Question:A long cable carries current in one direction uniformly distributed over its (circular) cross section. The current returns along the surface (there is a very thin insulating sheath separating the currents). Find the self-inductance per unit length.

Consider a long cable carries current in one direction uniformly distributed over its (circular) cross-section.

Write the formula ofthe energy stored in the toroidal coil.

$L=\frac{{\mu }_{\mathbf{0}}I}{\mathbf{8}\pi }$ …… (1)

Here, ${\mu }_{\mathbf{0}}$ is permeability and data-custom-editor="chemistry" $\mathrm{l}$is total current.

Determine theenergy in a current carrying wire is written as follows:

$\frac{1}{2}L{I}^2$

Here, L is the self-inductance of the wire and l is the current flowing through the wire.

According to Ampere’s law

$\oint \overrightarrow{B}·d\overrightarrow{l}={\mu }_0{I}_{enc}$

Here, $\overrightarrow{B}$is the magnetic field, $d\overrightarrow{l}$ is the length element, ${\mu }_0$is the permeability of the free space and$I_{enc}$ is the current flowing through the enclosed ampere’s loop.

Consider an Ampere’s loop of s radius.

$$\begin{gathered} \oint \overrightarrow{B}·d\overrightarrow{I}={\mu }_0{I}_{enc} \\ B·\left(2\pi s\right)={\mu }_0{I}_{enc} \end{gathered}$$

The current per unit area, is written as follows:

$J=\frac{I}{\pi R^2}$

Here, is the total current flowing the wire and is the radius of the wire.

From the above equation it follows that

$$\begin{gathered} I_{enc}=J\pi s^2 \\ =\left(\frac{1}{\pi R^2}\right)\pi s^2 \\ =I{\left(\frac{s}{r}\right)}^2 \end{gathered}$$

Substitute $I{\left(\frac{s}{r}\right)}^2$ for$I_{enc}$ into equation $B·\left(2\pi s\right)={\mu }_0{I}_{enc}$ .

$$\begin{gathered} B·\left(2\pi s\right)={\mu }_{0}I{\left(\frac{s}{r}\right)}^2 \\ B=\frac{{\mu }_{0}Is}{2\pi R^2} \end{gathered}$$

Determine the energy stored in the wire is calculated as follows:

$W=\frac{1}{2{\mu }_0}\underset{allspace}{\int }{B}^{2}d\tau$

Here, is the surface area element.

Substitute $\frac{{\mu }_{0}Is}{2\pi R^2}$for and for into equation $W=\frac{1}{2{\mu }_0}\underset{allspace}{\int }{B}^{2}d\tau$.

Here, l is the length of the wire.

role="math" localid="1658741154605" $$\begin{gathered} W=\frac{1}{2{\mu }_0}\underset{allspace}{\int }{\left(\frac{{\mu }_{0}Is}{2\pi R^2}\right)}^2\left(2\pi s\right)Ids \\ =\frac{1}{2{\mu }_0}{\left(\frac{{\mu }_{0}I}{2\pi R^2}\right)}^{2}2\pi \underset{0}{\overset{R}{\int }}{s}^{2}ds \end{gathered}$$

On further simplification

$$\begin{gathered} W=\frac{{\mu }_0{I}^{2}I}{4\pi R^4}\underset{0}{\overset{R}{\int }}{s}^{3}ds \\ =\frac{{\mu }_0{I}^{2}I}{4\pi R^4}{\left[\frac{s^4}{4}\right]}_0^R \\ =\frac{{\mu }_0{I}^{2}l}{16\pi } \end{gathered}$$

The work done is also equal to $\frac{1}{2}L{I}^{}$.

Equating, you have

$\frac{{\mu }_0{I}^{2}l}{16\pi }=\frac{1}{2}L{I}^2$

Determine the self-inductance per unit length.

$$\begin{gathered} L=\frac{{\mu }_{0}I}{8\pi } \\ \frac{L}{I}=\frac{{\mu }_0}{8\pi } \end{gathered}$$

Therefore, thevalue ofthe self-inductance per unit length is $\frac{{\mu }_0}{8\pi }$.