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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 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 ?
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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.
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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