A coil C of N turns is placed around a long solenoid S of radius R and n turns per unit length, as in Figure. (a) Show that the mutual inductance for the coil–solenoid combination is given by $\mathit{M}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}\mathbf{}\mathit{\pi }{\mathit{R}}^{\mathbf{2}}\mathit{n}\mathit{N}$. (b) Explain why M does not depend on the shape, size, or possible lack of close packing of the coil.

1. Number of turns in the coil are N.
2. Number of turns in the solenoid are n.
3. Radius of solenoid is R.

We have the relation between the mutual inductance and flux linked in the solenoid. We also have the formula for the flux linked in the solenoid. So, we can prove the required formula for M.

Formula:

Mutual inductance is given by,

$$\begin{gathered} M=\frac{N\varphi }{I} \\ \varphi ={\mu }_{0}InA \end{gathered}$$

We have mutual inductance given by,

$M=\frac{N\varphi }{I}$

But,

$$\begin{gathered} \varphi ={\mu }_{0}InA \\ M=\frac{N\left({\mu }_{0}InA\right)}{I} \end{gathered}$$

Here, the solenoid has cross – section as the circular cross – section.

$$\begin{gathered} A=\pi R^2 \\ M=\frac{N\left({\mu }_{0}In\pi R^2\right)}{I} \end{gathered}$$

We know that the magnetic field of the solenoid is entirely contained within the cross-section of the coil, which is given by,

$\varphi =BA$

The solenoid has cross – section as the circular cross – section.

$$\begin{gathered} A=\pi R^2 \\ \varphi =B\pi R^2 \end{gathered}$$

From this, we can say that, M does not depend upon the shape, size or possible lack of close packing of the coil.