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# hw 25 - nguyen(jmn727 homework 25 Turner(59070 This...

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nguyen (jmn727) – 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 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 z > V a (right) and V a > V z (left) 2. V a > V z (right) and V z > V a (left) 3. V a > V z (right) and V a > V z (left) 4. V a = V z (right) and V a = V z 5. V z > V a (right) and V z > V a (left) correct 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 37 turns, 0 . 24 m by 0 . 16 m, is rotated at 55 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 . 3 V. What is the magnitude of the field? Correct answer: 3 . 83907 mT. Explanation: Let : N = 37 turns , ω = 55 rad / s , ǫ max = 0 . 3 V , x = 0 . 24 m , and y = 0 . 16 m . From ǫ max = N A B ω , where the area A is A = (0 . 24 m)(0 . 16 m) = 0 . 0384 m 2 , so B = ǫ max N A ω = 0 . 3 V (37 turns)(0 . 0384 m 2 )(55 rad / s) = 0 . 00383907 T = 3 . 83907 mT .

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hw 25 - nguyen(jmn727 homework 25 Turner(59070 This...

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