CHAPTER
24,r
MagRetic Fields
Practice Problems
24.1
Magnets: Permanent and
Temporary
pages 643—65 1
page 647
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 mag
nets are held in the following ways?
a the two north poles are brought close
together
repulsive
a north pole and a south pole are
brought together
attractive
2
Figure 24-7 shows five disk magnets float
ing above each other. Iiie north pole of the
top-most disk faces up. Which poles are on
the top side of each of the other magnets?
4 Why do magnetic compasses sometimes
give false readings?
because Earth’s magnetic field is dis
toiled by objects made of fron, nickel, or
cobalt in the vicinity of the compass, and
by ore deposits of these same metals
page 650
5, A long, straight, current-carrying
from north to south,
ire runs
a. A ompass needle placed abo the wire
points with its north pole towacd the
east. In what direction is the urrent
flowing?
from south to north
b.
If a compass is put underneath the ii
in which direction will the wmpass
needle point?
west
6 flow does the strength of a magnetk field,
1 cm from a current carrying wire, compare
with each of the following?
a the strength of the field that s
from the wire
Because magnetic field strength
varies inversely with the distance
from the wire, the magnetic field at
1 cm will be twice as strong as the
magnetic field at 2 cm.
b the strength of the field that is 3 m
from the wire
*
Figure 24-7
south, north, south, north
Because magnetic field strength
varies inversely with the distance
from the wire, the magnetic field at
1 cm will be three times as strong as
the magnetic field at 3 cm.
3 A magnet attracts a nail, which, in turn, attracts
many small tacks, as shown in Figure 24-3 on
page 645. If the north pole of the permanent
magnet is the left end, as shown, which end
of the nail is the south pole?
the bottom (the point)
Physics: Principles and Problems
Solutions Manual
485
Chapter 24 continued
Chapter 24 continued
7. A student makes a magnet by winding wire
around a nail and connecting it to a battery,
as shown in Figure 24-13. Which end of
the nail, the pointed end or the head, will
be the north pole?
,
a
,
Figure 24-13
the pointed end
8, You have a spool of wire, a glass rod, an
iron rod, and an aluminum rod, Which rod
should you use to make an electromagnet
to pick up steel objects? Explain.
Use the iron rod. Iron would be attracted
to a permanent magnet and take on prop
erties of a magnet, whereas aluminum or
glass would not. This effect would sup
port the magnetic field in the wire coil and
thus make the strongest electromagnet.
9. The electromagnet in problem 8 works well,
but you decide that you would like to make its
strength adjustable by using a potentiometer
as a variable resistor. Is this possible? Explain.
Yes. Connect the potentiometer in
series with the power supply and the
coil. Adjusting the potentiometer for
more resistance will decrease the cur
rent flow and the strength of the field.
Section Review
24.1
Magnets: Permanent and
Temporary
pages 643—651
page 651
10. Magnetic Fields Is a magnetic field real, or
is it just a means of scientific modeling?
Field lines are not real. The field is real.
11. Magnetic Forces Identify some magnetic
forces around you. How could you demon
strate the cffects of those forces?
486
Sdution Manual
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.
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 material
that can be magnetized, nearby.
1 2. Magnetic Fields A current-carrying wire is
passed through a card on which iron filings
are sprinkled. The filings show the magnetic
field around the wire. A second wire is close
to and parallel to the first wire. There is an
identical current in the second wire, If the
two currents are in the same direction, how
will the first magnetic field be affected?
flow will it be affected if the two currents
are in opposite directions?
It would be approximately twice as
large; it would be approximately zero.
13. Direction of a Magnetic Field I)escribe
the right-hand rule used to determine the
direction of a magnetic field around a
straight, current-carrying wire.
lf 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 field.
‘14. Electromagnets A glass sheet is placed
over an active electromagnet, and iron fil
ings sprinkled on the sheet create a pattern
on it. lf this experiment is repeated with the
polarity of the power supply reversed, what
observable differences will result? Explain.
