Mutual Inductance question from JEE Main 2026 (2 Apr, Shift 2)

Answer: (d)

1. Field at the centre of a square loop (current $I$, side $L$): each side gives $\frac{\mu_0 I}{4\pi (L/2)}(\sin 45^\circ + \sin 45^\circ)$.
2. $B = 4 \times \frac{\mu_0 I}{2\pi L}\sqrt{2} = \frac{2\sqrt{2}\,\mu_0 I}{\pi L}$
3. Since $R \ll L$, $B$ is uniform over the small loop: $\phi = B\,\pi R^2 = \frac{2\sqrt{2}\,\mu_0 I R^2}{L}$
4. $M = \phi / I = \frac{2\sqrt{2}\,\mu_0 R^2}{L}$
5. So the answer is (d) $2\sqrt{2}\,\frac{\mu_0 R^2}{L}$.