91Ó°ÊÓ

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).

$\mathbf{E}\mathbf{(}\mathbf{s}\mathbf{,}\mathbf{Ï•}\mathbf{,}\mathbf{z}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{Acos}\mathbf{(}\mathbf{kz}\mathbf{-}\mathbf{Ó\neg ³Ù}\mathbf{)}\mathbf{}}{\mathbf{s}}\hat{\mathbf{s}}$

$\mathbf{B}\mathbf{(}\mathbf{s}\mathbf{,}\mathbf{Ï•}\mathbf{,}\mathbf{z}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{Acos}\mathbf{(}\mathbf{kz}\mathbf{-}\mathbf{Ó\neg ³Ù}\mathbf{)}\mathbf{}}{\mathbf{cs}}\widehat{\mathbf{Ï•}}$

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.

(a)

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

$\stackrel{¯}{\mathrm{V}}.E=\frac{1}{s}\frac{∂}{∂s}\left(s{E}_s\right)+\frac{1}{s}\frac{∂E_{\mathrm{ϕ}}}{∂\mathrm{ϕ}}+\frac{∂E_z}{∂\mathrm{z}}$ …… (1)

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

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

Substitute ${\mathrm{E}}_{\mathrm{s}}=\frac{\mathrm{Acos}\left(\mathrm{kx}-\mathrm{Ó\neg ³Ù}\right)}{\mathrm{s}}$,${\mathrm{E}}_{\mathrm{Ï•}}=0$ and${\mathrm{E}}_{\mathrm{z}}=0$ in equation (1).

$$\begin{gathered} \stackrel{¯}{\mathrm{V}}.E=\frac{1}{s}\frac{∂}{∂s}\left(s\frac{A\cos \left(kx-\mathrm{Ó\neg }t\right)}{s}\right)+0+0 \\ \stackrel{¯}{\mathrm{V}}.E=0 \end{gathered}$$

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

$\stackrel{¯}{\mathrm{V}}.\mathrm{B}=\frac{1}{\mathrm{s}}\frac{∂}{∂\mathrm{s}}\left({\mathrm{sB}}_{\mathrm{s}}\right)+\frac{1}{\mathrm{s}}\frac{∂{\mathrm{B}}_{\mathrm{ϕ}}}{∂\mathrm{ϕ}}+\frac{1}{\mathrm{s}}\frac{∂{\mathrm{B}}_{\mathrm{z}}}{∂\mathrm{z}}$ …… (2)

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

$$\begin{gathered} {\mathrm{B}}_{\mathrm{s}}=0 \\ {\mathrm{B}}_{\mathrm{Ï•}}=\frac{\mathrm{Acos}\left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)}{\mathrm{s}} \\ {\mathrm{B}}_{\mathrm{z}}=0 \end{gathered}$$

Substitute ${\mathrm{B}}_{\mathrm{s}}=0$, ${\mathrm{B}}_{\mathrm{Ï•}}=\frac{\mathrm{Acos}\left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)}{\mathrm{s}}$and ${\mathrm{B}}_{\mathrm{z}}=0$in equation (2).

$$\begin{gathered} \stackrel{¯}{\mathrm{V}}.\mathrm{B}=\frac{1}{\mathrm{s}}\frac{∂}{∂\mathrm{s}}\left(\mathrm{s}×0\right)+\frac{1}{\mathrm{s}}\frac{∂\left({\displaystyle \frac{\mathrm{Acos}\left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)}{\mathrm{s}}}\right)}{∂\mathrm{Ï•}}+0 \\ \stackrel{¯}{\mathrm{V}}.\mathrm{B}=0 \end{gathered}$$

Prove $\stackrel{¯}{\mathrm{V}}×E=-\frac{∂B}{∂t}$:

$$\begin{gathered} \stackrel{¯}{\mathrm{V}}×\mathrm{E}=\left(\frac{1}{\mathrm{s}}\left(\frac{∂{\mathrm{E}}_{\mathrm{z}}}{∂\mathrm{Ï•}}-\frac{∂{\mathrm{E}}_{\mathrm{Ï•}}}{∂\mathrm{z}}\right)\hat{\mathrm{s}}+\left(\frac{∂{\mathrm{E}}_{\mathrm{s}}}{∂\mathrm{z}}-\frac{∂{\mathrm{E}}_{\mathrm{z}}}{∂\mathrm{s}}\right)\widehat{\mathrm{Ï•}}\right)+\frac{1}{\mathrm{s}}\left[\frac{∂}{∂\mathrm{s}}\left({\mathrm{sE}}_{\mathrm{Ï•}}\right)-\frac{∂{\mathrm{E}}_{\mathrm{s}}}{∂\mathrm{Ï•}}\right]\hat{\mathrm{z}} \\ \stackrel{¯}{\mathrm{V}}×\mathrm{E}=0+\frac{∂{\mathrm{E}}_{\mathrm{s}}}{∂\mathrm{z}}\widehat{\mathrm{Ï•}}-\frac{1}{\mathrm{s}}\frac{∂{\mathrm{E}}_{\mathrm{s}}}{∂\mathrm{Ï•}}\hat{\mathrm{z}} \\ \stackrel{¯}{\mathrm{V}}×\mathrm{E}=\frac{-{\mathrm{E}}_0\mathrm{ksin}\left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)}{\mathrm{s}}\widehat{\mathrm{Ï•}} \\ \stackrel{¯}{\mathrm{V}}×\mathrm{E}=-\frac{∂\mathrm{B}}{∂\mathrm{t}} \end{gathered}$$

