Download Answers to Multiple-Choice Problems Solutions to Problems

20
MAGNETIC FIELD AND MAGNETIC FORCES
Answers to Multiple-Choice Problems
1. E 2. C 3. C
4. B 5. D 6. A 7. C
8. C
9. B
10. A 11. C
12. B
13. A 14. B
15. A
Solutions to Problems
S
S
S
20.1.S Set Up: The force F on the particle is in the direction of the deflection of the particle.
The directions of v, B
S
S
and F are shown in Figure 20.1. Apply the right-hand rule to the directions of v and B. See if your thumb is in the
S
direction of F, or opposite to that direction. Use F 5 0 q 0 vB sin f with f 5 90° to calculate F.
N
S
v
S
B
S
F
W
E
S
Figure 20.1
S
S
S
Solve: (a) When you apply the right-hand rule to v and B, your thumb points east. F is in this direction, so the charge
is positive.
(b) F 5 0 q 0 vB sin f 5 1 8.50 3 1026 C 2 1 4.75 3 103 m / s 2 1 1.25 T 2 sin 90° 5 0.0505 N
S
S
v and F are
20.2. Set Up: The force F on the ion is downward, in the direction of the deflection. The directions of S
S
S
S
shown in Figure 20.2a. Use the right-hand rule to relate the directions of v, B, and F. The charge is positive. Use
F 5 0 q 0 vB sin f with f 5 90° to calculate B. e 5 1.60 3 10219 C.
S
B
S
v
S
S
v
(a)
S
F
F
(b)
Figure 20.2
20-1
20-2
Chapter 20
S
S
S
S
Solve: B is perpendicular to both v and F, so B is either toward the top of the page or toward the bottom of the page.
S
If B is toward the top of the page, as shown in Figure 20.2b, then when you apply the right-hand rule your thumb
S
S
points into the page. This is the specified direction of F, so B is directed as shown in Figure 20.2b. F 5 0 q 0 vB sin f
gives
B5
F
6.94 3 10215 N
5
5 0.850 T
219
0 q 0 v sin f 6 1 1.60 3 10 C 2 1 8.50 3 103 m / s 2 sin 90°
S
20.3. Set Up: The directions of S
v and B are shown in Figure 20.3. A proton has charge e 5 11.60 3 10219 C.
F 5 0 q 0 vB sin f. In part (a), f 5 55.0°. An electron has charge 2e.
S
v
S
B
Figure 20.3
S
Solve: (a) The right-hand rule says F is into the page in Figure 20.3.
F 5 0 q 0 vB sin f 5 1 1.60 3 10219 C 2 1 3.60 3 103 m / s 2 1 0.750 T 2 sin 55.0° 5 3.54 3 10216 N.
(b) F is maximum when f 5 90°, when v is perpendicular to B. Fmax 5 0 q 0 vB 5 4.32 3 10216 N. F is minimum
S
S
when f 5 0° or 180°, when v is either parallel or antiparallel to B. Fmin 5 0.
(c) 0 q 0 is the same for an electron and a proton, so F 5 3.45 3 10216 N, the same as for a proton. Since the proton
and electron have charges of opposite sign, the forces on them are in opposite directions. The force on the electron is
directed out of the page in Figure 20.3.
S
S
Reflect: Only the component of v perpendicular to B contributes to the magnetic force. Therefore, this force is zero
S
when the charged particle moves along the direction of B and the force is maximum when the particle moves in a
S
direction perpendicular to B. When the sign of the charge changes, the force reverses direction. And, even though the
force magnitude is the same for the electron and proton, the effect of the force (the acceleration) for the electron
would be much greater, because of its smaller mass.
S
S
20.4. Set Up: The gravity force is downward so the force from the magnetic field must be upward. The charge’s
velocity and the forces are shown in Figure 20.4. Since the charge is negative, the magnetic force is opposite to the
S
right-hand rule direction. The minimum magnetic field is when the field is perpendicular to v. The force is also
S
S
perpendicular to B, so B is either eastward or westward.
N
W
S
E
v
S
S
w
S
FB
Figure 20.4
S
S
Solve: If B is eastward, the right-hand rule direction is into the page and FB is out of the page, as required. ThereS
fore, B is eastward. mg 5 0 q 0 vB sin f. f 5 90° and
B5
mg
v0q0
5
1 0.195 3 1023 kg 2 1 9.80 m / s2 2
5 1.91 T.
1 4.00 3 104 m / s 2 1 2.50 3 1028 C 2
Magnetic Field and Magnetic Forces
20-3
S
20.5.S Set
Up: The charge is positive, so the magnetic force is in the right-hand rule direction. Figure 20.5 shows v
S
S
and B. F 5 ma relates the force and acceleration.
y
S
v
x
z
S
B
Figure 20.5
S
S
S
Solve: (a) The right-hand rule applied to v and B gives that FB is in the 1z direction.
FB 5 0 q 0 vB sin f 5 1 3.50 3 1028 C 2 1 2.00 3 105 m / s 2 1 0.80 T 2 sin 90° 5 5.60 3 1023 N
5.60 3 1023 N
F
S
5
5 1.12 m / s2. a is in the 1z direction.
m
5.00 3 1023 kg
S
S
Reflect: The force and acceleration are perpendicular to the plane formed by the directions of v and B.
(b) a 5
S
20.6.
Set Up: Use F 5 0 q 0 vB sin f to relate the force F to the angle f between the direction of v and the direction
S
of B. Since 0 q 0 , v, and B are unknown and constant, set up a ratio that involves just F and f.
Solve:
F1
F2
F
5 0 q 0 vB, which is constant. Therefore,
5
.
sin f
sin f1
sin f 2
F2 5 F1
1
2 1
2
sin f 2
sin 30.0°
5F
5 1.93F.
sin f 1
sin 15.0°
20.7. Set Up: Use F 5 0 q 0 vB sin f to calculate the force
and then use a 5 F / m to calculate the acceleration. The
S
S
charge q of the nucleus is 4e. The directions of v and B are shown in Figure 20.7, where the plane of the page is
S
horizontal and upward is directed out of the page. The acceleration has the same direction as F.
N
W
E
S
S
S
v
14 e
B
Figure 20.7
Solve: F 5 0 q 0 vB sin f 5 4 1 1.60 3 10219 C 2 1 1.35 3 103 m / s 2 1 1.12 T 2 sin 90° 5 9.68 3 10216 N.
a5
9.68 3 10216 N
F
5 6.45 3 1010 m / s2.
5
m
1.50 3 10226 kg
S
The right-hand rule gives that F is north, so the acceleration is north.
20.8. Set Up: F 5 0 q 0 vB sin f. Since the particle has negative charge, the force is opposite to the right-hand rule
(rhr) direction.
S
S
Solve: (a) The directions of v and B are shown in Figure 20.8a. The angle f is 130°.
F 5 1 2.50 3 1028 C 2 1 40.0 3 103 m / s 2 1 2.00 T 2 sin 130° 5 1.53 3 1023 N.
The rhr direction is in the 1z direction so the force is in the 2z direction.
20-4
Chapter 20
y
y
S
v
S
v
S
F
140°
f
S
50°
B
50°
x
x
z
z
S
B
(a)
(b)
Figure 20.8
S
S
(b) The directions of v and B are shown in Figure 20.8b. f 5 90°.
F 5 0 q 0 vB sin f 5 1 2.50 3 1028 C 2 1 40.0 3 103 m / s 2 1 2.00 T 2 sin 90° 5 2.00 3 1023 N.
S
S
The right-hand rule direction and the direction of F are also shown in Figure 20.8b; F is 140° counterclockwise from
the 1x axis, so is 50° counterclockwise from the 1y axis.
S
20.9. Set Up: S
v and B at each lettered point are shown in Figure 20.9a–e. The direction of the magnetic force at
each point is given by the right-hand rule.
y
y
y
S
S
v
F
S
F
S
S
S
v
B
x
B
x
x
S
B
z
S
v
(b)
(a)
z
z
(c)
S
v
y
S
v
y
45.0°
x
S
x
B
S
F
z
S
B
z
S
F
(d)
(e)
Figure 20.9
S
S
Solve: Point a: F 5 qvBsinf 5 qvBsin90° 5 qvB. F is in the 2z direction. Point b: F 5 qvB. F is in the 1y
S
direction. Point c: f 5 180° and F 5 0. Point d: f 5 45° and F 5 qvB sin 45° 5 qvB / "2 . F is in the 2y direcS
tion. Point e: f 5 90° and F 5 qvB. F is in the yz plane, midway between the 2y and 2z axes.
S
S
S
Reflect: The force is largest when v is perpendicular to B and the force is zero when v is either parallel or antiS
parallel to B.
Magnetic Field and Magnetic Forces
20-5
20.10. Set Up: Eq. (20.3) relates v, B and E for a velocity selector. For the ions to pass through undeflected, the net
force must be zero and therefore the electric and magnetic forces must be in opposite directions. For positive charges
the electric force is in the same direction as the electric field and for negative charges the field and force are in
opposite directions.
Solve: (a) E 5 vB 5 1 8.75 3 103 m / s 2 1 0.550 T 2 5 4.81 3 103 N / C.
(b) Take the velocity to be to the right and let the electric field be downward, as shown in Figure 20.10. Since the
charge is positive, the electric force is downward. For an ion to pass through undeflected, the net force must be zero,
S
S
S
so the magnetic force must be upward. Using the right-hand rule for v and B we deduce that B must be directed into
the plane of the figure.