None. The filings would show the same
field pattern but a compass would show
the magnetic polarity reversal.
b. Assume that the top rod was lost and
replaced with another one. In this case,
the top rod falls on top ofthe bottom
rod 0(3 matter what its orientation is.
What type of replacement rod must
have been used?
If an ordinary iron bar is used on
top, it will be attracted to the bottom
magnet in any orientation.
Practice Problems
24.2
Forces Caused by
Magnetic Fields
pages 652—659
page 654
16. What is the name of the rule used to predict
the direction of kirce on a current-carrying
wire at right angles to a magnetic field?
ldenrih’ what must he known to use this rule.
Third right-hand rule; the direction of
the current and the direction of the field
must be known,
17. A wire that is 0.50 in long and carrying a
current of 8.0 A is at right angles to a 0.40-T
magnetic field, I-low strong is the force that
acts on the wire?
F = ILS = (8.0 A)(0.50 m)(0.40 N/Am)
=
15. Critical Thinking Imagine a toy contain
ing 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
The metal rods could be magnets
with their axes parallel. If the top
P1’ic: Pr’incipIe and J’robleins
1.6 N
F
IL
—
the force of gravity on the wire. What is the
strength of the magnetic field?
F
=
B
—
20.
ILB, where F = weight of the wire
F
0.35N
015T
IL
(6.0 A)(0.400 m)
—
—
[low much current will be required to pro
duce a force of 0.38 N on a 10.0 cm length
of wire at right angles to a 0.49-f field?
F=ILB
—
BL
—
0.38N
(0.49 T)(0.lOOm)
7.8 A
page 658
21. In what direction does the thumb point when
using the third right-hand rule for an electron
moving at right angles to a magnetic field?
opposite to the direction of the electron
motion
22. An electron passes through a magnetic field
at right angles to the field at a velocity of
4.0X 106 rn/s. The strength of the magnetic
field is 0.50 T. What is the magnitude of the
force acting on the electron?
F=qvB
=
=
10— C)(4.0
19
(1.60x
x10 m/s)(0.50 T)
6
3.2x10’ N
13
23. A stream of doubly ionized particles (miss
ing two electrons, and thus, carrying a net
charge of two elementary charges) moves at
a velocity of 3.OY
rn/s perpendicular to
a magnetic field of9,0x1O 2
magnitude of the force acting
What is the
OH each ion?
,
F=qvB
18. A wire that is 75 cm long, 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?
F=ILB
B
19. A 4
O.0-cm-long copper wire carries a cur
rent of 6.0 A and weighs 0.35 N. A certain
magnetic field is strong enough to balance
(6.0 A)(0.75 m)
Pli,i SIC’: Prim iIes and ProbIem.
013T
—
=
=
0x10’- C>
19
(2)(1.6
(3.0x10 2
4
mls)(9.OxlO’ T)
6’ N
8.6X10
24. Triply ionized particles in a beam carry a
net positive charge of three elementary
charge units. The beam enters a magnetic
fIeld of4.OXIO 2 T. The particles have a
speed of 9.OX 106 rn/s. What is the magni
tude of the force acting on each particle?
Solutions Manual
487
Chapter 24 continued
Chapter 24 continued
=
(3)(t60x10 C)
19
2 T)
(9JJX 106 mls)(4.0X10
=
13 N
t7X10
25. Doubly ionized helium atoms (alpha parth
km) are traveling at right angles to a mag
neth field at a speed of 4.0X 10 rn/s. The
held ‘trngth is 5.OX 10 2 What force
acts on each pankle?
