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CHAPTER 19
MAGNETISM
Multiple Choice Questions:
1.
(a), because unlike poles attract.
2.
(a).
3.
(c).
4.
(b). A calibrated compass points toward the Earth’s north geographic pole, which is really a south magnetic pole.
Therefore it would be attracted by the south pole and repelled by the north pole of the magnets shown, and hence
would point downward.
5.
(d).
6.
(a). By the right hand rule, the velocity is up and the force is to the left, so the magnetic field must be pointing away
from you.
7.
(a).
8.
(a). By the right hand rule, if the charge were positive the field would have to be into the page. But since the charge is
negative, the field must be out of the page.
9.
(d). When  = 0°, there is no magnetic force because F = qvB sin .
10.
(c), because r = mv/qB.
11.
(b), because r = mv/qB.
12.
(b). For no deflection, we must have v = E/B1. If the velocity is less than this value, the magnetic force is less than the
electrical force, so the charge will be deflected downward in the direction of the electric field.
13.
(d). The Earth’s magnetic field points from the south geographic pole to the north geographic pole and the
conventional current flows from east to west. By the right hand rule, the force is perpendicular to the wire pointing
downward.
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335
Chapter 19
14.
(a), according to the right-hand force rule.
15.
(b).
16.
(b). The torque is given by  = NIAB sin , which is a maximum when  = 90°.
17.
(c). By the right hand rule, with the thumb pointing from east to west, the curled fingers point to the south below the
Magnetism
wire.
18.
(a). By the right hand rule, if the thumb points along the current in a clockwise direction, the curled fingers point away
from you.
19.
(b), according to the right-hand source rule.
20.
(c). The point at the center is closest to the current and hence will have the largest field. The point 20 cm out from the
center is farthest and will therefore have the smallest field.
21.
(b).
22.
(d).
23.
(d). Heating the magnet increases the random thermal motion of its atoms. This will affect the domain orientation,
domain boundaries, and the electron spin orientations.
24.
(d).
25.
(b).
26.
(c). The calibrated compass is attracted to the Earth’s north geographic pole and repelled by its south geographic pole.
Since the magnetic points straight up, it is being repelled away from the ground, so you must be near the south
geographic pole.
27.
(a). In order to provide the centripetal force to keep the proton in orbit, the magnetic force must be pointing toward the
center of the Earth.

t
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College Physics Seventh Edition: Instructor Solutions Manual
336
Conceptual Questions:
1.
The magnet would attract the unmagnetized iron bar when a pole end is placed at the center of its long side. If the end
of the unmagnetized bar were placed at the center of the long side of the magnet, it would not be attracted.
2.
Near the north pole of a permanent bar magnet, the north pole of a compass will point away from the bar magnet so
field lines leave a north pole. Near the south pole of a permanent bar magnet, the south pole of a compass will point
toward the bar magnet so field lines enter a south pole.
3.
(a) The spacing of the iron filings gets farther apart . This tells us that the magnetic field strength decreases with
distance from the middle.
(b) The magnetic field points up-and-down (parallel to the line along which the magnets are aligned). But we cannot
tell if it points up or down from looking at the filing pattern alone.
4.
(a) The magnitudes are the same since their charge magnitudes are the same, but the directions are opposite
because their charges have opposite sign.
(b) The acceleration of the electron is about 1830 times the acceleration of the proton because the electron’s mass is
about 1/1830 times the proton’s mass.
5.
Not necessarily , because there still could be a magnetic field. If the magnetic field and the velocity of the charged
particle make an angle of either 0° or 180°, there is no magnetic force on the particle.
6.
The direction of bending tells us the sign of the charge. (a) (1) negative charge, (2) zero charge, (3) positive
charge. (b) The amount of curvature tells us the momentum. Since the particles have equal speeds, we can conclude
that m3  m1 .
7.
(a) The fields should be uniform and of equal magnitude but should point in opposite directions. The lower field
should point into the paper and the upper field should point out of the paper.
(b) The emerging kinetic energy is the same as the initial kinetic energy. Since the magnetic force is perpendicular
to the velocity, it changes only the direction of the velocity, not its magnitude, so the kinetic energy does not change.
8.
If the charges are electrons (negative), an excess negative charge accumulates on the left side of the strip, and if
positive, on the right side of the strip. The sign of the voltage across the strip would indicate the type of charge.
9.
The magnetic force on the electrons in the beam causes the deflection.
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337
10.
Chapter 19
Magnetism
No . The negative charge will actually experience a force that is in the same direction as the positive charges so they
also contribute to the propulsion. The negative charges will be accelerated by the electric field in opposite direction as
the positive charges; thus, the magnetic force on the charges (negative or positive) is backward according to the right
hand rule.
11.
The electric force is qE and the magnetic force is qvB. Since both forces depend linearly on the charge, the selected
speed, which we get by equating their magnitudes, is independent of the charge.
12.
The paths in the field B2 would be semicircles but they would curve upward instead of downward.
13.
(a) If the electric field is reduced, the magnetic force will be greater than the electric force. Therefore the positive
charges in the velocity selector will be deflected upward . Hence they will not enter the region of B2.
(b) If B1 is reduced, the electric force will be greater than the magnetic force. Thus the positive charges will be
deflected downward and will not enter the region of B2.
14.
The wires attract each other, and the magnitudes of the forces on each wire are the same .
15.
The spring should shorten because the coils of the spring attract each other due to the magnetic fields created in the
coils. (Parallel wires with current in same direction will attract each other.)
16.
(a) Orient the plane of its loop perpendicular to the magnetic field lines.
(b) Orient the plane of its loop parallel to the magnetic field lines.
(c) In both cases, the net force is zero because the forces on the segments of the loop are either zero or cancel.
17.
Pushing the button in both cases completes the circuit. The current in the wires activates the electromagnet, causing
the clapper to be attracted and ring the bell. However, this breaks the armature contact and opens the circuit. Holding
the button causes this to repeat, and the bell rings continuously. For the chimes, when the circuit is completed, the
electromagnet attracts the core and compresses the spring. Inertia causes it to hit one tone bar, and the spring force
then sends the core in the opposite direction to strike the other bar.
18.
For an electron, the right hand rule tells us that the magnetic force on the negative charges is to the south. The
conventional current is to the east, so the right hand rule tells us that the magnetic force on it is to the south . We get
the same answer both ways.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
College Physics Seventh Edition: Instructor Solutions Manual
19.
338
C
 N 
(a) µB  IAB  A  m2  T =   m2 
  Nm
s
 C  m/s 
 
