91Ó°ÊÓ

The radius of the solenoid is R .

Let’ assume the solenoid having the length l , number of the turns N and cross-sectional area is A . The current carried by the solenoid is $I$.

The area of the solenoid is given by,

$A=\mathrm{Ï€}{R}^2$

The magnetic field inside the solenoid is given by,

role="math" localid="1658135700537" $\mathbf{B}\mathbf{=}{\mathbf{μ}}_{\mathbf{0}}\mathbf{nI}$

Here $n$is the number of the turns per unit length.

The number of the turns per unit length.

$$\begin{gathered} n=\frac{N}{I} \\ N=nI \end{gathered}$$

The magnetic flux is given by,

$\mathrm{Ï•}=B.A$

Substitute ${\mathrm{μ}}_{0}nI$for B and$\mathrm{π}{R}^2$ for A into above equation.

$$\begin{gathered} \mathrm{ϕ}={\mathrm{μ}}_{0}nI.\mathrm{π}{R}^2 \\ \mathrm{ϕ}={\mathrm{μ}}_{0}nI.\mathrm{π}{R}^2 \end{gathered}$$

The inductance of the solenoid is given by,

$L=\frac{N\mathrm{Ï•}}{I}$

Substitute ${\mathrm{μ}}_{0}nI\mathrm{π}{R}^2$for $\mathrm{ϕ}$into above equation.

$$\begin{gathered} L=\frac{N{\mathrm{μ}}_{0}nL{\mathrm{π}}^2}{I} \\ L=\frac{N{\mathrm{μ}}_{0}nL{\mathrm{π}}^2}{I} \end{gathered}$$

Substitute $nI$for N into above equation.

$$\begin{gathered} L=\frac{nI{\mathrm{μ}}_{0}nI\mathrm{π}{R}^2}{I} \\ \frac{L}{I}=n^2{\mathrm{μ}}_0\mathrm{π}{R}^2 \end{gathered}$$

Hence the self-inductance per unit length of the solenoid is$n^2{\mathrm{μ}}_0\mathrm{π}{R}^2$