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A conductor of 3 m in length is moving perpendicularly to magnetic field of 10^-3 tesla with the speed of 10² m/s, then the e.m.f. produced across the ends of conductor will be

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0.03 volt

0.3 volt

$3 \times 10^{- 3} volt$

3 volt

answer is B.

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Detailed Solution

$e = Bvl = 3 \times 10^{- 3} \times 10^{2} = 0 .3 volt$

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