Question

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

• A. \(\frac{\text{R}_1}{\text{R}_2}\)
• B. \(\frac{R_2}{R_1}\)
• C. \(\frac{\text{R}_1^2}{\text{R}_2}\)
• D. \(\frac{R_2^2}{R_1}\)

Hint:

Flux depends on area → smaller loop area \(R^2\) is important.

Answer 1

The Correct Option is D

Concept:
Mutual inductance depends on magnetic flux linkage: \[ M = \frac{\Phi}{I} \] Step 1: Magnetic field of larger loop. Magnetic field at centre of loop of radius \(R_1\): \[ B = \frac{\mu_0 I}{2R_1} \]
Step 2: Flux through smaller loop. Area of smaller loop: \[ A = \pi R_2^2 \] \[ \Phi = B \cdot A = \frac{\mu_0 I}{2R_1} \cdot \pi R_2^2 \]
Step 3: Find M. \[ M = \frac{\Phi}{I} = \frac{\mu_0 \pi R_2^2}{2R_1} \]
Step 4: Conclusion. \[ M \propto \frac{R_2^2}{R_1} \]