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A circuit ABCD is held perpendicular to the uniform magnetic field of $B = 5 \times 10^{- 2} T$ extending over the region PQRS and B directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of 0.2 ms^-1 for 1.5 s. During this time, the current in the $5 Ω$ resistor is

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0.6 mA from B to C

0.9 mA from B to C

0.9 mA from C to B

0.6 mA from C to B

answer is A.

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Detailed Solution

$I = \frac{B l v}{R}; I = \frac{5 \times 10^{- 2} \times 0.3 \times 0.2}{5} A = 0.6 m A$

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