m)
4
= 9.4xlO- V 1m
[C] INDUCTANCE
Placing a capacitor in series with a resistor and battery in an electric circuit causes the current to drop
exponentially from E I R (when the switch is closed) to zero, as charge builds up on the capacitor and
causes the voltage across the capacitor to oppose the emf of the battery. The same thing happens
when an inductor is placed in series with a resistor and a seat of emf in an electric circuit.
An inductor is a circuit element that opposes changes in current. The prototypical inductor is a
coil (a solenoid) 'that looks like this in diagrams:
How does this coil oppose changes in current in a circuit? Well, when current exists in the coil,
a magnetic field is created, and magnetic flux passes through the loops. If the current changes, then
the magnetic field changes proportionately, as does the magnetic flux through the loops. But,
according to Faraday's Law of Induction, a changing magnetic flux induces an emf that opposes the
change that produced it. In other words, the changing magnetic flux through the coil, due to the
current in the coil itself, produces a self-induced emf. Since this self-induced emf opposes the
change in the current that produced it, it's also called back emf.
ELECTROMAGNETIC INDUCTION .
313
Assume that the inductor contains N turns and that cD B is the magnetic flux through each turn.
Then the total magnetic flux through the entire coil is N cD B; this is proportional to the current, so
N <PB = LI for some constant, L. The proportionality constant L is called the inductance (or selfinductance) of the coil:
L= N<PB
I
The SI unit for inductance is the weber per ampere, which is renamed a henry (H).
Example 14.8 An ideal solenoid of cross sectional area A and length t
contains n turns per unit length. What is its self-inductance?
Solution. The magnetic field inside an ideal solenoid is parallel to its central axis and has strength
B = l-lonI. Since the solenoid's length is t, the total number of turns, N, is nt. Therefore,
2A~
- N<PB _ N ·BA _ n£ ·BA _ n£'(/-lonI)A _
L- /-lon (,
I
I
I
I
n=N/l
Or, in terms of the total number of turns in the solenoid, L = I-lPA/ £.
The self-induced emf in an inductor now follows directly from the definition of L and Faraday's
Law. Since N <Ps = LI, we have N(d <Psi dt) = L(dIl dt). But by Faraday's Law, E = -N(d <Psi dt). So,
dI
E=- L-dt
The self-induced emf is proportional to the rate of change of the current. The faster the current tries
to change, the greater is the self-induced (back) emf. But if the current is steady, then E =o.
Rl Circuits
Consider the following circuit, which contains a battery, an inductor, and a resistor:
a S ~---------,
b
,---------~
R
L
314 •
CRACKING THE AP PHYSICS EXAM
Move the switch S to point a. If the inductor were absent, the current would rise abruptly to £ / R, but
the inductor opposes this rapid increase in current by producing a back emf. The result is a current
that increases gradually in time, according to the equation
E( _til)
I(t)=R 1-e
R
The ratio of L to R (in the exponent) is called the inductive time constant: r L = L/ R, and the equation
for the rising current in the RL circuit (a circuit that contains a resistor and an inductor) can be
written as:
which is similar to the equation for the gradually rising charge on the plates of a capacitor in an RC
circuit.
As current builds in the circuit, the source of emf provides electrical energy. Some of this energy
is dissipated as heat by the resistor, and the remainder is stored in the magnetic field of the inductor.
The amount of stored energy is
UB =.lLI2
. 2
where L is the inductance of the inductor carrying a current I.
Example 14.9 In a particular circuit, assume that £ = 12 V, R = 40 0,
and L =5 mHo How much energy is stored in the inductor's magnetic
field when the current reaches its maximum steady-state value?
Solution. When the current in the circuit reaches its maximum steady-state value, I
point, the energy stored in the magnetic field of the inductor is
= £ / R. At this
Once the current has reached its steady-state value of £ / R, imagine that you move the switch to
point b. Without the inductor, the current would drop abruptly to zero (because the source of emf has
been taken out of the circuit). But the inductor would oppose this abrupt decrease in the current by
producing a self-induced emf. The presence of this emf would cause the current to die out gradually,
according to the equation
I(t)= Ee -tlr ,
R
lC Circuits
The RC circuit we looked at in chapter 12 and the RL circuit above are similar in that they both
experience an exponential increase or decrease of current. However, an LC circuit, which is a circuit
that contains both an inductor and a capacitor, behaves quite differently.
ELECTROMAGNETIC INDUCTION. 315
The following figure shows a capacitor and an inductor in a simple series circuit.
