Question

Consider two long co-axial solenoids S₁ and S₂, each of length I (>> r₂) and of radius r₁ and r₂(r₂ > r₁). The number of turns per unit length are n₁ and n₂ respectively. Derive an expression for mutual inductance M₁₂ of solenoid S₁ with respect to solenoid S₂. Show that M₂₁ = M₁₂.

Answer 1

Let l = length of each solenoid

r₁, r₂ = radii of the two solenoids

A = `pi r_1^2`

A = area of cross-section of inner solenoid S₁

N₁, N₂ = number of turns in the two solenoids

First we pass a time varying current I₂ through S₂.

The magnetic field set up inside S₂ due to I₂ is:

B₂ = μ₀n₂I₂

Where n₂ = N₂/l = the number of turns per unit length of S₂

Total magnetic flux linked with the inner solenoid S₁ is:

Φ₁ = B₂AN₁ = μ₀n₂I₂AN₁

∴ Mutual inductance of coil 1 with respect to coil 2 is;

M₁₂ = `phi_1/l_2`

= μ₀n₂AN₁

= `(mu_0N_1N_2A)/l`

We now consider the flux connected to the outer solenoid S₂ as a result of the inner solenoid’s current I₁. S₁. The field B₁ due to I₁ is constant inside S₁ but zero in the annular region between the two solenoids.

Hence B₁ = μ₀n₁I₁

Where n₁ = N₁/l = the number of turns per unit length of S₁.

Total flux linked with the outer solenoid S₂ is:

Φ₂ =  B₁AN₂ = μ₀n₁I₁AN₂

= `(mu_0N_1N_2AI_1)/l`

∴ Mutual inductance of coil 2 with respect to coil 1 is:

M₂₁ = `(mu_0N_1N_2A)/l`

Clearly, M₁₂ = M₂₁ = M