JEE Main 2026 · 2 Apr · Shift 2 · Physics Q43

← JEE Main 2026 · 2 Apr · Shift 2

PhysicsSingle correct+4 / -1Question 43 of 75

A circular current loop of radius $R$ is placed inside square loop of side length $L$ ( $L\gg R$ ) such that they are co-planar and their centers coincide. The permeability of free space is $\mu_0$ . The mutual inductance between circular loop and square loop is ________.

2\sqrt2\dfrac{\mu_0L^2}R

\sqrt2\dfrac{\mu_0L^2}R

\sqrt2\dfrac{\mu_0R^2}L

2\sqrt2\dfrac{\mu_0R^2}L

Answer and worked solution

Official answer: D

Field at the centre of the square (four sides at distance $L/2$ ):
$B=4\cdot\frac{\mu_0I}{4\pi(L/2)}(\sin45^\circ+\sin45^\circ)=\frac{2\sqrt2\mu_0I}{\pi L}.$

Flux through the small circle (field nearly uniform since $L\gg R$ ): $\Phi=B\pi R^2$ .

Mutual inductance. $M=\dfrac\Phi I=\dfrac{2\sqrt2\mu_0R^2}L$ .

(D)

Topics:Mutual inductance Biot-Savart law

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JEE Main 2026 · 2 Apr · Shift 2, Physics question 43