Chapter 24 Practice Problems, Review, and Assessment
Section 1 Understanding Magnetism: Practice Problems
1. If you hold a bar magnet in each hand and bring your hands close together, will the force be attractive or repulsive
if the magnets are held in the following ways?
a. the two north poles are brought close together
b. a north pole and a south pole are brought together
SOLUTION:
a. repulsive
b. attractive
2. Figure 4 shows five disk magnets floating above one another. The north pole of the top-most disk faces up. Which
poles are on the top side of each of the other magnets?
SOLUTION:
south, north, south, north
3. The ends of a compass needle are marked N and S. How would you explain to someone why the pole marked N
points north? A complete answer should involve Earth’s magnetic poles.
SOLUTION:
Earth’s geographic north pole is actually its southern magnetic pole.
4. CHALLENGE When students use magnets and compasses, they often touch the magnets to the compasses. Then
they find that the compasses point south. Explain why this might occur.
SOLUTION:
When students bring compasses near magnets, the magnetization of the compasses flips. The iron in the
compass needle becomes a temporary magnet.
5. How does the strength of a magnetic field that is 1 cm from a current-carrying wire compare with each of the
following?
a. the strength of the field 2 cm from the wire
b. the strength of the field 3 cm from the wire
SOLUTION:
a. Because magnetic field strength varies inversely with distance from the wire, the magnetic field at 1
cm will be twice as strong as the magnetic field at 2 cm.
b. Because magnetic field strength varies inversely with distance from the wire, the magnetic field at 1
cm will be three times as strong as the magnetic field at 3 cm.
6. A long, straight current-carrying wire lies in a north-south direction.
a. The north pole of a compass needle placed above this wire points toward the east. In what direction is the
current?
b. If a compass were placed underneath this wire, in which direction would the compass needle point?
SOLUTION:
a. from south to north
b. west
7. A student makes a magnet by winding wire around a nail and connecting it to a battery, as in Figure 13. Which
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end or the head—is the north pole?
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of the- nail—the
b. If a compass were placed underneath this wire, in which direction would the compass needle point?
SOLUTION:
Chapter
24 Practice
Review, and Assessment
a. from
south toProblems,
north
b. west
7. A student makes a magnet by winding wire around a nail and connecting it to a battery, as in Figure 13. Which
end of the nail—the pointed end or the head—is the north pole?
SOLUTION:
the pointed end
8. You have a battery, a spool of wire, a glass rod, an iron rod, and an aluminum rod. Which rod could you use to
make an electromagnet that can pick up steel objects? Explain.
SOLUTION:
Use the iron rod. Iron would be attracted to a permanent magnet and take on properties of a magnet,
whereas aluminum or glass would not. This effect would support the magnetic field in the wire coil and
thus make the strongest electromagnet.
9. Challenge The electromagnet in the previous problem works well, but you would like to make the strength of the
electromagnet adjustable by using a potentiometer as a variable resistor. Is this possible? Explain.
SOLUTION:
Yes. Connect the potentiometer in series with the power supply and the coil. Adjusting the
potentiometer for more resistance will decrease the current and the field strength.
Section 1 Understanding Magnetism: Review
10. MAIN IDEA Explain how to construct an electromagnet.
SOLUTION:
You could connect either end of a wire to a source of current. The strongest electromagnets are
solenoids, where wire in a circuit is wrapped around a ferromagnetic rod, such as iron,
11. Magnetic Fields What two things about a magnetic field can magnetic field lines represent?
SOLUTION:
Field lines represent the strength and the direction of a magnetic field.
12. Magnetic Forces Identify some magnetic forces around you. How could you demonstrate the effects of those
forces?
SOLUTION:
Student answers may vary. Answers could include magnets on a refrigerator and Earth’s magnetic field.
The effects of these forces can be demonstrated by bringing another magnet, or a ferromagnetic
material, nearby.
13. Magnetic Fields Where on a bar magnet is the magnetic field the strongest?
SOLUTION:
at the poles
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14. Magnetic Fields Two current-carrying wires are close to and parallel to each other and have identical currents.
If
the two currents were in the same direction, how would the magnetic fields of the wires be affected? How would
the fields be affected if the two currents were in opposite directions?
material, nearby.
13. Magnetic Fields Where on a bar magnet is the magnetic field the strongest?
Chapter
24 Practice Problems, Review, and Assessment
SOLUTION:
at the poles
14. Magnetic Fields Two current-carrying wires are close to and parallel to each other and have identical currents. If
the two currents were in the same direction, how would the magnetic fields of the wires be affected? How would
the fields be affected if the two currents were in opposite directions?
SOLUTION:
If the currents were in the same direction, the magnetic field would be approximately twice as large; if
the currents were in opposite directions, the field would be approximately zero.
15. Direction of Field Describe how to use a right-hand rule to determine the direction of a magnetic field around a
straight, current-carrying wire.
SOLUTION:
If you grasp the wire with your right hand, with your thumb pointing in the direction of the conventional
current, your fingers curl in the direction of the magnetic field.
16. Electromagnets A glass sheet with iron filings sprinkled on it is placed over an active electromagnet. The iron
filings produce a pattern. If this scenario were repeated with the direction of current reversed, what observable
differences would result? Explain.
SOLUTION:
None; the filings would show the same field pattern. However, a compass would show that the magnetic
polarity had reversed.
17. Magnetic Domains Explain what happens to the domains of a temporary magnet when the temporary magnet is
removed from a magnetic field.
SOLUTION:
The domains return to a random arrangement because they no longer align with the domains of the field
of the permanent magnet.
18. Critical Thinking Imagine a toy containing two parallel, horizontal metal rods, one above the other. The top rod is
free to move up and down.
a. The top rod floats above the lower one If the top rod’s direction is reversed, however, it falls down onto the
lower rod. Explain why the rods could behave in this way.
b. Assume the toy’s top rod was lost and another rod replaced it. The new rod falls on top of the bottom rod no
matter its orientation. What type of material is in the replacement rod?
SOLUTION:
a. The metal rods could be magnets with their axes parallel. If the top magnet is positioned so that its
north and south poles are above the north and south poles of the bottom magnet, it will be repelled and
float above. If the top magnet is turned end for end, it will be attracted to the bottom magnet.
b. If the bar is made from a ferromagnetic material, such as iron, it will be attracted to the bottom
magnet in any orientation.
