192 Chapter 19 ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. The electron moves in a horizontal plane in a direction of 35° N or E, which is the same as 55° E of N. Since the magnetic fi eld at this location is horizontal and directed due north, the angle between the direction of the electron’s velocity and the direction of the magnetic fi eld is 55°. The magnitude of the magnetic force experienced by the electron is then F q B= = ×()×()×− −v sin . . .θ1 6 10 2 5 10 0 10 1019 6 C m s44 1855 3 3 10 T N()= ×−sin .°The right-hand rule number 1 predicts a force directed upward, away from the Earth’s surface for a positively charged particle moving in the direction of the electron. However, the negatively charged electron will experience a force in the opposite direction, downward toward the Earth’s surface. Thus, the correct choice is (d). 2. If the magnitude of the magnetic force on the wire equals the weight of the wire, then BI wl sinθ=, or B w I= l sinθ. The magnitude of the magnetic fi eld is a minimum when θ θ= =90 1° and sin. Thus, BwImin...= =×()()=−l1 0 100 500 202 N0.10 A m Tand (a) is the correct answer for this question. 3. The z-axis is perpendicular to the plane of the loop, and the angle between the direction of this normal line and the direction of the magnetic fi eld is θ= 30 0. °. Thus, the magnitude of the torque experienced by this coil containing N = 10turns is τ θ= =()()()()BIAN sin . . . .0 010 2 0 0 20 0 30 T A m m()= × ⋅−10 30 0 6 0 103sin . .° N mmeaning that (c) is the correct choice. 4. A charged particle moving perpendicular to a magnetic fi eld experiences a centripetal force of magnitude F m r q Bc= =v v2 and follows a circular path of radius r m qB= v. The speed of this proton must be v = =×()()×()− −qBrm1 6 10 0 050 1 0 10119 3. . .. C T m667 104 8 10273×= ×− kg m s.and choice (e) is the correct answer. 5. The magnitude of the magnetic fi eld at distance r from a long straight wire carrying current I is B I r=µ π02. Thus, for the described situation, B =× ⋅()()()= ×−−4 10 12 21 1077ππ T m A A m T/making (d) the correct response. 6. The force per unit length between this pair of wires is F I Idl= =× ⋅()()()= ×−µπππ0 1 2724 10 32 29 T m A A m2110 1 107 6− −× N N∼and (d) is the best choice for this question. 7. The magnitude of the magnetic fi eld inside the specifi ed solenoid is B nINI= == × ⋅()−µ µ π0 074 101200 50l T m A m.()= ×−2 0 6 0 104. . A Twhich is choice (e).56157_19_ch19_p171-218.indd 19256157_19_ch19_p171-218.indd 192 3/18/08 11:39:38 PM3/18/08 11:39:38 PMMagnetism 193 8. The magnitude of the magnetic force experienced by a charged particle in a magnetic fi eld is given by F q B= v sinθ, where v is the speed of the particle and θ is the angle between the direc-tion of the particle’s velocity and the direction of the magnetic fi eld. If either v = 0 [choice (e)] or sinθ= 0 [choice (c)], this force has zero magnitude. All other choices are false, so the correct answers are (c) and (e). 9. The force that a magnetic fi eld exerts on a moving charge is always perpendicular to both the direction of the fi eld and the direction of the particle’s motion. Since the force is perpendicular to the direction of motion, it does no work on the particle and hence does not alter its speed. Because the speed is unchanged, both the kinetic energy and the magnitude of the linear momen-tum will be constant. Correct answers among the list of choices are (d) and (e). All other choices are false.10. By the right-hand rule number 1, when the proton fi rst enters the fi eld, it experiences a force directed upward, toward the top of the page. This will defl ect the proton upward, and as the proton’s velocity changes direction, the force changes direction always staying perpendicular to the velocity. The force, being perpendicular to the motion, causes the particle to follow a circu-lar path, with no change in speed, as long as it is in the fi eld. After completing a half circle, the proton will exit the fi eld traveling toward the left. The correct answer is choice (d).11. The contribution made to the magnetic fi eld at point P by the lower wire is directed out of the page, while the contribution due to the upper wire is directed into the page. Since point P is equi-distant from the two wires, and the wires carry the same magnitude currents, these two oppositely directed contributions to the magnetic fi eld have equal magnitudes and cancel each other. There-fore, the total magnetic fi eld at point P is zero, making (a) the correct answer for this question. 12. The magnetic fi eld due to the current in the vertical wire is directed into the page on the right side of the wire and out of the page on the left side. The fi eld due to the current in the horizontal wire is out of the page above this wire and into the page below the wire. Thus, the two contributions to the total magnetic fi eld have the same directions at points B (both out of the page) and D (both contributions into the page), while the two contributions have opposite directions at points A and C. The magnitude of the total magnetic fi eld will be greatest at points B and D where the two contributions are in the same direction, and smallest at points A and C where the two contribu-tions are in opposite directions and tend to cancel. The correct choices for this question are (a) and (c).13. Any point in region I is closer to the upper wire which carries the larger current. At all points in this region, the outward directed fi eld due the upper wire will have a greater magnitude than will the inward directed fi eld due to the lower wire. Thus, the resultant fi eld in region I will be nonzero and out of the page, meaning that choice (d) is a true statement and choice (a) is false. In region II, the fi eld due to each wire is directed into the page, so their magnitudes add and the resultant fi eld cannot be zero at any point in this region. This means that choice (b) is false. In region III, the fi eld due to the upper wire is directed into the page while that due to the lower wire is out of the page. Since points in this region are closer to the wire carrying the smaller current, there are points in this region where the magnitudes of …