S
FB
S
B
S
v
S
E
S
FE
Figure 20.10
S
S
(c) For a negative charge and the E and B fields of Figure 20.10, the direction of each force is reversed. But they are
still in opposite directions from each other, so will cancel if their magnitudes are the same. The magnitude of the
charge divides out in the derivation of Eq. (20.3), so the same velocity selector works for negative ions and positive
ions of any charge.
S
20.11. Set Up: The relation between v, B and E is given by Eq. (20.3). E points from the positive plate to the
negative plate, so is directed upward in the figure. The electric field is related to the potential difference V between
the plates by V 5 Ed, where d is the separation between the plates.
Solve: (a) E 5
3.33 3 103 V / m
V
150 V
E
3
5 1.02 T.
5
V
m.
B
5
5
3.33
3
10
5
/
v
d
4.50 3 1022 m
3.25 3 103 m / s
(b) The forces and fields are shown in Figure 20.11. Assume a positive charge; the same field directions also work for
S
S
S
S
S
a negative charge. E is upward so the electric force FE is upward. In order for FB to be downward, to oppose FE , B
must be out of the page, as shown.
S
FE
S
B
S
v
S
E
S
FB
Figure 20.11
Reflect: Ions with speed v 5 E / B are undeflected no matter what is the sign or magnitude of their charge. If v is not
equal to E / B, then either FB or FE is larger than the other, there is a net force on the charge and the charge does not
travel in a straight line.
20-6
Chapter 20
1 2
20.12. Set Up: v 5 E / B and E is constant, so vB is constant.
Solve: v1B1 5 v2B2 and B2 5
v1
v
B 5
B 5 B / 2.
v2 1 2v
20.13. Set Up: Eq. (20.4) says R 5
Solve: v 5
mv
0q0B
. A proton has mass m 5 1.67 3 10227 kg and charge
q 5 1e 5 1.60 3 10219 C.
1 0.0613 m 2 1 1.60 3 10219 C 2 1 0.250 T 2
R0q0B
5
5 1.47 3 106 m / s.
m
1.67 3 10227 kg
20.14. Set Up: Since the particle moves perpendicular to the uniform magnetic field, the radius of its path is
R5
mv
0q0B
S
S
. The magnetic force is perpendicular to both v and B. The alpha particle has charge
q 5 12e 5 3.20 3 10219 C.
Solve: (a) R 5
1 6.64 3 10227 kg 2 1 35.6 3 103 m / s 2
5 6.73 3 1024 m 5 0.673 mm.
1 3.20 3 10219 C 2 1 1.10 T 2
The alpha particle moves in a circular arc of diameter 2R 5 1.35 mm.
(b) For a very short time interval the displacement of the particle is in the direction of the velocity. The magnetic
force is always perpendicular to this direction so it does no work. The work-energy theorem therefore says that the
kinetic energy of the particle, and hence its speed, is constant.
(c) The acceleration is
a5
0 q 0 vB sin f
1 3.20 3 10219 C 2 1 35.6 3 103 m / s 2 1 1.10 T 2 sin 90°
FB
5
5
5 1.88 3 1012 m / s2.
m
m
6.64 3 10227 kg
We can also use a 5
v2
and the result of part (a) to calculate
R
1 35.6 3 103 m / s 2 2
a5
5 1.88 3 1012 m / s2,
6.73 3 1024 m
S
S
the same result. The acceleration is perpendicular to v and B and so is horizontal, toward the center of curvature of
the particle’s path.
S
S
S
(d) The unbalanced force 1 FB 2 is perpendicular to v, so it changes the direction of v but not its magnitude, which is
the speed.
20.15. Set Up: A proton and a deuteron each have charge 1e 5 1.60 3 10219 C, but they have different masses.
The mass of a proton is 1.67 3 10227 kg. The radius of the circular arc that the particle moves in is R 5
mv
0q0B
.
1 0.556 m 2 1 1.60 3 10219 C 2 1 0.500 T 2
R0q0B
5
5 1.33 3 107 m / s.
m
3.34 3 10227 kg
The speed is not changed by the magnetic field so the deuteron has this speed both just before it enters the field and
just after it leaves the field.
R1
R2
v
R
5 constant and
5
.
(b) Only the mass changes. 5
m
m1
m2
0q0B
Solve: (a) v 5
R2 5 R1
1 2
1
2
1.67 3 10227 kg
m2
5 1 55.6 cm 2
5 27.7 cm.
m1
3.34 3 10227 kg
Reflect: The magnetic force does no work on the moving charge so the motion in a magnetic field is at constant
speed. The radius of the path decreases as m decreases.
Magnetic Field and Magnetic Forces
20-7
v2
and the net force is radially inward, toward the center of the
R
circle. The direction of the force is shown in Figure 20.16. The mass of a proton is 1.67 3 10227 kg.
20.16. Set Up: For motion in an arc of a circle, a 5
S
F
B
v
Figure 20.16
S
S
S
Solve: (a) F is opposite to the right-hand rule direction, so the charge is negative. F 5 ma gives
0 q 0 vB sin f 5 m
v2
. f 5 90° and
R
0 q 0 BR 3 1 1.60 3 10219 C 2 1 0.250 T 2 1 0.475 m 2
v5
5
5 2.84 3 106 m / s.
m
12 1 1.67 3 10227 kg 2
(b) FB 5 0 q 0 vB sin f 5 3 1 1.60 3 10219 C 2 1 2.84 3 106 m / s 2 1 0.250 T 2 sin 90° 5 3.41 3 10213 N.
w 5 mg 5 12 1 1.67 3 10227 kg 2 1 9.80 m / s2 2 5 1.96 3 10225 N. The magnetic force is much larger than the weight
of the particle, so it is a very good approximation to neglect gravity.
(c) The magnetic force is always perpendicular to the path and does no work. The particles move with constant speed.
20.17. Set Up: R 5
mv
.
0q0B
1 2 1 2
B1
B1
mv
5 constant. R1B1 5 R2B2 and R2 5 R1
5R
5 R / 3.
B2
3B1
0q0
Reflect: The radius of the circular path decreases when B increases.
Solve: RB 5
v2
and the net force is radially inward, toward the center of
R
the circle. The direction of the force is shown in Figure 20.18. Since the electron has negative charge, the force
on it is opposite to the right-hand rule direction. The magnetic force does no work and the electron moves at
constant speed. The distance from A to B along the path is pR, where R 5 5.00 cm. An electron has charge
q 5 2e 5 21.60 3 10219 C and mass 9.11 3 10231 kg. A proton has charge q 5 1e and mass 1.67 3 10227 kg.
20.18. Set Up: For motion in an arc of a circle, a 5
v0
v0
F
F
Figure 20.18
S
S
Solve: (a) To produce a force in the direction shown, the magnetic field must be directed into the page. F 5 ma
v2
gives 0 q 0 vB sin f 5 m . f 5 90° and
R
1 9.11 3 10231 kg 2 1 1.41 3 106 m / s 2
mv
B5
5 1.61 3 1024 T
5
1 1.60 3 10219 C 2 1 5.00 3 1022 m 2
0q0R
p 1 5.00 3 1022 m 2
pR
5
5 1.11 3 1027 s
v
1.41 3 106 m / s
S
(c) Since the proton has positive charge, B must be directed out of the page to produce a force in the required
direction.
1 1.67 3 10227 kg 2 1 1.41 3 106 m / s 2
mv
5 0.294 T.
5
B5
1 1.60 3 10219 C 2 1 5.00 3 1022 m 2
0q0R
(b) t 5
20-8
Chapter 20
20.19. Set Up: In the acceleration process the electric potential energy decrease, 0 q 0 V, equals the increase in kinetic
energy, 12 mv2. Use this to find the speed v of the protons after they have been accelerated. In the magnetic field,
mv
.
R5
0q0B
Protons have charge 1e and mass 1.67 3 10227 kg. Electrons have charge 2e and mass 9.11 3 10231 kg.
Solve: (a) 0 q 0 V 5 12 mv2. v 5
B5
(b) B 5
2 0q0 V
2 1 1.60 3 10219 C 2 1 0.745 3 103 V 2
mv
5
5 3.78 3 105 m / s. R 5
and
227
Å
Å m
0q0B
1.67 3 10 kg
1 1.67 3 10227 kg 2 1 3.78 3 105 m / s 2
mv
5 4.51 3 1023 T.
5
1 1.60 3 10219 C 2 1 0.875 m 2
0q0R
Bp
Be
v
B
mv
mv
. In
5
.
5 constant and
only m changes. 5
m
mp
me
0q0R
0q0R
0q0R
Be 5
1 2
1
2
9.11 3 10231 kg
me
1 4.51 3 1023 T 2 5 2.46 3 1026 T.
Bp 5
mp
1.67 3 10227 kg
Reflect: For lighter particles a smaller force and hence smaller B is needed to produce the same path. To produce
electrons of the same speed as the protons a smaller accelerating voltage would be required.
20.20. Set Up: 1 gauss 5 1 3 1024 T. The earth has radius 6.38 3 106 m. If the dust speck moves in orbit just
outside the earth’s surface, its orbit radius equals the radius of the earth. R 5
mv
0q0B
.
1 6.38 3 106 m 2 1 1.60 3 1029 C 2 1 0.50 3 1024 T 2
R0q0B
5
5 0.0227 m / s.
m
0.0225 3 1023 kg
(b) FB 5 qvB 5 1.8 3 10215 N. Fg 5 mg 5 1 0.0225 3 1023 kg 2 1 9.80 m / s2 2 5 2.21 3 1024 N. The gravity force
is much larger and should not be neglected.
