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).

$$\begin{gathered} \mathit{E}\mathbf{(}\mathit{s}\mathbf{,}\mathit{Ï•}\mathbf{,}\mathit{z}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\frac{\mathbf{A}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{k}\mathbf{z}\mathbf{-}\mathbf{Ó\neg }\mathbf{t}\mathbf{)}}{\mathbf{s}}\hat{\mathbf{s}} \\ \mathit{B}\mathbf{(}\mathit{s}\mathbf{,}\mathit{Ï•}\mathbf{,}\mathit{z}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\frac{\mathbf{A}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{k}\mathbf{z}\mathbf{-}\mathbf{Ó\neg }\mathbf{t}\mathbf{)}}{\mathbf{c}\mathbf{s}}\widehat{\mathbf{Ï•}} \end{gathered}$$

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,$\mathrm{Ï•}$is the phase andt is the time.

(a)

Prove $∇â‹\dots E=0$as follow:

$∇·E=\frac{1}{s}\frac{∂}{∂s}\left(s{E}_s\right)+\frac{1}{s}\frac{∂E_{\mathrm{ϕ}}}{∂\mathrm{ϕ}}+\frac{∂E_z}{∂z}$ …… (1)

Here, the value of , localid="1657501054364" $E_s,E_{\mathrm{Ï•}}$and localid="1657501101055" $E_z$is given as:

localid="1657501166011" $E_s=\frac{A\cos (kx-\mathrm{Ó\neg }t)}{s}$

$E_{\mathrm{Ï•}}=0$

$E_z=0$

Substitute localid="1657500893088" $E_s=\frac{A\cos (t\mathrm{Î\pm }-\mathrm{Ó\neg }t)}{s},E_{\mathrm{Ï•}}=0\text{and}{E}_z=0$, in equation (1).

Prove, $∇·B=0$

$∇·B=\frac{1}{s}\frac{∂}{∂s}\left(s{B}_s\right)+\frac{1}{s}\frac{∂B_{\mathrm{ϕ}}}{∂\mathrm{ϕ}}+\frac{∂B_z}{∂z}$ …… (2)

Here, the value of localid="1657502944630" $B_s,B_{\mathrm{Ï•}}$,and $B_z$is given as:

$$\begin{gathered} B_s=0 \\ {B}_{\mathrm{Ï•}}=\frac{A\cos (kz-\mathrm{Ó\neg }t)}{s} \\ {B}_z=0 \end{gathered}$$

Substitute $B_s=0,B_{\mathrm{Ï•}}=\frac{A\cos (kz-\mathrm{Ó\neg }t)}{s}\text{and}{B}_z=0$, in equation (2).

$∇·B=\frac{1}{s}\frac{∂}{∂s}(s×0)+\frac{1}{s}\frac{∂\left(\frac{A\cos (kz-\mathrm{Ó\neg }t)}{s}\right)}{∂\mathrm{Ï•}}+0$

$∇â‹\dots B=0$

Prove $∇×E=\left(\frac{1}{s}\left(\frac{∂E_z}{∂\mathrm{ϕ}}-\frac{∂E_{\mathrm{ϕ}}}{∂z}\right)\hat{s}+\left(\frac{∂E_s}{∂z}-\frac{∂E_z}{∂s}\right)\widehat{\mathrm{ϕ}}\right)+\frac{1}{s}\left[\frac{∂}{∂s}\left(s{E}_{\mathrm{ϕ}}\right)-\frac{∂E_s}{∂\mathrm{ϕ}}\right]\hat{z}$

$∇×E=0+\frac{∂E_s}{∂z}\widehat{\mathrm{ϕ}}-\frac{1}{s}\frac{∂E_s}{∂\mathrm{ϕ}}\hat{z}$

$$\begin{gathered} ∇×E=\frac{-E_{0}k\sin (kz-\mathrm{Ó\neg }t)}{s}\widehat{\mathrm{Ï•}} \\ ∇×E=-\frac{∂B}{∂t} \end{gathered}$$

Prove $∇×B=\frac{1}{c^2}\frac{∂E}{∂t}$

$$\begin{gathered} ∇×B=\left(\frac{1}{s}\left(\frac{∂B_z}{∂\mathrm{Ï•}}-\frac{∂B_{\mathrm{Ï•}}}{∂z}\right)\hat{s}+\left(\frac{∂B_s}{∂z}-\frac{∂B_z}{∂s}\right)\widehat{\mathrm{Ï•}}\right)+\frac{1}{s}\left[\frac{∂}{∂s}\left(s{B}_{\mathrm{Ï•}}\right)-\frac{∂B_s}{∂\mathrm{Ï•}}\right]\hat{z} \\ ∇×B=-\frac{∂B_{\mathrm{Ï•}}}{∂z}+\frac{1}{s}\frac{∂}{∂s}\left(s{B}_{\mathrm{Ï•}}\right)\hat{z} \\ ∇×B=\frac{E_{0}k}{c}\frac{\sin (kz-\mathrm{Ó\neg }t)}{s}\hat{s} \\ ∇×B=\frac{1}{c^2}\frac{∂E}{∂t} \end{gathered}$$

Satisfy the boundary conditions.

$$\begin{gathered} E=E_z=0 \\ {B}^{âŠ\yen }=B_5=0 \end{gathered}$$

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

$$\begin{gathered} ∫E·da=\frac{E_0\cos (kz-\mathrm{Ó\neg }t)}{s}\left(2\mathrm{Ï€}s\right)dz \\ \frac{E_0\cos (lz-\mathrm{Ó\neg }t)}{s}\left(2\mathrm{Ï€}s\right)dz=\frac{Q_{\text{enclaged}}}{{\mathrm{Î\mu }}_0}âˆ\pounds \\ {E}_0\cos (kz-\mathrm{Ó\neg }t)\left(2\mathrm{Ï€}\right)dz=\frac{\mathrm{λ}dz}{{\mathrm{Î\mu }}_0} \end{gathered}$$

Hence, the charge density will be,

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

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

$$\begin{gathered} ∫B·dl=\frac{E_0}{c}\frac{\cos (kz-\mathrm{Ó\neg }t)}{s}\left(2\mathrm{Ï€}s\right) \\ {∫}_{1}B·dl={\mathrm{μ}}_0{l}_{\text{er1dased}} \end{gathered}$$

Hence, the current density will be,

$I=\frac{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}{{\mathrm{μ}}_{0}C}\cos (kz-\mathrm{Ó\neg }t)$

Therefore, the charge density is $\mathrm{λ}(z,t)=2\mathrm{Ï€}{\mathrm{Î\mu }}_0{E}_0\cos (kz-\mathrm{Ó\neg }t)$, and the current density is $I=\frac{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}{{\mathrm{μ}}_{0}c}\cos (kz-\mathrm{Ó\neg }t)$.