Unformatted text preview: 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 moremassive 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 RightHand 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 1135 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 RightHand Rule No. 1 (Section 21.5), the magnetic force in drawing 1 points
north.
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 RightHand 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 RightHand 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) RightHand 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, 1136 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 × 10−6 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 1137 CHAPTER 21 MAGNETIC FORCES AND MAGNETIC FIELDS
PROBLEMS
1. SSM REASONING AND SOLUTION The magnitude of the force can be determined
using Equation 21.1, F = q vB sin θ, where θ is the angle between the velocity and the
magnetic field. The direction of the force is determined by using RightHand Rule No. 1.
a. F = q vB sin 30.0° = (8.4 × 10–6 C)(45 m/s)(0.30 T) sin 30.0° = 5.7 × 10 −5 N , directed into the paper .
b. F = q vB sin 90.0° = (8.4 × 10–6 C)(45 m/s)(0.30 T) sin 90.0° = 1.1 × 10 −4 N , directed into the paper .
c. F = q vB sin 150° = (8.4 × 10 –6 C)(45 m/s)(0.30 T) sin 150° = 5.7 × 10 −5 N , directed into the paper . 2. REASONING The electron’s acceleration is related to the net force ΣF acting on it by
Newton’s second law: a = ΣF/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 = q0 vB sin θ. Thus, the magnitude of the ( ) electron’s acceleration can be written as a = q0 vB sin θ / m . SOLUTION We note that θ = 90.0°, since the velocity of the electron is perpendicular to
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the magnetic field. Th...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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