Question

A long solenoid of length \( L \) and radius \( r_1 \) having \( N_1 \) turns is surrounded symmetrically by a coil of radius \( r_2 \, (r_2>r_1) \) having \( N_2 \) turns (\( N_2 \ll N_1 \)) around its mid-point. Derive an expression for the mutual inductance of solenoid and coil. Is \( M_{12} = M_{21} \) valid in this case?

Hint:

For solenoid–coil systems:

Flux area = cross-section of inner solenoid
Mutual inductance is always symmetric: \( M_{12} = M_{21} \)

Answer 1

Concept: Mutual inductance: \[ M = \frac{\text{Flux linked with secondary}}{\text{Current in primary}} \] Magnetic field inside a long solenoid: \[ B = \mu_0 n I = \mu_0 \frac{N_1}{L} I_1 \] Field is uniform inside solenoid and negligible outside.
Step 1: Flux through outer coil due to solenoid. Magnetic field exists only inside solenoid of radius \( r_1 \). Area contributing to flux: \[ A = \pi r_1^2 \] Flux through one turn of outer coil: \[ \phi = B A = \mu_0 \frac{N_1}{L} I_1 \cdot \pi r_1^2 \]
Step 2: Total flux linkage with outer coil. Outer coil has \( N_2 \) turns: \[ \Phi = N_2 \phi = N_2 \mu_0 \frac{N_1}{L} I_1 \pi r_1^2 \]
Step 3: Mutual inductance. \[ M_{12} = \frac{\Phi}{I_1} \] \[ M_{12} = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 \]
Step 4: Mutual inductance symmetry. In general: \[ M_{12} = M_{21} \] This is a fundamental property of mutual inductance, independent of geometry (as long as medium is linear and isotropic).
Step 5: Conclusion.

Mutual inductance: \[ M = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 \]
Yes, \( M_{12} = M_{21} \) is valid.