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Question

A square metallic wire loop of side 0.1m and resistance of 1Ω is moved with a constant velocity in a magnetic field of  2Wb-m2 as shown in the figure. The magnetic field is perpendicular to the plane of the loop which is connected to a network of resistances. If the steady state current in the loop is 1 mA, then the velocity of loop is (in   cm/sec)
 


 


Solution

Equivalent resistance of the given Wheatstone bridge circuit (balanced) is 3 Ω 
So total resistance in circuit is R=3+1=4 Ω. The emf induced in the loop is e=Bvl.
So induced current i=eR=BvlR
103=2×v×(10×102)4v= 2 cm/sec

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