(b) Curl your right hand fingers counterclockwise and your thumb points upward , toward you, which is the direction
of the magnetic moment.
20.
Both magnetic forces point toward the south, so the wire would accelerate toward the south . But the west side would
move faster than the east side, causing it to rotate counterclockwise as viewed from above.
21.
The compass points downward so the magnetic field is downward at the center of the loop. The current direction is
clockwise according to the right-hand source rule.
22.
23.
Because B  1 d you need to double the current and reverse its direction .
Not necessarily . The magnetic field in a solenoid depends on the current in it and the number of turns per unit
length (the density of turns), not just the number of turns. For example, if the 200 turns is over 0.20 m and the 100
turns is over 0.10 m, then they will have the same turns per unit length and the same magnetic field.
24.
There are two wires, one carrying the current into the appliance and the other carrying it out. These two currents are in
opposite directions. When the two wires are very close to each other, the magnetic fields created by the two opposite
currents essentially cancel.
25.
The direction of the current should be counterclockwise to cancel the magnetic field of the outer loop. Its current
should be smaller than 10 A, because the field created by a loop is inversely related to the radius of that loop. With
a smaller radius for the inner loop, its current must be less than 10 A.
26.
The direction of the magnetic field is away from you , according to the right-hand source rule (remember, the
electron has negative charge).
27.
The purpose of the iron core is to increase the magnetic permeability and magnetic field, because the magnetic field
is proportional to the magnetic permeability of the material.
28.
You can destroy or reduce the magnetic field of a permanent magnet by hitting or heating it. Also if placed in an
external magnetic field its direction can be reversed to match that field’s direction. In a rapidly oscillating external
magnetic field, the domains can be left randomized and the bar magnet could have zero residual permanent
magnetization.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
339
Chapter 19
29.
Hawaii is slightly north of the equator so the magnetic field there points northward and mostly parallel to the surface
Magnetism
of the Earth but slightly downward. This would be the direction of the remnant magnetism.
30.
Applying the right-hand rule and recalling that an electron is negative, we get (a) zero (because the velocity is
parallel to the magnetic field), (b) upward , and (c) eastward .
31.
It will be the north magnetic pole . Right now, the pole near the Earth’s geographical North Pole is actually a south
magnetic pole.
32.
Using Fig. 19.30, we have:
(a) Approximately 12.5°E since Phoenix is about midway between the 10°E and 15°E isogonic lines.
(b) About 2°W since Chicago is a little less than midway between the 0° and 5°W isogonic lines.
(c) About 2 12 °E since New Orleans is about midway between the 0° and 5°W isogonic lines.
Exercises:
1.
(a) The second magnet is directly below the first one, oriented vertically, with its north pole directly above its south
pole. The two north poles are 2.5 cm apart.
(b) Since the magnets are identical and the north poles of #2 and #3 are both 2.5 cm from the north pole of #1, they
will exert forces of equal magnitude on the north pole of #1. The force due to #2 is directly upward and the force due
to #3 is directly to the left, so the net force is
F
2.
(1.5 mN)2  (1.5 mN)2  2.1 mN at 45° above the horizontal toward the left
(a) The compass would be repelled equally by both north poles, so it would point into the third quadrant at
45° below the –x-axis . This is the direction of the net field at the origin.
(b) At the origin, the field from #1 would be downward and field from #2 would be to the right, so the net field (and
hence the compass) would point into the fourth quadrant at 45° below the +x-axis .
3.
(a) B1 is in the –x-direction and B2 is in the –y-direction, so the net field at the origin will point into the third
quadrant. If  is the angle it makes below the –x-axis, we have
tan  = B2/B1 = (B1/2)/B1 =
1
2

