Ch. 23 Solution
PHY 205
Dr. Hayel Shehadeh
Chapter 23: Magnetic Flux and Faraday’s Law of Induction
Answers Conceptual Questions
6.
Nothing. In this case, the break prevents a current from circulating around the ring. This, in turn,
prevents the ring from experiencing a magnetic force that would propel it into the air.
8.
As the penny begins to tip over, there is a large change in the magnetic flux through its surface, due
to the great intensity of the MRI magnetic field. This change in magnetic flux generates an induced
current in the penny that opposes its motion. As a result, the penny falls over slowly, as if it were
immersed in molasses.
12.
When the switch is opened in a circuit with an inductor, the inductor tries to maintain the original
current. (In general, an inductor acts to resist any change in the current—whether the current is
increasing or decreasing.) Therefore, the continuing current may cause a spark to jump the gap.
Solutions to Problems & Conceptual Exercises
2.
The image shows a box immersed in a vertical magnetic field.
Use equation 23-1 to calculate the flux through each side.
Solution: 1. The sides of the box are parallel to the field, so
the magnetic flux through the sides is zero .
2. Calculate the flux
through the bottom:
  BA cos   0.0250 T  0.325 m 0.120 m cos 0  9.75 104 Wb .
The height of the box is not important in this problem.
4.
A house has a floor of dimensions 22 m by 18 m. The local magnetic field due to Earth has a horizontal
component 2.6×10-5 T and a downward vertical component 4.2×10-5 T.
The horizontal component of the magnetic field is parallel to the floor, so it does not contribute to the flux. Use
equation 23-1 to calculate the flux using the vertical component.
Solution: Calculate the magnetic
flux:
  BA cos  B A   4.2 105 T   22 m 18 m   1.7 102 Wb
The flux through the vertical walls of the house is determined by the horizontal component of the magnetic
field instead of the vertical component.
10. The magnetic flux through a coil oscillates in time as indicated by the
graph at right.
The magnitude of the flux is greatest when the flux is at a maximum
or a minimum on the graph. The magnitude of the emf is greatest
when the flux has the greatest positive or negative slope.
Solution: 1. (a) The magnetic flux has its greatest magnitude at
t = 0 s, 0.2 s, 0.4 s, and 0.6 s .
2. (b) The magnitude of the induced emf is greatest at
1
Ch. 23 Solution
PHY 205
Dr. Hayel Shehadeh
t = 0.1 s, 0.3 s, and 0.5 s .
Note that the magnitude of the induced emf is zero when the magnitude of the flux is a maximum and the
magnitude of the induced emf is a maximum when the flux is zero.
12. A wire loop is placed in a magnetic field that is perpendicular to its
plane. The field varies with time as shown at right.
Faraday’s Law states that the magnitude of the induced emf is
proportional to the rate of change of the magnetic flux. In this case,
the magnetic flux is proportional to the field magnitude because the
area of the loop and its orientation remain constant. The rate of
change of the flux is therefore determined by the rate of change of the
field, which in turn is represented by the slope of the plot at the right.
Use the slope of the plot to determine the ranking of the magnitude of
the induced emf.
Solution: The graph region with the steepest slope (E) corresponds to
the greatest induced emf, and the regions with zero slope (D and F)
correspond to zero emf. Using similar reasoning we arrive at the
ranking:
D=F<A<B<C<E
Note that in regions D and F the field is nonzero (and so is the flux) but the induced emf is zero because it
depends on the rate of change of the flux, not the magnitude of the flux.
16. The image shows a single loop of area 7.4×10−2 m2 and resistance 110 Ω.
The loop is perpendicular to a magnetic field.
Solve Ohm’s Law (equation 21-2) for the necessary emf. Then insert the emf
into equation 23-4 to calculate the rate of change in the magnetic field.
Solution: 1. Calculate the emf :
2. Solve equation 23-4 for
the
change in magnetic field:


