Physics · Electromagnetic Induction

JEE Main 2024 — 5 April, Shift 1 — Question 41

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:

Option A:
$\frac{\mu_{0} \pi a^{2}}{2 b}$
Correct
Option B:
$\frac{\mu_{0}}{2 \pi} \cdot \frac{b^{2}}{a}$
Option C:
$\frac{\mu_{0} \pi b^{2}}{2 a}$
Option D:
$\frac{\mu_{0}}{2 \pi} \cdot \frac{a^{2}}{b}$

Answer: A

Step-by-step solution

$\phi=\mathrm{Mi}=\mathrm{BA}$

$\Rightarrow M i=\frac{\mu_{0} \mathrm{i}}{2 b} \pi \mathrm{a}^{2}$

$\therefore M=\frac{\mu_{0} \pi \mathrm{a}^{2}}{2 \mathrm{~b}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Self-Inductance and Mutual Inductance and Energy Density