Chapter 25 Practice Problems, Review, and Assessment
Section 1 Inducing Currents: Practice Problems
1. You move a straight wire that is 0.5 m long at a speed of 20 m/s vertically through a 0.4-T magnetic field pointed in
the horizontal direction.
a. What EMF is induced in the wire?
b. The wire is part of a circuit with a total resistance of 6.0 Ω. What is the current?
SOLUTION: a. EMF = BLv(sin θ )
= (0.4 T)(0.5 m)(20 m/s)(1)
= 4 V
b. 2. A straight wire that is 25 m long is mounted on an airplane flying at 125 m/s. The wire moves in a perpendicular direction through Earth’s magnetic field (B = 5.0×10
−5 T). What EMF is induced in the wire? SOLUTION: EMF = BLv(sin θ )
= (5.0×10−5 T)(25 m)(125 m/s)(1)
= 0.16 V
3. A straight wire segment in a circuit is 30.0 m long and moves at 2.0 m/s perpendicular to a magnetic field.
a. A 6.0-V EMF is induced. What is the magnetic field?
b. The total resistance of the circuit is 5.0 Ω. What is the current?
SOLUTION: a. EMF = BLv(sin θ )
6.0 V = B(30.0 m)(2.0 m/s)(1)
B = 0.10 T
b.
4. CHALLENGE A horseshoe magnet is mounted so that the magnetic field lines are vertical. You pass a straight
wire between the poles and pull it toward you. The current through the wire is from right to left. Which is the
magnet’s north pole? Explain. SOLUTION: Using a right-hand rule, the north pole is at the bottom.
5. A generator develops a maximum potential difference of 170 V.
a. What is the effective potential difference?
b. A 60-W lamp is placed across the generator with an Imax of 0.70. What is the effective current through the
lamp?
c. What is the resistance of the lamp? SOLUTION: a. eSolutions Manual - Powered by Cognero
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wire between the poles and pull it toward you. The current through the wire is from right to left. Which is the
magnet’s north pole? Explain. SOLUTION: Chapter
25 Practice Problems, Review, and Assessment
Using a right-hand rule, the north pole is at the bottom.
5. A generator develops a maximum potential difference of 170 V.
a. What is the effective potential difference?
b. A 60-W lamp is placed across the generator with an Imax of 0.70. What is the effective current through the
lamp?
c. What is the resistance of the lamp? SOLUTION: a. b. c. 6. The RMS potential difference of an AC household outlet is 117 V. What is the maximum potential difference across a lamp connected to the outlet? If the RMS current through the lamp is 5.5 A, what is the lamp’s maximum
current? SOLUTION: 7. If the average power used over time by an electric light is 75 W, what is the peak power? SOLUTION: 8. CHALLENGE An AC generator delivers a peak potential difference of 425 V.
a. What is the Veff in a circuit connected to the generator?
2 b. The resistance is 5.0×10 Ω. What is the effective current?
SOLUTION: a. b. eSolutions Manual - Powered by Cognero
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SOLUTION: Chapter 25 Practice Problems, Review, and Assessment
8. CHALLENGE An AC generator delivers a peak potential difference of 425 V.
a. What is the Veff in a circuit connected to the generator?
2 b. The resistance is 5.0×10 Ω. What is the effective current?
SOLUTION: a. b. Section 1 Inducing Currents: Review
9. MAIN IDEA Use the concept of electromagnetic induction to explain how an electric generator works.
SOLUTION: An EMF is induced in the armature of a generator at it is turned—by a mechanical force—in a magnetic
field. When the generator is in a circuit, the EMF induces a current. As the armature rotates through
180°, the induced EMF—and current—reverse direction.
10. Generator Could you make a generator by mounting permanent magnets on a rotating shaft and keeping the coil
stationary? Explain. SOLUTION: Yes, only relative motion between the coil and the magnetic field is important. Note, this generator
would not have much power as the relative velocities of the magnets and coil will be very small. 11. Bike Generator A small generator on your bicycle lights the bike’s headlight. What is the source of the energy for
the bulb when you ride along a flat road?
SOLUTION: You (the rider) provide the mechanical energy that turns the generator’s armature.
12. Microphone Consider the microphone shown in Figure 3. What happens when the diaphragm is pushed in?
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11. Bike Generator A small generator on your bicycle lights the bike’s headlight. What is the source of the energy for
the bulb when you ride along a flat road?
SOLUTION: Chapter
25 Practice Problems, Review, and Assessment
You (the rider) provide the mechanical energy that turns the generator’s armature.
12. Microphone Consider the microphone shown in Figure 3. What happens when the diaphragm is pushed in?
Figure 3 Microphone
SOLUTION: A current is induced in the coil.
13. Frequency What changes to an electric generator are required to increase frequency?
SOLUTION: You could increase the number of magnetic pole pairs or make the armature spin faster.
14. Output Potential Difference Explain why the output potential difference of an electric generator increases when
the magnetic field is made stronger. What is another way to increase the output potential difference? SOLUTION: The magnitude of the induced EMF is directly related to the strength of the magnetic field. A greater
potential difference is induced in the conductor(s) if the field strength is increased. Because EMF = BLv
(sin θ ), you can also increase output potential difference by increasing the length of the wire or the
velocity of the wire.
15. Critical Thinking A student asks, “Why does AC dissipate any power? The energy going into a lamp when the
current is positive is removed when the current is negative. The net is zero.” Explain why this reasoning is wrong. SOLUTION: Power is the rate at which energy is transferred. Power is the product of I and V. When I is positive, so
is V and therefore, P is positive. When I is negative, so is V; thus, P is positive again. Energy is always
transferred through the lamp. Section 2 Applications of Induced Currents: Practice Problems
For the following problems, effective currents and potential differences are indicated.