(2)(t6OX1O C)
19
2 T)
(4Ox 1O m/s)(5.0X10
—
6 N
6 4X10
Section Review
Forces Caused by
Magnetic Fields
pages 652—659
24.2
page 659
26. Magnetic Forces Imagine that a currentci r g iii is perpendicular to Faith’s
na t Ii Ic and runs east west If the cur
r t c t, i r hich direction is the force
up away from the surface of Earth
27. Deflection A beam of electrons in a
cathode-ray tube approaches the deflecting
magnets. l’he north pole is at the top of the
tube, the south pole is on the bottom. If you
are looking at the tube from the direction of
the phosphor screen, in which direction are
the electrons deflected?
to the left side of the screen
28
Galvanometers Compare the diagram of a
galvan rnetcr in I igure 24 18 on page 655
w t I electric motor in Figure 24-20 on
‘IS How is the galvannmeter similar
p,i
to an ci ctric motor? How are they different?
Both the galvanometer and the electric
motor use a ‘oop of wire positioned
between the poles of a permanent mag
net, When a current passes through the
488
S
S4anual
loop, the magnetic field of the perma
nent 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 com
mutator 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.
Chapter Assessment
Concept Mapping
page 664
32. Complete the following concept map using
the following: right-hand rule, F = qvB, and
F = ILB.
37,
29. Motors When the plane of the coil in a
motor is perpendicular to the magnetic
field, the forces do not exert a torque on the
coil. I)oes this mean that the coil does not
rotate? Explain.
Not necessarily; it 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.
30. Resistance A galvanometer requires 180 A
for full-scale deflection. What is the total
resistance of the meter and the multiplier
resistor for a 5.0-V full-scale deflection?
R==VA =28ki
31. Critical Thinking Ilow do you know that
the forces on parallel current-carrying wires
are a result of magnetic attraction between
wires, and not a result of electrostatics?
Hint: Consider what the clzarcfes are like when
the force is auractii’e. Then consider what the
forces are when three wires carry currents in the
same direction
If the currents are in the same direction,
the force is attractive. If it were due to
electrostatic forces, the like charges
would make the force repulsive. Three
wires would all attract each other,
which could never happen if the forces
were due to electrostatic charges.
PhysicS: Principles and Problems
Draw the magnetk field between two like
magnetic poles and then between two
unlike magnetic poles. Show the directions
ofthe fields. (24.1)
38. lfyou broke a magmwr in two, would you
have isolated north and south poles?
Explain. (24.1)
No, new poles would form on each of
the broken ends.
Mastering Concepts
page 664
33, State the rule for magnetic attraction and
repulsion. (24.1)
Like poles repel one another; opposite
poles attract.
34. I)escrihe how a temporary magnet differs
from a permanent magnet. (24.1)
A temporary magnet is like a magnet
only while under the influence of
another magnet. A permanent magnet
needs no outside influence.
35. Name the three most important common
magnetic elements. (24.1)
iron, cobalt, and nickel
36. Draw a small bar magnet and show the
magnetic field lines as they appear around
the magnet. Use arrows to show the direc
tion of the held lines. (24. 1)
Physics: Principles and Problern,5
39,
I)escribe how to use the first right hand rule
to determine he direction of a magnetic field
around a straight current carrying wire. (24. 1)
Grasp the wire with the right hand,
keeping thumb pointing in the direction
of the conventional current through the
wire. Fingers will encircle the wire and
point in the direction of the field.
40. If a current-carrying wire is bent into a
loop, why is the magnetic field inside the
loop stronger than the magnetic field out
side? (24.1)
The magnetic field lines are concentrat
ed inside the loop.
41. Describe how to use the second right-hand
rule to determine the polarity of an electro
magnet. (24.1)
Solutions Manual
489
Chapter 24 continued
Chapter 24 continued
Grasp the coil with the right hand,
keeping the fingers encircling the coil
in the direction of the conventional cur
rent flow through the loops. The thumb
of the right hand will point toward the
north pole of the electromagnet.
42.
Each electron in a piece of iron is like a tiny
magnet. The iron, however, may not be a
magnet. Explain. (24.1)
The electrons are not all oriented and
moving in the same direction; their
magnetic fields have random directions.
