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Question

# A square metal wire loop of side 10 cm and resistance 1 Ω is moved with a constant velocity v in a uniform magnetic field B = 2 T, as shown in Fig. 25.4. The loop is connected to 4 Ω resistor. What should be the speed v of the loop as to have a steady current of 1 mA in the loop ?

A

1 cms1

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B

2 cms1

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C

3 cms1

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D

4 cms1

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Solution

## The correct option is B 2 cms−1 The emf induced by a wire of length l moving a field B with speed v perpendicular to the field is given by e=B l v=2×0.1×v=0.2 v Current in the circuit is I=eR=0.2 v4. But I=1 mA=10−3 A. Therefore, 10−3=0.2 v4 which gives v=2×10−2ms−1=2cm s−1.

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