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Q.

A conducting rod moves with constant velocity $ϑ$ perpendicular to the long, straight wire carrying a current I as shown compute that the emf generated between the ends of the rod.

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a

\frac{μ_{0} υ I l}{4 π r}

b

\frac{μ_{0} υ I l}{π r}

c

\frac{μ_{0} υ I l}{2 π r}

d

\frac{2 μ_{0} υ I l}{π r}

answer is B.

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Detailed Solution

Question Image

induced emf $E = B l υ$

$= \frac{μ_{0} i}{2 π r} l υ$

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