91影视

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.

$L=\frac{{渭}_{\mathbf{0}}I}{\mathbf{8}蟺}$ 鈥︹�� (1)

Here, ${渭}_{\mathbf{0}}$ 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鈥檚 law

$鈭�\stackrel{鈫�}{B}路d\stackrel{鈫�}{l}={渭}_0{I}_{enc}$

Here, $\stackrel{鈫�}{B}$is the magnetic field, $d\stackrel{鈫�}{l}$ is the length element, ${渭}_0$is the permeability of the free space and$I_{enc}$ is the current flowing through the enclosed ampere鈥檚 loop.

Consider an Ampere鈥檚 loop of s radius.

$$\begin{gathered} 鈭�\stackrel{鈫�}{B}路d\stackrel{鈫�}{I}={渭}_0{I}_{enc} \\ B路\left(2蟺s\right)={渭}_0{I}_{enc} \end{gathered}$$

The current per unit area, is written as follows:

$J=\frac{I}{蟺R^2}$

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

From the above equation it follows that

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

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

$$\begin{gathered} B路\left(2蟺s\right)={渭}_{0}I{\left(\frac{s}{r}\right)}^2 \\ B=\frac{{渭}_{0}Is}{2蟺R^2} \end{gathered}$$

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

$W=\frac{1}{2{渭}_0}\underset{allspace}{鈭�}{B}^{2}d蟿$

Here, is the surface area element.

Substitute $\frac{{渭}_{0}Is}{2蟺R^2}$for and for into equation $W=\frac{1}{2{渭}_0}\underset{allspace}{鈭�}{B}^{2}d蟿$.

Here, l is the length of the wire.

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

On further simplification

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

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

Equating, you have

$\frac{{渭}_0{I}^{2}l}{16蟺}=\frac{1}{2}L{I}^2$

Determine the self-inductance per unit length.

$$\begin{gathered} L=\frac{{渭}_{0}I}{8蟺} \\ \frac{L}{I}=\frac{{渭}_0}{8蟺} \end{gathered}$$

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