48. A piece of metal is attracted to one pole of a
large magnet. Describe how you could tell
whether the metal is a temporary magnet or
Move it to the other pole. If the same
end is attracted, it is a temporary
magnet; it the same end is repelled and
the other end is attracted, it is a perma
nent magnet.
49, is the magnetic force that Earth exerts on a
44. Describe how to use the third righuhand
mie to determine the direction of force on a
current arrying wire plaed in a magnetic
field, (24,2)
Point the fingers of your right hand in
the direction of the magnetic field. Point
your thumb in the direction of the con
ventional current in the wire. The palm
of your hand then faces in the direction
of the force on the wire.
The forces are equal according to
Newton’s third law,
A strong current suddenly is switched on in a
wire, No force acts on the wire, however. Can
you conclude that there is no magnetic field
at the location of the wire? Explain. (242)
No, it a field is parallel to the wire, no
force would result.
46. What kind of meter is created when a shunt
is added to a galvanometer? (242)
an ammeter
Applying Concepts
pages 664-665
47 A small bar magnet is hidden in a fixed
position inside a tennis ball, Describe an
experiment that you could do to find the
location of the north pole and the south
pole of the magnet.
490
Solutions Manual
53. 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?
a permanent magnet.
43. Why will dropping or heating a magnet
weaken it? (24J)
The domains are jostled out of alignment.
45.
Use a compass to find the direction of
the magnetic field. Bring up a strong
magnet and find the direction of the force
on the wire, then use the right-hand rule.
Use a compass. The north pole of the
compass needle is attracted to the south
pole of the magnet and vice versa.
North pole
allel to each other.
a. lfthe two currents are in opposite direc
tions, where will the magnetic field
from the two wires be larger than the
field from either wire alone?
The magnetic field will be larger
anywhere between the two wires.
b. Wheme will the magnetic field from both
be exactly twice as large as from one wire?
The magnetic field will be twice as
large along a line directly between
the wires that is equal in distance
from each wire.
c. If the two currents are in the same direc
tion, where will the magnetic field be
exactly zero?
The magnetic field will be zero along
a line directly between the wires that
is equal in distance from each wire.
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 wireS By the
right-hand rule, the end of the compass
needle that points right is the north pole.
55.
51. A magnet can attract a piece of iron that is
not a permanent magnet. A charged rubber
110w is the range of a voltmeter changed
when the resistor’s resistance is increased?
The range of the voltmeter increases.
56. A magnetic field can exert a force on a
charged particle. Can the field change the
particle’s kinetic energy? Explain.
No, the force is always perpendicular
to the velocity. No work is done. The
energy is not changed.
rod can attract an uncharged insulator.
Describe the different microscopic processes
producing these similar phenomena.
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.
57. A beam of protons is moving from the
back to the front of a room. It is deflected
upward by a magnetic field. What is the
direction of the field causing the deflection?
52. A current-carrying wire nmns across a labora
tory bench. Describe at least two ways in
which you could find the direction of the
current.
and Problems
Magnetic pole
54. ‘I’wo wires carry equal currents and run par-
50 Compass Suppose you are lost in the
woods but have a compass with you.
Unfortunatel, the red paint marking the
north pole of the compass needle has worn
off, You have a flashlight with a battery and
a length of wire. How could you identifr
the north pole of the compass?
Principles
58. Earth’s magnetic field lines are shown in
Figure 24-23. At what location, poles or
equator, is the magnetic field strength
greatest? Explain.
Run the wire parallel to the magnetic field.
compass needle less than, equal to, or
greater than the force that the wmpass
needle exerts on Earth? Explain.
P/ivsicc:
Facing the front of the room, the
velocity is forward, the force is upward,
and therefore, using the third right-hand
rule, B is to the left.
j
Ph SICS: Principles and Problems
South pole /
Magnetic pole
S Figure 24-23
Earth’s magnetic field strength is
greatest at the poles. The field lines
are closer together at the poles.
