In one form of Tesla coil (a hugh voltage generator popular is science museums), a long solenoid with length $l$ and cross-sectional area $A$ is closely wound with $N_1$ turns of wire. A coil with $N_2$ turns surrounds it at its centre (shown in figure). Find the mutual inductance $M$ (in $\mu H$).

Identify and Set up: Mutual inductance occurs here because a current in either coil sets up a magnetic field that causes a flux through the other coil. 

We have an expression for the field magnitude $B_1$ at the center of the solenoid (coil $1$) in terms of the solenoid current $i_1$. This allows us to determine the flux through a cross section of the solenoid. Since there is no magnetic field outside a very long solenoid, this is also equal to the flux ${\varphi }_{B2}$ through each turn of the outer coil $\left(2\right)$. We then use 

$M=\frac{N_2{\varphi }_{B2}}{i_1}=\frac{N_1{\varphi }_{B1}}{i_2}$ (mutual inductance)              $...\left(i\right)$

in the form $M=N_2{\varphi }_{B2}/i_1$, to determine $M$.

Execute: The expression for the field magnitude is expressed in terms of the number of turns per unit length, which for solenoid $\left(1\right)$ is $n_1=N_1/L$. We then have 

$B_1={\mu }_0{n}_1{i}_1=\frac{{\mu }_0{N}_1{i}_1}{l}$

The flux through a cross section of the solenoid equals $B_{1}A.$ As we mentioned above, this also equals the flux ${\varphi }_{B2}$ through each turn of the outer coil. independent of its cross-sectional area. From equation $\left(i\right)$, the mutual inductance $M$ is then 

$M=\frac{N_2{\varphi }_{B2}}{i_1}=\frac{N_2{B}_{1}A}{i_1}=\frac{N_2}{i_1}\frac{{\mu }_0{N}_1{i}_i}{l}A=\frac{{\mu }_{0}A{N}_1{N}_2}{l}$

Evaluate: The mutual inductance $M$ of any two coils is proportional to the product $N_1{N}_2$ of their numbers of turnns. notice that $M$ depends only on the geometry of the two coils, not on the current. 

Here is a numerical example. Suppose $l=0.50m, A=10c{m}^2=1.0\times {10}^{-3}{m}^2, N_1=1000$ turns and $N_2=10$ turns then

$$\begin{gathered} M=\frac{\left(4\mathrm{\pi }\times {10}^{-7} \mathrm{Wb}/\mathrm{A}.\mathrm{m}\right)\left(1.0\times {10}^{-3}{m}^2\right)\left(1000\right)\left(10\right)}{0.50m} \\ =25\times {10}^{-6}Wb/A=25\times {10}^{-6}H=25\mu H \end{gathered}$$