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

(N/A) Consider two long coaxial solenoids,each of length $l$. Let the radius of the inner solenoid $S_{1}$ be $r_{1}$ and the number of turns per unit length be $n_{1}$. The corresponding quantities for the outer solenoid $S_{2}$ are $r_{2}$ and $n_{2}$ respectively. Let $N_{1}$ and $N_{2}$ be the total number of turns of coils $S_{1}$ and $S_{2}$ respectively.
When a current $I_{2}$ flows through $S_{2}$,it sets up a magnetic field $B_{2} = \mu_{0} n_{2} I_{2}$ inside it. This field passes through the inner solenoid $S_{1}$.
The magnetic flux $\Phi_{1}$ through each turn of $S_{1}$ is $\Phi_{1} = B_{2} A_{1} = (\mu_{0} n_{2} I_{2})(\pi r_{1}^{2})$.
The total flux linkage with $S_{1}$ is $N_{1} \Phi_{1} = (n_{1} l) (\mu_{0} n_{2} I_{2} \pi r_{1}^{2}) = (\mu_{0} n_{1} n_{2} \pi r_{1}^{2} l) I_{2}$.
By definition,$N_{1} \Phi_{1} = M_{12} I_{2}$,where $M_{12}$ is the mutual inductance of $S_{1}$ with respect to $S_{2}$.
Thus,$M_{12} = \mu_{0} n_{1} n_{2} \pi r_{1}^{2} l$.
Similarly,if a current $I_{1}$ flows through $S_{1}$,the flux linkage with $S_{2}$ is $N_{2} \Phi_{2} = M_{21} I_{1}$,where $M_{21} = \mu_{0} n_{1} n_{2} \pi r_{1}^{2} l$.
Since $M_{12} = M_{21} = M$,this demonstrates the reciprocity theorem,which states that the mutual inductance of two coils is the same regardless of which coil is used as the primary or secondary.