A small square loop of wire of side $l$ is placed inside a large square loop of wire $\mathrm{L}(\mathrm{L} \gg l)$. Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is :

(C) $\frac{2 \sqrt{2} \mu_{0} \ell^{2}}{\pi \mathrm{~L}}$

(A) $\frac{2 \sqrt{2} \mu_{0} \mathrm{~L}^{2}}{\pi \ell}$

(B) $\frac{\mu_{0} \ell^{2}}{2 \sqrt{2} \pi \mathrm{~L}}$

(D) $\frac{\mu_{0} \mathrm{~L}^{2}}{2 \sqrt{2} \pi \ell}$

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Electromagnetic Induction

A small square loop of wire of side

l

is placed inside a large square loop of wire

L (L ≫ l)

. Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is :

Answer: (C)

When current

I

flows in the outer loop, the magnetic field at the center is calculated by summing contributions from all four sides. Each side at distance

L /2

contributes

\frac{μ _{0} I}{4 π ( L /2 )} \times 2 sin 4 5^{\circ}

. The total field at center is

B = \frac{2 2 μ _{0} I}{π L}

. The flux through the inner loop of area

ℓ^{2}

Φ = B ℓ^{2} = \frac{2 2 μ _{0} I ℓ ^{2}}{π L}

. The mutual inductance is

M = \frac{Φ}{I} = \frac{2 2 μ _{0} ℓ ^{2}}{π L}

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option