Section 2 Applying Magnetic Forces: Practice Problems
19. Explain the method you could use to determine the direction of force on a current-carrying wire at right angles to a
magnetic field. Identify what must be known to use this method.
SOLUTION:
You would use the right-hand rule for magnetic force on a wire. When you point the fingers of your right
hand in the direction of the magnetic field and your thumb in the direction of the wire’s conventional
(positive) current, the palm of your hand will face in the direction of the force acting on the wire. To use
thisManual
method,
youbywould
need to know the direction of the current and the direction of the field.
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20. A wire that is 0.50 m long and carrying a current of 8.0 A is at right angles to a 0.40-T magnetic field. How strong
a. The metal rods could be magnets with their axes parallel. If the top magnet is positioned so that its
north and south poles are above the north and south poles of the bottom magnet, it will be repelled and
float above. If the top magnet is turned end for end, it will be attracted to the bottom magnet.
Chapter
Problems,
and Assessment
b. If24
thePractice
bar is made
from aReview,
ferromagnetic
material, such as iron, it will be attracted to the bottom
magnet in any orientation.
Section 2 Applying Magnetic Forces: Practice Problems
19. Explain the method you could use to determine the direction of force on a current-carrying wire at right angles to a
magnetic field. Identify what must be known to use this method.
SOLUTION:
You would use the right-hand rule for magnetic force on a wire. When you point the fingers of your right
hand in the direction of the magnetic field and your thumb in the direction of the wire’s conventional
(positive) current, the palm of your hand will face in the direction of the force acting on the wire. To use
this method, you would need to know the direction of the current and the direction of the field.
20. A wire that is 0.50 m long and carrying a current of 8.0 A is at right angles to a 0.40-T magnetic field. How strong
is the force that acts on the wire?
SOLUTION:
F = ILB = (8.0 A)(0.50 m)(0.40 N/Aim ) = 1.6 N
21. A wire that is 75 cm long and carrying a current of 6.0 A is at right angles to a uniform magnetic field. The
magnitude of the force acting on the wire is 0.60 N. What is the strength of the magnetic field?
SOLUTION:
F = ILB
22. A 40.0-cm-long copper wire carries a current of 6.0 A and weighs 0.35 N. A certain magnetic field is strong
enough to balance the force of gravity on the wire. What is the strength of the magnetic field?
SOLUTION:
F = ILB, where F = weight of the wire
23. How much current would be required to produce a force of 0.38 N on a 10.0-cm length of wire at right angles to a
0.49-T field?
SOLUTION:
24. CHALLENGE You are making your own loudspeaker. You make a 1-cm-diameter coil with 20 loops of thin wire.
You use hot glue to fasten the coil to an aluminum pie plate. The ends of the wire are connected to a plug that goes
into the earphone jack on an MP3 music player. You have a bar magnet to produce a magnetic field. How would
you orient the magnetic field to make the plate vibrate and produce sound?
SOLUTION:
One pole should be held as close to the coil as possible so that the field lines are perpendicular to both
the wires and the direction of motion of the plate.
25. In what direction is the force on an electron if that electron is moving east through a magnetic field that points
north?
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SOLUTION:
down
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you orient the magnetic field to make the plate vibrate and produce sound?
SOLUTION:
Chapter
Practice
Assessment
One24
pole
shouldProblems,
be held asReview,
close toand
the coil
as possible so that the field lines are perpendicular to both
the wires and the direction of motion of the plate.
25. In what direction is the force on an electron if that electron is moving east through a magnetic field that points
north?
SOLUTION:
down
26. What are the magnitude and direction of the force acting on the proton shown in Figure 20?
SOLUTION:
F = qvB
7
−19
= (+1.60×10
C)(4.0×10 m/s)(0.50 T)
−12
= 3.2×10
N
The direction of force is up.
27. A stream of doubly ionized particles (missing two electrons and thus carrying a net charge of two elementary
4
−2
charges) moves at a velocity of 3.0×10 m/s perpendicular to a magnetic field of 9.0×10
acting on each ion?
T. How large is the force
SOLUTION:
F = qvB
= (2)(1.60×10−19 C)
4
(3.0×10 m/s)(9.0×10
= 8.6×10−16 N
−2
T)
28. Triply ionized particles in a beam carry a net positive charge of three elementary charge units. The beam enters a
−2
6
magnetic field of 4.0×10 T. The particles have a speed of 9.0×10 m/s and move at right angles to the field. How
large is the force acting on each particle?
SOLUTION:
F = qvB
= (3)(1.60×10−19 C)(9.0×104 m/s)(4.0×10−2 T)
−13
= 1.7×10
N
29. A singly ionized particle experiences a force of 4.1×10
magnetic field. What is the particle’s velocity?
−13
N when it travels at right angles through a 0.61-T
SOLUTION:
F = qvB
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SOLUTION:
F = qvB
−19
4
−2
(3)(1.60×10
C)(9.0×10
m/s)(4.0×10
T)
Chapter=24
Practice Problems,
Review,
and Assessment
−13
= 1.7×10
N
29. A singly ionized particle experiences a force of 4.1×10
magnetic field. What is the particle’s velocity?
−13
N when it travels at right angles through a 0.61-T
SOLUTION:
F = qvB
30. CHALLENGE Doubly ionized helium atoms (alpha particles) are traveling at right angles to a magnetic field at a
4
speed of 4.0×10 m/s. The force on each particle is 6.4×10
−16
N. What is the magnetic field strength?
SOLUTION:
Section 2 Applying Magnetic Forces: Review
31. MAIN IDEA Explain how electric motors use magnets to convert electrical energy to mechanical energy.
SOLUTION:
An armature in a magnetic field rotates 360° as a split-ring commutator changes the direction of current,
producing mechanical energy.
32. Magnetic Forces Imagine that a current-carrying wire is perpendicular to Earth’s magnetic field and runs eastwest. If the current is east, in which direction is the force on the wire?
SOLUTION:
up, away from the surface of Earth
33. Synchrotrons In a synchrotron, magnetic fields bend particle beams into segments of a circle, and electric fields
accelerate the beams.
a. A beam of protons circulates in a clockwise direction. In what direction must the magnetic field be oriented? In
what direction must the electric fields be oriented?
b. If a beam of negatively charged antiprotons is to circulate in a counterclockwise direction, must the direction of
the magnetic field be changed? Must the direction of the electric fields be changed?