Mastering Problems
24,1 Magnets: Permanent and Temporary
pages 665—666
Level 1
59. As the magnet below in Figure 24-24
moves toward the suspended magnet, what
will the magnet suspended by the string do?
IL
• Figure 24-24
Move to the left or begin to turn. Like
poles repel.
60. As the magnet in Figure 24-25 moves
toward the suspended magnet, what will the
magnet that is suspended by the string do?
Solutions Manual
491
I
Chapter 24 continued
Chapter 24 continued
64. A conventional current flows through a
wire, as shown in Figure 24-28. Copy the
wire segment and sketch the magnetic field
that the current generates.
a. What is the direction of the magnetic
field inside the loops?
down into the page
b. What is the direction of the magnetic
field outside the loops?
up (out of the page)
24.2 Forces Caused by Magnetic Fields
pages 666—667
Level I
68. The arrangement shown in Figure 24-31 is
used to convert a galvanometer to what
type of device?
s tigure 24-25
Move to the right. Unlike poles attract.
R f ‘r to Figure 24-26 to answer the follow
I tg qt estions.
a Figure 24-28
Level 2
67, Ceramic Magnets The repulsive force
between two ceramic magnets was measured
and found to depend on distance, as given in
Table 24-1.
S
Ammeter, much of the current flows
through the resistor and allows the
measurement of higher currents.
Table 244
Where are the poles?
4 and 2, by definition
e is the north pole?
b 4%
2, by definition and field direction
c.
Where is
Force, F (N)
1.0
3.93
1.2
0.40
1.4
0.13
1 6
0.057
1.8
0.030
2.0
0.018
2.2
0.011
2.4
0.0076
2.6
0.0053
2.8
0.0038
3.0
0.0028
Figure 24-26
I
a.
Separation, d (cm)
the
65. The current is coming straight out of the
page in Figure 24-29. Copy the figure and
sketch the magnetix field that the current
generates.
69. What is the resistor shown in Figure 24 31
7
called
Shunt; by definition shunt is another
word for parallel.
70.
south pole?
4, by definition and field direction
62. Figure 24-27 shows the response of a compass
jn two different positions near a magnet.
I ci is the south pole of the magnet located?
a Figure 24-29
a. Plot the foice
as a function
Figure 24-31
of distance.
1 he arrangt ment sho yr n Figure 24-32 s
used to convert a gaivanotneter to what
type of device?
a Figure 24-32
.4.
1W
S
Voltmeter; the added resistance
decreases the current for any given
voltage.
40
3.0
Figure 24-27
71. What is the resistor shown in Figure 2 4-32
called?
z
At the right end, unlike poles attract.
63. A wire that is 1.50 m long and carrying a
r rren )f 10 0 A is at right angles to a uni
r )tm magnetic field, I he force acting on the
wire is 0.60 N. What is the strength of the
magnetic field?
u,. 20
66. Figure 24-30 shows the end view of an elec
tromagnet with current flowing through it.
Multiplier; by definition since it multi
plies the voltage range of the meter
1.0
72. A cunent-carrving win p1 ret I bctx ten the
poles of a magnet is shown n Figure 24-33.
What is the direction of the force on the wire?
‘
0.0
1.0
1.4
1.8
2.2
2.6
3.0
d(cm)
F=ILB
B
=
=
492
F
IL
6O
=
(l0.OA)(1.50m)
0040 T
S’iuti ‘us Manual
=
b. Does this force follow an
0.040 N/km
inverse square
law?
