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# Solution_pdf25 - tovar(jdt436 – homework 25 – Turner –(59070 1 This print-out should have 11 questions Multiple-choice questions may continue

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Unformatted text preview: tovar (jdt436) – homework 25 – Turner – (59070) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In the arrangement shown in the figure, the resistor is R and a B magnetic field is directed into the paper. The separation be- tween the rails is ℓ . Neglect the mass m of the bar. Assume the bar and rails have negligible resistance and friction. An applied force moves the bar to the left at a constant speed of v . m v R B B ℓ a z What is the relationship between the elec- tric potential at the ends of the resistor ( a or z ) while the bar is moving on the “right” side of the resistor (towards the resistor) and mov- ing on the “left” side of the resistor (after the bar moves past the resistor)? 1. V a = V z (right) and V a = V z 2. V z > V a (right) and V z > V a (left) correct 3. V a > V z (right) and V z > V a (left) 4. V a > V z (right) and V a > V z (left) 5. V z > V a (right) and V a > V z (left) Explanation: As the bar moves toward the resistor, the area of the current loop decreases, so the in- duced vector B ind is downward with I ind clockwise from above. Lenz’s law dictates that before moving past the resistor, current flows from z to a , so z is at a higher potential. After going past the resistor, Lenz’s law dictates that the induced vector B ind is now upward. This requires the current to reverse its rota- tional direction to be counter-clockwire from above. However, the direction z to a (through the resistor R ) also reverses its rotational di- rection. The emf across the bar does not change sign; i.e. , the current through the re- sistor R remains in the same direction. 002 10.0 points A rectangular coil of 64 turns, 0 . 12 m by . 28 m, is rotated at 77 rad / s in a magnetic field so that the axis of rotation is perpendicu- lar to the direction of the field. The maximum emf induced in the coil is 0 . 4 V. What is the magnitude of the field? Correct answer: 2 . 41574 mT. Explanation: Let : N = 64 turns , ω = 77 rad / s , ǫ max = 0 . 4 V , x = 0 . 12 m , and y = 0 . 28 m ....
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## This note was uploaded on 12/11/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Solution_pdf25 - tovar(jdt436 – homework 25 – Turner –(59070 1 This print-out should have 11 questions Multiple-choice questions may continue

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