Q 12-06-060JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
A small square loop of wire of side $l$ is placed inside a large square loop of wire of side $L$ $(L = l^2)$. The loops are coplanar and their centres coincide. The value of the mutual inductance of the system is $\sqrt x\times10^{-7}\ \text{H}$, where $x = $ ______.
Numerical value type. Enter your answer.
Answer: 128
A current $I$ in the large loop gives, at its centre,
$$B = 4\times\frac{\mu_0I}{4\pi(L/2)}\times2\sin45^\circ = \frac{2\sqrt2\,\mu_0I}{\pi L}$$
Treating this as uniform over the small loop, $\Phi = Bl^2$ and
$$M = \frac{2\sqrt2\,\mu_0 l^2}{\pi L} = \frac{2\sqrt2\,\mu_0}{\pi} = 2\sqrt2\times4\times10^{-7} = 8\sqrt2\times10^{-7} = \sqrt{128}\times10^{-7}\ \text{H}$$
$x = 128$.
Solution by Sreeraj P, M.Sc Physics