No.
a Figure 24 30
Plwsics: Principles and Problems
Pbrsics:
Principles and
Problems
Snlutons ;Viantuii
493
Chapter 24 continued
Chapter 24 continued
77. A wire that is 625 in long is perpendicular
to a 0.40-1’ magnetic field. A 1.8-N force acts
on the wire. What current is
the wire?
in
F=ILB
t8N
(0.40 l)(625m)
j=f,_
BL
81. Galvanometer A galvanometer deflects
full-scale for a 50.0-MA current.
a. What must he the total resistance of the
series resistor and the galvanometer to
make a voltmeter with 10.0-V full-scale
deflection?
0.0072 A
V=JR
7.2mA
R
73. A n ne that is 0.50 m long and carrying a
urrer t of 8 1 A is at right angles to a uni
form magnetic field. ihe three on the wire
is 0.4() N. What is the strength of the inag
nenc field?
F=ILB
8
F
IL
O.40N
(8.OA)(O.50m)
—o lOT
—
=
=3.OkA
F
=
ILB
=
2A N
(5.0 A)(O.$o m)(06O N/ACm)
3.6N
L=F
B!
(0.80 T)(7.5 A)
75. A wire that is 25
long
right angles
to a 0.30-f
magnetic field The cur
rent through the wire is 6 0 A. What is the
magnitude of the force on the wire?
cm
is
at
uniform
F
=
ILB
=
0A5 N
=
(6M A)(025 m)(03O N/Am)
wire
uniform
wire
wire?
If the wire is parallel to the field, no
cutting is taking place, so no force is
produced.
Se/ut arts \ianua/
line
west,
acting
b. If the galvanometer has a resistance of
1.0 kI 1, what should be the resistance of
the series (multiplier) resistor?
2.OOX 102 kil, so
Total resistance
2.00X10 kci
the series resistor is 2
1.0 kci = 199 kci.
0.60 m
225 A
parallel to the surface of Earth,
on
carries
a
current
from
meter
F=1LB
=
lB
=
0,011 N/rn
=
84. Subatomic Particle A muon (a particle
an electron) is
with the same charge
at right angles
m/s
X
1W
traveling at 4.21
to a magnetic field. The muon experiences
12 N.
a force of5.00X10
a. how strong is the magnetic field?
F=qvB
qv
5.OOX 10”12 N
C)(42ix10 mis)
x10 7
(1.609
1
(225 A)(5.OxlO”-5 T)
ammeter
b. What is the equivalent resistance of par
allel resistors having the potential differ
ence calculated in a circuit whir a total
current of 10 mA?
V=IR
V
I
R
c. In your judgn’ient, would this force he
important in designing towers to hold
this power line? Explain,
No; the force is much smaller than
the weight of the wires.
Physic’s’ Principles and Problems
0
2
5X1
V
0.OIA
1
so R
=
1,,,., 1
5i
R
2
—
5 ci
83. A beam of electrons moves at right angles to
a magnetic field of 6.0X 102 ‘L ihe electrons
have a velocity of 2,5X 106 rn/s. What is the
magnitude of the force on each electron?
Physics: Pi±ncip]es and Problems
ma
F
—
—
—
—
2
x10”
N
5.O0
1
:i’:iz28k9
2
2.66X 1&6 rn/s
85. A singly ionized particle experiences a force
of 4.1 X 10 13 N when it travels at right
angles through a 0.61 -T magnetic field.
What is the velocity of the particle?
F=qVB
—
=
1
1 .Oxl
=
=
—
1
RR
1
1i
R
a
V
c. What resistor should he placed parallel
with the galvanometer to make the
resistance calculated in part b?
—
b. What is the direction of the force?
The force would be downward.
F
a. What is the potential difference across the
galvanometer (1.0 k1 resistance) when a
current of 50 ,ccA passes through it?
A)(1.0X10 ci)
(50X10 3
V = JR = 6
005 V
0.742 T
b. What acceleration does the muon expe
28 kg?
rience if its mass is l.88X
10 mA.