 = 27° below the –x -axis
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
College Physics Seventh Edition: Instructor Solutions Manual
340
(b) Now B1 has the same magnitude as before but is in the +x-direction, and B2 is still in the –y-direction. So the net
field now points into the fourth quadrant at 27° below the +x-axis .
4.
(a) By the right-hand force rule, curl your fingers from v to B and your thumb points into the page, so the direction
of the magnetic force is (1) into the page .
(b) F = qvB sin 
20 N = (0.25 C)(200 m/s)(B) sin 90°
B = 0.40 T
F
10 N
=
= 3.5  103 m/s .
qB sin 
(0.050 C)(0.080 T) sin 45
5.
F = qvB sin ,
6.
sin  = F/qvB = (5.00 N)/[(0.250 C)(200 m/s)(0.500 T)] = 0.200

v=
 = 11.54° = 11.5°
But 180° 11.54° = 168.46° = 168° will also give the same magnitude force.
7.
The magnetic field has to be to the left, looking in the direction of the beam, so the magnetic force points upward.
F = qvB sin  = mg,
so
B=
mg
(1.67  10  27kg )(9.80m / s 2)
=
= 2.0  1014 T .
qvsin
(1.6  10  19C )(5.0  106m / s )sin90
The direction is to the left looking in the direction of the velocity , so the magnetic force points upward, opposite
to the gravitational force.
8.
(a) According to the right-hand force rule and the fact that electron carries negative charge, the magnetic field is
directed in the (4) z direction.
(b) From F = qvB sin ,
9.
B=
F
5.0  1019 N
=
= 1.0  106 T .
19
qv sin 
(1.6  10 C)(3.0  106 m/s) sin 90
(a) F = qvB sin  = (1.6  1019 C)(2.0  104 m/s)(1.2  103 T) sin 90 = 3.8  1018 N .
(b) F = (1.6  1019 C)(2.0  104 m/s)(1.2  103 T) sin 45 = 2.7  1018 N .
(c) F = 0 since sin 0 = 0.
(d) F = 0 since sin 180 = 0.
10.
(a) The maximum force occurs when the field is perpendicular to the velocity, in which case sin  = 0. So Fmax =
qvB. In this case, we have
F=
1
2
Fmax
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341
Chapter 19
Magnetism
qvB sin  =
sin  =
1
2
qvB

1
2
 = 30° or 150°
(b) F = qvB sin  = Fmax sin 
(i) F = Fmax sin 10° = 0.17Fmax
(ii) F = Fmax sin 80° = 0.98Fmax
(iii) F = Fmax sin 100° = 0.98Fmax
(c) sin(180° – 50°) = sin 50° = sin 130°, so 130° would give the same magnitude force. The right-hand force rule
tells us that the direction would be the same in both cases. Angles of 50° and 130° would also give the same
magnitude force, but the direction would be opposite.
11.
(a) a = F/m = (qvB sin )/m
a = (1.60  10–19 C)(3.0  105 m/s)(0.50 T)(sin 37°)/(1.67  10–27 kg)
a = 8.6  1012 m/s2
By the right-hand force rule, the direction is southward .
(b) The magnitude would be the same but the direction would be northward instead.
(c) The magnitude of the force would be the same because the electron and proton have the same magnitude charge,
but the direction of the force on the electron would be opposite to that on the proton because they have opposite
charges.
(d) Using a = F/m, we have
ae/ap = (F/me)/(F/mp) = mp/me = (1.67  10–27 kg)/(9.11  10–31 kg) = 1830
V
E
8.0  103 V/m
=
=
= 2.0  105 m/s .
Bd
B
0.040 T
12.
v=
13.
(a) v =
V
,
Bd

V = vBd = (8.0  104 m/s)(1.5 T)(0.015 m) = 1.8  103 V .
(b) It is the same voltage , 1.8  103 V, because it is independent of charge on the particle.
14.
(a) v =
V
E
3000 N/C
=
=
= 1.0  105 m/s .
Bd
B
0.030 T
(b) It is the same speed , 1.0  105 m/s, because it is independent of charge on the particle.
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College Physics Seventh Edition: Instructor Solutions Manual
15.
16.

