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Q.A conducting rod of length $l$ is moving with velocity $v$ parallel to a long horizontal wire carrying current $I$ . The induced emf is

a

\dfrac{\mu_0 I v}{2\pi} \ln \left(\dfrac{d_2}{d_1}\right)

b

\dfrac{\mu_0 I v}{2\pi} \ln \left(\dfrac{d_1}{d_2}\right)

c

\dfrac{\mu_0 I v l}{2\pi d}

d

zero

Correct Answer: Option D

The correct solution involves applying the fundamental concept to derive the final value step by step...

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