16. A step-down transformer has 7500 turns on its primary coil and 125 turns on its secondary coil. The potential difference across the primary circuit is 7.2 kV. What is the potential difference across the secondary circuit? If the current in the secondary circuit is 36 A, what is the current in the primary circuit?
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SOLUTION: Page 4
SOLUTION: Power is the rate at which energy is transferred. Power is the product of I and V. When I is positive, so
is V25
and
therefore,
P is positive.
When
is negative, so is V; thus, P is positive again. Energy is always
Chapter
Practice
Problems,
Review,
and IAssessment
transferred through the lamp. Section 2 Applications of Induced Currents: Practice Problems
For the following problems, effective currents and potential differences are indicated.
16. A step-down transformer has 7500 turns on its primary coil and 125 turns on its secondary coil. The potential difference across the primary circuit is 7.2 kV. What is the potential difference across the secondary circuit? If the current in the secondary circuit is 36 A, what is the current in the primary circuit?
SOLUTION: 17. CHALLENGE A step-up transformer has 300 turns on its primary coil and 90,000 turns on its secondary coil. The
potential difference of the generator to which the primary circuit is attached is 60.0 V. The transformer is 95 percent efficient. What is the potential difference across the secondary circuit? The current in the secondary circuit is 0.50 A. What current is in the primary circuit? SOLUTION: Section 2 Applications of Induced Currents: Review
18. MAIN IDEA You hang a coil of wire with its ends joined so it can swing easily. If you now plunge a magnet into
the coil, the coil will start to swing. Which way will it swing relative to the magnet and why? SOLUTION: Away from the magnet; the changing magnetic field induces a current in the coil, producing a magnetic
field. This field opposes the field of the magnet, and thus, the force between coil and magnet is
repulsive. 19. Motors If you unplugged a running vacuum cleaner from a wall outlet, you would be much more likely to see a
spark than you would be if you unplugged a lighted lamp from the wall. Why? eSolutions Manual - Powered by Cognero
SOLUTION: The inductance of the motor creates an EMF that causes the spark. The bulb has very low self-
Page 5
SOLUTION: Away from the magnet; the changing magnetic field induces a current in the coil, producing a magnetic
field. This field opposes the field of the magnet, and thus, the force between coil and magnet is
Chapter
25 Practice Problems, Review, and Assessment
repulsive. 19. Motors If you unplugged a running vacuum cleaner from a wall outlet, you would be much more likely to see a
spark than you would be if you unplugged a lighted lamp from the wall. Why? SOLUTION: The inductance of the motor creates an EMF that causes the spark. The bulb has very low selfinductance, so there is no EMF.
20. Transformers and Current Explain why a transformer may be operated only on alternating current.
SOLUTION: Transformers rely on changing currents to induce changing magnetic fields. DC always produces the
same magnetic field and thus can’t induce a current in another wire.
21. Transformers Frequently, transformer coils that have only a few turns are made of very thick (low-resistance)
wire, while those with many turns are made of thin wire. Why? SOLUTION: More current can go through the coil with fewer turns, so thick wires with capacity for large currents
are needed. Also, resistance must be kept low to prevent voltage drops and I2R power loss and heating.
22. Step-Up Transformers Refer to the step-up transformer shown in Figure 15. Explain what would happen to the
primary current if the secondary coil were short-circuited
Figure 15
SOLUTION: According to the transformer equations, the ratio of primary to secondary current is equal to the ratio of
turns and doesn’t change. Thus, if the secondary current increased, so would the primary.
23. Critical Thinking Would permanent magnets make good transformer cores? Explain.
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SOLUTION: No, induced EMF depends on a changing magnetic field through the core. Permanent magnets are
Page 6
Figure 15
SOLUTION: Chapter
25 Practice
Review,
and Assessment
According
to theProblems,
transformer
equations,
the ratio of primary to secondary current is equal to the ratio of
turns and doesn’t change. Thus, if the secondary current increased, so would the primary.
23. Critical Thinking Would permanent magnets make good transformer cores? Explain.
SOLUTION: No, induced EMF depends on a changing magnetic field through the core. Permanent magnets are
“permanent” because they are made of materials that resist such changes.
Chapter Assessment
Section 1 Inducing Currents: Mastering Concepts
24. BIG IDEA You have a wire connected in a circuit and two magnets. Describe how you could use them to
generate potential difference and current.
SOLUTION: You would set up a magnetic field (north and south poles facing). You could generate a potential
difference (EMF) by moving the wire within the field or moving the field and keeping the wire still. In
both cases, the wire and field are at right angles. Because the wire is connected in a circuit, the EMF
generates current.
25. Explain how AC and DC generators differ in the way they connect to electric circuits.
SOLUTION: DC generators connect using commutators so current is in one direction; AC generators use slip-ring
devices in which one ring is connected to one end of the wire on the armature, and the other ring is
connected to the other end of the wire. As the armature rotates in the magnetic field through 180°, the induced EMF reverses direction.
26. Why is iron used in an armature?
SOLUTION: Iron is used in an armature to increase the strength of the magnetic field.
27. A single conductor moves through a magnetic field and generates a potential difference. In what direction should
the wire be moved, relative to the magnetic field, to generate the minimum potential difference?
SOLUTION: The minimum amount of potential difference (0 V) is generated when the conductor is moving parallel to
the magnetic lines of force.