Solve: (a) v 5
20.21. Set Up: An electron has charge q 5 2e 5 21.60 3 10219 C and mass 9.11 3 10231 kg. When it moves in
v2
a circular path of radius R, its acceleration is . When a particle of charge 2e is accelerated through a potential
R
difference of magnitude V, it gains kinetic energy eV.
Solve: 12mv2 5 eV and
v5
2 1 1.60 3 10219 C 2 1 2.00 3 103 V 2
2eV
5
5 2.65 3 107 m / s.
Å m
Å
9.11 3 10231 kg
S
F 5 ma gives 0 q 0 vB sin f 5 m
v2
. f 5 90° and
R
1 9.11 3 10231 kg 2 1 2.65 3 107 m / s 2
mv
B5
5 8.38 3 1024 T.
5
1 1.60 3 10219 C 2 1 0.180 m 2
0q0R
S
20.22. Set Up: In a velocity selector, E 5 vB (Section 20.2). For motion in a circular arc in a magnetic field Br,
R5
mv
0 q 0 Br
(Section 20.3). The ion has charge 1e.
Solve: (a) E 5 vB 5 1 4.50 3 103 m / s 2 1 0.0250 T 2 5 112 V / m.
1 6.64 3 10226 kg 2 1 4.50 3 103 m / s 2
mv
(b) Br 5
5 1.49 3 1022 T.
5
1 1.60 3 10219 C 2 1 0.125 m 2
0q0R
20.23. Set Up: In a velocity selector, E 5 vB (Section 20.2). For motion in a circular arc in a magnetic field Br,
R5
mv
0 q 0 Br
(Section 20.3). The ions have charge 1e.
155 V / m
E
5
5 4.92 3 103 m / s
B
0.0315 T
1 0.175 m 2 1 1.60 3 10219 C 2 1 0.0175 T 2
R 0 q 0 Br
5
5 9.96 3 10226 kg
(b) m 5
v
4.92 3 103 m / s
Reflect: Ions with larger m / 0 q 0 move in a path of larger radius.
Solve: (a) v 5
Magnetic Field and Magnetic Forces
20-9
mv
. After completing one semicircle the separation between the ions is the difference in the
0q0B
diameters of their paths, or 2 1 R13 2 R12 2 . A singly ionized ion has charge 1e.
20.24. Set Up: R 5
1 1.99 3 10226 kg 2 1 8.50 3 103 m / s 2
mv
5 8.46 3 1023 T.
5
1 1.60 3 10219 C 2 1 0.125 m 2
0q0R
R13
R12
v
R
5
.
(b) The only difference between the two isotopes is their masses. 5
5 constant and
m
m 12
m13
0q0B
Solve: (a) B 5
R13 5 R12
1 2
1
2
2.16 3 10226 kg
m13
5 1 12.5 cm 2
5 13.6 cm.
m12
1.99 3 10226 kg
The diameter is 27.2 cm.
(c) The separation is 2 1 R13 2 R12 2 5 2 1 13.6 cm 2 12.5 cm 2 5 2.2 cm. This distance can be easily observed.
20.25. Set Up: 0 q 0 , v and B are the same as in Problem 20.24. For m 5 1.99 3 10226 kg 1 12C 2 , R12 5 12.5 cm.
The separation of the isotopes at the detector is 2 1 R15 2 R14 2 .
R14
R12
v
mv R
. 5
5
5 constant.
Solve: R 5
and
m14
m12
0q0B m
0q0B
R14 5 R12
R15 5 R12
1 2
1 2
1
21
2
2
2.32 3 10226 kg
m14
5 1 12.5 cm 2
5 14.6 cm.
m12
1.99 3 10226 kg
m15
5 1 12.5 cm
m12
2.49 3 10226 kg
1.99 3 10226 kg
5 15.6 cm.
The separation of the isotopes at the detector is 2 1 R15 2 R14 2 5 2 1 15.6 cm 2 14.6 cm 2 5 2.0 cm.
Reflect: The separation is large enough to be easily detectable. Since the diameter of the ion path is large, about
30 cm, the uniform magnetic field within the instrument must extend over a large area.
S
20.26.S Set Up: F 5 IlB sin f. The direction of F is given by applying the right-hand rule to the directions of
I and B.
S
Solve: (a) The current and field directions are shown in Figure 20.26a. The right-hand rule gives that F is directed to
the south, as shown. f 5 90° and F 5 1 1.20A 2 1 1.00 3 1022 m 2 1 0.588 T 2 5 7.06 3 1023 N.
N
W
E
F
S
B
60°
F
I
I
30°
I
B
B
F
(a)
(b)
(c)
Figure 20.26
S
(b) The right-hand rule gives that F is directed to the west, as shown in Figure 20.26b. f 5 90° and
F 5 7.06 3 1023 N, the same as in part (a).
S
(c) The current and field directions are shown in Figure 20.26c. The right-hand rule gives that F is 60.0° north of
23
west. f 5 90° so F 5 7.06 3 10 N, the same as in part (a).
20.27. Set Up: F 5 IlB sin f. The current and field are perpendicular, so f 5 90°. 1 gauss 5 1024 T.
Solve: F 5 1 50,000 A 2 1 1 m 2 1 0.5 3 1024 T 2 sin 90° 5 2.5 N
20.28. Set Up: F 5 IlB sin f. Since the field is perpendicular to the rod it is perpendicular to the current and
f 5 90°.
0.13 N
F
5
5 9.7A
Solve: I 5
1 0.200 m 2 1 0.067 T 2
lB
20-10
Chapter 20
S
20.29.S Set Up: F 5 IlB sin f. The direction of F is determined by applying the right-hand rule to the directions of
I and B. 1 gauss 5 1024 T.
S
Solve: (a) The directions of I and B are sketched in Figure 20.29a. f 5 90° and
F 5 1 1.5 A 2 1 2.5 m 2 1 0.55 3 1024 T 2 5 2.1 3 1024 N.
S
The right-hand rule says that F is directed out of the page, so is upward.
N
N
B
B
I
I
F
E
W
E
W
F
S
S
(b)
(a)
Figure 20.29
S
S
(b) The directions of I and B are sketched in Figure 20.29b. f 5 90° and F 5 2.1 3 1024 N. F is directed east to
west.
S
(c) B and the direction of the current are antiparallel. f 5 180° and F 5 0.
(d) The magnetic force of 2.1 3 1024 N is not large enough to cause significant effects.
S
Reflect: The magnetic force is a maximum when the directions of I and B are perpendicular and it is zero when the
current and magnetic field are either parallel or antiparallel.
20.30. Set Up: The magnetic force is F 5 IlB sin fSand is maximum when f 5 90°. For the wire to be completely
S
supported by the field requires that F 5 mg and that F and w are in opposite directions.
Solve: (a) IlB 5 mg.
I5
1 0.150 kg 2 1 9.80 m / s2 2
mg
5
5 1.34 3 104 A.
1 2.00 m 2 1 0.55 3 1024 T 2
lB
This is a very large current and ohmic heating due to the resistance of the wire would be severe; such a current isn’t
feasible.
S
S
(b) The magnetic force must be upward. The directions of I, B and F are shown in Figure 20.30, where we have
S
assumed that B is south to north. To produce an upward magnetic force, the current must be to the east. The wire
must be horizontal and perpendicular to the earth’s magnetic field.
N
B
W
I
F
S
Figure 20.30
E
Magnetic Field and Magnetic Forces
S
20-11
S
20.31. Set Up: F 5 IlB sin f. The direction of F is given by the right-hand rule applied to the directions of I and B.
Solve: (a) segment da: f 5 180° and F 5 0
S
S
segment ab: f 5 90° and F 5 IlB 5 1 8.0 A 2 1 0.200 m 2 1 0.750 T 2 5 1.2 N. The directions of I, B and F are
S
shown in Figure 20.31a. F is directed into the page.
segment bc: f 5 0° and F 5 0
S
S
S
segment cd: f 5 90° and F 5 1.2 N. The directions of I, B and F are shown in Figure 20.31b. F is directed out of
the page.
I
I
Fab
Fcd
B
B
(b)
(a)
Figure 20.31
S
S
(b) Fab and Fcd are equal in magnitude and opposite in direction. Their vector sum is zero and the net force on the
entire circuit is zero.
Reflect: The net force on any current loop in a uniform magnetic field is zero.
20.32. Set Up: The direction
of the magnetic force on each current segment in the field is shown in Figure 20.32.
S
S
By symmetry, Fa 5 Fb. Fa and Fb are in opposite directions so their vector sum is zero. The net force equals Fc.
F 5 IlB sin f. For Fc, f 5 90° and l 5 0.450 m.
B
I
Fa
Fb
I
Fc
I
Figure 20.32
Solve: Fc 5 IlB 5 1 6.00 A 2 1 0.450 m 2 1 0.666 T 2 5 1.80 N. The net force is 1.80 N, directed to the left.
20.33. Set Up: Label the three segments in the field as a, b, and c. Let x be the length of segment a. Segment b has
length 0.300 m and segment c has length 0.600 cm 2 x. Figure 20.33a shows the direction of the force on each
segment. For each segment, f 5 90°. The total force on the wire is the vector sum of the forces on each segment.