K = 12 mv2,
E
2.0  103 V/m
=
= 5.3  104 T .
v
3.77  106 m/s
v=
E
,
B
v=
V
E
1.0  103 V/m
=
=
= 1.0  104 m/s. Here the magnetic force provides centripetal force for the circular
Bd
B
0.10 T

B=
motion. Since F = qvB sin  = m
m=
17.
2(10  103 eV)(1.6  1019 J/eV)
= 3.77  106 m/s.
2.25  1028 kg
2K
=
m
v=
342
v2
,
R
qBR sin 
(1.6  1019 C)(0.10 T)(0.012 m) sin 90
=
= 1.9  1026 kg .
v
1.0  104 m/s
(a) v =
V
E
1.0  103 V/m
=
=
= 1.0  104 m/s.
Bd
B
0.100 T
In circular motion magnetic force provides centripetal force. F = qvB sin  = m
so
m=
v2
,
R
qBR sin 
(2  1.6  1019 C)(0.10 T)(0.015 m) sin 90
=
= 4.8  1026 kg .
v
1.0  104 m/s
(b) K = 12 mv2 = 12 (4.8  1026 kg)(1.0  104 m/s)2 = 2.4  1018 J .
(c) No , the kinetic does not increase. The centripetal force (magnetic force) is perpendicular to velocity so the
work equals zero .
18.
(a) The kinetic energy is related to the momentum by K = p2/2m, and the radius of the path is given by r = mv/qB.
Combining these equations gives
r = mv/qB = p/qB =
2mK
qB
Solving for B, we have
2mK

qr
B
2(1.67  10 27 kg)(10 4 eV)(1.60  19-19 J/eV)
 0.0289 T = 0.029 T
(1.60  10 -19 C)(0.50 m)
(b) Using mv  p 
2mK

m
v
2mK to find v gives
2K

m
2(10 4 ev)(1.60  10 -19 J/eV)
 1.384  106 m/s
1.67  10-27 kg
Finding t gives us
vt =
1
4
(2πr)
t = πr/2v = π(0.50 m)/[2(1.38  106 m/s)] = 5.7  107 s
(c) F = qvB = (1.60  10–19 C)(1.384  106 m/s)(0.0289 T) = 6.4  1015 N
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343
19.
Chapter 19
Magnetism
(a) The direction of the current is the direction that a positive charge would have to move to experience the given
force. Applying the right-hand force rule gives the following current directions.
(i) to the right
(ii) upward
(iii) into the page
(iv) to the left
(v) into or out of the page ; either would give zero force
(b) The force on a wire segment is F = ILB sin . In each case except (v),  = 90°, so the force is
F = (5.5 A)(0.15 m)(1.0  10–3 T)(sin 90°) = 8.3  104 N
In case (v) the force is zero .
20.
(a) The magnetic force is in the (3) +y- direction according to the right-hand force rule, (upward as +z).
(b) F = ILB sin  = (5.0 A)(1.0 m)(0.30 T) sin 90 = 1.5 N .
21.
F = ILB sin  = (20 A)(2.0 m)(0.050 T) sin 37 = 1.2 N perpendicular to the plane of B and I .
22.
(a) F = ILB sin  = (15 A)(0.75 m)(1.0  10–4 T) sin 30° = 5.6  104 N
(b) F =
1
2
Fa
ILB sin  =
sin  =
1
2
ILB sin 30°
1
4

Also, sin(180° – 14.5°) =
23.
 = 14.5°
1
4
, so 180° – 14.5° = 165.5° = 166° would also work.
(a) F/L = IB sin  = 0 N since  = 0°
(b) F/L = (10 A)(0.40 T) sin 90° = 4.0 N/m in the +z direction
(c) Using the same approach gives
F/L = 4.0 N/m in the  y direction
(d) F/L = 4.0 N/m in the z direction
(e) F/L = 4.0 N/m in the + y direction
(f) F/L = (10 A)(0.40 T) sin 45° = 2.8 N/m in the +z direction
24.
(a) F = ILB sin 
0.050 N = I (0.25 m)(0.30 T) sin 90°
I = 0.67 A in the z direction
(b) Using F = ILB sin , we have
0.025 N = I(0.25 m)(0.30 T) sin 90°
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College Physics Seventh Edition: Instructor Solutions Manual
344
I = 0.33 A in the + y direction
(c) This is not possible because the force must be perpendicular to the magnetic field.
25.
(a) First find the angle  between the wire (the +x-axis) and the magnetic field.
tan  = By/Bx = (0.040 T)/(0.020 T)

 = 63.43°
The magnitude of the magnetic field is
B
B x2  B y2 
(0.020 T)2 + (0.040 T)2  0.04472 T
F/L = IB sin  = (10 A)(0.04472 T) sin 63.43° = 0.400 N/m in the +z direction
(b) Since Bx is parallel to the wire, changing it will have no effect on the force, so the answers are the
same as in part (a) .
(c) Using the same approach as in part (a), we have
tan  = (0.050 T)/(0.020 T)
B
B x2  B y2 