28. In which direction would the electric field in the wire point if the wire were pulled to the right?
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SOLUTION: Chapter
Practiceamount
Problems,
Review, difference
and Assessment
The25
minimum
of potential
(0 V) is generated when the conductor is moving parallel to
the magnetic lines of force.
28. In which direction would the electric field in the wire point if the wire were pulled to the right?
SOLUTION: The magnetic field is out of the page, and the charges move to the right as the wire is pulled to the
right. According to a right-hand rule, this means that any current in the wire would be up, toward the top
of the page. In order for the current to be up, the electric field must also be up.
29. What is the effect of increasing the length of the wire of an armature in an electric generator?
SOLUTION: Increasing the wire length results in a net increase in induced potential difference because EMF = BLv
(sin θ ) .
30. How were Oersted’s and Faraday’s results similar? How were they different?
SOLUTION: They are similar in that each showed a relationship between electricity and magnetism. They are
different in that Oersted found that electric currents generated magnetic fields while Faraday found that
magnetic fields can generate EMFs and currents.
31. What does EMF stand for? Why is the name inaccurate?
SOLUTION: Electromotive force; it is not a force but an electric potential difference (energy per unit of charge). It
was named before related concepts were understood.
32. What is the difference between an electric generator and an electric motor?
SOLUTION: Generators and motors are constructed in similar ways, but in a generator, mechanical energy turns an
armature in a magnetic field. The induced potential difference produces current, thus producing electric
energy. In a motor, potential difference is placed across an armature in a magnetic field. The potential
difference produces current in the coil and the armature turns, producing mechanical energy. 33. List the major parts of an AC generator.
SOLUTION: An AC generator consists of a permanent magnet, an armature (made of wire loops), a set of brushes,
and a slip ring device to connect to the circuit.
34. Why is the effective value of an AC current less than its maximum value?
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SOLUTION: Page 8
As the armature turns in an alternating-current generator, the generated power varies between some
maximum value and zero. The average power is equal to one-half the maximum power. The effective
33. List the major parts of an AC generator.
SOLUTION: Chapter
25 Practice
Problems,
Review,
and Assessment
An AC
generator
consists of
a permanent
magnet, an armature (made of wire loops), a set of brushes,
and a slip ring device to connect to the circuit.
34. Why is the effective value of an AC current less than its maximum value?
SOLUTION: As the armature turns in an alternating-current generator, the generated power varies between some
maximum value and zero. The average power is equal to one-half the maximum power. The effective
current is the constant value of current that would cause the average power to be dissipated in the load,
R.
35. Hydroelectricity Water trapped behind a dam turns turbines that rotate generators. List all the forms of energy
that take part in the cycle that includes the stored water and the electricity produced.
SOLUTION: Stored water has gravitational potential energy. As it falls, it gains kinetic energy (KE). The KE is
transferred to the turbines and then to the generator’s armature. The generator transforms KE into
electrical energy. In addition, there are frictional losses in the turbine and generator resulting in
thermal energy.
Chapter Assessment
Section 1 Induced Currents: Mastering Problems
36. A 20.0-m-long wire moves perpendicularly through a magnetic field at 4.0 m/s. An EMF of 40 V is induced in the wire. What is the field’s strength? (Level 1)
SOLUTION: EMF = BLv(sin θ )
2 −5 37. Airplanes An airplane traveling at 9.50×10 km/h passes over a region where Earth’s magnetic field is 4.5×10 T
and is nearly vertical. What potential difference is induced between the plane’s wing tips, which are 75 m apart? (Level 1) SOLUTION: EMF = BLv(sin θ )
= (4.5×10−5 T)(75 m)(9.50×102 km/h)
(1000 m/km)(1 h/3600 s)(1)
= 0.89 V
38. A straight wire that is 0.75-m long moves upward through a horizontal 0.30-T magnetic field, as shown in
Figure 18, at a speed of 16 m/s. (Level 1)
a. What EMF is induced in the wire?
b. The wire is part of a circuit with a total resistance of 11 Ω. What is the current?
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SOLUTION: EMF = BLv(sin θ )
= (4.5×10−5 T)(75 m)(9.50×102 km/h)
Chapter
25 m/km)(1
Practice h/3600
Problems,
(1000
s)(1)Review, and Assessment
= 0.89 V
38. A straight wire that is 0.75-m long moves upward through a horizontal 0.30-T magnetic field, as shown in
Figure 18, at a speed of 16 m/s. (Level 1)
a. What EMF is induced in the wire?
b. The wire is part of a circuit with a total resistance of 11 Ω. What is the current?
SOLUTION: a. EMF = BLv(sin θ )
= (0.30 T)(0.75 m)(16 m/s)(1)
= 3.6 V
b. EMF = IR
39. At what speed would a 0.20-m length of wire have to move perpendicularly across a 2.5-T magnetic field to induce
an EMF of 10 V? (Level 1)
SOLUTION: 40. An AC generator has a maximum potential difference of 565 V. What effective potential difference does the generator deliver to an external circuit? (Level 1)
SOLUTION: 41. An AC generator develops a maximum potential difference of 150 V. It delivers a maximum current of 30.0 A to an external circuit. (Level 1)
a. What is the effective potential difference of the generator? b. What effective current does the generator deliver to the external circuit?
c. What is the average power dissipated in the circuit?
