Question

Obtain an expression for the coefficient of mutual induction of two long solenoids.

Answer 1

Mutual induction of two long solenoids:
$S_1$ and $S_2$ are two solenoids each of length $l$. The solenoid $S_2$ is wound closely over the solenoid $S_1$.
$N_1$ and $N_2$ are the number of turns in the solenoids $S_1$ and $S_2$ respectively.
Both the solenoids are considered to have the same area of cross section $A$ as they are closely wound together. $I_1$ is the current flowing through the solenoid $S_1$.
The magnetic field $B_1$ produced at any point inside the solenoid $S_1$ due to the current $I_1$ is
$B_i={\mu }_0\frac{N_1}{l}$ ......(1)
The magnetic flux linked with each turn of $S_2$ is equal to $B_{i}A$.
Total magnetic flux linked with solenoid $S_2$ having $N_2$ turns ${\varphi }_2=B_{1}A{N}_2$
Substituting for $B_1$ from equation (1)
${\varphi }_2=\left({\mu }_0\frac{N_1}{l}{I}_1\right)A{N}_2$
${\varphi }_2=\frac{\mu N_1{N}_{2}A{I}_1}{l}$ .....(2)
But, $\varphi =M{I}_1$ ......(3)
Where $M$ is the coefficient of mutual induction between $S_1$ and $S_2$.
From equations (2) and (3)
$M{I}_1=\frac{{\mu }_0{N}_1{N}_{2}A{I}_1}{l}$
$M{I}_i=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$
If the core is filled with a magnetic material of permeability $\mu$,
$M{I}_l=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$