91影视

The radius of the solenoid is R .

Let鈥� 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,

$A=蟺R^2$

The magnetic field inside the solenoid is given by,

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

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,

$蠒=B.A$

Substitute ${渭}_{0}nI$for B and$蟺R^2$ for A into above equation.

$$\begin{gathered} 蠒={渭}_{0}nI.蟺R^2 \\ 蠒={渭}_{0}nI.蟺R^2 \end{gathered}$$

The inductance of the solenoid is given by,

$L=\frac{N蠒}{I}$

Substitute ${渭}_{0}nI蟺R^2$for $蠒$into above equation.

$$\begin{gathered} L=\frac{N{渭}_{0}nL{蟺}^2}{I} \\ L=\frac{N{渭}_{0}nL{蟺}^2}{I} \end{gathered}$$

Substitute $nI$for N into above equation.

$$\begin{gathered} L=\frac{nI{渭}_{0}nI蟺R^2}{I} \\ \frac{L}{I}=n^2{渭}_0蟺R^2 \end{gathered}$$

Hence the self-inductance per unit length of the solenoid is$n^2{渭}_0蟺R^2$