Q65 / 90 · JEE Main 2024 (05 Apr Shift 1) · Physics

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

  • A
    $\frac{\mu_0 \pi b^2}{2 a}$
  • B
    $\frac{\mu_o}{2 \pi} \cdot \frac{b^2}{a}$
  • C
    $\frac{\mu_0}{2 \pi} \cdot \frac{a^2}{b}$
  • Correct answer
    D
    $\frac{\mu_0 \pi a^2}{2 b}$

Solution

$\begin{aligned} & \phi=\mathrm{Mi}=\mathrm{BA} \\ & \Rightarrow \mathrm{Mi}=\frac{\mu_0 \mathrm{i}}{2 \mathrm{~b}} \pi \mathrm{a}^2 \\ & \therefore \mathrm{M}=\frac{\mu_0 \pi \mathrm{a}^2}{2 \mathrm{~b}}\end{aligned}$

Hence, the correct answer is option: 'D'.

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