Mutual Inductance of Two Long Co-axial Solenoids of Equal Length
Consider two solenoids $S_1$ and $S_2$ such that the solenoid $S_2$ completely surrounds the solenoid $S_1$.
Let length of each solenoid be $l$, and the area of cross-section of each solenoid is $A$. Let $N_1$ and $N_2$ be the total number of turns of solenoid $S_1$ and $S_2$ respectively.
$\therefore$ Number of turns per unit length of solenoid $S_1$ is given by, $n_1 = \frac{N_1}{l}$.
Number of turns per unit length of solenoid $S_2$ is given by, $n_2 = \frac{N_2}{l}$.
Let current $I_1$ flow through solenoid $S_1$. Then magnetic field inside the solenoid $S_1$ is given by,
$ B_1 = \mu_0 n_1 I_1 = \mu_0 \frac{N_1}{l} I_1 $
Magnetic flux linked with each turn of solenoid $S_2$ is given by,
$d\phi_2 = B_1 A = \mu_0 \frac{N_1}{l} I_1 A$.
Then, total magnetic flux linked with $N_2$ turns of the solenoid $S_2$ is given by
$ \phi_2 = N_2 d\phi_2$
$ \phi_2= N_2 \mu_0 \frac{N_1}{l} I_1 A$
$ \phi_2= \mu_0 \frac{N_1 N_2}{l} A I_1 \quad \dots (1) $
But
$ \phi_2 = M_{12} I_1 \quad \dots (2) $
where $M_{12}$ is the mutual inductance of coil $S_2$ with respect to the coil $S_1$.
From (1) and (2), we get
$ M_{12} I_1 = \mu_0 \frac{N_1 N_2}{l} A I_1$
or
$M_{12} = \mu_0 \frac{N_1 N_2 A}{l} \quad \dots (3) $
Now, let current $I_2$ flows through solenoid $S_2$ and no current flows through the solenoid $S_1$.
The magnetic field inside solenoid $S_2$ is given by,
$ B_2 = \mu_0 n_2 I_2 = \mu_0 \frac{N_2}{l} I_2 $
Magnetic flux linked with each turn of solenoid $S_1$ is given by,
$ d\phi_1 = B_2 A = \mu_0 \frac{N_2}{l} I_2 A $
Total magnetic flux linked with all turns of solenoid $S_1$,
$ \phi_1 = N_1 d\phi_1 = \mu_0 \left(\frac{N_2}{l} I_2 A\right) N_1 = \mu_0 \frac{N_1 N_2}{l} A I_2 \quad \dots (4) $
Also,
$ \phi_1 = M_{21} I_2 \quad \dots (5) $
where $M_{21}$ is the mutual inductance of solenoid $S_1$ w.r.t. solenoid $S_2$.
From eqns. (4) and (5), we get
$ M_{21} I_2 = \mu_0 \frac{N_1 N_2 A}{l} I_2 $
$M_{12} = \mu_0 \frac{N_1 N_2 A}{l} \quad \dots (6) $
Comparing eqns. (3) and (6), we find
$M_{12} = M_{21} = M$(Theorem of reciprocity)
Thus,
$M = \mu_0 \frac{N_1 N_2 A}{l} \quad \dots (7) $
or
$ M = \mu_0 \left(\frac{N_1}{l}\right) \left(\frac{N_2}{l}\right) A l$
$M= \mu_0 n_1 n_2 Al \quad \dots (8)$
If the two solenoids are wound on a material having relative permeability $\mu_r$, then eqn. (8) can be written as
$M = \mu_0 \mu_r \frac{N_1 N_2 A}{l}$
$M= \mu N_1 N_2 \frac{A}{l} \quad (\because \mu = \mu_0 \mu_r) \quad \dots (9) $
Thus, mutual inductance of a pair of solenoids increases if they are wound on a material of relative permeability $\mu_r$.