Derive formula for mutual inductance for two very long coaxial solenoids. Also discuss reciprocity theorem.

Consider figure which shows two long coaxial solenoids each of length $l$. We denote the radius of the inner solenoid $\mathrm{S}_{1}$ by $r_{1}$ and the number of turns per unit length by $n_{1}$. The corresponding quantities for the outer solenoid $\mathrm{S}_{2}$ are $r_{2}$ and $n_{2}$ respectively. Let $\mathrm{N}_{1}$ and $\mathrm{N}_{2}$ be the total number of turns of coils $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ respectively.

When a current $\mathrm{I}_{2}$ is set up through $\mathrm{S}_{2}$, it in turn sets up a magnetic flux through $\mathrm{S}_{1}$. Let us denote it by $\Phi_{1}$.

The corresponding flux linkage with solenoid $\mathrm{S}_{1}$ is

$\therefore \mathrm{N}_{1} \Phi_{1}=\mathrm{M}_{12} \mathrm{I}_{2} \quad \ldots$ $(1)$

$\mathrm{M}_{12}$ is called the mutual inductance of solenoid $\mathrm{S}_{1}$ with respect to solenoid $\mathrm{S}_{2}$. It is also referred to as the coefficient of mutual induction.

$\mathrm{N}_{1} \Phi_{1}=\mathrm{N}_{1} \mathrm{~A}_{1} \mathrm{~B}_{2}$

$\mathrm{~N}_{1} \Phi_{1}=\left(n_{1} l\right)\left(\pi r_{1}{ }^{2}\right)\left(\mu_{0} n_{2} \mathrm{I}_{2}\right)$

$=\mu_{0} n_{1} n_{2} \pi r_{1}{ }^{2} l \mathrm{I}_{2}$

Where $n_{1} l$ is the total number of turns in solenoid $\mathrm{S}_{1}$. Thus, from equation $(1)$ and $(2)$, $\mathrm{M}_{12}=\mu_{0} n_{1} n_{2} \pi r_{1}{ }^{2} l$

... $(3)$

Note that we neglected the edge effects and considered the magnetic field $\mu_{0} n_{2} I_{2}$ to be uniform throughout the length and width of the solenoid $S_{2}$.

We now consider the reverse case. A current $\mathrm{I}_{1}$ is passed through the solenoid $\mathrm{S}_{1}$ and the flux linkage with coil $\mathrm{S}_{2}$ is,

$\mathrm{N}_{2} \Phi_{2}=\mathrm{M}_{21} \mathrm{I}_{1}$ $\mathrm{M}_{21}$ is called the mutual inductance of solenoid $\mathrm{S}_{2}$ with respect to solenoid $\mathrm{S}_{1}$.