B
I
Fb
c
I
u
Fc
B
Fb
I
B
b
a
Fac
F
Fa
(a)
Figure 20.33
(b)
20-12
Chapter 20
Solve: Fa 5 IlB 5 1 4.50 A 2 x 1 0.240 T 2 . Fc 5 1 4.50 A 2 1 0.600 m 2 x 2 1 0.240 T 2 . Since Fa and Fc are in the
same direction their vector sum has magnitude Fac 5 Fa 1 Fc 5 1 4.50 A 2 1 0.600 m 2 1 0.240 T 2 5 0.648 N and is
directed toward the bottom of the page in Figure 20.33a. Fb 5 1 4.50 A 2 1 0.300 m 2 1 0.240 T 2 5 0.324 N and is
S
S
directed to the right. The vector addition diagram for Fac and Fb is given in Figure 20.33b.
S
F 5 "Fac2 1 Fb2 5 " 1 0.648 N 2 2 1 1 0.324 N 2 2 5 0.724 N. tan u 5
S
Fac
0.648 N
5
and u 5 63.4°.
Fb
0.324 N
The net force has magnitude 0.724 N and its direction is specified by u 5 63.4° in Figure 20.33b.
Reflect: All three current segments are perpendicular to the magnetic field, so f 5 90° for each in the force equation. The direction of the force on a segment depends on the direction of the current for that segment.
20.34. Set Up: F S5 IlB sin f. The direction of the magnetic force is given by the right-hand rule applied to the
directions of I and B. The torque due to a force equals the force times the moment arm, the perpendicular distance
between the axis and the line of action of the force.
Solve: (a) The direction of the magnetic force on each segment of the circuit is shown in Figure 20.34. For segments
bc and da the current is parallel or antiparallel to the field and the force on these segments is zero.
I
c
b
B
I
B
B
Fab
Fcd
I
B
a
d
I
Figure 20.34
S
S
(b) Fab acts at the hinge and therefore produces no torque. Fcd tends to rotate the loop about the hinge so it does produce a torque about this axis. Fcd 5 IlB sin f 5 1 5.00 A 2 1 0.200 m 2 1 1.20 T 2 sin 90° 5 1.20 N
(c) t 5 Fl 5 1 1.20 N 2 1 0.350 m 2 5 0.420 N # m. The torque is directed so as to rotate side cd out of the plane of
the page in Figure 20.34.
20.35. Set Up: t 5 IAB sin f. Since the plane of the loop is parallel to the field, the field is perpendicular to the
normal to the loop and f 5 90°. The magnetic moment of the loop is m 5 IA.
Solve: (a) t 5 IAB 5 1 6.2 A 2 1 0.050 m 2 1 0.080 m 2 1 0.19 T 2 5 4.7 3 1023 N # m
(b) m 5 IA 5 1 6.2 A 2 1 0.050 m 2 1 0.080 m 2 5 0.025 A # m2
Reflect: The torque is a maximum when the field is in the plane of the loop and f 5 90°.
20.36. Set Up: t 5 IAB sin f. The maximum torque is for f 5 90°, where f is the angle between the field and the
normal to the plane of the loop. The coil has area A 5 pr 2 5 p 1 4.3 3 1022 m 2 2 5 5.81 3 1023 m2.
Solve: (a) t 5 1 15 2 1 2.7 A 2 1 5.81 3 1023 m2 2 1 0.56 T 2 5 0.132 N # m2. The plane of the coil is parallel to the
field.
(b) t 5 0.71tmax means sin f 5 0.71 and f 5 45°. The plane of the coil makes an angle of 45° with the field
direction.
Magnetic Field and Magnetic Forces
20-13
20.37. Set Up: t 5 IAB sin f. The coil as viewed along the axis of rotation is shown in Figure 20.37a for its
original position and in Figure 20.37b after it has rotated 30.0°.
F2
I
F2
I
axis
F3
F1
F3
I
F4
F4
B
30°
30° B
normal
I
F1
normal
(a)
(b)
Figure 20.37
S
S
S
S
Solve: (a) The forces on each side of the coil are shown in Figure 20.37a. F1 1 F2 5 0 and F3 1 F4 5 0. The net
force on the coil is zero. f 5 0° and sin f 5 0, so t 5 0. The forces on the coil produce no torque.
(b) The net force is still zero. f 5 30.0° and the net torque is
t 5 1 1 2 1 1.40 A 2 1 0.220 m 2 1 0.350 m 2 1 1.50 T 2 sin 30.0° 5 0.0808 N # m.
The net torque is clockwise in Figure 20.37b and is directed so as to increase the angle f.
Reflect: For any current loop in a uniform magnetic field the net force on the loop is zero. The torque on the loop
depends on the orientation of the plane of the loop relative to the magnetic field direction.
20.38. Set Up: The torque on a solenoid is t 5 NIAB sin f. The angle f is the angle between the axis of the
solenoid and the direction of the field.
Solve: (a) t 5 NIAB sin f 5 1 165 2 1 1.20 A 2 1 6.75 3 1024 m2 2 1 1.12 T 2 sin 90° 5 0.150 N # m
(b) f 5 0° and t 5 0.
(c) f 5 35.0° and t 5 1 0.150 N # m 2 sin 35.0° 5 0.0860 N # m
20.39. Set Up: t 5 NIAB sin f. The area A is related to the diameter D by A 5 14 pD 2.
Solve: t 5 NI 1 14 pD 2 2 B sin f. t is proportional to D 2. Increasing D by a factor of 3 increases t by a factor of 32 5 9.
m0 I
. m 5 4p 3 1027 T # m / A.
2pr 0
1 2p 2 1 0.040 m 2 1 5.50 3 1024 T 2
2prB
Solve: (a) I 5
5
5 110 A
m0
4p 3 1027 T # m / A
20.40. Set Up: For a long, straight wire, B 5
(b) B 5
part (a).
1 4p 3 1027 T # m / A 2 1 110 A 2
5 2.75 3 1024 T. B is proportional to 1 / r so the field is half what it is in
2p 1 8.00 3 1022 m 2
20.41. Set Up: For a long straight wire, B 5
m 0 5 4p 3 1027 T # m / A.
Solve: B 5
field.
m0 I
. The magnetic field of the earth is of the order of 1024 T.
2pr
1 4p 3 1027 T # m / A 2 1 10 A 2
5 4.0 3 1025 T. This field is the same order of magnitude as the earth’s
2p 1 0.050 m 2
m0 I
. m 5 4p 3 1027 T # m / A. 1 gauss 5 1024 T.
2pr 0
1 4p 3 1027 T # m / A 2 1 50,000 A 2
5 2.0 3 1023 T
Solve: (a) B 5
2p 1 5.0 m 2
20.42. Set Up: For a long straight wire, B 5
(b) Bearth 5 0.5 3 1024 T, so
B
Bearth
5
2.0 3 1023 T
5 40
0.5 3 1024 T
20-14
Chapter 20
20.43. Set Up: 1 gauss 5 1024 T. For a long straight wire, B 5
Solve: I 5
m0 I
. m 0 5 4p 3 1027 T # m / A.
2pr
2p 1 0.050 m 2 1 1.0 3 10210 T 2
2prB
5 2.5 3 1025 A 5 25 mA.
5
m0
4p 3 1027 T # m / A
m0 I
. m 0 5 4p 3 1027 T # m / A.
2pr
1 4p 3 1027 T # m / A 2 1 12 3 1023 A 2
Solve: B 5
5 1.2 3 1027 T 5 0.12 mT
2p 1 0.020 m 2
20.44. Set Up: B 5
m0 I
2pr
Solve: (a) B as a function of r is sketched in Figure 20.45.
20.45. Set Up: B 5
B
r
Figure 20.45
(b) B is proportional to 1 / r. B approaches zero as r increases but becomes zero only at an infinite distance from the wire.
m0 I
Reflect: B 5
applies precisely only to an infinitely long straight wire. Any real wire is finite in length. But this
2pr
expression is accurate for a wire of finite length at points much closer to the wire than their distance from the closest
end of the wire.
m0 I
. B1 5 B when r1 5 R. Calculate r2 when B2 5 B / 3.
2pr
m0 I
B1
B
5 constant and B1r1 5 B2r2. r2 5 r1
5R
5 3R.
Solve: Br 5
2p
B2
B/3
1 2 1 2
20.46. Set Up: B 5
20.47. Set Up: B 5
m0 I
. B 5 8.0 mT when r1 5 2.0 cm.
2pr 1
12
1
2
m0 I
r1
2.0 cm
5 constant and B1r1 5 B2r2. B2 5 B1
5 1 8.0 mT 2
5 4.0 mT.
r2
2p
4.0 cm
8.0 mT
B1
5 1 2.0 cm 2
5 16.0 cm.
(b) r2 5 r1
B2
1.0 mT
m0
I2
B1
B2
B
2I
5 constant and
5 . B2 5 B1
5 1 8.0 mT 2
5 16.0 mT
(c) 5
I
2pr
I1
I2
I1
I
1 2
Solve: (a) Br 5
1
2
12
1 2
Reflect: B is proportional to I and to 1 / r. When I is doubled, B is doubled. When r is doubled, B is halved.
m0 I
. m 5 4p 3 1027 T # m / A.
2pr 0
Solve: Assume that the lines are 10 m off the ground. At this distance
20.48. Set Up: B 5
1 4p 3 1027 T # m / A 2 1 100 A 2
5 2 3 1026 T.
2p 1 10 m 2
This is about 4% of the earth’s magnetic field. The magnetic field is too small to be of concern.
B5
Magnetic Field and Magnetic Forces
20-15
m0 I
. If the currents are in the same direction, their fields add. If a pair
2pr
of wires carry current in opposite directions, their fields are in opposite directions and cancel.
