91Ó°ÊÓ

Write the expression for the total energy stored in electromagnetic fields.

$\mathit{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{Î\mu }}_{\mathbf{0}}{\mathit{E}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}{\mathit{B}}^{\mathbf{2}}$…… (1)

Here, E is the electric field,${\mathrm{Î\mu }}_0$ is the permittivity of free space, ${\mathrm{μ}}_0$is the permeability of free space, and B is the magnetic field.

(a)

Write the expression for the electric field.

$E=\frac{1}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\mathrm{λ}}{s}\hat{s}$

Here,$\mathrm{λ}$ is the linear charge density, and s is the surface charge density.

Write the expression for the magnetic field.

$B=\frac{{\mathrm{μ}}_0}{2\mathrm{π}s}\widehat{\mathrm{ϕ}}$

Here, I is the current, which is given as:

$l=\mathrm{λ}v$

Substitute the value of E, B and I in equation (1).

$$\begin{gathered} u=\frac{1}{2}\left[{\mathrm{Î\mu }}_0{\left(\frac{1}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)}^2\frac{{\mathrm{λ}}^2}{s^2}+\frac{1}{{\mathrm{μ}}_0}{\left(\frac{{\mathrm{μ}}_0}{2\mathrm{Ï€}}\right)}^2\frac{{\mathrm{λ}}^2{v}^2}{s^2}\right] \\ u=\frac{{\mathrm{λ}}^2}{8{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\left(1+{\mathrm{Î\mu }}_0{\mathrm{μ}}_0{v}^2\right)\frac{1}{s^2} \end{gathered}$$

Substitute${\mathrm{Î\mu }}_0{\mathrm{μ}}_0=\frac{1}{c^2}$in the above expression and integrate over the volume between the cylinders.

localid="1657524792153" $$\begin{gathered} \frac{w}{l}=\frac{{\mathrm{λ}}^2}{8{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\left(1+\frac{v^2}{c^2}\right){∫}_a^b\frac{1}{s^2}2\mathrm{Ï€}sds \\ \frac{w}{l}=\frac{{\mathrm{λ}}^2}{8{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\left(1+\frac{v^2}{c^2}\right)\left(2\mathrm{Ï€}\right){∫}_a^b\frac{1}{s^2}2.ds \\ \frac{w}{l}=\frac{{\mathrm{λ}}^2}{8{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\left(1+\frac{v^2}{c^2}\right)\left(2\mathrm{Ï€}\right){∫}_a^b\frac{1}{s^2}.ds \\ \frac{w}{l}=\frac{{\mathrm{λ}}^2}{8{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\left(1+\frac{v^2}{c^2}\right)\left(2\mathrm{Ï€}\right){∫}_a^b\frac{1}{s^2}s.ds \\ \frac{w}{l}=\frac{{\mathrm{λ}}^2}{8{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\left(1+\frac{v^2}{c^2}\right)\left(2\mathrm{Ï€}\right){∫}_a^b\frac{1}{s^{}}.ds \\ \frac{w}{l}=\frac{{\mathrm{λ}}^2}{8{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\left(1+\frac{v^2}{c^2}\right)\left(2\mathrm{Ï€}\right){\left[\mathrm{In}\left(\mathrm{s}\right)\right]}_a^b \end{gathered}$$

On further solving,

$$\begin{gathered} \frac{w}{l}=\frac{{\mathrm{λ}}^2}{8{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\left(1+\frac{v^2}{c^2}\right)\left[\left(2\mathrm{Ï€}\right)\mathrm{In}\left(\mathrm{b}\right)-\mathrm{In}\left(\mathrm{a}\right)\right] \\ \frac{w}{l}=\frac{{\mathrm{λ}}^2}{4{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\left(1+\frac{v^2}{c^2}\right)\mathrm{In}\left(\frac{\mathrm{b}}{\mathrm{a}}\right) \end{gathered}$$

Therefore, the energy per unit length stored in the fields is .

$\frac{w}{l}=\frac{{\mathrm{λ}}^2}{4{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\left(1+\frac{v^2}{c^2}\right)\mathrm{In}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)$.

(b)

Write the expression for the momentum density.

$$\begin{gathered} g={\mathrm{μ}}_0{\mathrm{Î\mu }}_{0}S \\ G={\mathrm{Î\mu }}_0(E×B) \end{gathered}$$

Substitute the value of E and B in the above expression.

$$\begin{gathered} g={\mathrm{Î\mu }}_0\left(\frac{1}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\mathrm{λ}}{s}\hat{s}×\frac{{\mathrm{μ}}_0}{2\mathrm{Ï€}}\frac{\mathrm{l}}{s}\widehat{\mathrm{Ï•}}\right) \\ g={\mathrm{Î\mu }}_0\left(\frac{\mathrm{λ}}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\left(\frac{{\mathrm{μ}}_0\mathrm{λ}v}{2\mathrm{Ï€}s}\right)\hat{z} \\ \frac{p}{l}=\frac{{\mathrm{μ}}_0{\mathrm{λ}}^{2}v}{4{\mathrm{Ï€}}^2}\widehat{\hat{z}{∫}_a^b}\frac{1}{s^2}2\mathrm{Ï€}sds \\ \frac{p}{l}=\frac{{\mathrm{μ}}_0{\mathrm{λ}}^{2}v}{4{\mathrm{Ï€}}^2}\widehat{\hat{z}\left(2\mathrm{Ï€}\right){∫}_a^b}\frac{1}{s}ds \end{gathered}$$

On further solving,

$$\begin{gathered} \frac{p}{l}=\frac{{\mathrm{μ}}_0{\mathrm{λ}}^{2}v}{4\mathrm{π}}\hat{z}\left(2\mathrm{π}\right){\left[\ln \left(s\right)\right]}_a^b \\ \frac{p}{l}=\frac{{\mathrm{μ}}_0{\mathrm{λ}}^{2}v}{4\mathrm{π}}\ln \left(\frac{b}{a}\right)\hat{z}\left(2\mathrm{π}\right)\left[In\left(b\right)-in\left(a\right)\right] \\ \frac{p}{l}=\frac{{\mathrm{μ}}_0{\mathrm{λ}}^{2}v}{4\mathrm{π}}\ln \left(\frac{b}{a}\right)\hat{z} \end{gathered}$$

Therefore, the momentum per unit length in the fields is$\frac{p}{l}=\frac{{\mathrm{μ}}_0{\mathrm{λ}}^{2}v}{2\mathrm{π}}\ln \left(\frac{b}{a}\right)\hat{z}$ .