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(D) The magnetic field $B$ produced by a square loop of side $L$ carrying current $I$ at its centre is given by the sum of fields due to four wires of

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$\frac{l^2}{L}$

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(D) The magnetic field $B$ produced by a square loop of side $L$ carrying current $I$ at its centre is given by the sum of fields due to four wires of length $L$ at a distance $d = L/2$ from the centre.For one wire,the field is $B_{wire} = \frac{\mu_0 I}{4 \pi d} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{4 \pi (L/2)} (2 	imes \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{\sqrt{2} \pi L}$.Since there are four such wires,the total magnetic field at the centre is $B = 4 	imes \frac{\mu_0 I}{\sqrt{2} \pi L} = \frac{2 \sqrt{2} \mu_0 I}{\pi L}$.Since $L \gg l$,we can assume the magnetic field $B$ is uniform over the area of the smaller loop $S_2 = l^2$.The magnetic flux linked with the smaller loop is $\phi_2 = B 	imes S_2 = \frac{2 \sqrt{2} \mu_0 I}{\pi L} 	imes l^2$.The mutual inductance $M$ is defined as $M = \frac{\phi_2}{I} = \frac{2 \sqrt{2} \mu_0}{\pi} \frac{l^2}{L}$.Therefore,$M \propto \frac{l^2}{L}$.

$A$ small square loop of wire of side $l$ is placed inside a large square loop of side $L$ $(L \gg l)$. If the loops are coplanar and their centres coincide,the mutual inductance of the system is directly proportional to :

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