SOLUTION: a. Veff = (0.707)Vmax = (0.707)(150 V) = 110 V
b. Ieff = (0.707)Imax = (0.707)(30.0 V) = 21.2 A
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c. Page 10
generator deliver to an external circuit? (Level 1)
SOLUTION: Chapter 25 Practice Problems, Review, and Assessment
41. An AC generator develops a maximum potential difference of 150 V. It delivers a maximum current of 30.0 A to an external circuit. (Level 1)
a. What is the effective potential difference of the generator? b. What effective current does the generator deliver to the external circuit?
c. What is the average power dissipated in the circuit?
SOLUTION: a. Veff = (0.707)Vmax = (0.707)(150 V) = 110 V
b. Ieff = (0.707)Imax = (0.707)(30.0 V) = 21.2 A
c. 42. Stove An electric stove connects to an AC source with an effective potential difference of 240 V. (Level 1)
a. Find the maximum potential difference across one of the stove’s elements when it is operating.
b. The resistance of the operating element is 11 Ω. What is the effective current?
SOLUTION: a. b. 43. You wish to generate an EMF of 4.5 V by moving a wire at 4.0 m/s through a 0.050-T magnetic field. You want to
use the shortest length of wire you can. How long must the wire be and what should the angle be between the field
and the velocity? (Level 1)
SOLUTION: EMF = BLv(sin θ )
This is the shortest length of wire assuming that the wire and the direction of motion are each
perpendicular to the field (sin θ = 90°).
44. You connect both ends of a 0.10-Ω copper wire to the terminals of a 875-Ω galvanometer. You then move a 10.0−2
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cm segment of the wire upward at 1.0 m/s through a 2.0×10
galvanometer indicate? (Level 2)
SOLUTION: -T magnetic field. What current will the
Page 11
This25is Practice
the shortest
length of
wire assuming
that the wire and the direction of motion are each
Chapter
Problems,
Review,
and Assessment
perpendicular to the field (sin θ = 90°).
44. You connect both ends of a 0.10-Ω copper wire to the terminals of a 875-Ω galvanometer. You then move a 10.0−2
cm segment of the wire upward at 1.0 m/s through a 2.0×10 -T magnetic field. What current will the
galvanometer indicate? (Level 2)
SOLUTION: 45. Refer to Example Problem 1 and Figure 19 to determine the following. (Level 2)
a. induced potential difference in the conductor
b. current (I)
c. polarity of point A relative to point B
SOLUTION: a. EMFind = BLv(sin θ )
−2
= (7.0×10
= 0.13 V
b. T)(0.50 m)(3.6 m/s)(1)
c. Point A is negative relative to point B.
46. The direction of a 0.045-T magnetic field is 60.0º above the horizontal. A wire, 2.5-m long, moves horizontally at
2.4 m/s. (Level 2)
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a. What
the vertical
component of the magnetic field?
b. What EMF is induced in the wire?
SOLUTION: c. Point A is negative relative to point B.
Chapter
25 Practice Problems, Review, and Assessment
46. The direction of a 0.045-T magnetic field is 60.0º above the horizontal. A wire, 2.5-m long, moves horizontally at
2.4 m/s. (Level 2)
a. What is the vertical component of the magnetic field?
b. What EMF is induced in the wire?
SOLUTION: a. The vertical component of the magnetic field is B sin 60.0° = (0.045 T)(sin 60.0°) = 0.039 T
b. EMF = ByLv = (0.039 T)(2.5 m)(2.4 m/s) = 0.23 V
6 47. Dams A generator at a dam can supply 375 MW (375×10 W) of electrical power. Assume that the turbine and
generator are 85 percent efficient. (Level 2) a. Find the rate at which falling water must supply energy to the turbine. b. The energy of the water comes from a change in gravitational potential energy, GPE = mgΔh. What is the
change in GPE needed each second?
c. If the water falls 22 m, what is the mass of the water that must pass through the turbine each second to supply this power?
SOLUTION: a. b. 440 MW = 440 MJ/s = 4.4×108 J each second
c. 48. A 20-cm conductor is moving with a constant velocity of 1 m/s in a magnetic field of 4.0 T. What is the induced potential difference when the conductor moves perpendicular to the line of force? (Level 2)
SOLUTION: When the conductor is moving perpendicular to the line of force
EMF = BLv(sin θ )
= (4.0 T)(0.20 m)(1 m/s)(1)
= 0.8 V
Chapter Assessment
Section 2 Applications of Currents: Mastering Concepts
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49. State Lenz’s law and explain why it is consistent with the law of conservation of energy.
Page 13
SOLUTION: When the conductor is moving perpendicular to the line of force
EMF = BLv(sin θ )
Chapter
25 T)(0.20
Practicem)(1
Problems,
= (4.0
m/s)(1)Review, and Assessment
= 0.8 V
Chapter Assessment
Section 2 Applications of Currents: Mastering Concepts
49. State Lenz’s law and explain why it is consistent with the law of conservation of energy.
SOLUTION: When a change in a magnetic field induces a current, the current is in a direction that opposes the
change in the original magnetic field. It is consistent with the law of conservation of energy because it
prevents a changing magnetic field to grow without limits, and thus it prevents current from increasing
to infinity. Otherwise, energy would be created.
50. Why does current through a motor decrease when the motor speeds up?
SOLUTION: This is Lenz’s law. Once the motor starts turning, there is more and more current through its wires,
strengthening the wires’ magnetic field. According to Lenz’s law, the wires generate an EMF to oppose
the strengthening current.
51. Power Saws When a power saw is turned on in a garage, why do the lights dim in the house?
SOLUTION: When a saw and lightbulbs are in parallel circuit, a large current in the wires of the saw increases the
potential difference across the wires but decreases the potential difference across the lightbulbs. Thus,
they dim.