1 4p 3 1027 T # m / A 2 1 0.300 A 2
m0 I
Solve: (a) B 5 6
56
5 1.44 3 1027 T
2pr
2p 1 2.50 m 2
(b) The fields of two pairs cancel and the net field is due to the remaining two wires.
20.49. Set Up: For a long, straight wire B 5
m0 I
5 1 1 1.44 3 1027 T 2 5 4.80 3 1028 T
2pr 3
Reflect: In a typical household appliance cord there are two closely spaced conductors carrying current in opposite
directions and the currents in the cord produce very little net magnetic field.
B52
20.50. Set Up: The net magnetic field at each point is the vector sum of the fields due to each wire. For each wire
m0 I
S
and the direction of B is determined by the right-hand rule described in Section 20.7. Let the wire with
2pr
12.0 A be wire 1 and the wire with 10.0 A be wire 2.
1 4p 3 1027 T # m / A 2 1 12.0 A 2
m 0 I1
S
Solve: (a) Point Q: B1 5
5
5 1.6 3 1025 T. The direction of B1 is out of
2pr1
2p 1 0.15 m 2
the page.
1 4p 3 1027 T # m / A 2 1 10.0 A 2
m 0 I2
B2 5
5
5 2.5 3 1025 T.
2pr2
2p 1 0.080 m 2
B5
S
S
S
The direction of B2 is out of the page. Since B1 and B2 are in the same direction, B 5 B1 1 B2 5 4.1 3 1025 T and
S
B is directed out of the page.
Point P: B1 5 1.6 3 1025 T, directed into the page. B2 5 2.5 3 1025 T, directed into the page.
B 5 B1 1 B2 5 4.1 3 1025 T
S
and B is directed into the page.
S
S
(b) B1 is the same as in part (a), out of the page at Q and into the page at P. The direction of B2 is reversed from what
it was in (a) so is into the page at Q and out of the page at P.
S
S
Point Q: B1 and B2 are in opposite directions so B 5 B2 2 B1 5 2.5 3 1025 T 2 1.6 3 1025 T 5 9.0 3 1026 T
S
and B is directed into the page.
S
S
S
Point P: B1 and B2 are in opposite directions so B 5 B2 2 B1 5 9.0 3 1026 T and B is directed out of the page.
m0 I
S
. The direction of B is given by the right-hand rule in Section 20.7. Call the wires a and b,
2pr
as indicated in Figure 20.51. The magnetic fields of each wire at points P1 and P2 are shown in Figure 20.51.
20.51. Set Up: B 5
Ba
P1
a
Bb
Ba
b
P2
Bb
5.0 cm
5.0 cm
20.0 cm
25.0 cm
Figure 20.51
Solve: (a) At P1, Ba 5 Bb and the two fields are in opposite directions, so the net field is zero.
m0 I
m0 I S
S
(b) Ba 5
. Bb 5
. Ba and Bb are in the same direction so
2pra
2prb
B 5 Ba 1 Bb 5
S
1
2
1 4p 3 1027 T # m / A 2 1 4.00 A 2
m0 I 1
1
1
1
1
5
1
5 6.67 3 1026 T
S
r
r
2p a
2p
0.300 m
0.200 m T
b
B has magnitude 6.67 mT and is directed toward the top of the page.
Reflect: At points directly to the left of both wires the net field is directed toward the bottom of the page.
20-16
Chapter 20
m0 I
S
. The direction of B is given by the right-hand rule in Section 20.7.
2pr
Solve: (a) The magnetic fields of the two wires are in opposite directions at points between the wires, as shown in
Figure 20.52a. Consider a point that is a distance x from wire 1 and hence 0.400 m 2 x from wire 2. B1 5 B2 gives
20.52. Set Up: B 5
m 0 I1
m 0 I2
75.0 A
25.0 A
5
.
. 3.00x 5 0.400 m 2 x and x 5 0.100 m.
5
x
1 0.400 m 2 x 2
2px
2p 1 0.400 m 2 x 2
The net field is zero at a point between the wires, 10.0 cm from the wire carrying 25.0 A and 30.0 cm from the wire
carrying 75.0 A.
B2
B1
x
1
I1 5 25A
2
I2 5 75A
P
I2
I1
0.4 m 2 x
x
0.4 m
B1
B2
(a)
(b)
Figure 20.52
(b) The magnetic fields of the two wires are in opposite directions at points along the extension of the line connecting
the wires and either to the right or to the left of both wires. But for the fields to cancel, they must have equal magnitudes and this happens only at points closer to the wire carrying the smaller current. Therefore, consider a point P
that is a distance x to the left of wire 1 and hence 0.400 m 1 x to the left of wire 2, as shown in Figure 20.52b.
m 0 I1
m 0 I2
25.0 A
75.0 A
5
.
.
5
x
1 0.400 m 1 x 2
2px
2p 1 0.400 m 1 x 2
3.00x 5 0.400 m 1 x and x 5 0.200 m. The net field is zero at a point 20.0 cm from the wire carrying 25.0 A and
60.0 cm from the wire carrying 75.0 A.
m 0 lIIr
. m 0 5 4p 3 1027 N / A2. Parallel conductors carrying currents in opposite directions
2pr
repel each other. Parallel conductors carrying currents in the same direction attract each other.
1 4p 3 1027 N / A2 2 1 15 m 2 1 25 A 2 1 75 A 2
Solve: F 5
5 0.0161 N. Since the currents are in the same direction
2p 1 0.35 m 2
the force is attractive.
Reflect: The currents are large but the force per meter on each wire is very small.
20.53. Set Up: F 5
20.54. Set Up: Label wires 1 and 2. Find the direction of the magnetic field of one wire at the location of the other.
Then use the right-hand rule to find the direction of the force on the second wire due to this magnetic field.
Solve: Figure 20.54a shows the magnetic fields at the wires for the case where the currents are both to the left. The
directions of the forces on the wires exerted by these fields are also shown. The wires attract.
F1
I1
B2
1
I1
B2
1
F1
F2
2
I2
(a)
Figure 20.54
B1
2
B1
I2
F2
(b)
Magnetic Field and Magnetic Forces
20-17
Figure 20.54b shows the magnetic fields at the wires for the case where one current is to the left and the other is to
the right. The directions of the forces on the wires exerted by these fields are also shown. The wires repel.
20.55. Set Up: Measurements on a typical extension cord give that the centers of the two wires are separated by
m 0lIIr
. 4p 3 1027 N / A2.
2pr
1 4p 3 1027 N / A2 2 1 2.0 m 2 1 5.0 A 2 2
5 2.5 3 1023 N. Since the currents are in opposite directions the
Solve: F 5
2p 1 4.0 3 1023 m 2
wires repel.
about 4 mm. F 5
20.56. Set Up: F 5
m 0lI1I2
.
2pr
Solve: The force is proportional to each current. Fnew 5
m 0l 1 3I1 2 1 2I2 2
m 0lI1I2
56
5 6F.
2pr
2pr
m0 I
S
. 4p 3 1027 T # m / A. The direction of B can be found by
2R
the right-hand rule: with the thumb of your right hand in the direction of the current your fingers curl around the wire
in the direction of the magnetic field.
2 1 0.11 m 2 1 0.50 3 1024 T 2
2RB
Solve: (a) I 5
5
5 8.8 A.
m0
4p 3 1027 T # m / A
(b) The plane of the loop should be perpendicular to the earth’s field and the current should be directed as shown in
Figure 20.57.
20.57. Set Up: At the center of a circular loop B 5
I
Bearth
I
Figure 20.57
m0 I
only applies at the center of the circular loop. The current
2R
produces magnetic field at other points but only at the center of the loop do we have a simple expression for the magnitude of the field.
Reflect: A large current is required. Note that B 5
20.58. Set Up: The magnetic field at the center of N circular loops is B 5
Solve: B 5
1 4p 3 1027 T # m / A 2 1 600 2 1 0.500 A 2
5 9.42 3 1023 T
2 1 2.00 3 1022 m 2
20.59. Set Up: The magnetic field at the center of N circular loops is B 5
Solve: N 5
Nm 0 I
.
2R
Nm 0 I
.
2R
2 1 6.00 3 1022 m 2 1 6.39 3 1024 T 2
2RB
5
5 24.4 Therefore, 24 turns are required.
m0 I
1 4p 3 1027 T # m / A 2 1 2.50 A 2
m0 I
. R 5 8 cm. 1 gauss 5 1 3 1024 T.
2R
2 1 8 3 1022 m 2 1 3.0 3 10212 T 2
2RB
5
5 3.8 3 1027 A.
Solve: I 5
m0
4p 3 1027 T # m / A
20.60. Set Up: B 5
20-18
Chapter 20
m0 I
. The radius R of the circle is related to the circumference C 5 0.360 m by C 5 2pR. The
2R
S
direction of B is given by applying the right-hand rule to the current direction.
1 4p 3 1027 T # m / A 2 1 15.0 A 2
0.360 m
C
Solve: R 5
5
5 5.73 3 1022 m. B 5
5 1.64 3 1024 T. The rela2p
2p
2 1 5.73 3 1022 m 2
S
tion between the direction of I and the direction of B is illustrated in Figure 20.61.
20.61. Set Up: B 5
I
B
I
Figure 20.61
Reflect: The field at the center of the circular turn is larger than the field at a point a distance R from a long straight
wire that carries the same current.
