JEE Main 2026 — Physics Electromagnetism

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

Magnetic field at the centre of the square loop:

For a square of side $L$L, the perpendicular distance from the centre to any side is $d = \dfrac{L}{2}$d=2L​, and each side subtends angles $\theta_1 = \theta_2 = 45^\circ$θ1​=θ2​=45∘ at the centre.

The field due to one finite straight segment is:

$$B_{\text{segment}} = \dfrac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)$$Bsegment​=4πdμ0​I​(sinθ1​+sinθ2​)

$$B_{\text{segment}} = \dfrac{\mu_0 I}{4\pi \cdot \frac{L}{2}}\left(\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\right) = \dfrac{\mu_0 I}{2\pi L} \cdot \sqrt{2} = \dfrac{\sqrt{2}\,\mu_0 I}{2\pi L}$$Bsegment​=4π⋅2L​μ0​I​(2​1​+2​1​)=2πLμ0​I​⋅2​=2πL2​μ0​I​

All four sides contribute equally in the same direction at the centre, so:

$$B_{\text{total}} = 4 \cdot \dfrac{\sqrt{2}\,\mu_0 I}{2\pi L} = \dfrac{2\sqrt{2}\,\mu_0 I}{\pi L}$$Btotal​=4⋅2πL2​μ0​I​=πL22​μ0​I​

Flux through the circular loop:

Since $L \gg R$L≫R, the field $B_{\text{total}}$Btotal​ can be treated as uniform over the area $A = \pi R^2$A=πR2 of the circular loop.

$$\Phi = B_{\text{total}} \cdot A = \dfrac{2\sqrt{2}\,\mu_0 I}{\pi L} \cdot \pi R^2 = \dfrac{2\sqrt{2}\,\mu_0 R^2 I}{L}$$Φ=Btotal​⋅A=πL22​μ0​I​⋅πR2=L22​μ0​R2I​

Mutual inductance:

$$M = \dfrac{\Phi}{I} = \dfrac{2\sqrt{2}\,\mu_0 R^2}{L}$$M=IΦ​=L22​μ0​R2​

Hence, the correct option is $(4)\ \dfrac{2\sqrt{2}\,\mu_0 R^2}{L}$(4) L22​μ0​R2​.