# HW 25 - terry (ect328) homework 25 Turner (59130) 1 This...

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Unformatted text preview: terry (ect328) homework 25 Turner (59130) 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 out of the paper. The separation between the rails is . 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 ( z or a ) 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 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 a > V z (left) 5. V z = V a (right) and V z = V a 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. Lenzs 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, Lenzs 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 10.0 points A rectangular coil of 50 turns, 0 . 26 m by . 2 m, is rotated at 141 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: 0 . 818331 mT. Explanation: Let : N = 50 turns , = 141 rad / s , max = 0 . 3 V , x = 0 . 26 m , and y = 0 . 2 m ....
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## This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas at Austin.

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HW 25 - terry (ect328) homework 25 Turner (59130) 1 This...

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