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(B) Let the current in the large loop be $I_1$. The magnetic field $B_1$ produced by the large loop at its center is given by $B_1 = \frac{4\mu_0 I_1}

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$l^2/L$

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(B) Let the current in the large loop be $I_1$. The magnetic field $B_1$ produced by the large loop at its center is given by $B_1 = \frac{4\mu_0 I_1}{\sqrt{2}\pi L} = \frac{2\sqrt{2}\mu_0 I_1}{\pi L}$.Since $L \gg l$,we can assume the magnetic field $B_1$ is approximately uniform over the area of the small loop.The magnetic flux $\phi_2$ linked with the small loop is $\phi_2 = B_1 	imes A_2$,where $A_2 = l^2$ is the area of the small loop.$\phi_2 = \left( \frac{2\sqrt{2}\mu_0 I_1}{\pi L} ight) l^2$.The mutual inductance $M$ is defined as $M = \frac{\phi_2}{I_1}$.$M = \frac{2\sqrt{2}\mu_0 l^2}{\pi L}$.Thus,$M \propto \frac{l^2}{L}$.

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

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