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Figure shows a conducting frame having battery and a resistance on which a movable conductor of length 0.5 m can slide. The whole arrangement is placed in a uniform magnetic field of B = 0.4 T directed perpendicular and into the plane of frame. Initially the circuit is open. When the key is inserted, the conductor begins to move. It is found that a force 0.5 N has to be applied on the conductor to the left to keep it moving at constant speed to the right. Current flowing in the conductor is:

a
5A
b
2.5A
c
1.25A
d
1A

detailed solution

Correct option is B

F=Bil⇒i=FBl=0.50.4×0.5=2.5A

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