Chapter 23: Faraday’s Law and Inductance
1. A circular loop of wire is held in a uniform magnetic field, with the plane of the loop
perpendicular to the field lines. Which of the following will not cause a current to be
induced in the loop? (a) crushing the loop (b) rotating the loop about an axis
perpendicular to the field lines (c) keeping the orientation of the loop fixed and moving it
along the field lines (d) pulling the loop out of the field
Answer: (c). In all cases except this one, there is a change in the magnetic flux through
2. Figure 23.7 shows a graphical representation of the field magnitude versus time for a
magnetic field that passes through a fixed loop and that is oriented perpendicular to the
plane of the loop. The magnitude of the magnetic field at any time is uniform over the
area of the loop. Rank the magnitudes of the emf generated in the loop at the five instants
indicated, from largest to smallest.
Figure 23.7 The time behavior of a magnetic field through a loop.
Answer: c, d = e, b, a. The magnitude of the emf is proportional to the rate of change of
the magnetic flux. For the situation described, the rate of change of magnetic flux is
proportional to the rate of change of the magnetic field. This rate of change is the slope of
the graph in Figure 23.7. The magnitude of the slope is largest at c. Points d and e are on
a straight line, so the slope is the same at each point. Point b represents a point of
relatively small slope, and a is at a point of zero slope because the curve is horizontal at
3. You wish to move a rectangular loop of wire into a region of uniform magnetic field at
a given speed so as to induce an emf in the loop. The plane of the loop must remain
perpendicular to the magnetic field lines. In which orientation should you hold the loop
while you move it into the region of magnetic field so as to generate the largest emf? (a)
with the long dimension of the loop parallel to the velocity vector (b) with the short
dimension of the loop parallel to the velocity vector (c) either way because the emf is the
same regardless of orientation
Answer: (b). According to Equation 23.5, because B and v are fixed, the emf depends
only on the length of the wire moving in the magnetic field. Thus, you want the long
dimension moving through the magnetic field lines so that it is perpendicular to the
velocity vector. In this case, the short dimension is parallel to the velocity vector.
( Bx) B
Bv [Eq. 23.5]
4. In Active Figure 23.11, a given applied force of magnitude Fapp results in a constant
speed v and a power input . Imagine that the force is increased so that the constant
speed of the bar is doubled to 2v. Under these conditions, what are the new force and the
new power input? (a) 2F and 2 (b) 4F and 2 (c) 2F and 4 (d) 4F and 4
Active Figure 23.11 (a) A conducting bar sliding with a velocity v along two conducting rails under the
action of an applied force Fapp . (b) The equivalent circuit diagram for the pictorial representation in (a).
Answer: (c). The force on the wire is of magnitude Fapp = FB = IℓB, with I given by
Equation 23.6. Thus, the force is proportional to the speed and the force doubles. Because
= Fappv, the doubling of the force and the speed results in the power being four times
5. In equal-arm balances from the early 20th century (Fig. 23.18), it is sometimes
observed that an aluminum sheet hangs from one of the arms and passes between the
poles of a magnet, which causes the oscillations of the equal arm balance to decay
rapidly. In the absence of such magnetic braking, the oscillation might continue for a very
long time and the experimenter would have to wait to take a reading. Why do the
oscillations decay? (a) The aluminum sheet is attracted to the magnet. (b) Currents in the
aluminum sheet set up a magnetic field that opposes the oscillations. (c) Aluminum is
(Photos by John Jewett)
Figure 23.18 In an old-fashioned equal-arm balance, an aluminum sheet hangs between the poles of a
Answer: (b). When the aluminum sheet moves between the poles of the magnet, circular
currents called eddy currents are established in the aluminum. According to Lenz’s law,
these currents are in a direction so as to oppose the original change, which is the
movement of the aluminum sheet in the magnetic field. Thus, the effect of the eddy
currents is create a force opposite to the velocity. This magnetic braking causes the
oscillations of the equal arm balance to settle down, and a reading of the mass can take
place. Magnetic damping has an advantage over frictional damping in that the magnetic
damping force goes exactly to zero as the speed goes to zero. On the other hand, if
mechanical friction were used to damp the oscillation of the balance beam, the speed
might go to zero at a final position other than zero.
6. In a region of space, a magnetic field is uniform over space but increases at a constant
rate. This changing magnetic field induces an electric field that (a) increases in time, (b)
is conservative, (c) is in the direction of the magnetic field, or (d) has a constant
Answer: (d). The constant rate of change of B will result in a constant rate of change of
the magnetic flux. According to Equation 23.9, if dΦB/dt is constant, E is constant in
E ds ddt
7. The circuit in Figure 23.28 includes a power source that provides a sinusoidal voltage.
Thus, the magnetic field in the inductor is constantly changing. The inductor is a simple
air-core solenoid. The switch in the circuit is closed and the lightbulb glows steadily. An
iron rod is inserted into the interior of the solenoid, which increases the magnitude of the
magnetic field in the solenoid. As that happens, the brightness of the lightbulb (a)
increases, (b) decreases, or (c) is unaffected.
Figure 23.28 A lightbulb is powered by an AC source with an inductor in the circuit. When the iron bar is
inserted into the coil, what happens to the brightness of the lightbulb?
Answer: (b). When the iron rod is inserted into the solenoid, the inductance of the coil
increases. As a result, more potential difference appears across the coil than before.
Consequently, less potential difference appears across the lightbulb, so the bulb is
8. Two circuits like the one shown in Active Figure 23.26 are identical except for the
value of L. In circuit A, the inductance of the inductor is LA, and in circuit B, it is LB. The
switch has been in position b for both circuits for a long time. At t = 0, the switch is
thrown to a in both circuits. At t = 10 s, the switch is thrown to b in both circuits. The
resulting graphical representation of the current as a function of time is shown in Figure
23.29. Assuming that the time constant of each circuit is much less than 10 s, which of
the following is true? (a) LA > LB. (b) LA < LB. (c) There is not enough information to
determine the relative values.
Active Figure 23.26 An RL circuit. When the switch S is in position a, the battery is in the circuit. When
the switch is thrown to position b, the battery is no longer part of the circuit. The switch is designed so that
it is never open, which would cause the current to stop.
Figure 23.29 Current–time graphs for two circuits with different inductances.
Answer: (b). Figure 23.29 shows that circuit B has the larger time constant because in
this circuit it takes longer for the current to reach its maximum value and then longer for
this current to drop back down to zero after the switch is thrown to b. Equation 23.15
indicates that, for equal resistances RA and RB, the condition τB > τA means that LA < LB.
9. You are performing an experiment that requires the highest possible energy density in
the interior of a very long solenoid. Which of the following increases the energy density?
(More than one choice may be correct.) (a) increasing the number of turns per unit length
on the solenoid (b) increasing the cross-sectional area of the solenoid (c) increasing only
the length of the solenoid while keeping the number of turns per unit length fixed (d)
increasing the current in the solenoid
Answer: (a), (d). Because the energy density depends on the magnitude of the magnetic
field, we must increase the magnetic field to increase the energy density. For a solenoid,
B = μ0nI, where n is the number of turns per unit length. In (a), we increase n to increase
the magnetic field. In (b), the change in cross-sectional area has no effect on the magnetic
field. In (c), increasing the length but keeping n fixed has no effect on the magnetic field.
Increasing the current in (d) increases the magnetic field in the solenoid.