SOLUTION:
a. The magnetic field must be up, at a right angle to the protons’ velocity. This is found by using a righthand rule. The electric fields should be in the direction of the velocity— clockwise.
b. Neither field needs to be changed. Both the sign of the charge and the direction of motion have been
reversed, so the magnetic field is in the same direction. The electric fields remain the same because
negatively charged particles are accelerated in the direction opposite that of the electric field.
34. Galvanometers Compare the diagram of a galvanometer in the left part of Figure 17 with the electric motor in
Figure 19. How is the galvanometer similar to an electric motor? How is it different?
SOLUTION:
Both
the- galvanometer
and the electric motor use a loop of wire positioned between the poles of a Page 6
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permanent magnet. When a current passes through the loop, the magnetic field of the permanent
magnet exerts a force on the loop. The loop in a galvanometer cannot rotate more than 180°. The loop
a. The magnetic field must be up, at a right angle to the protons’ velocity. This is found by using a righthand rule. The electric fields should be in the direction of the velocity— clockwise.
b. Neither field needs to be changed. Both the sign of the charge and the direction of motion have been
Chapter
24 Practice
Review,
reversed,
so theProblems,
magnetic field
is inand
the Assessment
same direction. The electric fields remain the same because
negatively charged particles are accelerated in the direction opposite that of the electric field.
34. Galvanometers Compare the diagram of a galvanometer in the left part of Figure 17 with the electric motor in
Figure 19. How is the galvanometer similar to an electric motor? How is it different?
SOLUTION:
Both the galvanometer and the electric motor use a loop of wire positioned between the poles of a
permanent magnet. When a current passes through the loop, the magnetic field of the permanent
magnet exerts a force on the loop. The loop in a galvanometer cannot rotate more than 180°. The loop
in an electric motor rotates through many 360° turns. The motor’s split-ring commutator allows the
current in the loop to reverse as the loop becomes vertical in the magnetic field, enabling the loop to
spin in the magnetic field. The galvanometer measures unknown currents; the electric motor has many
uses.
35. Motors When the plane of an armature in a motor is perpendicular to the magnetic field, the forces do not exert a
torque on the coil. Does this mean that the coil does not rotate? Explain.
SOLUTION:
Not necessarily; if the coil is already in rotation, then rotational inertia will carry it past the point of zero
torque. It is the coil’s acceleration that is zero, not the velocity.
36. Resistance A galvanometer requires 180 μA for full-scale deflection. When it is used as a voltmeter, what total
resistance of the meter and the multiplier resistor is needed for a 5.0-V full-scale deflection?
SOLUTION:
37. Critical Thinking Two current-carrying wires move toward each other when they are placed parallel to each
other. Compare the directions of the two currents. Explain your reasoning.
SOLUTION:
Because the force is attractive, the currents are in the same direction. That is, an up current in the first
wire creates a magnetic field that intersects the second wire. If the current in the second wire is in the
same direction, the force on it will pull the wires together.
Chapter Assessment
Section 1 Understanding Magnetism: Mastering Concepts
38. State a rule describing magnetic attraction and repulsion.
SOLUTION:
Like poles repel one another; opposite poles attract.
39. Describe how a temporary magnet differs from a permanent magnet.
SOLUTION:
A temporary magnet is like a magnet only while under the influence of another magnet. Its domains will
revert to a random arrangement when the magnet is removed. The domains of a permanent magnet are
permanently aligned.
40. Name three common ferromagnetic elements.
SOLUTION:
iron, cobalt, and nickel
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41. BIG IDEA Draw a small bar magnet with field lines around it. Use arrows to show the direction of the field. Draw
a small nail in this magnetic field with the induced north and south poles indicated. The bar magnet's field exerts an
attractive force on one pole of the nail and a repulsive force on the other. Why is the nail pulled toward the bar
SOLUTION:
A temporary magnet is like a magnet only while under the influence of another magnet. Its domains will
Chapter
24 Practice
Problems,
Review,
andthe
Assessment
revert
to a random
arrangement
when
magnet is removed. The domains of a permanent magnet are
permanently aligned.
40. Name three common ferromagnetic elements.
SOLUTION:
iron, cobalt, and nickel
41. BIG IDEA Draw a small bar magnet with field lines around it. Use arrows to show the direction of the field. Draw
a small nail in this magnetic field with the induced north and south poles indicated. The bar magnet's field exerts an
attractive force on one pole of the nail and a repulsive force on the other. Why is the nail pulled toward the bar
magnet?
SOLUTION:
The bar magnet's field is stronger closer to the bar magnet, so the attractive force on the closer pole is
stronger than the repulsive force on the further pole, making a net attractive force.
42. Draw the magnetic field between two like magnetic poles and then between two unlike magnetic poles. Show the
directions of the fields.
SOLUTION:
43. If you broke a magnet in two, would you have isolated north and south poles? Explain.
SOLUTION:
No; new poles would form on each of the broken ends.
44. Describe how to use a right-hand rule to determine the direction of a magnetic field around a straight, currentcarrying wire.
SOLUTION:
Grasp the wire with your right hand, keeping your thumb pointing in the direction of the conventional
current through the wire. Your fingers will encircle the wire and point in the direction of the magnetic
field.
45. If a current-carrying wire were bent into a loop, why would the magnetic field inside the loop be stronger than the
magnetic field outside?
SOLUTION:
The magnetic field is concentrated inside the loop because the direction of the fields from the individual
loops is always the same, and the fields add together.
46. Describe how to use a right-hand rule to determine the north and south ends of an electromagnet.
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SOLUTION:
Grasp the electromagnet with your right hand, keeping your fingers encircling the electromagnet in the
direction of the conventional current flow through the loops. The thumb of your right hand will point
magnetic field outside?
SOLUTION:
Chapter
Practicefield
Problems,
Review, inside
and Assessment
The24
magnetic
is concentrated
the loop because the direction of the fields from the individual
loops is always the same, and the fields add together.
46. Describe how to use a right-hand rule to determine the north and south ends of an electromagnet.
SOLUTION:
Grasp the electromagnet with your right hand, keeping your fingers encircling the electromagnet in the
direction of the conventional current flow through the loops. The thumb of your right hand will point
toward the north pole of the electromagnet.
