Question

(a) Show directly that Eqs. 9.197 satisfy Maxwell’s equations (Eq. 9.177) and the boundary conditions (Eq. 9.175).

(b) Find the charge density, $\mathit{\lambda }\left(z,t\right)$, and the current, $\mathit{I}\left(z,t\right)$, on the inner conductor.

Short answer

(a) The equation 9.197 satisfies Maxwell’s equation and the boundary conditions.

(b) The charge density is $\mathrm{\lambda }\left(\mathrm{z},\mathrm{t}\right)=2{\mathrm{\pi \epsilon }}_0{\mathrm{E}}_0\cos \left(\mathrm{kz}-\mathrm{\omega t}\right)$, and the current density is $\mathrm{I}=\frac{2{\mathrm{\pi \epsilon }}_0}{{\mathrm{\mu }}_0\mathrm{c}}\cos \left(\mathrm{kz}-\mathrm{\omega t}\right)$.

Step-by-step solution

Expression for the electric and magnetic fields at a distance s from the axis of the co-axial transmission line:

Write the expression for the electric and magnetic fields at a distance s from the axis of the co-axial transmission line (Eqs 9.197).

$\mathbf{E}\mathbf{(}\mathbf{s}\mathbf{,}\mathbf{\varphi }\mathbf{,}\mathbf{z}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{Acos}\mathbf{(}\mathbf{kz}\mathbf{-}\mathbf{\omega t}\mathbf{)}\mathbf{}}{\mathbf{s}}\hat{\mathbf{s}}$

$\mathbf{B}\mathbf{(}\mathbf{s}\mathbf{,}\mathbf{\varphi }\mathbf{,}\mathbf{z}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{Acos}\mathbf{(}\mathbf{kz}\mathbf{-}\mathbf{\omega t}\mathbf{)}\mathbf{}}{\mathbf{cs}}\hat{\mathbf{\varphi }}$

Here, E is the electric field, B is the magnetic field, s is the Poynting vector, k is the wave number, c is the speed of light, is the phase and t is the time.

Satisfy Maxwell’s equation and V¯×E=0 and V¯×B=0 :

(a)

Prove$\overline{\mathrm{V}}.\mathrm{E}=0$ as follow:

$\overline{\mathrm{V}}.E=\frac{1}{s}\frac{\partial }{\partial s}\left(s{E}_s\right)+\frac{1}{s}\frac{\partial E_{\varphi }}{\partial \varphi }+\frac{\partial E_z}{\partial \mathrm{z}}$ …… (1)

Here, the value of ${\mathrm{E}}_{\mathrm{s}}$,${\mathrm{E}}_{\mathrm{\varphi }}$ and${\mathrm{E}}_{\mathrm{z}}$ is given as:

$$\begin{gathered} E_{\mathrm{s}}=\frac{A\cos \left(kx-\omega t\right)}{s} \\ {E}_{\varphi }=0 \\ {E}_z=0 \end{gathered}$$

Substitute ${\mathrm{E}}_{\mathrm{s}}=\frac{\mathrm{Acos}\left(\mathrm{kx}-\mathrm{\omega t}\right)}{\mathrm{s}}$,${\mathrm{E}}_{\mathrm{\varphi }}=0$ and${\mathrm{E}}_{\mathrm{z}}=0$ in equation (1).

$$\begin{gathered} \overline{\mathrm{V}}.E=\frac{1}{s}\frac{\partial }{\partial s}\left(s\frac{A\cos \left(kx-\omega t\right)}{s}\right)+0+0 \\ \overline{\mathrm{V}}.E=0 \end{gathered}$$

Prove $\overline{\mathrm{V}}.\mathrm{B}=0$:

$\overline{\mathrm{V}}.\mathrm{B}=\frac{1}{\mathrm{s}}\frac{\partial }{\partial \mathrm{s}}\left({\mathrm{sB}}_{\mathrm{s}}\right)+\frac{1}{\mathrm{s}}\frac{\partial {\mathrm{B}}_{\mathrm{\varphi }}}{\partial \mathrm{\varphi }}+\frac{1}{\mathrm{s}}\frac{\partial {\mathrm{B}}_{\mathrm{z}}}{\partial \mathrm{z}}$ …… (2)