20.62. Set Up: Let wire 1 be the inner wire Swith diameter
20.0 cm and let wire 2 be the outer wire with diameter
S
30.0 cm. To produce zero net field, the fields B1 and B2 of the two wires must have equal magnitudes and opposite
m0 I
S
. The direction of B is given by the right-hand rule applied to the
directions. At the center of a wire loop B 5
2R
current direction.
m0 I
m0 I
m 0 I1
m 0 I2
R2
15.0 cm
1 12.0 A 2 5 18.0 A.
, B2 5
. B1 5 B2 gives
5
I1 5
Solve: B1 5
and I2 5
2R1
2R2
2R1
2R2
R1
10.0 cm
S
S
The directions of I1 and of its field are shown in Figure 20.62. Since B1 is directed into the page, B2 must be directed
out of the page and I2 is counterclockwise.
1 2 1
2
I2
I1
2
1
B1
B2
Figure 20.62
m0 I
. By symmetry each segment of the
2R
loop that has length Dl contributes equally to the field, so the field at the center of a semicircle is 12 that of a full loop.
m0 I
Since the straight sections produce no field at P, the field at P is B 5
4R
m0 I
S
S
Solve: B 5
. The direction of B is given by the right-hand rule: B is directed into the page.
4R
m0 I
Reflect: For a quarter-circle section of wire the magnetic field at its center of curvature is B 5
8R
20.63. Set Up: The magnetic field at the center of a circular loop is B 5
Magnetic Field and Magnetic Forces
20-19
750 coils
N
5
5 5000 coils / m.
L
0.150 m
Solve: B 5 1 4p 3 1027 T # m / A 2 1 5000 coils / m 2 1 7.00 A 2 5 0.0440 T.
20.64. Set Up: B 5 m 0 nI. n 5
N
L
20.65. Set Up: At the center of a long solenoid B 5 m 0 nI 5 m 0 I.
Solve: I 5
1 0.150 T 2 1 1.40 m 2
BL
5 41.8 A
5
m0 N
1 4p 3 1027 T # m / A 2 1 4000 2
20.66. Set Up: At the center of a long solenoid B 5 m 0 I. The total length of wire is N 1 2pr 2 .
N
L
1 0.0279 T 2 1 0.400 m 2
BL
5
5 740 turns
m0 I
1 4p 3 1027 T # m / A 2 1 12.0 A 2
(b) The total length of wire is N 1 2pr 2 5 1 740 2 1 2p 2 1 1.40 3 1022 m 2 5 65.1 m.
Solve: (a) N 5
20.67. Set Up: The magnetic field at the center of a circular loop is Bloop 5
m0 I
. The magnetic field at the center of
2R
N
is the number of turns per meter.
L
1 4p 3 1027 T # m / A 2 1 2.00 A 2
m0 I
5
5 2.51 3 1025 T.
Solve: (a) Bloop 5
2R
2 1 0.050 m 2
N
1000
5 200 m21. Bsolenoid 5 m 0nI 5 1 4p 3 1027 T # m / A 2 1 200 m21 2 1 2.00 A 2 5 5.03 3 1024 T.
(b) n 5 5
L
5.00 m
Bsolenoid 5 20Bloop. The field at the center of a circular loop depends on the radius of the loop. The field at the center
of a solenoid depends on the length of the solenoid, not on its radius.
Reflect: Eq. (20.14) for the field at the center of a solenoid is only correct for a very long solenoid, one whose length
L is much greater than its radius R. We cannot consider the limit that L gets small and expect the expression for the
solenoid to go over to the expression for N circular loops.
a solenoid is Bsolenoid 5 m 0nI, where n 5
N
L
20.68. Set Up: At the center of a solenoid, Bsolenoid 5 m 0 nI 5 m 0 I. A distance R from a long straight wire,
Solve: (a) Bsolenoid 5 1 4p 3 1027 T # m / A 2
2
m0 I
.
2pR
450
1 1.75 A 2 5 2.83 3 1023 T.
0.35 m
1 4p 3 10
5 3.50 3 1025 T.
2p 1 1.0 3 1022 m 2
The magnetic field due to the wire is much less than the field at the center of the solenoid. For the solenoid, the fields
of all the wires add to give a much larger field.
27
(b) Bwire 5
T # m / A 2 1 1.75 A 2
1
Bwire 5
m 0 NI
. r is the distance from the
2pr
center of the torus and must lie between the inner and outer radii of the solenoid.
1 4p 3 1027 T # m / A 2 1 375 2 1 1.20 A 2
Solve: B 5
5 1.50 3 1023 T.
2p 1 0.0600 m 2
Reflect: The magnetic field varies somewhat across the cross section of the interior of the solenoid. At the inner
edge of the windings it is 1.80 3 1023 T and at the outer edge it is 1.2 3 1023 T.
20.69. Set Up: The magnetic field inside a toroidal solenoid is given by B 5
m 0 NI
. r 5 0.140 m is the distance from the center of the torus to the point where B is to be
2pr
calculated. This point must be between the inner and outer radii of the solenoid, but otherwise the field doesn’t
depend on those radii.
2p 1 0.140 m 2 1 3.75 3 1023 T 2
2prB
5
5 1750 turns
Solve: N 5
m0 I
1 4p 3 1027 T # m / A 2 1 1.50 A 2
20.70. Set Up: B 5
20-20
Chapter 20
m 0 IDl sin u
S
. The wire and the vector r from each short segment
4p
r2
5.00 cm
to point P are shown in Figure 20.71. For segment A, tan uA 5
and uA 5 35.5°. rA 5 8.60 cm. For segment C,
7.00 cm
uC 5 90° and rC 5 5.00 cm.
20.71. Set Up: The Biot and Savart law is DB 5
P
rA
rB
uC
uA
A
1
C
I
2
Figure 20.71
4p 3 1027 T # m / A 1 10.0 A 2 1 2.00 3 1023 m 2 sin 35.5°
5 1.57 3 1027 T.
1 8.60 3 1022 m 2 2
4p
S
The right-hand rule says that DB is directed out of the page.
4p 3 1027 T # m / A 1 10.0 A 2 1 2.00 3 1023 m 2 sin 90°
(b) DB 5
5 8.00 3 1027 T.
1 5.00 3 1022 m 2 2
4p
S
The right-hand rule says that DB is directed out of the page.
Reflect: Both segments produce magnetic field at P that is in the same direction. At P the field due to segment C is
larger than that due to segment A.
Solve: (a) DB 5
1
2
m 0 IDl sin u
S
. The wire and the vector r from the short segment to
4p
r2
point P are shown in Figure 20.72. r 5 0.150 m and u 5 90°.
20.72. Set Up: The Biot and Savart law is DB 5
I
r
D,
1
P
2
Figure 20.72
4p 3 1027 T # m / A 1 5.00 A 2 1 1.50 3 1022 m 2 sin 90°
5 3.33 3 1027 T.
1 0.150 m 2 2
4p
S
The right-hand rule says that DB is directed into the page.
Solve: DB 5
m 0 IDl sin u
S
. The wire and the vector r from the short segment to
4p r 2
point P are shown in Figure 20.73. r 5 0.150 m and u 5 150°.
20.73. Set Up: The Biot and Savart law is DB 5
150°
I
r
P
30°
30°
1
2
Figure 20.73
4p 3 1027 T # m / A 1 5.00 A 2 1 1.50 3 1022 m 2 sin 150°
5 1.67 3 1027 T.
1 0.150 m 2 2
4p
S
The right-hand rule says that DB is directed into the page.
Reflect: The current in the long straight segments also produces magnetic field at point P.
Solve: DB 5
Magnetic Field and Magnetic Forces
20-21
20.74. Set Up: Ampere’s law says that the sum of B Ds around any closed curve equals m 0 Iencl , where Iencl is the
net current encircled by the curve.
2.80 3 1025 T # m
5 22.3 A.
Solve: m 0 I 5 2.80 3 1025 T # m so I 5
4p 3 1027 T # m / A
i
20.75. Set Up: The field lines are circles centered on the conductor. By symmetry, the magnitude of the field
depends only on the distance from the wire. Ampere’s law says a B Ds 5 m 0 Iencl . Apply this law to a circular path
with radius r and with the wire at its center.
S
Solve: On the path B is tangent to the path and constant, so a B Ds 5 B a Ds 5 B 1 2pr 2 . Iencl 5 I, the current in
m0 I
the wire. Ampere’s law therefore gives B 1 2pr 2 5 m 0 I and B 5
.
2pr
i
i
20.76. SetS Up: Apply Ampere’s law to a circle of radius r that has the axis of the conductors at its center. By
symmetry B is constant on this circle and is tangent to it, so a B Ds 5 B 1 2pr 2 .
Solve: (a) For R1 , r , R2 , Iencl 5 I, the current in the inner conductor. Ampere’s law says B 1 2pr 2 5 m 0 I.
m0 I
B5
.
2pr
(b) For r . R3 , Iencl 5 0, Since the circle encloses both conductors and they carry currents in opposite directions.
Ampere’s law says B 1 2pr 2 5 0 and B 5 0.
i
20.77. Set Up: A sketch of a cross section in a plane perpendicular to the common axis of the conductors is given
in Figure 20.77. Let the inner conductor have current I into the page and let the outer conductor have current I out of
the page. Apply Ampere’s law to a circular path of radius r . By symmetry the magnetic field is constant on the path
and tangent to it, so B 5 B.
i
I
I
R3
R1
R2
Figure 20.77
Solve: (a) R1 , r , R2: a B Ds 5 B 1 2pr 2 . The current enclosed by the path is I, in the inner conductor.
m0 I
.
