JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Electromagnetic Induction

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Two conducting circular loops A and B are placed in the same plane with their centres coinciding as shown in figure. The mutual inductance between them is:

(A) $\frac{μ _{0} π a ^{2}}{2 b}$ ✓ Correct
(B) $\frac{μ _{0}}{2 π} \cdot \frac{b ^{2}}{a}$
(C) $\frac{μ _{0} π b ^{2}}{2 a}$
(D) $\frac{μ _{0}}{2 π} \cdot \frac{a ^{2}}{b}$

Answer: (A)

Solution

Let loop

B

(radius

b

) carry current

i

. Since

b ≫ a

, the field at the common centre due to

B

is uniform over the small loop

A

B = \frac{μ _{0} i}{2 b} .

Flux through

A

(area

π a^{2}

ϕ = B \cdot π a^{2} = \frac{μ _{0} i π a ^{2}}{2 b}

. Therefore

M = \frac{ϕ}{i} = \frac{μ _{0} π a ^{2}}{2 b} .

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Single correct option

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