Here, the value of ${\mathrm{B}}_{\mathrm{s}}$,${\mathrm{B}}_{\mathrm{\varphi }}$ and${\mathrm{B}}_{\mathrm{z}}$ is given as:

$$\begin{gathered} {\mathrm{B}}_{\mathrm{s}}=0 \\ {\mathrm{B}}_{\mathrm{\varphi }}=\frac{\mathrm{Acos}\left(\mathrm{kz}-\mathrm{\omega t}\right)}{\mathrm{s}} \\ {\mathrm{B}}_{\mathrm{z}}=0 \end{gathered}$$

Substitute ${\mathrm{B}}_{\mathrm{s}}=0$, ${\mathrm{B}}_{\mathrm{\varphi }}=\frac{\mathrm{Acos}\left(\mathrm{kz}-\mathrm{\omega t}\right)}{\mathrm{s}}$and ${\mathrm{B}}_{\mathrm{z}}=0$in equation (2).

$$\begin{gathered} \overline{\mathrm{V}}.\mathrm{B}=\frac{1}{\mathrm{s}}\frac{\partial }{\partial \mathrm{s}}\left(\mathrm{s}\times 0\right)+\frac{1}{\mathrm{s}}\frac{\partial \left({\displaystyle \frac{\mathrm{Acos}\left(\mathrm{kz}-\mathrm{\omega t}\right)}{\mathrm{s}}}\right)}{\partial \mathrm{\varphi }}+0 \\ \overline{\mathrm{V}}.\mathrm{B}=0 \end{gathered}$$

Prove $\overline{\mathrm{V}}\times E=-\frac{\partial B}{\partial t}$:

$$\begin{gathered} \overline{\mathrm{V}}\times \mathrm{E}=\left(\frac{1}{\mathrm{s}}\left(\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{\varphi }}-\frac{\partial {\mathrm{E}}_{\mathrm{\varphi }}}{\partial \mathrm{z}}\right)\hat{\mathrm{s}}+\left(\frac{\partial {\mathrm{E}}_{\mathrm{s}}}{\partial \mathrm{z}}-\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{s}}\right)\hat{\mathrm{\varphi }}\right)+\frac{1}{\mathrm{s}}\left[\frac{\partial }{\partial \mathrm{s}}\left({\mathrm{sE}}_{\mathrm{\varphi }}\right)-\frac{\partial {\mathrm{E}}_{\mathrm{s}}}{\partial \mathrm{\varphi }}\right]\hat{\mathrm{z}} \\ \overline{\mathrm{V}}\times \mathrm{E}=0+\frac{\partial {\mathrm{E}}_{\mathrm{s}}}{\partial \mathrm{z}}\hat{\mathrm{\varphi }}-\frac{1}{\mathrm{s}}\frac{\partial {\mathrm{E}}_{\mathrm{s}}}{\partial \mathrm{\varphi }}\hat{\mathrm{z}} \\ \overline{\mathrm{V}}\times \mathrm{E}=\frac{-{\mathrm{E}}_0\mathrm{ksin}\left(\mathrm{kz}-\mathrm{\omega t}\right)}{\mathrm{s}}\hat{\mathrm{\varphi }} \\ \overline{\mathrm{V}}\times \mathrm{E}=-\frac{\partial \mathrm{B}}{\partial \mathrm{t}} \end{gathered}$$

