Q 12-06-096JEE MainJEE Main 2022 (25 Jul, Shift 1)Medium
A small square loop of wire of side $l$ is placed inside a large square loop of wire of side $L$ $(L\gg l)$. Both loops are coplanar and their centres coincide at point $O$ as shown in figure. The mutual inductance of the system is
Answer: (D) $\dfrac{2\sqrt2\mu_0l^2}{\pi L}$
Current $I$ in the large loop. Each side (distance $\frac L2$ from the centre, ends at $45^\circ$) gives
$$B_1 = \frac{\mu_0I}{4\pi(L/2)}(\sin45^\circ + \sin45^\circ) = \frac{\sqrt2\mu_0I}{2\pi L}$$
Four sides: $B = \dfrac{2\sqrt2\mu_0I}{\pi L}$, nearly uniform over the small loop since $l\ll L$.
$$M = \frac{\Phi}{I} = \frac{Bl^2}{I} = \frac{2\sqrt2\mu_0l^2}{\pi L}$$
Solution by Sreeraj P, M.Sc Physics