 = 68.20°
(0.020 T)2 + (0.050 T)2  0.05385 T
F/L = (10 A)(0.05385 T) sin 68.20° = 0.500 N/m in the z direction
26.
F = ILB sin  = (1000 A)(15 m)(5.0  105 T) sin 45 = 0.53 N .
The direction of the force is northward at an angle of 45 above the horizontal , according to the right-hand rule.
27.
F = ILB sin  = (5.0 A)(3  0.50 m)(1.0 T) sin 90 = 7.5 N upward in the plane of the paper .
28.
(a) For the magnetic torque to be at maximum, the plane of the coil should be (1) parallel to the magnetic field.
 = IAB sin  and the angle  here is between the magnetic field and a direction perpendicular to the plane of the
coil. When  = 90,  is maximum.
(b) m = IA = (1.5 A)(0.20 m)(0.30 m) = 9.0  102 A·m2 .
 = IAB sin  = mB sin  = (9.0  104 Am2)(1.6 T) sin 90 = 0.14 mN .
(c) sin  = 20% = 0.20,
29.

 =. 12o or 168o .
(a)  = NIAB sin  = (1)(0.25 A)(0.20 m2)(0.30 T) sin 60° = 0.013 m N
(b) Since the torque is proportional to the magnetic field strength, doubling the field would double the torque. If
you are allowed to also change the direction of the field, any combination of B and  such that B sin  was doubled
would also double the torque.
(c) Doubling the current will double the torque.
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345
Chapter 19
Magnetism
(d) Doubling the area will double the torque.
(e) Since sin 60° = 0.866 and the maximum that the sine can be is 1, you cannot double the sine by increasing the
angle.
μ oI
,
2r

B=N
31.
At the center of a circular coil, the magnetic field is B = µ0NI/2r. To double the field, r would have to be reduced by
a factor of
1
4
32.
1
2
I=
2(0.15m)(0.80  10 3 T)
2rB
=
= 3.8 A .
(50)(4  107 T·m / A)
Nμo
30.
. Since the area is proportional to the square of the radius, the area would decrease to
of its initial value .
(a) Putting the SI units into the off-center axial field formula (Eq. 19.13b), we have
B

µ0 NIr 2
2 r2  x 2

3/2

 T m/A (A)(m 2 )
m 
2 3/2

T m 3
 T
m3
(b) At the center of the coil, x = 0 so the above equation becomes
B

µ0 NIr 2
2 r 2  x2

3/ 2

µ0 NIr 2 µ0 NI

2r
2r 3
which is Eq. 19.13a.
33.
The field is given by B 
0.80  10 3 T =

µ0 NIr 2
2 r2  x 2

3/2
. Putting in numbers and solving for I gives
(4  10 7 T m/A )(25)(0.15 m)2 I
2 (0.075 m)2 + (0.15 m)2 
3/2
I = 11 A
34.
B=
o I
(4  107 T·m/A)(2.5 A)
=
= 2.0  106 T .
2 d
2 (0.25 m)
35.
B=
o I
,
2 d
36.
n=
100 turns
= 500 turns/m.
0.20 m

d=
o I
(4  107 Tm/A)(5.0 A)
=
= 0.25 m .
2 B
2 (4.0  106 T)
Since B = o nI,
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346
College Physics Seventh Edition: Instructor Solutions Manual
I=
37.
B
1.5  103 T
=
= 2.4 A .
o n
(4  107 T·m/A)(500 /m)
(a) The directions of the fields by the two wires are opposite. B =
B1 =
(4  107 T·m/A)(8.0 A)
= 2.67  105 T,
2 (0.060 m)
B2 =
o I
.
2 d
(4  107 T·m/A)(2.0 A)
= 6.67  106 T.
2 (0.060 m)
So the net field is B = B1  B2 = 2.67  105 T  6.67  106 T = 2.0  105 T .
(b) Assume the net field is zero at x from wire 1.
B1 =
o (8.0 A)
o (2.0 A)
=
,
2 x
2 (0.12 m  x)
Solve for
38.
x = 4(0.12 m  x).
or
x = 9.6  102 m = 9.6 cm from wire 1 .
(a) The fields by the two wires (right and left) are equal in magnitude and
opposite in direction. So the net field is 0 .
I
I
I
I
BR
BR
BL
BL
(b) The fields by the two wires (right and left) are equal in magnitude and in
the same direction. So the net field is
B = 2BR = 2
39.
o I
o I
(4  107 T·m/A)(4.0 A)
=
=
= 6.4  106 T .
2 d
d
 (0.25 m)
At the right-hand point, the two fields are both into the paper, so their magnitudes add algebraically. The field due to
a single wire is B = µ0I/2πd, so we have
B = B1 + B2 =
 (1.5 A)  1
0 I  1 1 
1 

