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(D) The magnetic field $B$ produced by the square loop at its center is $B = \frac{\mu_0 I}{\pi} \frac{4 \sin(45^\circ)}{L} = \frac{2\sqrt{2}\mu_0 I}{

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$2\sqrt{2} \frac{\mu_0 R^2}{L}$

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(D) The magnetic field $B$ produced by the square loop at its center is $B = \frac{\mu_0 I}{\pi} \frac{4 \sin(45^\circ)}{L} = \frac{2\sqrt{2}\mu_0 I}{\pi L}$.Since the circular loop is very small $(L >> R)$, we assume the magnetic field $B$ is uniform over the area of the circular loop.The magnetic flux $\phi$ linked with the circular loop of radius $R$ is $\phi = B 	imes A = B 	imes (\pi R^2)$.Substituting the value of $B$, we get $\phi = \left( \frac{2\sqrt{2}\mu_0 I}{\pi L} ight) 	imes (\pi R^2) = \frac{2\sqrt{2}\mu_0 I R^2}{L}$.The mutual inductance $M$ is defined as $M = \phi / I$.Therefore, $M = \frac{2\sqrt{2}\mu_0 R^2}{L}$.

$A$ circular current loop of radius $R$ is placed inside a square loop of side length $L$ $(L >> R)$ such that they are co-planar and their centers coincide. The permeability of free space is $\mu_0$. The mutual inductance between the circular loop and the square loop is . . . . . . .

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