Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres…

Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coinciding. If $R_1>R_2$, the mutual inductance $M$ between them will be directly proportional to
  1. $\frac{\mathrm{R}_1}{\mathrm{R}_2}$
  2. $\frac{\mathrm{R}_2}{\mathrm{R}_1}$
  3. $\frac{\mathrm{R}_1^2}{\mathrm{R}_2}$
  4. $\frac{\mathrm{R}_2^2}{\mathrm{R}_1}$

Solution

Mutual inductance between concentric coplanar loops

Let $R_1$ be the radius of the outer loop and $R_2$ the radius of the inner loop, with $R_1 > R_2$. The magnetic field at the center of the outer loop carrying current $I_1$ is $B = \frac{\mu_0 I_1}{2R_1}$.

Since $R_1 > R_2$, the field is approximately uniform over the area of the inner loop. The flux through the inner loop is $\Phi_{21} = B \cdot \pi R_2^2 = \frac{\mu_0 \pi I_1 R_2^2}{2R_1}$.

The mutual inductance is $M = \frac{\Phi_{21}}{I_1} = \frac{\mu_0 \pi R_2^2}{2R_1}$, which is proportional to $\frac{R_2^2}{R_1}$.

Comparing with the given options, the mutual inductance varies as $\frac{R_2^2}{R_1}$, corresponding to option D.

Asked in: MHT CET 2025 (19 April Shift 1)

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