52. A 150-W transformer has an input potential difference of 9.0 V and an output current of 5.0 A. What is the ratio of
Voutput to Vinput?
SOLUTION: 53. Why is the self-inductance of a coil a major factor when the coil is in an AC circuit but a minor factor when the coil
is in a DC circuit?
SOLUTION: Self-inductance only plays a role when current (and its induced magnetic field) are changing. An
alternating current is always changing in magnitude and direction. A direct current eventually becomes steady, and thus, after a short time, there is no changing magnetic field.
54. Explain why the word change appears so often in this chapter.
SOLUTION: As Faraday discovered, only a changing magnetic field induces EMF and current. And only a changing
current generates a changing magnetic field.
55. Upon what does the ratio of the potential difference in the primary circuit of a transformer to the potential
difference in the secondary circuit of the transformer depend?
SOLUTION: eSolutions
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The ratio of EMFs in a transformer is set by the ratio of the turns of wire in the primary coil to the
number of turns of wire in the secondary coil.
Page 14
54. Explain why the word change appears so often in this chapter.
SOLUTION: As Faraday
discovered,
only
a changing
magnetic field induces EMF and current. And only a changing
Chapter
25 Practice
Problems,
Review,
and Assessment
current generates a changing magnetic field.
55. Upon what does the ratio of the potential difference in the primary circuit of a transformer to the potential
difference in the secondary circuit of the transformer depend?
SOLUTION: The ratio of EMFs in a transformer is set by the ratio of the turns of wire in the primary coil to the
number of turns of wire in the secondary coil.
56. Trains The train in Figure 20 makes no contact with the rails. There are electromagnets in the train but not in the
rails. Explain how this train is able to levitate above the rails as long as it is moving.
SOLUTION: The changing magnetic field in the track induces eddy currents that create a magnetic field that exerts a
repulsive force on the coils in the train. When the train stops, the magnetic field no longer changes so
there is no longer a repulsive force.
Chapter Assessment
Section 2 Applications of Induced Currents: Mastering Problems
57. Doorbell A doorbell requires an effective potential difference of 18 V from a 120-V line. (Level 1) a. If the primary coil has 475 turns, how many does the secondary coil have?
b. The doorbell draws 125-mA of current. What current is in the primary circuit?
SOLUTION: a. b. 58. Scott connects a transformer to a 24-V source and measures 8.0 V at the secondary circuit. If the primary and secondary circuits were reversed, what would the new output potential difference be? (Level 1)
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Chapter
25 Practice Problems, Review, and Assessment
58. Scott connects a transformer to a 24-V source and measures 8.0 V at the secondary circuit. If the primary and secondary circuits were reversed, what would the new output potential difference be? (Level 1)
SOLUTION: 59. Hair Dryer A hair dryer manufactured for use in the United States uses 10 A at 120 V. In a country where the line potential difference is 240 V, it must be used with a transformer. (Level 2)
a. What turns ratio should the transformer have?
b. What current will the hair dryer draw from the 240-V line?
SOLUTION: a. b. 60. A step-up transformer has 80 turns on its primary coil and 1200 turns on its secondary coil. The primary circuit is supplied with an alternating current at 120 V. (Level 1)
a. What potential difference is being applied across the secondary circuit?
b. The current in the secondary circuit is 2.0 A. What current is in the primary circuit?
c. What are the power input and the power output of the transformer? SOLUTION: a. eSolutions Manual - Powered by Cognero
Page 16
b. Chapter 25 Practice Problems, Review, and Assessment
60. A step-up transformer has 80 turns on its primary coil and 1200 turns on its secondary coil. The primary circuit is supplied with an alternating current at 120 V. (Level 1)
a. What potential difference is being applied across the secondary circuit?
b. The current in the secondary circuit is 2.0 A. What current is in the primary circuit?
c. What are the power input and the power output of the transformer? SOLUTION: a. b. c. Chapter Assessment: Applying Concepts
61. A wire is moved horizontally between the poles of a magnet, as shown in Figure 21. What is the direction of the
induced current?
SOLUTION: No current is induced because the direction of the velocity is parallel to the magnetic field.
62. You wind wire around a large nail, as shown in Figure 22. You connect the wire to a battery to make an
electromagnet. Is the current larger just after you make the connection or several hundredths of a second after you
make the connection? Or, is it always the same? Explain. eSolutions Manual - Powered by Cognero
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SOLUTION: Chapter
25 Practice
Problems,
Review,
and Assessment
No current
is induced
because
the direction
of the velocity is parallel to the magnetic field.
62. You wind wire around a large nail, as shown in Figure 22. You connect the wire to a battery to make an
electromagnet. Is the current larger just after you make the connection or several hundredths of a second after you
make the connection? Or, is it always the same? Explain. SOLUTION: The current is larger several hundredths of a second after the connection is made. A decreasing selfinduced EMF will cause the current to grow gradually to its maximum value.
63. You move a length of copper wire down through a magnetic field, as shown in Figure 23.
a. Will the induced current move to the right or to the left in the wire segment?
b. What will be the direction of the force acting on the wire as a result of the induced current?
SOLUTION: a. A right-hand rule will show the current moving left.
b. The force will act in an upward direction.
64. Use unit substitution to show that the units of BLv are volts.
SOLUTION: 65. Ranking Consider the transformers described below. Rank them according to the current induced in the secondary
coil, from least to greatest. Indicate whether any have equal currents.