Ampere’s law therefore gives B 5
2pr
m0 I
.
(b) r . R2: The net current enclosed by the path is 2I. a B Ds 5 B 1 2pr 2 . Ampere’s law therefore gives B 5
pr
Reflect: Outside a long solid cylindrical conductor of radius R the magnetic field is the same as that due to a long
straight wire and is independent of the value of R. In the air space inside a long hollow cylindrical conductor the
magnetic field is zero. The magnetic field for the two conductors in this problem is the superposition of the fields due
to each conductor.
i
i
20-22
Chapter 20
20.78. Set Up: The toroidal solenoid is sketched in Figure 20.78. A few representative turns of wire are shown in
the sketch. These turns carry current into the page at the inner edge of the solenoid and carry current out of the page
at the outer edge of the solenoid. Apply Ampere’s law to a circular path of radius r. By symmetry the magnetic field
is constant on the path and tangent to it, so B 5 B.
i
I
I
I
I
r
I
I
Figure 20.78
Solve: (a) r , a: a B Ds 5 B 1 2pr 2 . No current passes through the path so the enclosed current is zero. Ampere’s
law gives B 1 2pr 2 5 0 and B 5 0.
r . b: a B Ds 5 B 1 2pr 2 . Each wire passes through the path twice, once carrying current into the page and once
carrying current out of the page. Therefore, the net current through the path is zero. Ampere’s law gives B 1 2pr 2 5 0
and B 5 0.
(b) a , r , b: a B Ds 5 B 1 2pr 2 . Each of the N turns carries current through the area enclosed by the path and the
m 0 IN
.
enclosed current is NI. Ampere’s law gives B 1 2pr 2 5 m 0 NI and B 5
2pr
i
i
i
20.79. Set Up: m 5 Km m 0 , with m 0 5 4p 3 1027 T # m / A. The magnetic field in the material is a factor of Km
greater than it is in vacuum.
Solve: (a) m 5 1 1.00026 2 1 4p 3 1027 T # m / A 2 5 1.257 3 1026 T # m / A
(b) Binside 5 KmBexternal 5 1 1.00026 2 1 1.3500 T 2 5 1.3504 T
Reflect: The magnetic field inside the paramagnetic material is larger than the external field, but only slightly larger.
20.80. Set Up: Km 5 Binside / Boutside . m 5 Km m 0
1.5023 T
5 1.0015.
1.5000 T
(b) m 5 Km m 0 5 1 1.0015 2 1 4p 3 1027 T # m / A 2 5 1.259 3 1026 T # m / A
Solve: (a) K 5
Magnetic Field and Magnetic Forces
20-23
20.81. Set Up: Use vy2 5 v0y2 1 2ayy to calculate vy . With 1y downward, v0y 5 0, ay 5 19.80 m / s2 and
y 5 125 m. The direction of the force is given by the right-hand rule and the magnitude is given by F 5 0 q 0 vB sin f.
The charge of the ball is 2 1 4.00 3 108 2 e.
S
S
Solve: The directions of v and B are shown in Figure 20.81, where the downward direction is into the page. The
right-hand rule (rhr) direction is north, but since the charge is negative the force is opposite the rhr direction and is to
the south. At the bottom of the shaft the speed of the ball is
v 5 "2g 1 125 m 2 5 "2 1 9.80 m / s2 2 1 125 m 2 5 49.5 m / s.
0 q 0 5 1 4.00 3 108 2 1 1.60 3 10219 C 2 5 6.40 3 10211 C.
f 5 90°, so F 5 0 q 0 vB 5 1 6.40 3 10211 C 2 1 49.5 m / s 2 1 0.250 T 2 5 7.92 3 10210 N.
N
W
E
rhr direction
S
B
v
F
Figure 20.81
Reflect: The magnetic force is much less than the gravity force on the ball.
S
S
20.82. Set Up: The magnetic force FB must be upward and equal to mg. The direction of FB is determined by the
V
direction of I in the circuit. FB 5 IlB sin f, with f 5 90°. I 5 , where V is the battery voltage.
R
S
Solve: (a) The forces are shown in Figure 20.82. The current I in the bar must be to the right to produce FB upward.
To produce current in this direction, point a must be the positive terminal of the battery.
FB
B
I
mg
Figure 20.82
(b) FB 5 mg. IlB 5 mg. m 5
1 175 V 2 1 0.600 m 2 1 1.50 T 2
VlB
IlB
5
5
5 3.21 kg.
g
Rg
1 5.00 V 2 1 9.80 m / s2 2
20.83. Set Up: The maximum current I is the current that results in an upward magnetic force FB that equals the
V
. F 5 IlB sin f, with f 5 90°.
R
1 0.750 kg 2 1 9.80 m / s2 2 1 25.0 V 2
mgR
VlB
5 mg. V 5
5
5 817 V.
Solve: FB 5 mg. IlB 5 mg.
1 0.500 m 2 1 0.450 T 2
R
lB
Reflect: The current required to lift the bar is quite large, about 33 A. For a bar of less mass, less current would be
required to lift the bar off the supports and break the circuit.
weight mg of the bar. I 5
20-24
Chapter 20
20.84. Set Up: Segments bc and cd each have Slength l 5 "2 1 0.750 m 2 2 5 1.06 m. Draw each segment in the
plane of the page. F 5 IlB sin f. The direction of F is given by the right-hand rule.
S
Solve: (a) segment ab (Figure 20.84a): F is in the 2z direction. f 5 90° so
F 5 IlB 5 1 6.58 A 2 1 0.750 m 2 1 0.860 T 2 5 4.24 N.
S
segment bc (Figure 20.84b): F is in the 2y direction.
F 5 IlB sin f 5 1 6.58 A 2 1 1.06 m 2 1 0.860 T 2 sin 45° 5 4.24 N.
S
segment cd (Figure 20.84c): F is parallel to the yz plane, midway between the 1y and 1z axes.
F 5 IlB sin f 5 1 6.58 A 2 1 1.06 m 2 1 0.860 T 2 sin 90° 5 6.00 N.
S
segment de (Figure 20.84d): F is in the 2y direction.
F 5 IlB sin f 5 1 6.58 A 2 1 0.750 m 2 1 0.860 T 2 sin 90° 5 4.24 N.
segment ef (Figure 20.84e): f 5 180° so F 5 0.
(b) The force on segment cd has a component of 4.24 N in the 1z direction and a component of 4.24 N in the 1y
direction. Adding the forces on the segments therefore gives a net force of 4.24 N, in the 2y direction.
y
y
x
y
f
I
F
B
I
I
F
x
z
F
B
x
(b)
(a)
B
z
z
(c)
y
y
I
x
z
B
B
x
I
F
(d)
(e)
Figure 20.84
20.85. Set Up: The ion has charge q 5 1e. After being accelerated through a potential difference V the ion has
kinetic energy qV. The acceleration in the circular path is v2 / R.
Solve: K 5 qV 5 1eV. 12 mv2 5 eV and
v5
2 1 1.60 3 10219 C 2 1 220 V 2
2eV
5
5 7.79 3 104 m / s. FB 5 0 q 0 vB sin f. f 5 90°.
Å m
Å
1.16 3 10226 kg
S
F 5 ma gives 0 q 0 vB 5 m
S
R5
v2
.
R
1 1.16 3 10226 kg 2 1 7.79 3 104 m / s 2
mv
5 7.81 3 1023 m 5 7.81 mm.
5
1 1.60 3 10219 C 2 1 0.723 T 2
0q0B
Magnetic Field and Magnetic Forces
20-25
m0 I
. The magnetic field at the center of
2pr
20.86. Set Up: The magnetic field of a long straight wire is given by B 5
m0 I
. Let 1 refer to the wire and let 2 refer to the loop. The center of the loop is a distance
2R
S
r 5 3.25 cm from the wire. By the right-hand rule, at the center of the loop the magnetic field B1 due to the wire is
S
directed into the page and the magnetic field B2 due to the loop is directed out of the page.
1 4p 3 1027 T # m / A 2 1 10.0 A 2
m 0 I1
5
5 6.15 3 1025 T.
Solve: B1 5
2pr
2p 1 3.25 3 1022 m 2
1 4p 3 1027 T # m / A 2 1 12.0 A 2
m 0 I2
B2 5
5
5 2.32 3 1024 T.
2R
2 1 3.25 3 1022 m 2
a circular loop is B 5
S
S
S
S
Since B1 and B2 are in opposite directions, B 5 B2 2 B1 5 1.70 3 1024 T. B2 is larger than B1 , so the net field is in
S
the direction of B2 , out of the page.
S
20.87. Set Up: The direction of the current is shown in Figure 20.87. The direction of B is given by the right-hand
rule. B 5
m0 I
.
2pr
N
W
E
I
S
Figure 20.87
Solve: (a) Below the line the magnetic field is toward the east.
1 4p 3 1027 T # m / A 2 1 800 A 2
5 2.91 3 1025 T.
B5
2p 1 5.50 m 2
(b) The field due to the current is the same order of magnitude as the earth’s field so is a problem.
20.88. Set Up: Calculate the magnetic field of the wire at the location of the electron. Then calculate the force this
field exerts on the electron. The electron has negative charge so the force on it is opposite to the right-hand rule
direction.
