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(C) Let a current $I$ flow through the outer square loop of side $L$. The magnetic field $B$ produced by this loop at its centre $O$ is given by the s

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$\frac{2 \sqrt{2} \mu_{0} \ell^{2}}{\pi L}$

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(C) Let a current $I$ flow through the outer square loop of side $L$. The magnetic field $B$ produced by this loop at its centre $O$ is given by the sum of the fields due to its four sides:$B = 4 	imes \left( \frac{\mu_{0} I}{4 \pi (L/2)} 	imes (\sin 45^{\circ} + \sin 45^{\circ}) ight) = 4 	imes \left( \frac{\mu_{0} I}{2 \pi L} 	imes 2 	imes \frac{1}{\sqrt{2}} ight) = \frac{2 \sqrt{2} \mu_{0} I}{\pi L}$.Since $L \gg l$,we can assume this magnetic field is uniform over the area of the small inner loop.The magnetic flux $\phi$ linked with the inner loop of side $l$ is:$\phi = B 	imes 	ext{Area} = \left( \frac{2 \sqrt{2} \mu_{0} I}{\pi L} ight) 	imes l^{2}$.The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.Therefore,$M = \frac{2 \sqrt{2} \mu_{0} l^{2}}{\pi L}$.

$A$ small square loop of wire of side $l$ is placed inside a large square loop of wire $L$ $(L \gg l)$. Both loops are coplanar and their centres coincide at point $O$ as shown in the figure. The mutual inductance of the system is:

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