47. Each atom in a piece of iron is like a tiny magnet. The iron, however, may not be a magnet. Explain.
SOLUTION:
The poles of the atoms do not necessarily align in the same direction. When not in the presence of a
strong magnet, the iron is not magnetized; the poles of its atoms point in random directions. If the iron
were placed near a strong magnet, however, the poles of the atoms would align.
48. Why will heating a magnet weaken it?
SOLUTION:
Its domains are jostled out of alignment.
Chapter Assessment
Section 1 Understanding Magnetism: Mastering Problems
49. Refer to Figure 22 to answer the following questions. (Level 1)
a. Where are the poles?
b. Where is the north pole?
c. Where is the south pole?
SOLUTION:
a. 4 and 2, by definition
b. 2, by definition and field direction
c. 4, by definition and field direction
50. As the magnet on the right in Figure 23 moves toward the magnet suspended on a string, what will the suspended
magnet do? (Level 1)
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SOLUTION:
a. 4 and 2, by definition
b. 2, by definition and field direction
Chapter
Problems,
c. 4,24
byPractice
definition
and field Review,
directionand Assessment
50. As the magnet on the right in Figure 23 moves toward the magnet suspended on a string, what will the suspended
magnet do? (Level 1)
SOLUTION:
Move to the left or begin to turn. Like poles repel.
51. As the magnet on the right in Figure 24 moves toward the magnet suspended on a string, what will the suspended
magnet do? (Level 1)
SOLUTION:
Move to the right. Unlike poles attract.
52. Figure 25 shows the response of a compass in two different positions near a magnet. Where is the south pole of
the magnet located? (Level 1)
SOLUTION:
At the right end. Unlike poles attract.
53. Copy the wire segment in Figure 26 and sketch the magnetic field that its current generates. (Level 1)
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SOLUTION:
Chapter
24 Practice Problems, Review, and Assessment
At the right end. Unlike poles attract.
53. Copy the wire segment in Figure 26 and sketch the magnetic field that its current generates. (Level 1)
SOLUTION:
54. The current is coming straight out of the page in the wire in Figure 27. Copy the figure and sketch the magnetic
field that the current generates. (Level 1)
SOLUTION:
55. Figure 28 shows the end view of an electromagnet that is carrying current. (Level 1)
a. What is the direction of the magnetic field inside the loops?
b. What is the direction of the magnetic field outside the loops?
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Chapter 24 Practice Problems, Review, and Assessment
55. Figure 28 shows the end view of an electromagnet that is carrying current. (Level 1)
a. What is the direction of the magnetic field inside the loops?
b. What is the direction of the magnetic field outside the loops?
SOLUTION:
a. down (into the page)
b. up (out of the page)
56. The repulsive force between two magnets was measured and found to depend on distance, as given in Table 1.
(Level 2)
a. Plot the force as a function of distance.
b. Does this force follow an inverse square law?
SOLUTION:
a.
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SOLUTION:
Chapter
24 Practice
Review, and Assessment
a. down
(into theProblems,
page)
b. up (out of the page)
56. The repulsive force between two magnets was measured and found to depend on distance, as given in Table 1.
(Level 2)
a. Plot the force as a function of distance.
b. Does this force follow an inverse square law?
SOLUTION:
a.
b. no
Chapter Assessment
Section 2 Applying Magnetic Forces: Mastering Concepts
57. What kind of meter is created when a shunt is added to a galvanometer?
SOLUTION:
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an Manual
ammeter
58. Electric Motor The torque on the armature of an electric motor depends on the length of the wires along the
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Chapter 24 Practice Problems, Review, and Assessment
b. no
Chapter Assessment
Section 2 Applying Magnetic Forces: Mastering Concepts
57. What kind of meter is created when a shunt is added to a galvanometer?
SOLUTION:
an ammeter
58. Electric Motor The torque on the armature of an electric motor depends on the length of the wires along the
armature’s axis of rotation. Are there forces on the wires perpendicular to that axis? If so, why do these forces not
contribute to the torque?
SOLUTION:
The forces on these wires are parallel to the axis of rotation. Only forces perpendicular to the axis can
contribute to the torque that rotates the armature.
59. A strong current suddenly is switched on in a wire. No force, however, acts on the wire. Can you conclude that
there is no magnetic field at the location of the wire? Explain.
SOLUTION:
No, if a magnetic field is parallel to a wire, there is no force on the wire.
60. Earbud You are able to listen to music through an earbud because current in a wire coil in the earbud changes
direction. What happens to the force on the wire when current changes direction?
SOLUTION:
The force from the magnetic field on the coil pushes the coil into or out of the field, depending on the
direction of current. When current is in one direction, the forces pushes the coil into the field; when
current is in the other direction, the force pushes the coil out of the field. The motion causes the cone
to vibrate, thereby creating sound waves in the air.
61. Synchrotron A synchrotron uses magnets to direct charged particles in a circular path. Explain how the magnetic
force on a charged particle makes it move in a circle.
SOLUTION:
Because a particle’s velocity is always at right angles to a magnetic field, a moving particle must travel
in a circle to maintain that angle.
62. The arrangement shown in Figure 29 is used to convert a galvanometer to what type of device? (Level 1)
SOLUTION:
Ammeter; much of the current is through the resistor and allows the measurement of higher currents.
63. What is the resistor shown in Figure 29 called? (Level 1)
SOLUTION:
Shunt; by definition, shunt is another word for parallel.
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62. The arrangement shown in Figure 29 is used to convert a galvanometer to what type of device? (Level 1)
SOLUTION:
Chapter
24 Practice
Review,
and Assessment
Ammeter;
muchProblems,
of the current
is through
the resistor and allows the measurement of higher currents.
63. What is the resistor shown in Figure 29 called? (Level 1)
SOLUTION:
Shunt; by definition, shunt is another word for parallel.
64. The arrangement shown in Figure 30 is used to convert a galvanometer to what type of device? (Level 1)
SOLUTION:
Voltmeter; the added resistance decreases the current for any given voltage.
65. What is the resistor shown in Figure 30 called? (Level 1)
SOLUTION:
Multiplier; it multiplies the voltage range of the meter.
