Question

Find the self-inductance per unit length of a long solenoid, of radius R , carrying n turns per unit length.

Short answer

The self-inductance per unit length of the solenoid is ${\mathrm{n}}^2{\mathrm{\mu }}_0{\mathrm{\pi R}}^2$.

Step-by-step solution

Write the given data from the question.

The radius of the solenoid is R .

Determine the self-inductance per unit length of a solenoid.

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,

$A=\pi R^2$

The magnetic field inside the solenoid is given by,

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

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,

$\varphi =B.A$

Substitute ${\mu }_{0}nI$for B and$\pi R^2$ for A into above equation.

$$\begin{gathered} \varphi ={\mu }_{0}nI.\pi R^2 \\ \varphi ={\mu }_{0}nI.\pi R^2 \end{gathered}$$

The inductance of the solenoid is given by,

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

Substitute ${\mu }_{0}nI\pi R^2$for $\varphi$into above equation.

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

Substitute $nI$for N into above equation.

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

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