 IR   0.32 A 110   35 V
N

AB
N
t
t
B
 
35 V

 4.9  102 T/s
t
 1 7.2 10 2 m 2 
The magnitude of the magnetic field (0.48 T) is not important, only the rate of change in the field.
20. A metal ring is dropped into a localized region of constant magnetic field, as indicated
in the figure at the right. The magnetic field is zero above and below the indicated
region.
Use Lenz’s Law to determine the direction that the induced current must flow in order
to oppose the change in the magnetic flux through the ring.
Solution: 1. (a) At location 1 the magnetic flux through the ring is increasing in the out-of-the-page direction.
The induced current will flow clockwise in order to produce into-the-page flux to oppose this change. At
location 2 the flux through the ring is not changing so that the induced current is zero. At location 3 the out-ofthe-page flux through the ring is decreasing, so the induced current will flow counterclockwise to oppose this
change by producing out-of-the-page flux. In summary: Location 1, clockwise; location 2, zero; location 3,
counterclockwise.
2
Ch. 23 Solution
PHY 205
Dr. Hayel Shehadeh
2. The best explanation is I. Clockwise at 1 to oppose the field; zero at 2 because the field is uniform;
counterclockwise at 3 to try to maintain the field. Statement II has the current directions reversed, and
statement III does not properly recognize the rate of change of the magnetic flux through the loop.
Statement II would be correct if the magnetic field pointed into the page.
22. The figure at the right shows two metal disks of the same size and material
oscillating in and out of a region with a magnetic field. One disk is solid;
the other has a series of slots.
Note that the action of the slots will be to suppress the induced currents in
the disk that result from the changing magnetic field through the disk.
The disk with slots will therefore experience a much smaller magnetic
force than will the solid disk.
Solution: 1. (a) The slots in the second disk will tend to suppress the clockwise and counterclockwise currents
that are induced by the change in the magnetic field through the disk. The diminished currents through the disk
will result in a diminished magnetic force on the disk, so we conclude that the retarding effect of eddy currents
on the solid disk is greater than the retarding effect on the slotted disk.
2. The best explanation is I. The solid disk experiences a greater retarding force because eddy currents in it
flow freely and are not interrupted by the slots. Statement II does not recognize that the field penetration
through the disk is not affected by the slots, and statement III ignores the importance of the eddy currents in
explaining the motion of the disks.
In this common classroom demonstration the motion of the two disks are dramatically different. The motion of
the solid disk is quickly damped to zero while the slotted disk continues to swing nearly unimpeded.
26. The image shows a loop of wire dropping between the poles of a magnet.
Use Lenz’s Law to determine the direction of the induced current.
Solution: 1. (a) When the loop is above the magnet, the magnetic field is
increasing and directed out of the page. The current in the loop will oppose
the increasing field by flowing clockwise.
2. (b) When the loop is below the magnet, the magnetic field is decreasing
and is directed out of the page. The current in the loop will oppose the
decreasing field by flowing counterclockwise.
When the loop is directly between the two poles the flux is a maximum, and
therefore momentarily not changing. At this point the induced current is zero.
29. The image shows a loop with resistance R to the left of an upward current.
Use Lenz’s Law to determine the direction of the induced current in the
loop.
Solution: 1. (a) Because the current in the wire is constant, the magnetic
field through the circuit does not vary with time, so the induced current is
zero.
2. (b) The magnetic field through the circuit is increasing because the current in the wire is increasing. And,
since the magnetic field is directed out of the page, the induced current in the circuit will induce a magnetic
field into the page. So, the current in the circuit flows clockwise.
If the current were decreasing, the outward magnetic field would decrease, inducing a counterclockwise
current.
3
Ch. 23 Solution
PHY 205
Dr. Hayel Shehadeh
32. The image shows a circuit containing a resistor and capacitor. A
magnetic field initially pointing into the page reverses to point out of
the page.
Use Lenz’s Law to determine the direction of current flow. As the
current flows onto the capacitor plate it becomes positively charged.
Solution: Because the field changes from in to out of the page, the
induced current in the circuit will flow clockwise to generate a field
directed into the page. Therefore, the top plate will become positively
charged.
As the capacitor plate becomes charged it opposes the induced current. The emf around the loop is still the
same as given by equation 23-3, but the voltage across the capacitor ( V  Q C ) must be subtracted to calculate
the current through the resistor.
38. The image shows a frictionless rod sliding across two rails
separated by 0.45 m. A magnetic field of 0.750 T points out
of the page.
Calculate the emf using Ohm’s Law (equation 21-2). Insert
the emf into equation 23-3 and solve for the speed of the rod.
Solution: 1. (a) Calculate
:


2. Solve for v:
v
IR
  0.155 A 12.5  
 1.94 V

B

1.94 V
 5.75 m/s
 0.750 T  0.450 m 
3. (b) No, the answer to part (a) would not change because the equation for v is independent of the direction of
motion of the bar.
Lenz’s Law says that as the bar moves to the right, the current flows upward in the resistor. If the bar moved at
the same speed to the left, the same current would flow downward through the resistor.
42. The emf produced by a rotating coil is proportional to the speed of the coil rotation.
Use equation 23-11 to determine the maximum emf as a function of the rotation speed. Place all of the
constants on the right side of the equation and develop a ratio for the maximum emf and rotation speed. Solve
the resulting ratio for the unknown rotation speed.
Solution: 1. Create a ratio using equation 23-11:

2. Solve for the final rotation speed:
2 
max
 NBA 


max 2
max1
1 

max1
1
 NBA 

max 2
2
55 V
 210 rpm   260 rpm
45 V
The emf is linearly proportional to the rotation speed. Doubling the rotation speed will double the emf.
4
Ch. 23 Solution
PHY 205
Dr. Hayel Shehadeh
46. A set of coils with area 0.016 m2 rotates at 3600 rpm in a 0.050-T magnetic field. The coils produce an emf of
170 V.
Solve equation 23-11 for the number of turns N.
Solution:
Calculate
the number of
turns:
N

max
B A

 rev  60 s 


  560 turns
 0.050 T   0.016 m   3600 rev/min   2 rad  min 
170 V
2
Note that the magnetic field and number of turns are inversely proportional. To produce the same emf at the
same frequency with half the number of coils it would be necessary to double the magnetic field.
54. The four electrical circuits shown in the figure at the
right have identical batteries, resistors, and inductors.
A long time after the switch is closed each inductor
will act like an ideal wire. Use this principle and the
rules of adding resistors in series and in parallel to
determine the ranking of the currents supplied by the
batteries.
Solution: Let each battery have emf  and each
resistor have resistance R. For circuit A the inductor
will act like an ideal wire when the switch has been
closed for a long time, and the current supplied by
the battery will be  R . For circuits B and C the
resistors in parallel with the inductor are shorted out,
and make no contribution to the current in the
circuit. The current supplied by the battery in those
circuits is also  R . For circuit D the equivalent
resistance of the two resistors in parallel is R 2, so
the current supplied by the
battery after a long time, when the inductor acts as an ideal wire, is 2 R . The ranking of the currents,
therefore, is
A = B = C < D.
While inductors act like ideal wires in DC circuits, capacitors act like open switches (see Chapter 21). If each
inductor were replaced by a capacitor, the currents in each circuit would be A, zero; B,  2R ; C, 2 3R ; and
D, zero.
62. A circuit is constructed with a 62.0-mH inductor in series with two
parallel resistors and a 12-V battery. A long time after the switch is
closed the inductor stores 0.110 J of energy.
Solve equation 23-19 for the current through the battery. Then use
Ohm’s Law (equation 21-2) to calculate the current through the 7.50Ω resistor and subtract that current from the total to calculate the
current through the unknown resistor. Divide the voltage across the
battery by the current to calculate the resistance.
5
Ch. 23 Solution
PHY 205
Solution: 1. (a) Calculate the
current through the inductor:
Dr. Hayel Shehadeh
U
I
1 2
LI
2
2U

L
2  0.110 J 
62.0 mH
 1.884 A
V
12 V

 1.60 A
R 7.50 
2. Calculate the current through the 7.50-Ω resistor:
I
3. Calculate the current through the unknown resistor:
I R  I L  I 7.5  1.884 A 1.600 A= 0.284 A
4. Divide the voltage by the current:
R
V
12 V

 42 
I 0.284 A
5. (b) The energy stored in the inductor is proportional to the square of the total current. Decreasing the
resistance will increase the current, thereby increasing the energy stored. Therefore, in order to store more
energy in the inductor the value of R should be less than the value found in part (a).
If the resistance is decreased to 20 Ω the energy stored in the inductor increases to 0.150 J.
6