Chapter Assessment
Section 2 Applying Magnetic Forces: Mastering Problems
66. A wire that is 0.50 m long and carrying a current of 8.0 A is at right angles to a uniform magnetic field. The force
on the wire is 0.40 N. What is the strength of the magnetic field? (Level 1)
SOLUTION:
67. A wire that is 25 cm long is at right angles to a 0.30-T uniform magnetic field. The current through the wire is
6.0 A. What is the magnitude of the force on the wire? (Level 1)
SOLUTION:
F = ILB = (6.0 A)(0.25 m)(0.30 N/A · m)
= 0.45 N
68. A wire that is 35 cm long is parallel to a 0.53-T uniform magnetic field. The current through the wire is 4.5 A. What
force acts on the wire? (Level 1)
SOLUTION:
If the wire is parallel to the field, no force is produced.
69. The force acting on a wire that is at right angles to a 0.80-T magnetic field is 3.6 N. The current in the wire is
7.5 A. How long is the wire? (Level 1)
SOLUTION:
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force acts on the wire? (Level 1)
SOLUTION:
Chapter
24 Practice Problems, Review, and Assessment
If the wire is parallel to the field, no force is produced.
69. The force acting on a wire that is at right angles to a 0.80-T magnetic field is 3.6 N. The current in the wire is
7.5 A. How long is the wire? (Level 1)
SOLUTION:
70. A current-carrying wire is placed between the poles of a magnet, as shown in Figure 31. What is the direction of
the force on the wire? (Level 1)
SOLUTION:
The force is down.
71. The force on a 0.80-m wire that is perpendicular to Earth’s magnetic field is 0.12 N. What is the current in the
−5
wire? Use 5.0×10 T for Earth’s magnetic field. (Level 1)
SOLUTION:
72. Problem Posing Complete this problem so that it must be solved using the concept of magnetic force: “A proton
has a velocity of 600 m/s….” (Level 2)
SOLUTION:
Answers will vary. A possible form of the correct answer would be: “…if it is traveling through a
magnetic field of strength 750 mT, what magnetic force does it experience?”
73. A power line carries a 225-A current from east to west, parallel to the surface of Earth. (Level 2)
a. What are the magnitude and direction of the force resulting from Earth’s magnetic field acting on each meter of
−5
the wire? Use BEarth = 5.0×10 T.
b. In your judgment, would this force be important in designing towers to hold this power line? Explain.
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SOLUTION:
a. The force would be downward.
Page 16
SOLUTION:
Answers
will vary.
A possible
form and
of the
correct answer would be: “…if it is traveling through a
Chapter
24 Practice
Problems,
Review,
Assessment
magnetic field of strength 750 mT, what magnetic force does it experience?”
73. A power line carries a 225-A current from east to west, parallel to the surface of Earth. (Level 2)
a. What are the magnitude and direction of the force resulting from Earth’s magnetic field acting on each meter of
−5
the wire? Use BEarth = 5.0×10 T.
b. In your judgment, would this force be important in designing towers to hold this power line? Explain.
SOLUTION:
a. The force would be downward.
b. No; the force is much smaller than the weight of the wires.
74. Galvanometer A galvanometer deflects full-scale for a 50.0-μA current. (Level 2)
a. What must be the total resistance of the series resistor and the galvanometer to make a voltmeter with 10.0-V
full-scale deflection?
b. If the galvanometer has a resistance of 1.0 kΩ, what should be the resistance of the series (multiplier) resistor?
SOLUTION:
a.
2
2
b. Total resistance = 2.00×10 kΩ, so the series resistor is 2.00×10 kΩ − 1.0 kΩ = 199 kΩ.
75. The galvanometer in the previous problem is used to make an ammeter that deflects full-scale for 10 mA. (Level 2)
a. What is the potential difference across the galvanometer (1.0 kΩ resistance) when a current of 50 μA passes
through it?
b. What is the equivalent resistance of parallel resistors having the potential difference calculated in a circuit with a
total current of 10 mA?
c. What resistor should be placed parallel with the galvanometer to make the resistance calculated in part b?
SOLUTION:
a. V = IR = (5.00×10−5 A)(1.0×103 Ω) = 0.05 V
b.
c.
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Page 17
Chapter 24 Practice Problems, Review, and Assessment
2
2
b. Total resistance = 2.00×10 kΩ, so the series resistor is 2.00×10 kΩ − 1.0 kΩ = 199 kΩ.
75. The galvanometer in the previous problem is used to make an ammeter that deflects full-scale for 10 mA. (Level 2)
a. What is the potential difference across the galvanometer (1.0 kΩ resistance) when a current of 50 μA passes
through it?
b. What is the equivalent resistance of parallel resistors having the potential difference calculated in a circuit with a
total current of 10 mA?
c. What resistor should be placed parallel with the galvanometer to make the resistance calculated in part b?
SOLUTION:
a. V = IR = (5.00×10−5 A)(1.0×103 Ω) = 0.05 V
b.
c.
−2
76. A beam of electrons moves at right angles to a magnetic field of 6.0×10 T. The electrons have a velocity of
6
2.5×10 m/s. What is the magnitude of the force on each electron? (Level 1)
SOLUTION:
77. A room contains a strong, uniform magnetic field. A loop of fine wire in the room has current in it. The loop is
rotated until there is no torque on it as a result of the field. What is the direction of the magnetic field relative to the
plane of the coil? (Level 2)
SOLUTION:
The magnetic field is perpendicular to the plane of the coil. You would use a right-hand rule to find the
direction of the field produced by the coil. The field in the room is in the same direction.
−28
78. Subatomic Particle A muon (a particle with the same charge as an electron but with a mass of 1.88×10 kg) is
7
−12
traveling at 4.21×10 m/s at right angles to a magnetic field. The muon experiences a force of −5.00×10 N.
(Level 2)
a. How strong is the magnetic field?
b. What acceleration does the muon experience?
SOLUTION:
a.
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plane of the coil? (Level 2)
SOLUTION:
The24
magnetic
is perpendicular
to the
plane of the coil. You would use a right-hand rule to find the
Chapter
Practicefield
Problems,
Review, and
Assessment
direction of the field produced by the coil. The field in the room is in the same direction.
−28
78. Subatomic Particle A muon (a particle with the same charge as an electron but with a mass of 1.88×10 kg) is
7
−12
traveling at 4.21×10 m/s at right angles to a magnetic field. The muon experiences a force of −5.00×10 N.
