JEE Main 2026 — Electromagnetic Induction Question with Solution

Using the reciprocity theorem, assume a current $I$ flows through the larger square loop and compute the flux through the smaller circular loop. The mutual inductance is given by $M=\frac{\Phi }{I}$.

Magnetic field at the centre of the square loop:
For a square of side $L$, the perpendicular distance from the centre to any side is $d=\frac{L}{2}$, and each side subtends angles ${\theta }_1={\theta }_2=45^{\circ }$ at the centre.

The field due to one finite straight segment is:
$B_{\text{segment}}=\frac{{\mu }_{0}I}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

$B_{\text{segment}}=\frac{{\mu }_{0}I}{4\pi \cdot \frac{L}{2}}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=\frac{{\mu }_{0}I}{2\pi L}\cdot \sqrt{2}=\frac{\sqrt{2}{\mu }_{0}I}{2\pi L}$

All four sides contribute equally in the same direction at the centre, so:
$B_{\text{total}}=4\cdot \frac{\sqrt{2}{\mu }_{0}I}{2\pi L}=\frac{2\sqrt{2}{\mu }_{0}I}{\pi L}$

Flux through the circular loop:
Since $L\gg R$, the field $B_{\text{total}}$ can be treated as uniform over the area $A=\pi R^2$ of the circular loop.

$\Phi =B_{\text{total}}\cdot A=\frac{2\sqrt{2}{\mu }_{0}I}{\pi L}\cdot \pi R^2=\frac{2\sqrt{2}{\mu }_0{R}^{2}I}{L}$

Mutual inductance:
$M=\frac{\Phi }{I}=\frac{2\sqrt{2}{\mu }_0{R}^2}{L}$

Hence, the correct option is $(4)\frac{2\sqrt{2}{\mu }_0{R}^2}{L}$.