9 Solution Set 9 on chapter 21

# 9 Solution Set 9 on chapter 21 - On Chapter 21 Magnetic...

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Unformatted text preview: On Chapter 21 Magnetic Forces and Magnetic Fields Conceptual Problems pp. 672: # 6, 8, 17 Problems pp. 673: # 8, 11, 19, 23, 33, 38, 50, 58, 59, 67, 78 Conceptual Problems pp. 672: # 6, 8, 17 6. (e) In this situation, the centripetal force, F c = 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, ( 29 / 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 q 1 is larger than q 2 , which is larger than q 3 . 8. (d) According to Equation 21.2, the radius r of the circular path is given by ( 29 / 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 F c 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 must travel counterclockwise around the circle in order that the magnetic force point toward the center of the circle. 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 F BA + F BC , where F BA and F BC are the magnitudes of the forces exerted on wire B by A and on wire B by C. However, F BA = F BC , since the wires A and C are equidistance from B. Therefore, the net force on wire B has a magnitude of 2 F BA . The net force exerted on wire A is less than this, because wire A is attracted to B and repelled by C, 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. 1 Problems pp. 673: # 8, 11, 19, 23, 33, 38, 50, 58, 59, 67, 78 8. REASONING According to Equation 21.1, the magnetic force has a magnitude of F = q vB sin θ . The field B and the directional angle θ are the same for each particle. Particle 1, however, travels faster than particle 2. By itself, a faster speed v would lead to a greater force magnitude F . But the force on each particle is the same. Therefore, particle 1 must have a smaller charge to counteract the effect of its greater speed....
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9 Solution Set 9 on chapter 21 - On Chapter 21 Magnetic...

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