Question

An infinitely long cylindrical tube, of radius a, moves at constant speed v along its axis. It carries a net charge per unit length $\lambda$, uniformly distributed over its surface. Surrounding it, at radius b, is another cylinder, moving with the same velocity but carrying the opposite charge $-\lambda$. Find:

(a) The energy per unit length stored in the fields.

(b) The momentum per unit length in the fields.

(c) The energy per unit time transported by the fields across a plane perpendicular to the cylinders.

Short answer

(a)The energy per unit length stored in the fields is$\frac{w}{l}=\frac{{\lambda }^2}{4\pi {\epsilon }_0}\left(1+\frac{v^2}{c^2}\right)$$In\left(\frac{b}{a}\right)$.

(b) The momentum per unit length in the fields is$\frac{p}{l}=\frac{{{\mu }_{0\lambda }}^{2}v}{2\pi }In\left(\frac{b}{a}\right)\hat{z}$ .

(c) The energy per unit time transported by the fields across a plane perpendicular to the cylinders is$dW=\frac{{\mu }_0{\lambda }^2{C}^2{V}^2}{2\pi }In\left(\frac{b}{a}\right)$ .

Step-by-step solution

Expression for the total energy stored in electromagnetic fields:

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

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

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

Determine the energy per unit length stored in the fields:

(a)

Write the expression for the electric field.

$E=\frac{1}{2\pi {\epsilon }_0}\frac{\lambda }{s}\hat{s}$

Here,$\lambda$ is the linear charge density, and s is the surface charge density.

Write the expression for the magnetic field.

$B=\frac{{\mu }_0}{2\pi s}\hat{\varphi }$

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

$l=\lambda v$

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

$$\begin{gathered} u=\frac{1}{2}\left[{\epsilon }_0{\left(\frac{1}{2\pi {\epsilon }_0}\right)}^2\frac{{\lambda }^2}{s^2}+\frac{1}{{\mu }_0}{\left(\frac{{\mu }_0}{2\pi }\right)}^2\frac{{\lambda }^2{v}^2}{s^2}\right] \\ u=\frac{{\lambda }^2}{8{\pi }^2{\epsilon }_0}\left(1+{\epsilon }_0{\mu }_0{v}^2\right)\frac{1}{s^2} \end{gathered}$$

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

localid="1657524792153" $$\begin{gathered} \frac{w}{l}=\frac{{\lambda }^2}{8{\pi }^2{\epsilon }_0}\left(1+\frac{v^2}{c^2}\right){\int }_a^b\frac{1}{s^2}2\pi sds \\ \frac{w}{l}=\frac{{\lambda }^2}{8{\pi }^2{\epsilon }_0}\left(1+\frac{v^2}{c^2}\right)\left(2\pi \right){\int }_a^b\frac{1}{s^2}2.ds \\ \frac{w}{l}=\frac{{\lambda }^2}{8{\pi }^2{\epsilon }_0}\left(1+\frac{v^2}{c^2}\right)\left(2\pi \right){\int }_a^b\frac{1}{s^2}.ds \\ \frac{w}{l}=\frac{{\lambda }^2}{8{\pi }^2{\epsilon }_0}\left(1+\frac{v^2}{c^2}\right)\left(2\pi \right){\int }_a^b\frac{1}{s^2}s.ds \\ \frac{w}{l}=\frac{{\lambda }^2}{8{\pi }^2{\epsilon }_0}\left(1+\frac{v^2}{c^2}\right)\left(2\pi \right){\int }_a^b\frac{1}{s^{}}.ds \\ \frac{w}{l}=\frac{{\lambda }^2}{8{\pi }^2{\epsilon }_0}\left(1+\frac{v^2}{c^2}\right)\left(2\pi \right){\left[\mathrm{In}\left(\mathrm{s}\right)\right]}_a^b \end{gathered}$$

On further solving,

$$\begin{gathered} \frac{w}{l}=\frac{{\lambda }^2}{8{\pi }^2{\epsilon }_0}\left(1+\frac{v^2}{c^2}\right)\left[\left(2\mathrm{\pi }\right)\mathrm{In}\left(\mathrm{b}\right)-\mathrm{In}\left(\mathrm{a}\right)\right] \\ \frac{w}{l}=\frac{{\lambda }^2}{4{\pi }^2{\epsilon }_0}\left(1+\frac{v^2}{c^2}\right)\mathrm{In}\left(\frac{\mathrm{b}}{\mathrm{a}}\right) \end{gathered}$$

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

$\frac{w}{l}=\frac{{\lambda }^2}{4{\pi }^2{\epsilon }_0}\left(1+\frac{v^2}{c^2}\right)\mathrm{In}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)$.

Determine the momentum per unit length in the fields:

(b)

Write the expression for the momentum density.

$$\begin{gathered} g={\mu }_0{\epsilon }_{0}S \\ G={\epsilon }_0(E\times B) \end{gathered}$$

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

$$\begin{gathered} g={\epsilon }_0\left(\frac{1}{2\pi {\epsilon }_0}\frac{\lambda }{s}\hat{s}\times \frac{{\mu }_0}{2\pi }\frac{\mathrm{l}}{s}\hat{\varphi }\right) \\ g={\epsilon }_0\left(\frac{\lambda }{2\pi {\epsilon }_0}\right)\left(\frac{{\mu }_0\lambda v}{2\pi s}\right)\hat{z} \\ \frac{p}{l}=\frac{{\mu }_0{\lambda }^{2}v}{4{\pi }^2}\widehat{\hat{z}{\int }_a^b}\frac{1}{s^2}2\pi sds \\ \frac{p}{l}=\frac{{\mu }_0{\lambda }^{2}v}{4{\pi }^2}\widehat{\hat{z}\left(2\pi \right){\int }_a^b}\frac{1}{s}ds \end{gathered}$$

On further solving,

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

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