A circular current loop of radius $R$ is placed inside square loop of side length $L\ (L\gg R)$ such that they are co-planar and their centers coincide. The permeability of free space is $\mu_0$. The mutual inductance between circular loop and square loop is ______.
Answer: (D) $2\sqrt2\dfrac{\mu_0R^2}{L}$
Pass current $I$ through the square loop. Each side is at distance $L/2$ from the centre and subtends $45^\circ$ on each side:
$$B_{side}=\frac{\mu_0I}{4\pi(L/2)}(\sin45^\circ+\sin45^\circ)=\frac{\sqrt2\mu_0I}{2\pi L}$$
All four sides add: $B=\dfrac{2\sqrt2\mu_0I}{\pi L}$. Since $R\ll L$, this field is uniform over the small circle.
Flux through the circle: $\phi=B\pi R^2=\dfrac{2\sqrt2\mu_0IR^2}{L}$
$$M=\frac{\phi}{I}=2\sqrt2\frac{\mu_0R^2}{L}$$
Solution by Sreeraj P, M.Sc Physics