A. primary coil: 50 turns; secondary coil: 25 turns; primary current: 2 A
B. primary coil: 50 turns; secondary coil: 60 turns; primary current: 2 A
C. primary coil: 20 turns; secondary coil: 10 turns; primary current: 4 A
D. primary coil: 20 turns; secondary coil: 40 turns; primary current: 8 A
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E. primary coil: 10 turns; secondary coil: 100 turns; primary current: 1 A
SOLUTION: Chapter 25 Practice Problems, Review, and Assessment
65. Ranking Consider the transformers described below. Rank them according to the current induced in the secondary
coil, from least to greatest. Indicate whether any have equal currents.
A. primary coil: 50 turns; secondary coil: 25 turns; primary current: 2 A
B. primary coil: 50 turns; secondary coil: 60 turns; primary current: 2 A
C. primary coil: 20 turns; secondary coil: 10 turns; primary current: 4 A
D. primary coil: 20 turns; secondary coil: 40 turns; primary current: 8 A
E. primary coil: 10 turns; secondary coil: 100 turns; primary current: 1 A
SOLUTION: 66. When a wire is moved through a magnetic field, does the resistance of the closed circuit affect current only, EMF
only, both, or neither?
SOLUTION: current only
67. A transformer is connected to a battery through a switch, as shown in Figure 24. The secondary circuit contains a
lightbulb. Will the bulb be lighted as long as the switch is closed, only at the moment the switch is closed, or only at
the moment the switch is opened? Explain. SOLUTION: The bulb will light because there is a current in the secondary circuit. This will happen whenever the
primary current changes, so the bulb will glow either when the switch is closed or when it is opened.
However, it will glow only for a moment. The secondary coil will only generate current when current in
the primary coil is changing. 68. Explain why the initial start-up current is so high in an electric motor. Also explain how Lenz’s law applies at the
instant the electric motor starts.
SOLUTION: If the armature (conductors) are not rotating, no lines of force are being cut, and no potential difference
is induced. Therefore, the induced EMF is zero. Since there is no current in the armature, no magnetic
field is formed around the stationary conductor. It should be noted that this explanation only holds true
at the instant of startup, at time just greater than 0. The instant the armature begins to rotate, it will be
cutting the lines of force and will have an induced potential difference. This potential difference, the
induced EMF, will have a polarity such that it produces a magnetic field opposing the field that created
it. This reduces the current in the motor. Therefore, the motion of the motor increases its apparent
resistance.
69. A transformer is constructed with a laminated core that is not a superconductor. Because the eddy currents cannot
be completely eliminated, there is always a small transformation of electrical to thermal energy. This results in a net
reduction of power available to the secondary circuit. As long as the core has eddy currents, what fundamental law
makes it impossible for the output power to equal the input power? eSolutions
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SOLUTION: Second law of thermodynamics
at the instant of startup, at time just greater than 0. The instant the armature begins to rotate, it will be
cutting the lines of force and will have an induced potential difference. This potential difference, the
induced EMF, will have a polarity such that it produces a magnetic field opposing the field that created
it. This
reduces Problems,
the currentReview,
in the motor.
Therefore, the motion of the motor increases its apparent
Chapter
25 Practice
and Assessment
resistance.
69. A transformer is constructed with a laminated core that is not a superconductor. Because the eddy currents cannot
be completely eliminated, there is always a small transformation of electrical to thermal energy. This results in a net
reduction of power available to the secondary circuit. As long as the core has eddy currents, what fundamental law
makes it impossible for the output power to equal the input power? SOLUTION: Second law of thermodynamics
70. Earth’s Magnetic Field The direction of Earth’s magnetic field in the northern hemisphere is downward and to
the north, as shown in Figure 25. If an east-west wire moves from north to south, in which direction is the current?
SOLUTION: The current is from west to east.
71. A physics instructor drops a magnet through a copper pipe, as illustrated in Figure 26. The magnet falls very
slowly, and the students in the class conclude that there must be some force opposing gravity. a. The magnetic field in the pipe changes. What is induced in the pipe?
b. How does what is induced in the pipe affect the falling magnet? Explain in terms of energy.
c. If the instructor used a plastic pipe, would the falling magnet slow down?
SOLUTION: a. The changing magnetic field in the pipe induces eddy currents. No matter whether the field is
increasing or decreasing, there are always eddy currents produced.
b. The eddy currents cause the pipe to get warmer. As the magnet falls, its gravitational potential
energy decreases but its kinetic energy does not. The thermal energy of the pipe increases and as it
gets hotter than the surrounding air, heat is transferred to the air. c. A non-conducting (or poorly conducting) pipe would have no or very small eddy currents, and so the
magnet wouldn't slow down.
72. Reverse Problem Write a physics problem with real-life objects for which the following equation would be part of
theManual
solution:
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b. The eddy currents cause the pipe to get warmer. As the magnet falls, its gravitational potential
energy decreases but its kinetic energy does not. The thermal energy of the pipe increases and as it
gets hotter than the surrounding air, heat is transferred to the air. c. A25
non-conducting
(or poorly
conducting)
pipe would have no or very small eddy currents, and so the
Chapter
Practice Problems,
Review,
and Assessment
magnet wouldn't slow down.
72. Reverse Problem Write a physics problem with real-life objects for which the following equation would be part of
the solution:
SOLUTION: Answers will vary, but a correct form of the answer is: “A metal bar of length 0.80 m is moving to the
left through a 0.15-T magnetic field directed straight up. If the EMF induced between the ends is 6.5 V,
what is the speed at which the wire is moving?”
