JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Electromagnetic Induction

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Find the mutual inductance in the arrangement, when a small circular loop of wire of radius ' R ' is placed inside a large square loop of wire of side L

(L ≫ R)

. The loops are coplanar and their centres coincide :

(A) $M = \frac{2 μ _{0} R ^{2}}{L}$
(B) $M = \frac{2 2 μ _{0} R}{L ^{2}}$
(C) $M = \frac{2 2 μ _{0} R ^{2}}{L}$ ✓ Correct
(D) $M = \frac{2 μ _{0} R}{L ^{2}}$

Answer: (C)

Solution

The magnetic field at the centre of a square loop of side

L

carrying current

i

B = \frac{2 2 μ _{0} i}{π L}

. Since the inner loop is very small (

L ≫ R

), this field is essentially uniform over it, so the flux linked is

Φ = B π R^{2} = \frac{2 2 μ _{0} R ^{2} i}{L}

. With

Φ = M i

, the mutual inductance is

M = \frac{2 2 μ _{0} R ^{2}}{L}

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Electromagnetic Induction

Solution

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