 2.9  10 6 T
  = 0


 0.15 m 0.35 m 
2
2  d1 d 2 
into the page . The field at the left-hand point will have the same magnitude (by symmetry), but it will point
out of the page .
40.
d=
(0.12 m)2 + (0.090 m)2 = 0.15 m,
0.12 m
 = tan1 0.090 m = 53.1.
B2 =
o I2
(4  107 T·m/A)(2.0 A)
=
= 4.44  106 T,
2 d2
2 (0.090 m)
so
B 2 = (4.44  106 T) ^x.
B1 =
(4  107 T·m/A)(8.0 A)
= 1.07  105 T.
2 (0.15 m)
B 1 = (1.07  105 T)(cos 53.1 ^x  sin 53.1 y ) = (6.42  106 T) ^x + (8.56  106 T) y .
I1 0.12 m
So the net field is B = B 1 + B 2 = (1.09  105 T) ^x + (8.56  106 T) y ,

d
B2

B1
I2
0.090 m
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347
Chapter 19
or
Magnetism
(1.09  105 T)2 + (8.56  106 T)2 = 1.4  105 T . The angle between the field and a horizontal line
B=
to the left is  = tan1
o I
4(4  107 T·m/A)(2.0 A)
= 1.0  104 T‚ away from the observer .
2(0.050 m)
41.
B=N
42.
(a) It is (1) toward the observer.
2r
(b) B =
43.
o I
2r
=
(4  107 T·m/A)(1.8 A)
= 1.9  105 T .
0.12 m
The currents are opposite, and so the fields produced by the two loops are also opposite.
B1 =
44.
=
8.56  106 T
= 38 below a horizontal line to the left .
1.09  105 T
o I1
2r1
and
(a) B = o nI,
B2 =
o I2
2r2
.
B2 = 2 B1,
B2
n2
N/L2
L1
=
=
=
.
B1
n1
N/L1
L2


I2 = 2
L2 =
r2
10 cm
I =2
 (1.0 A) = 4.0 A .
r1 1
5.0 cm
B1
4.0  104 T
 L1 =
 (10 cm) = 6.7 cm .
B2
6.0  104 T
(b) Twice as many windings are required to double the field from 4.0  104 T to 8.0  104 T. So the number is
N2 = 2000 turns .
(c) B1 = o n1 I1,

I1 =
B1
4.0  104 T
=
= 0.0318 A.
o n1 (4  107 T·m/A)[1000/(0.10 m)]
Therefore the current should be
I2 =
45.
B2
9.0  104 T
 I1 =
 (0.0318 A) = 0.072 A opposite the original current direction .
B1
4.0  104 T
(a) n1 = 200 /cm = 200  102 /m and n2 = 180 /cm = 180  102 /m.
The fields by the two coils are opposite. So the net field is
B = B2  B1 = o n2 I2  o n1 I1 = (4  107 T·m/A)[(180  102 /m)(15 A)  (200  102 /m)(10 A)] = 8.8  102 T .
(b) The direction of the magnetic field at the center is to the right .
46.
(a) One cable carries current into dead battery and the other carries it out, so the currents in the cables run in
opposite directions. Therefore the two wires repel each other.
(b) B = µ0I/2πd = µ0(15 A)/[2π(0.15 m) = 2.0  10 5 T
(c) F/L = IB sin  = (15 A)(2.0  10–5 T) sin 90° = 3.0  104 N/m
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College Physics Seventh Edition: Instructor Solutions Manual
47.
348
(a) Assume the currents both run to the right with wire 1 being the upper wire and wire 2 being the lower wire. The
magnetic field at wire 2 due to wire 1 points into the paper. By the right-hand force rule, the force on wire 2 is
upward (toward wire 1), so wire 1 attracts wire 2. By Newton’s third law, wire 2 also attracts wire 1. Therefore the
forces on the wires are (1) attractive . Or you could apply the same reasoning as we just did to show that the force
on wire 1 is toward wire 2.
(b) F/L = IB sin  = I(µ0I/2πd) sin 90°
Solving for I gives
I
2 d (F / L )

µ0
2 (0.24 m)(0.024 N/m)
 170 A
4  10-7 T m/A
(c) Bnet = 0 T because the two magnetic fields are equal and in opposite directions.
48.
d=
a
o I
o I
. B1 = B2 = B3 = B4 =
=
=
2

d
2
2 (a/ 2 )
(a/2)2 + (a/2)2 =
o I
.
2a
1
2
So the net field is
B2
o I
=
2a
B = B1 + B4 = 2
2 o I
, at 45 toward the lower left wire (wire 3).
a
B1
d
B3
B4
3
49.
50.
B
4
0.71 T
(a) B =  n I ,

 = n I = [(100)/(0.50 m)](0.95 A) = 3.74  103 T·m/A .
(b)  = m o ,

m =
3.74  103 T·m/A

=
= 3.0  103 .
o
4  107 T·m/A
(a) With the magnetic material, the field is B = mµ0NI/2r. Substituting numbers gives
B = (2200)(4π  10–7 T  m/A )(100)(7.5 A)/[2(0.025 m)] = 42 T
(b) B 
51.
 m µ0 NIr 2