Prove $\stackrel{¯}{\mathrm{V}}×\mathrm{B}=\frac{1}{{\mathrm{c}}^2}\frac{∂\mathrm{E}}{∂\mathrm{t}}$:

$$\begin{gathered} \stackrel{¯}{\mathrm{V}}×B=\left(\frac{1}{s}\left(\frac{∂B_z}{∂\mathrm{Ï•}}-\frac{∂B_{\mathrm{Ï•}}}{∂z}\right)\hat{s}+\left(\frac{∂B_s}{∂\mathrm{z}}-\frac{∂B_z}{∂\mathrm{s}}\right)\widehat{\mathrm{Ï•}}\right)+\frac{1}{s}\left[\frac{∂}{∂s}\left(s{B}_{\mathrm{Ï•}}\right)-\frac{∂B_s}{∂\mathrm{Ï•}}\right]\hat{z} \\ \stackrel{¯}{\mathrm{V}}×B=-\frac{∂B_{\mathrm{Ï•}}}{∂z}+\frac{1}{s}\frac{∂}{∂s}\left(s{B}_{\mathrm{Ï•}}\right)\hat{z} \\ \stackrel{¯}{\mathrm{V}}×B=\frac{{\mathrm{E}}_0\mathrm{k}}{\mathrm{c}}\frac{\sin \left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)}{\mathrm{s}}\hat{\mathrm{s}} \\ \stackrel{¯}{\mathrm{V}}×B\mathit{=}\frac{\mathit{1}}{{\mathit{c}}^{\mathit{2}}}\frac{\mathit{∂}\mathit{E}}{\mathit{∂}\mathit{t}} \end{gathered}$$

Satisfy the boundary conditions.

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

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

(b)

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

$$\begin{gathered} \widehat{\mathrm{â„•}}E.da=\frac{E_0\cos \left(kz-\mathrm{Ó\neg }t\right)}{s}\left(2\mathrm{Ï€}s\right)dz \\ \frac{E_0\cos \left(kz-\mathrm{Ó\neg }t\right)}{s}\left(2\mathrm{Ï€}s\right)dz\mathit{=}\frac{{\mathit{Q}}_{\mathit{e}\mathit{n}\mathit{c}\mathit{l}\mathit{o}\mathit{s}\mathit{e}\mathit{d}}}{{\mathit{Î\mu }}_{\mathit{0}}} \\ {E}_{\mathit{0}}\cos \left(kz\mathrm{-}\mathrm{Ó\neg }t\right)\left(\mathrm{2}\mathrm{Ï€}\right)dz\mathit{=}\frac{\mathit{λ}\mathit{d}\mathit{z}}{{\mathit{Î\mu }}_{\mathit{0}}} \end{gathered}$$

Hence, the charge density will be,

$\mathrm{λ}\left(\mathrm{z},\mathrm{t}\right)=2\mathrm{Ï€}{\mathrm{Î\mu }}_0{E}_0\cos \left(kz-\mathrm{Ó\neg }t\right)$

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

$$\begin{gathered} \widehat{\mathrm{â„•}}B.dl=\frac{E_0}{c}\frac{\cos \left(kz-\mathrm{Ó\neg }t\right)}{s}\left(2\mathrm{Ï€}s\right) \\ \widehat{\mathrm{â„•}}B.dl={\mathrm{μ}}_0{\mathrm{I}}_{\mathrm{enclosed}} \end{gathered}$$

Hence, the current density will be,

$\mathrm{I}=\frac{2{\mathrm{Ï€Î\mu }}_0}{{\mathrm{μ}}_0\mathrm{c}}\cos \left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)$

Therefore, the charge density is $\mathrm{λ}\left(\mathrm{z},\mathrm{t}\right)=2{\mathrm{Ï€Î\mu }}_0{\mathrm{E}}_0\cos \left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)$, and the current density is $\mathrm{I}=\frac{2{\mathrm{Ï€Î\mu }}_0}{{\mathrm{μ}}_0\mathrm{c}}\cos \left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)$.