=
=
a. What is the magnitude of the force
resulting from Earth’s magnetic field
each
of the wire? Use
Bfh
f
76. A
that is 35 cm long is parallel to a
0.53-I
magnetic field. The current
through the
is 4.5 A. What lone acts
on the
494
power
X10 kfl
2
2.00
an
79. The force acting on a wire that is at right
angles to a 0.80-f magnetic field is 3.6 N.
The current in the wire is 7.5 A. flow long
is the wire?
Level 2
80. A
east to
2.OOX l0 ci
82. The galvanorneter in l)rOblern 81 is used to
that deflects full-scale for
make
niagni
wirer
10’ N
14
2.4x
=
lO
3
3.Ox
A
F=ILB
74. 1 he current through a wiie that is 0.80 in
long is 5.0 A. The wire is perpendir ular to a
0.60:1 magnetk field What is th
tude of the force on the
=
C)(2.5x10 m/s)
10’’ 6
19
(1.6X
(6.0x10 T)
2
—
O.12N
i0 m)
0xj0
)(6
(5.
T
5
BL
=
=
=
=
as
W Figure 24-33
78, The force on a 0.80-in wire that is perpen
dicular to Earth’s magnetic field is 0.12 N.
What is the current in the wire? Use
5.0 X 10
T for Earth’s magnetic field.
F=ILB
F=qvB
F
Bq
—
—
10
3
4.1x
N
’ C)
Ox1O’
T)(1.6
(0.61 9
10 mis
6
4.2X
86. A room contains a strong, uniform magnetic
field, A loop of fine wire in the room has
current flowing through it. Assume that you
rotate the loop until there is no tendency for
it to rotate as a result of the field. What is the
direction of the magnetic field relative to the
plane of the coil?
The magnetic field is perpendicular to
the plane of the coil. The right-hand rule
would be used to find the direction of
the field produced by the coil. The field
in the room is in the same direction.
Solutions Manual
495
Chapter 24 continued
Chapter 24 continued
6
Afor
act5
ceof5787
0na
87 N
n
l0’
unknown particle traveling at a 900 angle
through a magnetic field If the velocity of
rn/s and the field is
the particle is 565X
2
320X 10
‘i’, how many elementary
charges does the particle carry?
F
qvB
=
F
By
—
q
16
578X1
N
O
(320X10_2 T)(565X10
4 mis)
—
19
32OX1
O C
N
=
1 charge
9 x
(3.2OX1O’
6
C)(
jj19
f
=
2 charges
=
)
0.62 N
—
c I)etermine the force on the wire (direc
tion and magnitude) when the switch is
closed and the battery is reversed.
Level 2
88W. A opper wile of insignificant resistance
is placed in the cc nter of an air gap
betwun two niagnetic poles as shown
ii Figure 24 34. 1 he field is confined
to the gap and has a strength of 1 .9 ‘1.
5,5
I
=
!1
15,A
24 V
Zr
—
Figure 24-34
Determinr the force on the wire (direc
tijn and magnitude) when the switch is
open
o N. With no current, there is no mag
Up, O62 N The direction of the force
is given by the third right-hand rule.
=
(24 Vj(O.075 m)(1 .9 T
5.511+5.5a
91.
0.31 N
89. Two galvanometers are available. One has
500-/Lr\ full-scale sensitivity and tire other
has 300.0 rA full sale sensitivity Both
have the same coil resistance of 835 IL Your
challenge is to ronert them to measure a
current of tOO.() mA, full-scale.
a. Determine the shunt resistor for the
30.0-hA meter.
Find the voltage across the meter
coil at full scale.
V
=
If?
V
—
=
=
(50.0 A)(855 !)
=
I
—
—
O.0428V
100.0 mA 50.0 ,tA
496
a
—
—
=
92.
—
—
F
18 rn/s
2.6X10
2
95,
If?