Solve: The wire and electron are shown in Figure 20.88a. At the electron, the field of the wire has magnitude
1 4p 3 1027 T # m / A 2 1 2.50 A 2
m0 I
5
5 1.11 3 1025 T
B1 5
2pr
2p 1 4.50 3 1022 m 2
and is directed out of the page. By the right-hand rule the force on the electron is to the right. The force has magnitude F 5 0 q 0 vB sin f 5 1 1.60 3 10219 C 2 1 6.00 3 104 m / s 2 1 1.11 3 1025 T 2 sin 90° 5 1.07 3 10219 N.
B
F
F
B
v
v
I
I
(a)
(b)
Figure 20.88
(b) The wire and electron are shown in Figure 20.88b. The magnitude of the force is the same as in part (a),
1.07 3 10219 N. The force is directed away from the wire.
20-26
Chapter 20
m0 I
S
. The direction of B is given by the right-hand rule.
2pr
Solve: (a) The currents are the same so points where the two fields are equal in magnitude are equidistant from the
two wires. The net field is zero along the dashed line shown in Figure 20.89a.
20.89. Set Up: B 5
1
slope 5 12
3
10.0 A
10.0 A
10.0 A
slope 5 21.00
3.0 A
(a)
(b)
slope 5 11.00
20.0 A
20.0 A
(c)
Figure 20.89
(b) For the magnitudes of the two fields to be the same at a point, the point must be 3 times closer to the wire with the
smaller current. The net field is zero along the dashed line shown in Figure 20.89b.
(c) As in (a), the points are equidistant from both wires. The net field is zero along the dashed line shown in Figure 20.89c.
Reflect: The lines of zero net field consist of points at which the fields of the two wires have opposite directions and
equal magnitudes.
20.90. Set Up: Since the magnetic force is always perpendicular to the velocity of the object, a charged particle
mv
.
0q0B
Solve: (a) The direction of curvature of the path tells us the direction of the force on the particle. If this is the same
S
S
as the right-hand direction for v and B, then the particle has positive charge it is opposite to this direction then the
particle has negative charge.
(b) At any point the momentum of the electron is given by mv 5 R 0 q 0 B. A measurement of the radius of curvature,
R, of the path gives the momentum of the particle. If the mass is known then the speed is momentum divided by
mass.
mv
(c) The electron is slowing down and by R 5
this causes the radius of curvature of the path to decrease. The
0q0B
electron loses kinetic energy as it moves through the liquid.
(d) A neutron has no charge. The magnetic field exerts no force on it, so it moves in a straight line.
moving perpendicular to the field travels in an arc of a circle with radius R 5
20.91. Set Up: Since the magnetic force is always perpendicular to the velocity of the object, a charged particle
moving perpendicular to the field travels in an arc of a circle with radius R 5
S
S
mv
0q0B
. The force on a positive charge is
in the right-hand rule direction as applied to v and B, and the force on a negative charge is opposite to this direction.
Magnetic Field and Magnetic Forces
20-27
Solve: (a) Since the paths curve in opposite directions, the forces on the two particles are in opposite directions. One
particle has positive charge and one has negative charge.
(b) The particle with larger momentum mv moves in a path of larger radius, and this is the particle on the right.
(c) The curvature of the path decreases as the particle loses energy.
2p 1 2.00 3 1022 m 2 1 37.2 T 2
m0 I
2pB
so I 5
5
5 3.72 3 106A.
m0
2pr
4p 3 1027 T # m / A
2 1 0.210 m 2 1 37.2 T 2
Nm 0 I
2RB
5
5 1.24 3 105 A.
(b) B 5
so I 5
2R
Nm 0
1 100 2 1 4p 3 1027 T # m / A 2
1 37.2 T 2 1 0.320 m 2
N
BL
(c) B 5 m 0 I so I 5
5 237 A.
5
L
m0 N
1 4p 3 1027 T # m / A 2 1 40,000 2
20.92. Set Up and Solve: (a) B 5
mv
0q0B
1 2
20.93. Set Up: R 5
. The time T is given by T 5
2pR
.
v
2p mv
2pm
5
.
v 0q0B
0q0B
(b) The time for one revolution is the same for both. An ion moving in a larger circle is traveling faster.
(c) To complete one revolution in the same time, an ion moving in a larger circle must travel faster. We can also see
mv
this from R 5
.
0q0B
0 q 0 BR
mv
.
(d) From R 5
we get v 5
m
0q0B
Reflect: Since the orbital period depends only on B and on the nature of the ion 1 m / 0 q 0 2 and not on the speed (and
radius of the path) of the ion, the rate at which the electric field is pulsed does not have to change as the ions are
accelerated.
Solve: (a) T 5
20.94. Set Up: 1 eV 5 1.60 3 10219 J. The energy of the ions and their speed are related by K 5 12 mv2. The mass
of a proton is 1.67 3 10227 kg. From Problem 20.93, the speed v of the particle and the magnetic field B are related
qBR
. The radius R is related to the circumference C by C 5 2pR.
by v 5
m
1.60 3 10219 J
5 2.00 3 10213 J. K 5 12 mv2 gives
Solve: (a) K 5 1.25 MeV 5 1 1.25 3 106 eV 2
1 eV
1
v5
(b) v 5
2
2 1 2.00 3 10213 J 2
2K
5
5 1.55 3 107 m / s
Åm
Å 1.67 3 10227 kg
qBR
C
6.4 3 103 m
mv
.R5
5
5 1.02 3 103 m.
gives B 5
m
qR
2p
2p
B5
1 1.67 3 10227 kg 2 1 1.55 3 107 m / s 2
5 1.59 3 1024 T.
1 1.60 3 10219 C 2 1 1.02 3 103 m 2
20.95. Set Up: The torque on a current loop is t 5 IAB sin f. We can use the magnetic field of the ring, B 5
m0 I
,
2R
to calculate the current in the ring.
2RBring
2 1 2.50 3 1022 m 2 1 75.4 3 1026 T 2
5 3.00 A. The torque is a maximum when f 5 90°
5
Solve: I 5
m0
4p 3 1027 T # m / A
and the plane of the ring is parallel to the field.
tmax 5 IAB 5 1 3.00 A 2 1 0.375 T 2 p 1 2.50 3 1022 m 2 2 5 2.21 3 1023 N # m.
Reflect: When the external field is perpendicular to the plane of the ring the torque on the ring is zero.
20-28
Chapter 20
20.96. Set Up: Label the three segments in the magnetic field 1, 2, and 3, as shown in Figure 20.96a. The force on
a current carrying conductor is F 5 IlB sin f, where f is the angle between the direction of the current and the
direction of the magnetic field. The direction of the force on each segment is given by the right-hand rule and is
S
S
S
shown in the figure. The sum of F1 and F3 is the same as the force F13 on a wire 0.307 m long. Section 2 has length
0.800 m. The current in each segment is perpendicular to the magnetic field, so f 5 90°.
F3
F2
3
B
2
I
60°
0.400 m
B
F1
I
30°
B
1
0.307 m2x
0.693 m
I
x
(a)
y
F2
y
F
F2 sin 60°
Fy
F13
60°
u
x
F2 cos 60°
x
Fx
(b)
(c)
Figure 20.96
Solve: F13 5 IlB sin f 5 1 6.50 A 2 1 0.307 m 2 1 0.280 T 2 sin 90° 5 0.559 N.
F2 5 IlB sin f 5 1 6.50 A 2 1 0.800 m 2 1 0.280 T 2 sin 90° 5 1.46 N. The forces and a coordinate system are shown in
S
Figure 20.96b. F2 has been resolved into its x and y components.
Fx 5 F2x 1 F13x 5 2F2 cos 60.0° 5 2 1 1.46 N 2 1 cos 60.0° 2 5 20.730 N.
Fy 5 F2y 1 F13y 5 F2 sin 60.0° 1 F13 5 1 1 1.46 N 2 1 sin 60.0° 2 1 0.559 N 5 11.83 N.
S
Fx , Fy and the resultant total force F are shown in Figure 20.96c. The resultant force has magnitude 1.97 N and is at
68.3° clockwise from the left-hand straight segment.
20.97. Set Up: Conservation of energy relates the accelerating potential difference V to the final speed of the ions.
In the magnetic field region the ions travel in an arc of a circle that has radius R 5
mv
0q0B
. The quarter-circle paths of
the two ions are shown in Figure 20.97. The separation at the detector is Dr 5 R18 2 R16 . Each ion has charge
q 5 1e.
R18
Dr
R16
Figure 20.97
Magnetic Field and Magnetic Forces
Solve: (a) Conservation of energy gives 0 q 0 V 5 12 mv2 and v 5
0 q 0 5 e for each ion. Dr 5 R18 2 R16 5
(b) V 5
1 DreB 2 2
2e 1 "m 18 2 "m 16 2 2
V 5 3.32 3 103 V.
5
20-29
20q0V
20q0V
"2 0 q 0 mV
m
.R5
5
.
Å m
0q0BÅ m
0q0B
"2eV
1 "m18 2 "m16 2 .
eB
e 1 Dr 2 2B 2
2 1 "m 18 2 "m16 2 2
5
1 1.60 3 10219 C 2 1 4.00 3 1022 m 2 2 1 0.050 T 2 2
2 1 "2.99 3 10226 kg 2 "2.66 3 10226 kg 2 2
Reflect: The speed of the 16O ion after it has been accelerated through a potential difference of V 5 3.32 3 103 V is
2.00 3 105 m / s. Increasing the accelerating voltage increases the separation of the two isotopes at the detector. But it
does this by increasing the radius of the path for each ion, and this increases the required size of the magnetic field
region.