Chapter Assessment: Mixed Review
73. Problem Posing Complete this problem so it can be solved using the concept of electromagnetic inductance: “A
copper wire of length 1.25 m….” (Level 2)
SOLUTION: Answers will vary. A possible form of the correct answer would be: “…moves perpendicularly to a 0.06T magnetic field at a speed of 5.0 m/s. What is the EMF induced between its ends?”
74. A step-up transformer’s primary coil has 500 turns. Its secondary coil has 15,000 turns. The primary circuit is connected to an AC generator having an EMF of 120 V. (Level 1)
a. Calculate the EMF of the secondary circuit.
b. Find the current in the primary circuit if the current in the secondary circuit is 3.0 A.
c. What power is drawn by the primary circuit? What power is supplied by the secondary circuit?
SOLUTION: a. b. c. 75. With what speed must a 0.20-m-long wire cut across a magnetic field for which B is 2.5 T if it is to have an EMF
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of 10 V induced in it? (Level 1)
SOLUTION: Page 21
Chapter 25 Practice Problems, Review, and Assessment
75. With what speed must a 0.20-m-long wire cut across a magnetic field for which B is 2.5 T if it is to have an EMF
of 10 V induced in it? (Level 1)
SOLUTION: 76. At what speed must a wire conductor 50-cm long be moved at right angles to a magnetic field of induction 0.20 T to
induce an EMF of 1.0 V? (Level 1)
SOLUTION: 77. A house lighting circuit is rated at 120-V effective potential difference. What is the peak potential difference that
can be expected in this circuit? (Level 1)
SOLUTION: 78. Toaster A toaster draws 2.5 A of alternating current. What is the peak current through this toaster? (Level 1)
SOLUTION: 79. The insulation of a capacitor will break down if the instantaneous potential difference exceeds 575 V. What is the largest effective alternating potential difference that can be applied to the capacitor? (Level 1)
SOLUTION: 80. Circuit Breaker A magnetic circuit breaker will open its circuit if the instantaneous current reaches 21.25 A. What is the largest effective current the circuit will carry? (Level 1)
SOLUTION: 81. The electricity received at an electrical substation has a potential difference of 240,000 V. What should the turns ratio of the step-down transformer be to have an output of 440 V? (Level 1)
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SOLUTION: Page 22
What is the largest effective current the circuit will carry? (Level 1)
SOLUTION: Chapter 25 Practice Problems, Review, and Assessment
81. The electricity received at an electrical substation has a potential difference of 240,000 V. What should the turns ratio of the step-down transformer be to have an output of 440 V? (Level 1)
SOLUTION: 82. An AC generator supplies a 45-kW industrial electric heater. If the effective potential difference is 660 V, what is the peak current supplied? (Level 1)
SOLUTION: 83. Hybrid autos use regenerative braking. When the driver applies the brakes, electronic circuits connect the electric
motor to the battery in such a way that the battery is charged. How would this connection slow the automobile?
(Level 1)
SOLUTION: When the driver applies the brakes of an electric or hybrid vehicle, the motor goes into generator
mode. The kinetic energy of the car is converted into electricity and charges the batteries. This in turn
slows the wheels of the vehicle, bringing it to a stop. These vehicles have a back up braking system for
high-speed, emergency situations.
84. Would you expect a real transformer to become hotter if it supplied a small or a large current? Explain. (Level 2) SOLUTION: Large current. The efficiency doesn't change, so if the output power is increased, so is the difference
between input and output power, and thus the rate of heating. 85. A step-down transformer has 100 turns on the primary coil and 10 turns on the secondary coil. If a 2.0-kW resistive
load is connected to the transformer, what is the effective primary current? Assume the peak secondary potential
difference is 60.0 V. (Level 1) SOLUTION: 86. A transformer rated at 100 kW has an efficiency of 98 percent. (Level 1)
a. If the connected load consumes 98 kW power, what is the value of the input kW to the transformer?
b. What is the maximum primary current when the transformer delivers its rated power? Assume that VP = 600 V.
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SOLUTION: a. Page 23
Chapter 25 Practice Problems, Review, and Assessment
86. A transformer rated at 100 kW has an efficiency of 98 percent. (Level 1)
a. If the connected load consumes 98 kW power, what is the value of the input kW to the transformer?
b. What is the maximum primary current when the transformer delivers its rated power? Assume that VP = 600 V.
SOLUTION: a. b. 87. A wire, 0.40-m long, cuts perpendicularly at a velocity of 8.0 m/s across a magnetic field for which B is 2.0 T. (Level 2)
a. What EMF is induced in the wire?
b. If the wire is in a circuit with a resistance of 6.4 Ω, what is the size of the current in the wire?
SOLUTION: a. b. 88. A 7.50-m-long wire is moved perpendicularly to Earth’s magnetic field at 5.50 m/s. What is the size of the current −2 −5 in the wire if the total resistance of the wire is 5.0×10 Ω? Assume Earth’s magnetic field is 5×10 T. (Level 2)
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2 b. Chapter 25 Practice Problems, Review, and Assessment
88. A 7.50-m-long wire is moved perpendicularly to Earth’s magnetic field at 5.50 m/s. What is the size of the current −2 −5 in the wire if the total resistance of the wire is 5.0×10 Ω? Assume Earth’s magnetic field is 5×10 T. (Level 2)
SOLUTION: 2 89. The peak value of the alternating potential difference applied to a 144-V resistor is 1.00×10 Ω. What maximum
power can the resistor handle? (Level 2)
SOLUTION: 90. X-rays A dental X-ray machine uses a step-up transformer to change 240 V to 96,000 V. The secondary side of the transformer has 200,000 turns and an output of 1.0 mA. (Level 2)
a. How many turns does the primary side have?
b. What is the input current?
