homework 22a-solutions.pdf

# homework 22a-solutions.pdf - agha(msa2448 homework 22a...

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agha (msa2448) – homework 22a – turner – (90130) 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 10.0points In the arrangement shown in the figure, the resistor is R and a B magnetic field is directed out of the paper. The separation between the rails is ℓ . 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 ( z or a ) while the bar is on the right moving to- ward the resistor, then while the bar is moving away from the resistor on the left? Neglect the mass m of the bar. Assume the bar and rails have negligible resistance and friction. 1. V a > V z (right) and V z > V a (left) 2. V z > V a (right) and V z > V a (left) 3. V a > V z (right) and V a > V z (left) correct 4. V z = V a (right) and V z = V a (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 induced vector B ind is upward with I ind counter- clockwire from above. Lenz’s law dictates that before moving past the resistor, current flows from a to z , so a is at a higher potential. After going past the resistor, Lenz’s law dictates that the induced vector B ind is now down- ward. This requires the current to reverse its ro- tational direction to be clockwise from above. However, the direction a to z (through the resistor R ) also reverses its rotational direc- tion. The emf across the bar does not change sign; i.e. , the current through the resistor R remains in the same direction. 002 10.0points A rectangular coil of 23 turns, 0 . 14 m by 0 . 39 m, is rotated at 114 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 . 5 V. What is the magnitude of the field? Correct answer: 3 . 49257 mT. Explanation: Let : N = 23 turns , ω = 114 rad / s , ǫ max = 0 . 5 V , x = 0 . 14 m , and y = 0 . 39 m . From ǫ max = N A B ω , where the area A is A = (0 . 14 m)(0 . 39 m) = 0 . 0546 m 2 , so B = ǫ max N A ω = 0 . 5 V (23 turns)(0 . 0546 m 2 )(114 rad / s) = 0 . 00349257 T = 3 . 49257 mT . 003 10.0points A 763-turn circular-loop coil 19 . 9 cm in di- ameter is initially aligned so that its axis is parallel to Earth’s magnetic field. In 2 . 32 ms the coil is flipped so that its axis is perpendic- ular to Earth’s magnetic field.

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• Spring '16
• dr burat

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