solutions_chapter21

# Solve a the field is into the page and is increasing

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Solve: (a) The field is into the page and is increasing so the flux is increasing. The field of the induced current is out of the page. To produce field out of the page the induced current is counterclockwise. (b) The field is into the page and is decreasing so the flux is decreasing. The field of the induced current is into the page. To produce field into the page the induced current is clockwise. (c) The field is constant so the flux is constant and there is no induced emf and no induced curent. v v 5 E max NBL 2 5 1 20.0 3 10 2 3 V 2 1 15 21 0.0250 T 21 0.120 m 2 2 5 3.70 rad / s. E max 5 NBL 2 v . 2 NBL v 5 NBL 2 v . BL v . v 5 v 1 L / 2 2 . F B E 5 2 DF B D t . F B A B C D E t ( a ) t ( b ) E F B F B 5 BA cos f 5 BA cos v t . v f 5 v t , t 5 0. f 5 t 5 0 Electromagnetic Induction 21-3

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21.14. Set Up: The field of the induced current is directed to oppose the change in flux in coil B. When S is closed the magnetic field of coil A is directed to the right. The field of A is stronger at points closer to A. The field of A is proportional to the current in A. Solve: (a) The flux due to field directed to the right decreases (goes to zero) when S is opened. The field due to the induced current is to the right inside of coil B. To produce field in this direction, the induced current passes through the resistor from a to b. (b) The flux through coil B due to the field of A increases, so the field due to the induced current is to the left inside of coil B. To produce field in this direction, the induced current passes through the resistor from b to a. (c) Decreasing R causes the current in coil A to increase, so the field of A increases. The flux through coil B due to the field of A increases, so the field due to the induced current is to the left inside of coil B. To produce field in this direc- tion, the induced current passes through the resistor from b to a. 21.15. Set Up: When S is closed, the current in the large loop is counterclockwise and its field is directed out of the page inside the small ring. The field of the induced current is directed to oppose the change in flux in the small ring. Solve: (a) The field is initially zero and then is directed out of the page. The flux through the ring increases, so the field of the induced current is directed into the page inside the ring. To produce field in this direction, the induced current is clockwise. (b) The current in the large loop is constant so its field is constant and the flux through the small ring isn’t changing. The induced current is zero. (c) The field is initially out of the page inside the ring and then becomes zero. The flux through the ring decreases so the field of the induced current is directed out of the page inside the ring. To produce field in this direction, the induced current is counterclockwise. Reflect: When the current in the outer loop is increasing, the current in the loop and the induced current in the small ring are in opposite directions. When the current in the loop is decreasing, the two currents are in the same direction.
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