2 r2  x 2

3/2
=
(2200)(4  107 T  m/A)(100)(7.5 A)(0.025 m)2
2 (0.025 m)2 + (0.050 m)2 
3/ 2
= 3.7 T
(a) According to Fig. 19.30, the magnetic field at Washington, D.C. points about 10° west of true north. For a
maximum vertical force, the proton’s velocity should be perpendicular to this field. Therefore the proton should be
directed at 10° north of east .
(b) a = F/m = qvB/m
a = (1.60  10–19 C)(2000 m/s)(0.05  10–3 T)/(1.67  10–27 kg) = 9.6  106 m/s2
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349
52.
Chapter 19
Magnetism
(a) Since the magnetic field is 0.10 mT at the Earth’s surface, we need to use the formula for an off-center point on
the axis of a current loop, Eq. 19.13b, which is B 

µ0 NIr 2
2 r2  x 2

3/2
. In this case, N = 1, r = 500 km, and x = 6357 km
(the radius of the Earth. Substituting these numbers gives
0.10  103 T =
(4  10 7 T  m/A)(500,000 m)2 I
2 (500,000 m)2 + (6,357,000 m)2 
3/ 2
I = 1.7  1011 A
(b) m = IA = Iπr2 = (1.7  1011 A)π(500,000 m)2 = 1.3  1023 Am2
53.
(a) Applying Newton’s second law gives
qvB = mv2/r
Using v = 2πr/T gives
qB = m(2πr/T)/r = 2πm/T
T = 2πm/qB
The frequency is
f = 1/T =
qB
2 m
Since the radius r and speed v do not appear in these results, and the results depend only on constant quantities, T
and f are independent of r and v.
(b) r = mv/qB = (9.11  10–31 kg)(1.0  105 m/s)/[(1.60  10–19 C)(0.00010 T)
r = 5.7  10–3 m = 5.7 mm
f = qB/2πm = (1.60  10–19 C)(0.00010 T)/[2π(9.11  10–31 kg)]
f = 2.8  106 Hz = 2.8 MHz
54.
(a) Applying Newton’s second law to the electron and using v = 2πr/t gives
ke2/r2 = mv2/r = m(2πr/t)2/r
Solving for t gives
t
4 2 r 3 m

ke 2
4 2 (0.0529  109 m)3 (9.11  1031 kg)
= 1.52  1016 s
( 9.00  109 Nm 2 /C2 )(1.60  1019 C)2
In this time, a charge e goes around once, so the current is
I = e/t = (1.60  1019 C)/(1.52  1016 s) = 1.05  103 A = 1.05 mA
(b) B = µ0NI/2r
B = (4π  107 Tm/A)(1)(1.05  103 A)/[2(0.0529  109 m)] = 12.5 T
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College Physics Seventh Edition: Instructor Solutions Manual
350
(c) An electron (which is negative) orbiting clockwise is equivalent to a proton orbiting counterclockwise. Therefore,
by the right-hand source rule, the magnetic field points out of the paper, or toward the viewer .
55.
(a) Let the wires be parallel, one above the other with the upper current (I1) going to the right. The magnetic field at
wire 2 (the lower wire) due to wire 1 points into the paper, by the right-hand source rule. By the right-hand force
rule, the force on the lower wire is away from wire 1, so wire 1 repels wire 2. Similar reasoning shows that wire 2
also repels wire 1, so the forces on the wires are (2) repulsive . Or, since wire 1 repels wire 2, wire 2 also repels
wire 1, by Newton’s third law.
(b) B = µ0I/2πd
B = (4π  107 Tm/A) (3.0 A)/[2π(0.10 m)] = 6.0  106 T = 6.0 µT
(c) F1/L = I1B2 = (3.0 A)(6.0 µT) = 18 µT/m
56.
(a) By the right-hand source rule, the magnetic field is perpendicular to the plane containing the two wires,
pointing out of the figure .
(b) Since I1 > I2, B1 > B2 midway between them. So the net field is perpendicular to the plane containing the two
wires and points out of the figure .
(c) F1/L = I1B2 = I1(µ0I/2πd) =
F1/L = (4π  107 Tm/A)(8.0 A)(2.0 A)/[2π(0.12 m)] = 2.7  10 5 N/m
Since the currents are in the same direction, the wires attract so the force is toward wire 2 .
57.
(a) The top wire must attract the lower wire so it can stay in equilibrium. For the forces to attract, the currents
should be in (1) the same direction.
(b) Magnetic force cancels gravity.
For a 1.0-m length, I = I1 = I2 =
58.
F = mg.
2 dmg
o
So
=
o I1 I2 mg
F
=
=
.
L
L
2 d
2 (0.020 m)(1.5  103 kg)(9.80 m/s2)
= 38 A .
4  107 T·m/A
(a) Energy conservation gives
eV =
1
2
mv2
Solving for v we get
v
2eV

