Consider two long co-axial solenoids $\displaystyle \mathrm{S}_{1}$ and $\displaystyle \mathrm{S}_{2}$, each of length $\displaystyle l\left(\gg \mathrm{r}_{2}\right)$ and of radius $\displaystyle \mathrm{r}_{1}$ and $\displaystyle \mathrm{r}_{2}\left(\mathrm{r}_{2}>\mathrm{r}_{1}\right)$. The number of turns per unit length are $\displaystyle \mathrm{n}_{1}$ and $\displaystyle \mathrm{n}_{2}$ respectively. Derive an expression for mutual inductance $\displaystyle \mathrm{M}_{12}$ of solenoid $\displaystyle \mathrm{S}_{1}$ with respect to solenoid $\displaystyle \mathrm{S}_{2}$. Show that $\displaystyle \mathrm{M}_{21}=\mathrm{M}_{12}$.

Let $\displaystyle N_{1}$ and $\displaystyle N_{2}$ be the total number of turns of coils $\displaystyle S_{1}$ and $\displaystyle S_{2}$ respectively. When current $\displaystyle I_{2}$ is set up in $\displaystyle S_{2}$, flux linkage with solenoid $\displaystyle S_{1}$ is: $\displaystyle N_{1} \phi_{1}=M_{12} I_{2}$ --(i) $\displaystyle N_{1} \phi_{1}=\left(n_{1} l\right)\left(\pi r_{1}^{2}\right)\left(\mu_{0} n_{2} I_{2}\right)$ $\displaystyle N_{1} \phi_{1}=\mu_{0} n_{1} n_{2} \pi r_{1}^{2} l I_{2}$ --(ii) From (i) and (ii): $\displaystyle M_{12}=\mu_{0} n_{1} n_{2} \pi r_{1}^{2} l$ Considering the reverse case, when $\displaystyle I_{1}$ current is set up in $\displaystyle S_{1}$, flux linkage with $\displaystyle S_{2}$ is: $\displaystyle N_{2} \phi_{2}=M_{21} I_{1}$ --(iii) $\displaystyle N_{2} \phi_{2}=\left(n_{2} l\right)\left(\pi r_{1}^{2}\right)\left(\mu_{0} n_{1} I_{1}\right)$ --(iv) From (iii) and (iv): $\displaystyle M_{21}=\mu_{0} n_{1} n_{2} \pi r_{1}^{2} l$ $\displaystyle \therefore M_{12}=M_{21}$