Questions 31 through 33 pertain to the situation described below: A loop is pull
ID: 2066837 • Letter: Q
Question
Questions 31 through 33 pertain to the situation described below:
A loop is pulled with a force F to the right to maintain a constant speed of 8.0 m/s. The loop has a length of 0.15 m, a width of 0.080 m, and a resistance of 200.0 W. At the instant shown, the loop is partially in and partially out of a uniform magnetic field that is directed into the paper. The magnitude of the field is 1.2 T.
31. What is the magnitude of the emf induced in the loop?
(a) zero volts (c) 1.4 V (e) 9.6 V
(b) 0.77 V (d) 4.9 V
32. What is the induced current in the loop?
(a) 0.048 A (c) 7.2 ´ 10-3 A (e) zero amperes
(b) 0.024 A (d) 3.8 ´ 10-3 A
33. Determine the magnitude of the force F required to pull the loop.
(a) 1.3 ´ 10-4 N (c) 3.7 ´ 10-4 N (e) 9.0 ´ 10-4 N
(b) 2.1 ´ 10-4 N (d) 6.8 ´ 10-4 N
Explanation / Answer
EMF is gathered by multiplying magnetic field, lenth of the device and the speed at which it is being pulled out. So (1.2T times 8m/s times .15m) this equals 1.44 V or answer (c) In order to get the current we must divide the EMF by the total resistance in the device. You gave us 200 ohms for resistance. So we would divide 1.44V/200 ohms this equals 0.0072 or 7.2x10^-3, answer (c) Finally, in order to get the force we will use and multiply Current by length and magnetic field. So (1.2T times .15m times 0.0072Amps) = 0.001296 N which rounds to 1.3x10^-3 N which is none of those answers but is closest to (a)
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