(Level 2)
a. How strong is the magnetic field?
b. What acceleration does the muon experience?
SOLUTION:
a.
b.
−16
79. A force of 5.78×10 N acts on an unknown particle traveling at a 90° angle through a magnetic field. If the
4
−2
velocity of the particle is 5.6×10 m/s and the field is 3.20×10 T, how many elementary charges does the particle
carry? (Level 2)
SOLUTION:
80. Ranking Task Rank the following situations according to the magnitude of magnetic force experienced, from
greatest to least. Indicate whether any of them are the same. (Level 2)
A. stationary charge of 8.0 nC in a magnetic field of 3.0 T
B. a charge of 3.0 nC moving with a speed of 9.0 m/s in a magnetic field of 1.0 T
C. a charge of 1.5 nC moving with a speed of 4.0 m/s in a magnetic field of 1.0 T
D. a charge of 6.0 nC moving with a speed of 9.0 m/s in a magnetic field of 0.5 T
E. a charge of 2.0 nC moving with a speed of 2.0 m/s in a magnetic field of 3.0 T
SOLUTION:
Use F = qvB. So, B = D > E > C > A
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Chapter Assessment: Applying Concepts
Page 19
81. A small bar magnet is hidden in a fixed position inside a tennis ball. Describe an experiment you could do to find the
location of the magnet’s poles.
Chapter 24 Practice Problems, Review, and Assessment
80. Ranking Task Rank the following situations according to the magnitude of magnetic force experienced, from
greatest to least. Indicate whether any of them are the same. (Level 2)
A. stationary charge of 8.0 nC in a magnetic field of 3.0 T
B. a charge of 3.0 nC moving with a speed of 9.0 m/s in a magnetic field of 1.0 T
C. a charge of 1.5 nC moving with a speed of 4.0 m/s in a magnetic field of 1.0 T
D. a charge of 6.0 nC moving with a speed of 9.0 m/s in a magnetic field of 0.5 T
E. a charge of 2.0 nC moving with a speed of 2.0 m/s in a magnetic field of 3.0 T
SOLUTION:
Use F = qvB. So, B = D > E > C > A
Chapter Assessment: Applying Concepts
81. A small bar magnet is hidden in a fixed position inside a tennis ball. Describe an experiment you could do to find the
location of the magnet’s poles.
SOLUTION:
Use a compass. The north pole of the compass needle will attract the south pole of the magnet and vice
versa.
82. Compass Suppose you are taking a walk in the woods and realize you are lost. You have a compass, but the red
paint marking the north pole of the compass needle has worn off. You also have a length of wire and a flashlight
with a battery. How could you identify the north pole of the compass?
SOLUTION:
Connect the wire to the battery so that the current is away from you in one section. Hold the compass
directly above and close to that section of the wire. By a right-hand rule, the end of the compass needle
that points right is the north pole.
83. Two parallel wires carry equal currents
a. If the two currents are in opposite directions, where will the magnetic field from the two wires be larger than the
field from either wire alone?
b. Where will the magnetic field from both wires be exactly twice as large as from one wire?
SOLUTION:
a. The magnetic field will be larger anywhere between the two wires.
b. The magnetic field will be twice as large along a line directly between the wires that is equal in
distance from each wire.
84. A nail is attracted to one pole of a permanent magnet. Describe how you could tell whether the metal is
magnetized.
SOLUTION:
If not magnetized, either end of the nail would be attracted to either pole of the magnet. If magnetized,
each end of the nail would be attracted to one pole and repelled from the other.
85. Is the magnetic force that Earth exerts on a compass needle less than, equal to, or greater than the force that the
needle exerts on Earth? Explain.
SOLUTION:
The strength of the forces are equal according to Newton’s third law.
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attract
a piece
86. A magnet
of iron that is not a permanent magnet. A charged rubber rod can attract an
uncharged insulator. What different microscopic processes produce these similar phenomena?
SOLUTION:
Page 20
85. Is the magnetic force that Earth exerts on a compass needle less than, equal to, or greater than the force that the
needle exerts on Earth? Explain.
SOLUTION:
Chapter
24 Practice Problems, Review, and Assessment
The strength of the forces are equal according to Newton’s third law.
86. A magnet can attract a piece of iron that is not a permanent magnet. A charged rubber rod can attract an
uncharged insulator. What different microscopic processes produce these similar phenomena?
SOLUTION:
The magnet causes the domains in the iron to point in the same direction. The charged rod separates
the positive and negative charges in the insulator.
87. A current-carrying wire runs across a laboratory bench. Describe at least two ways in which you could find the
direction of the current.
SOLUTION:
You could use a compass to find the direction of the magnetic field. You could also bring up a strong
magnet and find the direction of the force on the wire, then use a right-hand rule.
88. In which direction, in relation to a magnetic field, would you run a current-carrying wire so that the force on it,
resulting from the field, is minimized, or even made to be zero?
SOLUTION:
Run the wire parallel to the magnetic field.
89. A magnetic field can exert a force on a charged particle. Can the field change the particle’s kinetic energy?
Explain.
SOLUTION:
No, the force is always perpendicular to the velocity. No work is done. The energy is not changed.
90. As a beam of protons moves from the back to the front of a room, it is deflected upward by a magnetic field. What
is the direction of the field?
SOLUTION:
Facing the front of the room, the velocity is forward, the force is upward, and therefore, using a righthand rule, B is to the left.
91. Field lines representing Earth’s magnetic field lines are shown in Figure 32. At what location, poles or equator, is
the magnetic field strength greatest? Explain.
SOLUTION:
Earth’s magnetic field strength is greatest at the poles. The field lines are closer together at the poles.
Chapter Assessment: Mixed Review
5
92. A magnetic field of 16 T acts in a direction due west. An electron is traveling due south at 8.1×10 m/s. What are
Page 21
the magnitude and direction of the force acting on the electron? (Level 2)
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SOLUTION:
SOLUTION:
Chapter
24 Practice Problems, Review, and Assessment
Earth’s magnetic field strength is greatest at the poles. The field lines are closer together at the poles.