=
=
(15 A)(0.25 m)(0,85 T)(sin 90°)
=
3.2 N
F=qvB
19
(1.6x1
0” 5
C)(8.1X10 m/s)(16 T)
=
12 N, upward (right-hand
2.1X10
ILE3 sin (1
—
=
(15 A)(.25 m)(0,85 T)(sin 45<)
=
2.3 N
sin 0°
0
so F
0 N
=
\n electror is a
ft and i
st t m
a potential differen c ot 20.000 V X\hiiil
exists between plates P i rd P) showm
Figure 24-35. The clccton then passes
through a small opening into a magnctft
field of uniform field sn igth, B. As md
cated, the magnetic field is directed into
the page.
due west. An electron is traveling due south
at 8.1 X 10 m
s, What arc the magnitude
1
and the direction of the force acting on the
electron?
=
0
c. 0°
2.4x10
N
12
9.11x10
kg
31
A magnetic field of 16 ‘1 acts in a direction
sm
b. 13
ma
F
m
ILB
1
P
2
P
-
(500.0 i1
A)(855 )
0.428 V
93. Loudspeaker ‘the magnetic field in a loud
speaker is 0. 1 5 ‘I’. The wire consists of 250
turns wound on a 2.5-cm-diameter cylindri
cal form. The resistance of the wire is 8.0 .Q.
Find the force exerted oii the wire when
15 V is placed across the wire.
I
R
VLB
S’!utwtn \Fznual
94. A wire carrying 1 5 .\ ot current has a length
of 25 m in a magnetic field )f 0,S5 F. The
force on a crirment carrying wire in a
form magnetic field can be found using the
ILB sin 8. timid thc fom
ecluation F
on
the wire when it makes the following angles
with tire magnen fidd lines I
a. 90°
0,428 i
Find the voltage across the meter
coil at full scale.
=
=
5.5 N
m/s)(O.60 T)
rule—remembering that electron flow is
opposite to current flow)
—
R
F=ILB=
F
0.0428 V
b. Determine the shunt resistor for tire
500.0-hA meter.
V
=
lire mass of an electron is 9.11 X 10 ‘ kg.
What is the magnitude of the acceleration of
the beta iarticle described in problem 90?
Calculate the shunt resistor.
netic field produced by the wire and
copper is not a magnetic materiaL
b. l)etermine the force on the wire (direc
non and magnitude) nhen the switch is
closed.
VLB
fl5V)(250)(n)(0.025 m)(0.15 T)
8.0 Il
—
F
1
(1.6X1
0 C)(2.5X
2.4X10’ N
2
nrrd
Vn,-dB
F?
F—
4.30 II
90. Subatomic Particle A beta particle (highspeed electron) is traveling at right angles to
a 0.60-T magnetic field, It has a speed of
2.5X i0
7 rn/s. What size force acts on the
particle?
=
F=ILB
F=qvB
F=ILB=
(# of turns)(circumference)
=
—
The 50.0-pA meter is better. Its
much lower shunt resistance will do
less to alter the total resistance of
the circuit being measured. An ideal
ammeter has a resistance of 0 fi.
R
—
L
c. Determine which of the two is better for
actual use, Explain.
=
—
—
=
Down, 0.62 N. The direction of the
force is given by the third right-hand
rule and the magnitude of the force
is the same as in part b.
Up, 0.31 N. The direction of the force
is given by the third right-hand rule.
pages 667-668
I
Calculate the shunt resistor.
V
0.428V
I
100,0 mA 500.0 pA
4y){ppmX1 9 T)
5.5 !l
d. l)eterrnine the Iotce on the wire (direc
tion and magnitude) when the switch is
closed and the wile S replaced with a dif
ferent piece having a resistance of 3.5 IL
Mixed Review
a
—
Electron
a. State the direction of the electrft field
between the plates as miher P to j) r
.
1
12 to P
from P
2 to P
1
5:
Princip!n and Problems
J’hvsics: Priiiciplcc and Prohkms
3ai11110115 $.ialluaj
497