CHAPTER 21 MAGNETIC FORCES AND MAGNETIC Chapter 21 Problems 1079 CHAPTER 21 MAGNETIC FORCES AND MAGNETIC

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  • CHAPTER 21 MAGNETIC FORCES AND

    MAGNETIC FIELDS

    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

    zero.

    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

    screen.

    4. B = 1.1  101 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 2 /r (Equation 5.3), is provided by the

    magnetic force, F = qvB sin 90.0 (Equation 21.1), so mv 2 /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

    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 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  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 1079

    CHAPTER 21 MAGNETIC FORCES AND MAGNETIC FIELDS

    PROBLEMS

    1. SSM REASONING The electron’s acceleration is related to the net forceF 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 19

    C, and the

    electron’s mass is 9.11  10 31

    kg (see the inside of the front cover), so

       

    0

    19 6

    12 2

    31

    sin

    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

    q vB a

    m

     

          

    2. REASONING The magnitude B of the magnetic field is  0 sin

    F B

    q v   (Equation 21.1),

    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 19

    0 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   .

    SOLUTION a. Using Equation 21.1, we find that the magnitude of the magnetic field for the proton is

        

    14

    19 6 0

    8.0 10 N 0.11 T

    sin 1.60 10 C 4.5 10 m / s sin90.0

    F B

    q v 

       

      

  • 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 0

    sinF q vB  . Since the magnetic field points due north and the