Chapter Assessment: Mixed Review
5
92. A magnetic field of 16 T acts in a direction due west. An electron is traveling due south at 8.1×10 m/s. What are
the magnitude and direction of the force acting on the electron? (Level 2)
SOLUTION:
upward (right-hand rule—remembering that direction of force is reversed with negative charge)
93. Subatomic Particle A beta particle (high-speed electron) is traveling at right angles to a 0.60-T magnetic field. It
7
has a speed of 2.5×10 m/s. What force acts on the particle? (Level 2)
SOLUTION:
F = qvB
−19
7
= (1.6×10
C)(2.5×10 m/s)(0.60 T)
−12
= −2.4×10
N
94. A copper wire of insignificant resistance is placed in the center of an air gap between two magnetic poles, as
shown in Figure 33. The field is confined to the gap and has a strength of 1.9 T. (Level 2)
a. Determine the force on the wire (direction and magnitude) when the switch is open.
b. Determine the force on the wire (direction and magnitude) when the switch is closed.
c. Determine the force on the wire (direction and magnitude) when the switch is closed and the battery is reversed.
d. Determine the force on the wire (direction and magnitude) when the switch is closed and the wire has two 5.5-Ω
resistors in series.
SOLUTION:
a. 0 N. With no current, no magnetic field is produced by the wire. Copper is not a magnetic material.
b.
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Page 22
SOLUTION:
F = qvB
−19
7
= (1.6×10
C)(2.5×10
Chapter
24 Practice
Problems,m/s)(0.60
Review, T)
and Assessment
−12
= −2.4×10
N
94. A copper wire of insignificant resistance is placed in the center of an air gap between two magnetic poles, as
shown in Figure 33. The field is confined to the gap and has a strength of 1.9 T. (Level 2)
a. Determine the force on the wire (direction and magnitude) when the switch is open.
b. Determine the force on the wire (direction and magnitude) when the switch is closed.
c. Determine the force on the wire (direction and magnitude) when the switch is closed and the battery is reversed.
d. Determine the force on the wire (direction and magnitude) when the switch is closed and the wire has two 5.5-Ω
resistors in series.
SOLUTION:
a. 0 N. With no current, no magnetic field is produced by the wire. Copper is not a magnetic material.
b.
c. Down, 0.62 N. The direction of the force is given by a right-hand rule, and the magnitude of the force
is the same as in part b.
d. Up, 0.31 N. The direction of the force is given by a right-hand rule.
95. Two galvanometers are available. One has 50.0-μ A full-scale sensitivity and the other has 500.0-μ A full-scale
sensitivity. Both have the same coil resistance of 855 Ω. Your challenge is to convert them to measure a current of
100.0 mA, full-scale. (Level 2)
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a. Determine the shunt resistor for the 50.0-μA meter.
b. Determine the shunt resistor for the 500.0-μA meter.
c. Determine which of the two is better for actual use. Explain.
Page 23
Chapter 24 Practice Problems, Review, and Assessment
95. Two galvanometers are available. One has 50.0-μ A full-scale sensitivity and the other has 500.0-μ A full-scale
sensitivity. Both have the same coil resistance of 855 Ω. Your challenge is to convert them to measure a current of
100.0 mA, full-scale. (Level 2)
a. Determine the shunt resistor for the 50.0-μA meter.
b. Determine the shunt resistor for the 500.0-μA meter.
c. Determine which of the two is better for actual use. Explain.
SOLUTION:
a. Find the voltage across the meter coil at full scale.
V = IR = (50.0 μ A)(855 Ω) = 0.0428 V
Calculate the shunt resistor.
b. Find the voltage across the meter coil at full scale.
V = IR = (500.0 μ A)(855 Ω) = 0.428 V
Calculate the shunt resistor.
c. The 50.0-μ A meter is better. Its lower shunt resistance will do less to alter the total resistance of the
circuit being measured. An ideal ammeter has 0-Ω resistance.
96. Loudspeaker The magnetic field in a loudspeaker is 0.15 T. The wire makes 250 turns around a cylindrical form
that is 2.5 cm in diameter. The resistance of the wire is 8.0 Ω. Find the force on the wire when 15 V is placed
across the wire. (Level 2)
SOLUTION:
97. A wire carrying 15 A of current has a length of 25 cm in a uniform magnetic field of 0.85 T. Using the equation
F = ILB(sin θ ), find the force on the wire when it makes the following angles with the magnetic field lines.
(Level 2)
a. 90°
b. 45°
c. 0°
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SOLUTION:
a. F = ILB(sin θ)
Page 24
Chapter 24 Practice Problems, Review, and Assessment
97. A wire carrying 15 A of current has a length of 25 cm in a uniform magnetic field of 0.85 T. Using the equation
F = ILB(sin θ ), find the force on the wire when it makes the following angles with the magnetic field lines.
(Level 2)
a. 90°
b. 45°
c. 0°
SOLUTION:
a. F = ILB(sin θ)
= (15 A)(0.25 m)(0.85 T)(sin 90°)
= 3.2 N
b. F = ILB(sin θ )
= (15 A)(0.25 m)(0.85 T)(sin 45°)
= 2.3 N
c. sin 0° = 0
so F = 0 N
4
98. An electron is accelerated from rest through a potential difference of 2.0×10 V, which exists between plates P1
and P2, shown in Figure 34. The electron then passes through a small opening into a magnetic field of uniform field
strength. As indicated, the magnetic field is directed into the page. (Level 2)
a. State the direction of the electric field between the plates as either P1 to P2 or P2 to P1.
b. In terms of the information given, calculate the electron’s speed at plate P2.
c. Describe the motion of the electron through the magnetic field.
SOLUTION:
a. from P2 to P1
b.
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c. cManual
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Chapter Assessment: Thinking Critically
Page 25
b. F = ILB(sin θ )
= (15 A)(0.25 m)(0.85 T)(sin 45°)
= 2.3 N
c. sin
=0
Chapter
240°
Practice
Problems, Review, and Assessment
so F = 0 N
4
98. An electron is accelerated from rest through a potential difference of 2.0×10 V, which exists between plates P1
and P2, shown in Figure 34. The electron then passes through a small opening into a magnetic field of uniform field
strength. As indicated, the magnetic field is directed into the page. (Level 2)
a. State the direction of the electric field between the plates as either P1 to P2 or P2 to P1.
b. In terms of the information given, calculate the electron’s speed at plate P2.
c. Describe the motion of the electron through the magnetic field.
