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Question

A square metal wire loop of side 10 cm is moved with a constant velocity v in a uniform magnetic field B=2 Wbm2 as shown in figure. The magnetic field lines are perpendicular to the plane of the loop and directed into the plane. The loop is connected to the network of resistances, each of value 3 Ω. The resistance of the other wires are negligible. The speed of the loop so as to have a steady current of 1 mA in the loop is

A
1 cms1
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B
4 cms1
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C
1.5 cms1
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D
0.5 cms1
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Solution

The correct option is C 1.5 cms1
Let the speed of the rod be v
ϵ=Bvl=2×0.1v
ϵ=0.2v

Since the above network is a wheatstone bridge,
Therefore total resistance is R=3 Ω
Now current is given as 103 mA
103=0.2vR
103=0.2v3
v=1.5 cms1
For detailed solution watch the next video.

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