CHAPTER 21 MAGNETIC FORCES AND
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (d) Right-Hand Rule No. 1 gives the direction of the magnetic force as x for both drawings
A and B. In drawing C, the velocity is parallel to the magnetic field, so the magnetic force is
2. (b) Using Right-Hand Rule No. 1 (see Section 21.2), we find that the direction of the
magnetic force on a positively charged particle is to the west. Reversing this direction
because the particle is a negative electron, we see that the magnetic force acting on it points
to the east.
3. (a) Using Right-Hand Rule No. 1 (see Section 21.2), we find that the direction of the
magnetic force on a positively charged particle is straight down toward the bottom of the
4. B = 1.1 101 T, south
5. (c) The electric force points out of the screen, in the direction of the electric field. An
application of Right-Hand Rule No. 1 shows that the magnetic force also points out of the
screen, parallel to the electric force. When two forces have the same direction, the
magnitude of their sum has the largest possible value.
6. (e) In this situation, the centripetal force, Fc = mv
/r (Equation 5.3), is provided by the
magnetic force, F = qvB sin 90.0 (Equation 21.1), so mv
/r = qvB sin 90.0. Thus,
/q mv rB , and the charge magnitude q is inversely proportional to the radius r. Since the
radius of curve 1 is smaller than that of curve 2, and the radius of curve 2 is smaller than
that of curve 3, we conclude that q1 is larger than q2, which is larger than q3.
7. (a) The magnetic force that acts on the electron in regions 1 and 2 is always perpendicular
to its path, so the force does no work. According to the work-energy theorem, Equation 6.3,
the kinetic energy, and hence speed, of the electron does not change when no work is done.
8. (d) According to Equation 21.2, the radius r of the circular path is given by /r mv qB .
Since v, q, and B are the same for the proton and the electron, the more-massive proton
travels on the circle with the greater radius. The centripetal force Fc acting on the proton
must point toward the center of the circle. In this case, the centripetal force is provided by
the magnetic force F. According to Right-Hand Rule No. 1, the direction of F is related to
the velocity v and the magnetic field B. An application of this rule shows that the proton
Chapter 21 Answers to Focus on Concepts Questions 1077
must travel counterclockwise around the circle in order that the magnetic force point toward
the center of the circle.
9. rproton/relectron = 1835
10. (c) When, for example, a particle moves perpendicular to a magnetic field, the field exerts a
force that causes the particle to move on a circular path. Any object moving on a circular
path experiences a centripetal acceleration.
11. F = 3.0 N, along the y axis
12. (e) The magnetic field is directed from the north pole to the south pole (Section 21.1).
According to Right-Hand Rule No. 1 (Section 21.5), the magnetic force in drawing 1 points
13. (c) There is no net force. No force is exerted on the top and bottom wires, because the
current is either in the same or opposite direction as the magnetic field. According to Right-
Hand Rule No. 1 (Section 21.5), the left side of the loop experiences a force that is directed
into the screen, and the right side experiences a force that is directed out of the screen
(toward the reader). The two forces have the same magnitude, so the net force is zero. The
two forces on the left and right sides, however, do exert a net torque on the loop with respect
to the axis.
14. (d) According to Right-Hand Rule No. 1 (Section 21.5), all four sides of the loop are
subject to forces that are directed perpendicularly toward the opposite side of the square. In
addition, the forces have the same magnitude, so the net force is zero. A torque consists of a
force and a lever arm. For the axis of rotation through the center of the loop, the lever arm
for each of the four forces is zero, so the net torque is also zero.
15. N = 86 turns
16. (a) Right-Hand Rule No. 2 (Section 21.7) indicates that the magnetic field from the top wire
in 2 points into the screen and that from the bottom wire points out of the screen. Thus, the
net magnetic field in 2 is zero. Also, the magnetic field from the horizontal wire in 4 points
into the screen and that from the vertical wire points out of the screen. Thus, the net
magnetic field in 4 is also zero.
17. (b) Two wires attract each other when the currents are in the same direction and repel each
other when the currents are in the opposite direction (see Section 21.7). Wire B is attracted
to A and repelled by C, but the forces reinforce one another. Therefore, the net force has a
magnitude of FBA + FBC, where FBA and FBC are the magnitudes of the forces exerted on
wire B by A and on wire B by C. However, FBA = FBC, since the wires A and C are
equidistance from B. Therefore, the net force on wire B has a magnitude of 2FBA. The net
force exerted on wire A is less than this, because wire A is attracted to B and repelled by C,
1078 MAGNETIC FORCES AND MAGNETIC FIELDS
the forces partially canceling. The net force expected on wire C is also less than that on A. It
is repelled by both A and B, but A is twice as far away as B.
18. (a) The magnetic field in the region inside a solenoid is constant, both in magnitude and in
direction (see Section 21.7).
19. B = 4.7 106 T, out of the screen
20. (d) According to Ampere’s law, I is the net current passing through the surface bounded by
the path. The net current is 3 A + 4 A 5 A = 2 A.
Chapter 21 Problems 1079
CHAPTER 21 MAGNETIC FORCES
AND MAGNETIC FIELDS
1. SSM REASONING The electron’s acceleration is related to the net forceF acting on it
by Newton’s second law: a = F/m (Equation 4.1), where m is the electron’s mass. Since
we are ignoring the gravitational force, the net force is that caused by the magnetic force,
whose magnitude is expressed by Equation 21.1 as F = 0q vB sin . Thus, the magnitude of
the electron’s acceleration can be written as 0 sin /a q vB m .
SOLUTION We note that = 90.0, since the velocity of the electron is perpendicular to
the magnetic field. The magnitude of the electron’s charge is 1.60 10
C, and the
electron’s mass is 9.11 10
kg (see the inside of the front cover), so
1.60 10 C 2.1 10 m/s 1.6 10 T sin 90.0
5.9 10 m/s
9.11 10 kg
2. REASONING The magnitude B of the magnetic field is
where F is the magnitude of the magnetic force on the charge, whose magnitude is 0q and
whose velocity has a magnitude v and makes an angle θ with the direction of the field. Both
the proton in part a and the electron in part b have the same charge magnitude of
1.60 10 Cq . Therefore, the magnetic field has the same magnitude in both parts
of the problem. However, the direction of the field is different for the proton and the
electron. This is because the proton charge is positive, whereas the electron charge is
negative. Finally, we note that the magnitude of the magnetic force is a maximum, which
means that the velocity is perpendicular to the magnetic field, so that 90.0 .
a. Using Equation 21.1, we find that the magnitude of the magnetic field for the proton is
8.0 10 N
sin 1.60 10 C 4.5 10 m / s sin90.0
1080 MAGNETIC FORCES AND MAGNETIC FIELDS
Since the proton is traveling due east and the force points due south, we find from right hand
rule no. 1 that the magnetic field points upward, perpendicular to the earth’s surface .
b. For the electron, the magnitude of the field is the same as for the proton, since the two
charges have the same magnitude. Thus, 0.11 TB . Since the electron is a negative
charge, however, right-hand rule no. 1 reveals that the field direction is
downward, perpendicular to the earth’s surface .
3. SSM REASONING According to Equation 21.1, the magnitude of the magnetic force on
a moving charge is
sinF q vB . Since the magnetic field points due north and the