Electromagnetic Induction and Alternating Current 2 Question 12

13. A square metal wire loop of side 10cm and resistance 1Ω is moved with a constant velocity v0 in a uniform magnetic field of induction B=2Wb/m2 as shown in the figure. The magnetic field lines are perpendicular to the plane of the loop (directed into the paper). The loop is connected to a network of resistors each of value 3Ω.

What should be the speed of the loop so as to have a steady current of 1mA in the loop? Give the direction of current in the loop.

(1983,6 M)

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Answer:

Correct Answer: 13. 0.02m/s, direction of induced current is clockwise

Solution:

  1. Given network forms a balanced Wheatstone’s bridge. The net resistance of the circuit is therefore 3Ω+1Ω=4Ω. Emf of the circuit is Bv0l. Therefore, current in the circuit would be

i=Bv0lR or v0=iRBl=(1×103)(4)2×0.1=0.02m/s

Cross magnetic field passing through the loop is decreasing. Therefore, induced current will produce magnetic field in cross direction. Or direction of induced current is clockwise.



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