SOLUTION: a. b. eSolutions Manual - Powered by Cognero
Chapter Assessment: Thinking Critically
Page 25
Chapter 25 Practice Problems, Review, and Assessment
90. X-rays A dental X-ray machine uses a step-up transformer to change 240 V to 96,000 V. The secondary side of the transformer has 200,000 turns and an output of 1.0 mA. (Level 2)
a. How many turns does the primary side have?
b. What is the input current?
SOLUTION: a. b. Chapter Assessment: Thinking Critically
91. Apply Concepts Suppose an “anti-Lenz’s law” existed so that an induced current resulted in a force that
increased the change in a magnetic field. Thus, when more energy was demanded, the force needed to turn a
generator was reduced. What conservation law would be violated by this “anti-law”? Explain.
SOLUTION: It would violate the law of conservation of energy. It would allow a changing magnetic field to grow
without limits. The current would increase without any work done. A generator would create energy, not
just change it from one form to another. 92. Analyze A step-down transformer that has an efficiency of 92.5 percent is used to obtain 28.0 V from a 125-V
household potential difference. The current in the secondary circuit is 25.0 A. a. Write an expression for transformer efficiency in percent, using power relationships.
b. What is the current in the primary circuit?
c. At what rate is thermal energy generated?
SOLUTION: a. b. Manual - Powered by Cognero
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SOLUTION: It would violate the law of conservation of energy. It would allow a changing magnetic field to grow
without limits. The current would increase without any work done. A generator would create energy, not
Chapter
25 Practice Problems, Review, and Assessment
just change it from one form to another. 92. Analyze A step-down transformer that has an efficiency of 92.5 percent is used to obtain 28.0 V from a 125-V
household potential difference. The current in the secondary circuit is 25.0 A. a. Write an expression for transformer efficiency in percent, using power relationships.
b. What is the current in the primary circuit?
c. At what rate is thermal energy generated?
SOLUTION: a. b. c. Thermal energy will be produced at the rate of 57 J/s. This is the difference between the input power
(757 W) and the output power (700 W).
93. Analyze and Conclude A transformer that supplies eight homes has an efficiency of 95 percent. Each home contains an electric oven that draws 35 A from 240-V lines.
a. How much power is supplied to the ovens in the eight homes?
b. How much power is wasted as heat in the transformer?
SOLUTION: a. Secondary power: Ps = (# of homes)VsIs = (8)(240 V)(35 A) = 67 kW 67 kW is supplied to the ovens in the eight homes.
b. Primary power:
The difference between the secondary and primary power is the power that heats up the transformer, 4
kW.
Chapter Assessment: Writing in Physics
94. Common tools, such as an electric drills and circular saws, are typically constructed using universal motors. Using
your local library and other sources, explain how this type of motor may operate on either AC or DC current. SOLUTION: eSolutions
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A series DC motor uses both an armature and series coil. When operated on alternating current, the
polarity on both fields changes simultaneously. Therefore, the polarity of the magnetic field remains
The25
difference
theReview,
secondary
primary power is the power that heats up the transformer, 4
Chapter
Practicebetween
Problems,
andand
Assessment
kW.
Chapter Assessment: Writing in Physics
94. Common tools, such as an electric drills and circular saws, are typically constructed using universal motors. Using
your local library and other sources, explain how this type of motor may operate on either AC or DC current. SOLUTION: A series DC motor uses both an armature and series coil. When operated on alternating current, the
polarity on both fields changes simultaneously. Therefore, the polarity of the magnetic field remains
unchanged, and hence the direction of rotation is constant. Chapter Assessment: Cumulative Review
14 95. Astronomers find that a distant star emits light at a frequency of 4.56×10 Hz. If the star is moving toward Earth
at a speed of 2750 km/s, what frequency light will observers on Earth detect?
SOLUTION: 96. A distant galaxy emits light at a frequency of 7.29×10
6.14×10
14 Hz. Observers on Earth receive the light at a frequency of
14 Hz. How fast is the galaxy moving, and in what direction?
SOLUTION: Because the observed light has a lower frequency, the galaxy must be moving away from Earth. So, use
the negative form of the equation for the observed light frequency.
97. How much charge is on a 22-μF capacitor with 48 V applied to it?
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Chapter 25 Practice Problems, Review, and Assessment
97. How much charge is on a 22-μF capacitor with 48 V applied to it?
SOLUTION: 98. Find the potential difference across a 22-Ω, 5.0-W resistor operating at half of its rating.
SOLUTION: 99. What is the total resistance of three 85-Ω resistors connected in parallel and then series-connected to two 85-Ω resistors in parallel, as in Figure 27?
SOLUTION: 6 100. An electron with a velocity of 2.1×10 m/s is at right angles to a 0.81-T magnetic field. What is the force on the
electron produced by the magnetic field? What is the electron’s acceleration? The mass of an electron is
9.11×10
−31
kg.
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Chapter 25 Practice Problems, Review, and Assessment
6 100. An electron with a velocity of 2.1×10 m/s is at right angles to a 0.81-T magnetic field. What is the force on the
electron produced by the magnetic field? What is the electron’s acceleration? The mass of an electron is
9.11×10
−31
kg.
SOLUTION: eSolutions Manual - Powered by Cognero
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