An infinitely long straight wire PQ, carrying a current of 10 A lies in the plane of a square loop ABCD as shown in the figure. The loop is made to move with a constant velocity of 1m/s away from the wire PQ. Then the emf induced in the loop, at the instant shown in figure, is

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$10^{- 4} volt$

$10^{- 6} volt$

$10^{- 3} volt$

$10^{- 5} volt$

answer is A.

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Detailed Solution

Net emf = Emf induced in AD - Emf induced in BC

$= (\frac{μ_{0} x 10}{2 π x 1} x 1 x 1 - \frac{μ_{0} x 10}{2 π x 2} x 1 x 1) volt$

$= \frac{5 μ_{0}}{2 π} volt = \frac{5 x 4 π x 10^{- 7}}{2 π} volt = 10^{- 6} volt$

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