m
2(1.60  10 19 C)(3000 V)
 7.58  105 m/s = 7.6  105 m/s
–27
1.67  10 kg
(b) To balance the forces, we must have
eE = evB
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351
Chapter 19
Magnetism
B = E/v = (V/d)/v = (250 V)/[(0.10 m)(7.58  105 m/s)] = 3.3  103 T = 3.3 mT
The magnetic force on the protons must be downward, so the magnetic field must point toward the south .
(c) In this case, the magnetic force will exceed the electric force, so the protons will be deflected downward .
59.
(a) Since the currents are in the same direction, the magnitudes of the magnetic fields at the center add algebraically.
Bnet = B1 + B2 = (µ0/L)(N1I1 + N2I2)
Bnet = [(4π  107 Tm/A)/(0.10 m)][(3000)(5.0 A) + (2000)(10 A)] = 0.44 T
(b) In order to double the magnetic field at the center, the fields due to the two solenoids should point in the same
direction and have equal magnitudes.
B2 = B1
µ0N2I2/L = µ0N1I1/L
I2 = N1I1/N2 = (3000)(5.0 A)/(2000) = 7.5 A
The direction of I2 should be the same as that of I1.
(c) The currents should be in opposite directions. The fields oppose, so they will cancel if their magnitudes are
equal.
µ0N2I2/L = µ0N1I1/L
I2 = N1I1/N2 = (3000)(5.0 A)/(2000) = 7.5 A
60.
(a) Using r = mv/qB = p/qB, we get
p = qBr = (1.60  10–19 C)(0.80 T)(0.046 m) = 5.89  10–21 kg m/s
p = 5.9  1021 kgm/s
(b) K = p2/2m = (5.89  10–21 kg m/s )2/[2(1.67  10–27 kg)] = 1.04  10–14 J
K = 1.0  1014 J
(c) Since p = mv, doubling the speed would double the momentum to 2(5.89  10–21 kgm/s ) = 12 × 10–21 kg m/s .
Since K = p2/2m, doubling the momentum would increase the kinetic energy by a factor of 4 to 4.2  1014 J . The
radius is proportional to the linear momentum, so it would be doubled to 9.2 cm .
61.
The current is I =
B=
Q
(1.75  1013)(1.60  1019 C)
=
= 2.8  106 A.
1.00 s
t
o I
(4  107 T·m/A)(2.8  106 A)
=
= 2.3  1013 T down .
2 d
2 (2.40 m)
No , it does not seem likely this would interfere with a magnetic strip on an ATM card, because it is much smaller
than the Earth’s field in the range of 105 T.
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College Physics Seventh Edition: Instructor Solutions Manual
62.
352
(a) First find the current in the coil using B = µ0NI/2r.
7.45  10–3 T = (4π  10–7 Tm/A)(200)I/[2(0.100 m)]
I = 5.929 A
Applying Ohm’s law give us
V = RI = (0.115 Ω)(5.929 A) = 0.682 V
(b) B 

µ0 NIr 2
2 r2  x

2 3/2
=
(4  10 7 T m/A)(200)(5.929 A)(0.10 m)2
2 (0.100 m)2 + (0.045 m)2 
3/2
B = 5.65  10–3 T = 5.65 mT
63.
(a) The magnitude of the magnetic field created by the coil is
B=N
o I
2r
= (100) 
(4  107 T·m/A)(10.0 A)
= 1.26  104 T.
2(0.200 m)
Its direction is along the east-west direction. The direction of the Earth’s field in the horizontal direction is due
BEh
,
1.26  104 T
north. Tan 60 =
(b) B =
64.
BEh = (1.26  104 T) tan 60 = 2.18  104 T .
BEh
2.18  104 T
=
= 3.80  104 T .
cos 60
cos 55
(a) The magnetic field could be zero (4) inside the smaller one or outside the larger one .
(b) B1 = N1
I2 =
65.

o I1
2r1
= B2 = N2
o I2
2r2
.
N1 I1 r2
(200)(11.5 A)(2.50 cm)
=
= 6.05 A .
N2 I2 r1
(100)(9.50 cm)
(a) At the center of each coil, the fields are in the same direction and hence add algebraically.
Bnet = B1 + B2 =
Bnet =
0 IN  1
2

r2

 r (r 2  x 2 )3 / 2 


(4  10 7 T  m/A)(7.5 A)(100)  1
(0.10 m)2



3/2 
2
2
2
0.10 m (0.10 m) + (0.10 m)  


 

Bnet = 6.4  10–3 T = 6.4 mT
(b) The fields are of equal magnitude and in the same direction.


µ0 NIr 2

Bnet = 2B1 = 2 
 2 r2  x 2 3 / 2 




Using x = 0.050 m, this equation gives
Bnet = 6.7  10–3 T = 6.7 mT
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353
Chapter 19
Magnetism
(c) Applying the off-center formula with x1 = 10 cm and x2 = 20 cm, we get
B net 
µ0 NIr 2 
1
2
2  r  x2
1



3/2



 = 2.1  10–3 T = 2.1 mT
3/2

r 2  x 22

1

If we call x the distance from the midpoint between the two coils, a sketch of the graph of the magnetic field as a
function of location would look like the following.
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