Solution 26 - homework 26 KANUNGO ANIL Due Nov 6 2007 4:00...

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homework 26 – KANUNGO, ANIL – Due: Nov 6 2007, 4:00 am 1 Question 1, chap 31, sect 3. part 1 of 2 10 points A bar magnet is held above the center of a conducting wire loop in a horizontal plane with the south end of the magnet toward the loop. The magnet is dropped. S N counter- clockwise clockwise bar magnet Find the direction of the current, as viewed from above, around the wire loop while the magnet is falling toward the loop. 1. clockwise correct 2. Unable to determine the direction. 3. counter-clockwise Explanation: To oppose the field due to the south end of the magnet as it approaches the wire loop, the magnetic field should be directed downward along the axis of the loop, so the current must be clockwise when viewed from above the loop. Question 2, chap 31, sect 3. part 2 of 2 10 points Find the direction of the current around the loop after the magnet has passed through the wire loop and moves away from it. 1. Unable to determine the direction. 2. counter-clockwise correct 3. clockwise Explanation: To enhanse the field due to the north end of the magnet as it receeds from the wire loop, the magnetic field should be directed upward along the axis of the loop, so the current must be counter-clockwise when viewed from above the loop. Question 3, chap 32, sect 4. part 1 of 3 10 points A capacitor of 46 μ F is connected in series to an inductor of 21 mH and to an open switch. The capacitor is first charged to a voltage of 133 V. The charging battery is then removed, and the switch is closed. What is the maximum charge on the capac- itor? Correct answer: 0 . 006118 C (tolerance ± 1 %). Explanation: Let : C = 46 μ F = 4 . 6 × 10 - 5 F , and V 0 = 133 V . When the switch of the LC circuit is first closed, the voltage across the capacitor is V and the initial current, I (0), in the cir- cuit is zero. The voltage will oscillate and never be larger than its initial voltage, so that the change across the capacitor will never be larger than its initial charge which is Q = C V 0 = (4 . 6 × 10 - 5 F) (133 V) = 0 . 006118 C .
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