SOLUTION:
a. from P2 to P1
b.
c. clockwise
Chapter Assessment: Thinking Critically
99. Synchrotron The LHC synchrotron has a 27-km circumference and 1232 bending magnets. Protons of mass
–27
–19
8
1.67×10 kg and charge 1.602×10 C are all traveling at speeds close to the speed of light, 3.00×10 m/s. What
magnetic field would be required to bend their paths into a circle? Hint: Recall that centripetal force is given by
2
F = mv /r. The forces come from the particle moving through the magnetic field. The radius of a circle of
circumference C is r = C/2π. Note that due to the effects of special relativity on the protons, the actual magnetic
field needed is 8.36 T.
SOLUTION:
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Page 26
Chapter 24 Practice Problems, Review, and Assessment
c. clockwise
Chapter Assessment: Thinking Critically
99. Synchrotron The LHC synchrotron has a 27-km circumference and 1232 bending magnets. Protons of mass
–27
–19
8
1.67×10 kg and charge 1.602×10 C are all traveling at speeds close to the speed of light, 3.00×10 m/s. What
magnetic field would be required to bend their paths into a circle? Hint: Recall that centripetal force is given by
2
F = mv /r. The forces come from the particle moving through the magnetic field. The radius of a circle of
circumference C is r = C/2π. Note that due to the effects of special relativity on the protons, the actual magnetic
field needed is 8.36 T.
SOLUTION:
This is less than 1/10,000 the strength actually needed.
100. Apply Concepts A current is sent through a vertical spring, as shown in Figure 35. The end of the spring is in a
cup filled with mercury. What will happen? Why?
SOLUTION:
When the current passes through the coil, the magnetic field increases and forces cause the spring to
compress. The wire comes out of the mercury, the circuit opens, the magnetic field decreases, and the
spring drops down. The spring will oscillate up and down.
101. 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 proton moves through a magnetic field of 2.3
T. If it is to experience a force of magnitude 60 N, how fast must it be moving?”
102. Apply Concepts The magnetic field produced by a long, current-carrying wire is represented by
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−7
Page 27
B = (2×10 T · m/A)(I/d ), where B is the field strength in teslas, I is the current in amps, and d is the distance
from the wire in meters. Use this equation to estimate the magnetic fields in the following scenarios.
SOLUTION:
Answers
will vary,
but a correct
form
the answer is: “A proton moves through a magnetic field of 2.3
Chapter
24 Practice
Problems,
Review,
andofAssessment
T. If it is to experience a force of magnitude 60 N, how fast must it be moving?”
102. Apply Concepts The magnetic field produced by a long, current-carrying wire is represented by
−7
B = (2×10 T · m/A)(I/d ), where B is the field strength in teslas, I is the current in amps, and d is the distance
from the wire in meters. Use this equation to estimate the magnetic fields in the following scenarios.
a. The wires in your home seldom carry more than 10 A. How does the magnetic field that is 0.5 m from such a
wire compare to Earth’s magnetic field?
b. High-voltage power transmission lines often carry 200 A at potential differences as high as 765 kV. Estimate the
magnetic field on the ground under such a line, assuming that it is about 20 m high. How does this field compare
with a magnetic field in your home?
SOLUTION:
a.
Earth’s field is 5×10−5 T, so Earth’s field is about 12 times stronger than that of the wire.
b.
This is half as strong as the field in part a.
103. Add Vectors In the scenarios described in the previous problem, a second wire carries the same current in the
opposite direction. Find the net magnetic field that is a distance of 0.10 m from each wire carrying 10 A. The wires
are 0.01 m apart. Make a scale drawing of the situation. Calculate the magnitude of the field from each wire and
use a right-hand rule to draw vectors showing the directions of the fields. Finally, find the vector sum of the two
fields. State its magnitude and direction.
SOLUTION:
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Page 28
Chapter 24 Practice Problems, Review, and Assessment
This is half as strong as the field in part a.
103. Add Vectors In the scenarios described in the previous problem, a second wire carries the same current in the
opposite direction. Find the net magnetic field that is a distance of 0.10 m from each wire carrying 10 A. The wires
are 0.01 m apart. Make a scale drawing of the situation. Calculate the magnitude of the field from each wire and
use a right-hand rule to draw vectors showing the directions of the fields. Finally, find the vector sum of the two
fields. State its magnitude and direction.
SOLUTION:
Chapter Assessment: Writing in Physics
104. Research superconducting magnets and write a one-page summary of present and proposed uses for such magnets.
Describe any hurdles that stand in the way of their practical application.
SOLUTION:
Student answers may vary. Super-conducting magnets currently are used in magnetic resonance imaging
(MRI), a medical technology. They are being tested for use in magnetically levitated high-speed trains,
and it is hoped that superconducting magnets will help to make nuclear fusion energy practical. A
drawback of superconducting magnets is that they require extremely low temperatures (near absolute
zero). Scientists are trying to develop materials that are superconductive at higher temperatures.
Chapter Assessment: Cumulative Review
105. How much work is required to move a charge of 6.40×10
−3
C through a potential difference of 2500 V?
SOLUTION:
W = qV = (6.40×10−3 C)(2500 V) = 16 J
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106. The current in a 120-V circuit increases from 1.3 A to 2.3 A. Calculate the change in power.
SOLUTION:
Page 29
Student answers may vary. Super-conducting magnets currently are used in magnetic resonance imaging
(MRI), a medical technology. They are being tested for use in magnetically levitated high-speed trains,
and it is hoped that superconducting magnets will help to make nuclear fusion energy practical. A
drawback of superconducting magnets is that they require extremely low temperatures (near absolute
Chapter
24 Scientists
Practice Problems,
andmaterials
Assessment
zero).
are trying Review,
to develop
that are superconductive at higher temperatures.
Chapter Assessment: Cumulative Review
105. How much work is required to move a charge of 6.40×10
−3
C through a potential difference of 2500 V?
SOLUTION:
W = qV = (6.40×10−3 C)(2500 V) = 16 J
106. The current in a 120-V circuit increases from 1.3 A to 2.3 A. Calculate the change in power.
SOLUTION:
P = IV
P1 = I1V, P2 = I2V
ΔP = P2 − P1 = I2V − I1V
= V(I2 − I1)
= (120 V)(2.3 A − 1.3 A)
= 120 W
107. Determine the total resistance of three 55-Ω resistors connected in parallel and then series-connected to two 55-Ω
resistors connected in series.
SOLUTION:
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Page 30