JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

A circular current loop of radius $R$ is placed inside square loop of side length $L$ ( $L ≫ R$ ) such that they are co-planar and their centers coincide. The permeability of free space is $μ_{0}$ . The mutual inductance between circular loop and square loop is____.

A
$22 \frac{μ _{0} L ^{2}}{R}$
B
$2 \frac{μ _{0} L ^{2}}{R}$
C
$2 \frac{μ _{0} R ^{2}}{L}$
D
$22 \frac{μ _{0} R ^{2}}{L}$

Solution

Mutual inductance is symmetric, so compute the flux through the small circle due to current $I$ in the square loop.

The field at the centre of a square of side $L$ carrying current $I$ is $B = 4 \cdot \frac{μ _{0} I}{4 π ( L /2 )} \cdot 2 sin 4 5^{\circ} = \frac{2 2 μ _{0} I}{π L}$ .

Since $L ≫ R$ , this field is essentially uniform over the small circle. Flux: $Φ = B \cdot π R^{2} = \frac{2 2 μ _{0} R ^{2} I}{L}$ .

Mutual inductance: $M = \frac{Φ}{I} = \frac{2 2 μ _{0} R ^{2}}{L}$ .

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