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A current loop A of radius R is placed as shown in figure. Another circular loop B of radius r(<<R) is placed coaxially at a distance x(>>R) from the centre of A. The current in A is in anticlock wise direction as seen from B. Now the loop B is given a velocity v towards A as shown in figure. The emf induced in B as well as current direction as seen from the side of A is [Assume A to be fixed]

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$\frac{3}{2} \frac{μ_{0} I R^{2} π r^{2} v}{x^{4}}$ , anticlockwise

$\frac{5 μ_{0} I R^{2} π r^{2} v}{2 x^{4}}$ , anticlockwise

$\frac{3}{2} \frac{μ_{0} I R^{2} π r^{2} v}{x^{4}}$ , clock wise

$\frac{5 μ_{0} I R^{2} π r^{2} v}{2 x^{4}}$ , clock wise

answer is A.

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Detailed Solution

$ϕ_{B} = [\frac{μ_{0} i R^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}}] π r^{2} e_{B} = \frac{- d ϕ_{B}}{d t}$

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