5 110se at time t =0
Qrnax
++++++
__-_...,_-_-_-_ C
L
Let's assume that initially there is no current and that the capacitor is charged; at time t = 0, the
switch 5 is closed and the circuit is complete. The presence of the inductor prevents the capacitor
from discharging abruptly, so the current rises gradually and the energy in the electric field of the
capacitor is transferred to energy in the magnetic field of the inductor. Once the capacitor has
discharged, all the energy is in the inductor's magnetic field. The current in the inductor (now at its
maximum), delivers charge to the capacitor, but in the opposite direction from its original charge
configuration. The current gradually returns to zero as the magnetic-field energy in the inductor is
transformed back to electric-field energy in the capacitor.
The capacitor starts to discharge again, but this time it sends current in the opposite direction. The
current rises gradually, reaching a maximum value as the charge on the capacitor reaches zero. The
inductor continues to deliver charge to the capacitor as the current gradually returns to zero, and it
finds itself right back where it started. The circuit oscillates, and this defines one cycle of oscillation.
Frequency of oscillation is related to inductance and capacitance by the equation
1
f = 2rr{[C
The charge on the capacitor, as a function of time (assuming that it possessed its maximum charge,
Qmax,at time t = 0) is
Q(t)
where
OJ
= Qmax cos (OJ t)
= 21tf
Example 14.10 Find an equation for the current in the LC circuit as a
function of time. How long does it take for the circuit to complete a full
oscillation?
Solution. By definition, the current is the time derivative of the charge. So,
I(t) = ~; =:t [Qmax cos(mt)] =- coQmax sin(mt)
The time required to complete a cycle (the period) would be the reciprocal of the frequency:
T =!:.. =2rr{[C
f
316 •
CRACKING THE AP PHYSICS EXAM
[C] MAXWELL'S EQUATIONS
We'll finish up this chapter with a set of four equations that embody the subject of electromagnetism.
1.
GAUSS'S LAW
This law, first studied in Chapter 10, gives us a method for calculating electric fields and is particularly useful when the system we're working with possesses symmetry.
J
E . dA = Qenc10sed
Eo
closed
surface
2.
GAUSS'S LAW FOR MAGNETIC FIELDS
We mentioned (chapter 13) that magnetic field lines always form closed loops that encircle the
current that generates them. (Remember: This property is not shared by electrostatic fields, which
radiate away from positive source charges and toward negative ones.) The fact that magnetic field
lines always close upon themselves tells us that the magnetic flux through any closed surface must
be zero; as much magnetic flux will enter the closed surface as will exit it. Mathematically, this says:
fB·dA=O
closed
surface
Another statement of this law is that there are no magnetic monopoles. If there were, a single
isolated magnetic north pole would generate a magnetic field that radiates away, and an isolated
magnetic south pole would generate a magnetic field that radiates inward, toward the pole. A closed
surface surrounding such a magnetic charge (if one existed) would have a nonzero flux through it.
No magnetic monopoles like this have ever been observed.
3.
FARADAY'S LAW
As we discussed in this chapter, a changing magnetic flux induces an emf; this also means that a
changing magnetic field produces an electric field. The equation is
,( E . dl =_ d<I>B
j
4.
dt
THE AMPERE-MAXWELL LAW
The final equation in this set begins with Ampere's Law, which provides us with a method for
calculating magnetic fields and is particularly useful when systems are symmetrical. It reads
~ B · ds = 1l0Ienclosed
You may notice that the first two equations listed in this section possess a sort of symmetry;
there's a Gauss's Law for E-fields and one for B-fields. The third equation says that a changing B-field
produces an E-field. One question we haven't asked yet, which is of interest here is, Does a changing
E-field produce a B-field? The answer is yes, and the equation can be amended in the following way
[Maxwell added the missing piece, which is proportional to the rate of change of the electric flux,
flo E o(d <I> / dt)]:
ElECTROMAGNETIC INDUCTION .
317
This shows that a changing electric field, which appears on the right-hand side of the equation [in
the term Ilo to(d <1>/ dt)], will produce a magnetic field, which appears on the left-hand side of the
equation.
The quantity to(d <1>/ dt) has units of current and is called displacement current, ID • It differs from
the current I, which is conduction current (Ic)' because while Ie is composed of moving electric
charge, ID is not. The Ampere- Maxwell equation can be written in terms of conduction current and
displacement current as follows:
318 •
CRACKING THE AP PHYSICS EXAM