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

Step 1: Flux through the outer coil due to the solenoid.
Magnetic field exists only inside the solenoid of radius \( r_1 \). Area of solenoid cross-section: \[ A = \pi r_1^2 \] Flux through one turn of the outer coil: \[ \phi = B A = \mu_0 \frac{N_1}{L} I_1 \cdot \pi r_1^2 \]
Step 2: Total flux linkage with the outer coil.
For \( N_2 \) turns in the outer coil: \[ \Phi = N_2 \phi = N_2 \mu_0 \frac{N_1}{L} I_1 \pi r_1^2 \]
Step 3: Calculate mutual inductance. \[ M_{12} = \frac{\Phi}{I_1} = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 \]
Step 4: Symmetry of mutual inductance. \[ M_{12} = M_{21} \] Mutual inductance is symmetric, independent of the coil geometry, as long as the medium is linear and isotropic.
Step 5: Final Expression. \[ M = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 \] Yes, \( M_{12} = M_{21} \) holds true.