Prove $\overline{\mathrm{V}}\times \mathrm{B}=\frac{1}{{\mathrm{c}}^2}\frac{\partial \mathrm{E}}{\partial \mathrm{t}}$:

$$\begin{gathered} \overline{\mathrm{V}}\times B=\left(\frac{1}{s}\left(\frac{\partial B_z}{\partial \varphi }-\frac{\partial B_{\varphi }}{\partial z}\right)\hat{s}+\left(\frac{\partial B_s}{\partial \mathrm{z}}-\frac{\partial B_z}{\partial \mathrm{s}}\right)\hat{\varphi }\right)+\frac{1}{s}\left[\frac{\partial }{\partial s}\left(s{B}_{\varphi }\right)-\frac{\partial B_s}{\partial \varphi }\right]\hat{z} \\ \overline{\mathrm{V}}\times B=-\frac{\partial B_{\varphi }}{\partial z}+\frac{1}{s}\frac{\partial }{\partial s}\left(s{B}_{\varphi }\right)\hat{z} \\ \overline{\mathrm{V}}\times B=\frac{{\mathrm{E}}_0\mathrm{k}}{\mathrm{c}}\frac{\sin \left(\mathrm{kz}-\mathrm{\omega t}\right)}{\mathrm{s}}\hat{\mathrm{s}} \\ \overline{\mathrm{V}}\times B\mathit{=}\frac{\mathit{1}}{{\mathit{c}}^{\mathit{2}}}\frac{\mathit{\partial }\mathit{E}}{\mathit{\partial }\mathit{t}} \end{gathered}$$

Satisfy the boundary conditions.

$$\begin{gathered} {\mathrm{E}}^{\mathrm{P}}={\mathrm{E}}_{\mathrm{z}}=0 \\ {\mathrm{B}}^{\perp }={\mathrm{B}}_{\mathrm{s}}=0 \end{gathered}$$

Therefore, equation 9.197 satisfies Maxwell’s equation and the boundary conditions.

Determine the charge density and current density on the inner conductor:

(b)

Apply Gauss law for a cylinder of radius s and length dz to determine the charge density.

$$\begin{gathered} \widehat{\mathrm{\mathbb{N}}}E.da=\frac{E_0\cos \left(kz-\omega t\right)}{s}\left(2\mathrm{\pi }s\right)dz \\ \frac{E_0\cos \left(kz-\omega t\right)}{s}\left(2\mathrm{\pi }s\right)dz\mathit{=}\frac{{\mathit{Q}}_{\mathit{e}\mathit{n}\mathit{c}\mathit{l}\mathit{o}\mathit{s}\mathit{e}\mathit{d}}}{{\mathit{\epsilon }}_{\mathit{0}}} \\ {E}_{\mathit{0}}\cos \left(kz\mathrm{-}\omega t\right)\left(\mathrm{2}\mathrm{\pi }\right)dz\mathit{=}\frac{\mathit{\lambda }\mathit{d}\mathit{z}}{{\mathit{\epsilon }}_{\mathit{0}}} \end{gathered}$$

Hence, the charge density will be,

$\lambda \left(\mathrm{z},\mathrm{t}\right)=2\pi {\epsilon }_0{E}_0\cos \left(kz-\omega t\right)$

Apply Ampere’s law for a circle of radius s to determine the current density.

$$\begin{gathered} \widehat{\mathrm{\mathbb{N}}}B.dl=\frac{E_0}{c}\frac{\cos \left(kz-\omega t\right)}{s}\left(2\pi s\right) \\ \widehat{\mathrm{\mathbb{N}}}B.dl={\mathrm{\mu }}_0{\mathrm{I}}_{\mathrm{enclosed}} \end{gathered}$$

Hence, the current density will be,

$\mathrm{I}=\frac{2{\mathrm{\pi \epsilon }}_0}{{\mathrm{\mu }}_0\mathrm{c}}\cos \left(\mathrm{kz}-\mathrm{\omega t}\right)$

Therefore, the charge density is $\mathrm{\lambda }\left(\mathrm{z},\mathrm{t}\right)=2{\mathrm{\pi \epsilon }}_0{\mathrm{E}}_0\cos \left(\mathrm{kz}-\mathrm{\omega t}\right)$, and the current density is $\mathrm{I}=\frac{2{\mathrm{\pi \epsilon }}_0}{{\mathrm{\mu }}_0\mathrm{c}}\cos \left(\mathrm{kz}-\mathrm{\omega t}\right)$.