Question

Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other,

Obtain the expression for the the mutual inductance of two long co-axial solenoids S₁ and S₂ wound one over the other, each of length L and radii r₁ and r₂ and n₁ and n₂ number of turns per unit length, when a current I is set up in the outer solenoid S₂.

Answer 1

Consider two long solenoids S₁ and S₂ of same length (l) such that solenoid S₂ surrounds solenoid S₁ completely.

Let:

n₁ = Number of turns per unit length of S₁

n₂ = Number of turns per unit length of S₂

I₁ = Current passed through solenoid S₁

Φ₂₁ = Flux linked with S2 due to current flowing through S₁

Φ₂₁ ∝ I₁

Φ₂₁ = M₂₁I₁

_Where

M₂₁ = Coefficient of mutual induction of the two solenoids

When current is passed through solenoid S₁, an emf is induced in solenoid S₂.

Magnetic field produced inside solenoid S₁ on passing current through it is given by 

B₁ = μ₀n₁I₁

Magnetic flux linked with each turn of solenoid S₂ will be equal to B₁ times the area of cross-section of solenoid S₁

Magnetic flux linked with each turn of the solenoid S₂ = B₁A

Therefore, total magnetic flux linked with the solenoid S₂ is given by 

Φ₂₁ = B₁A × n₂l = μ₀n₁I₁ × A× n₂l

Φ₂₁ = μ₀n₁n₂AI₁

∴ M₂₁ = μ₀n₁n₂Al

Similarly, the mutual inductance between the two solenoids, when current is passed through solenoid S₂ and induced emf is produced in solenoid S₁, is given by

M₁₂ = μ₀n₁n₂Al

∴ M₁₂ = M₂₁ = M (say)

Hence, coefficient of mutual induction between the two long solenoids is given by 

M = `μ (n_1n_2A)/l`