Question

Derive an expression for the mutual inductance of two coaxial solenoids of the same length wound one over the other. Define mutual inductance and give its S.I. unit.

Accepted Answer

Mutual inductance, between a pair of coils, equals the magnetic flux, linked with one of them due to a change of unit current flowing in the other.
Let $r_{1}$ and $r_{2}$ be radii of inner (let it be primary P) and outer (let it be secondary S) co-axial solenoids respectively, and $n_{1}$ and $n_{2}$ be number of turns per unit length of the two solenoids. Let $N_{1}$ and $N_{2}$ be total number of turns in two solenoids and each of length l.
Let the secondary solenoid carry current $I_{2}$ . This current sets up magnetic flux $ϕ_{1}$ through the primary (inner) solenoid.

The total flux linkages with the primary solenoid are given by $N_{1} ϕ_{1} = M_{12} I_{2}$
Where, $M_{12}$ is the mutual inductance of the primary (inner) solenoid with respect to the secondary (outer) solenoid.
Magnetic field at the centre of the secondary solenoid due to a current $I_{2}$ is given by $B 2 = μ_{0} n_{2} I_{2}$ .
The total flux linkages with the primary solenoid are given by

$N_{1} ϕ_{1} = (n_{1} l) B_{2} I_{1} N_{1} ϕ_{1} = (μ_{0} {πn}_{1} n_{2} {lr}_{1}^{2}) I_{2} M_{12} = \frac{N_{1} ϕ_{1}}{l_{2}} = μ_{0} {πn}_{1} n_{2} {lr}_{1}^{2} Similarly, M_{21} = \frac{N_{2} ϕ_{2}}{l_{2}} = μ_{0} {πn}_{1} n_{2} {lr}_{1}^{2}$

Derive an expression for the mutual inductance of two coaxial solenoids of the same length wound one over the other. Define mutual inductance and give its S.I. unit.

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