PHY
solutions_chapter21

# C both objects fall at the same rate the distance

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(c) Both objects fall at the same rate. The distance between them doesn’t change. The flux through the coil doesn’t change and there is no current induced in the coil. B Electromagnetic Induction 21-5

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21.19. Set Up: The magnetic field of the solenoid is directed toward the top of the page in the figure in the problem. Loop A and the magnetic field, as viewed from above the loop, are shown in Figure 21.19. Lenz’s law says that the magnetic field of the induced current is directed to oppose the change in flux through the circuit. Figure 21.19 Solve: (a) The magnetic field of the solenoid decreases with distance from the solenoid so the flux through the loop decreases. To oppose this decrease, the flux through the loop due to the induced current is directed out of the page in Figure 21.19. To produce magnetic field inside the loop that is out of the page the induced current must be counter- clockwise. (b) Increasing I increases the magnetic field of the solenoid and this increases the flux through the loop. To oppose this increase, the flux through the loop due to the induced current is directed into the page in Figure 21.19. To pro- duce magnetic field inside the loop that is into the page, the induced current must be clockwise. Reflect: When loop A is moved upward, away from the solenoid, the force on the loop due to the induced current is downward, toward the solenoid. When the field of the solenoid is increased, the force on the loop due to the induced current is upward, away from the solenoid. Both these results agree with Lenz’s law. 21.20. Set Up: For a bar magnet the magnetic field is directed into the south pole and out of the north pole. The loop and the current induced in it, as viewed by the observer, are shown in Figure 21.20. Lenz’s law says that the magnetic field of the induced current is directed to oppose the change in flux through the loop. Figure 21.20 Solve: The magnetic field of the induced current is directed out of the page inside the loop. The magnitude of the flux decreases when the bar magnet moves away from the loop. Since the induced flux tends to oppose this decrease, the flux due to the bar magnet is directed out of the page. The magnetic field of the bar magnet is away from pole C, so pole C is a north pole and pole A is a south pole. B I induced B A 21-6 Chapter 21
21.21. Set Up: The bar and magnetic field are shown in Figure 21.21. To determine which end is at higher potential consider the direction of the magnetic force on a positive charge in the moving rod. Accumulation of positive charge produces higher potential. Figure 21.21 Solve: (a) (b) The magnetic force on a positive charge in the bar is eastward. Positive charge accumulates at end a of the bar and that end (the east end) is at higher potential. (c) If the bar is moving east to west the force is to the north. The force is perpendicular to the bar and there is no charge accumulation at its ends; the potential difference between its ends is zero.

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