samp07 - counterclockwise, the same direction as the emf of...

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SAMPLE PROBLEM (Class 7) ANS. a) The induced emf must be calculated and added to the emf of the battery. The induced emf is given by dt d B i Φ - = E , so first, Φ B must be calculated. Since only half the area of the loop is in the field, Φ B = ½ L ² B , where L is the length of one side of the square loop. Then, dt dB L dt B L d dt d B i 2 ) ( 2 2 2 1 - = - = Φ - = E . B = 0.0420 – 0.870 t , so T/s 870 . 0 - = dt dB . B is decreasing . ) T/s 870 . 0 ( 2 ) m 00 . 2 ( 2 2 2 - - = - = dt dB L i E = 1.74 V Since B points out of the page, and it is decreasing, the induced emf is
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Unformatted text preview: counterclockwise, the same direction as the emf of the battery. Therefore, they add. E net = E i + E bat = 1.74 V + 20 V E net = 21.7 V, counterclockwise b) The current is in the direction of the net emf, counterclockwise. The current goes through the battery from negative terminal to positive terminal....
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This note was uploaded on 09/17/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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