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, $\mathit{\lambda }\mathbf{(}\mathit{z}\mathbf{,}\mathit{t}\mathbf{)}$, and the current,$\mathit{I}\mathbf{(}\mathit{z}\mathbf{,}\mathit{t}\mathbf{)}$ , on the inner conductor.

Short answer

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

The charge density is $\lambda (z,t)=2\pi {\epsilon }_0{E}_0\cos (kz-\omega t)$,and the current density is$I=\frac{2\pi {\epsilon }_0}{{\mu }_{0}C}\cos (kz-\omega t)$

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).

$$\begin{gathered} \mathit{E}\mathbf{(}\mathit{s}\mathbf{,}\mathit{\varphi }\mathbf{,}\mathit{z}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\frac{\mathbf{A}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{k}\mathbf{z}\mathbf{-}\mathbf{\omega }\mathbf{t}\mathbf{)}}{\mathbf{s}}\hat{\mathbf{s}} \\ \mathit{B}\mathbf{(}\mathit{s}\mathbf{,}\mathit{\varphi }\mathbf{,}\mathit{z}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\frac{\mathbf{A}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{k}\mathbf{z}\mathbf{-}\mathbf{\omega }\mathbf{t}\mathbf{)}}{\mathbf{c}\mathbf{s}}\hat{\mathbf{\varphi }} \end{gathered}$$

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,$\varphi$is the phase andt is the time.

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

(a)

Prove $\nabla \cdot E=0$as follow:

$\nabla ·E=\frac{1}{s}\frac{\partial }{\partial s}\left(s{E}_s\right)+\frac{1}{s}\frac{\partial E_{\varphi }}{\partial \varphi }+\frac{\partial E_z}{\partial z}$ …… (1)

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

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

$E_{\varphi }=0$

$E_z=0$

Substitute localid="1657500893088" $E_s=\frac{A\cos (t\alpha -\omega t)}{s},E_{\varphi }=0\text{and}{E}_z=0$, in equation (1).

Prove, $\nabla ·B=0$

$\nabla ·B=\frac{1}{s}\frac{\partial }{\partial s}\left(s{B}_s\right)+\frac{1}{s}\frac{\partial B_{\varphi }}{\partial \varphi }+\frac{\partial B_z}{\partial z}$ …… (2)

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

$$\begin{gathered} B_s=0 \\ {B}_{\varphi }=\frac{A\cos (kz-\omega t)}{s} \\ {B}_z=0 \end{gathered}$$

Substitute $B_s=0,B_{\varphi }=\frac{A\cos (kz-\omega t)}{s}\text{and}{B}_z=0$, in equation (2).

$\nabla ·B=\frac{1}{s}\frac{\partial }{\partial s}(s\times 0)+\frac{1}{s}\frac{\partial \left(\frac{A\cos (kz-\omega t)}{s}\right)}{\partial \varphi }+0$

$\nabla \cdot B=0$

Prove $\nabla \times E=\left(\frac{1}{s}\left(\frac{\partial E_z}{\partial \varphi }-\frac{\partial E_{\varphi }}{\partial z}\right)\hat{s}+\left(\frac{\partial E_s}{\partial z}-\frac{\partial E_z}{\partial s}\right)\hat{\varphi }\right)+\frac{1}{s}\left[\frac{\partial }{\partial s}\left(s{E}_{\varphi }\right)-\frac{\partial E_s}{\partial \varphi }\right]\hat{z}$

$\nabla \times E=0+\frac{\partial E_s}{\partial z}\hat{\varphi }-\frac{1}{s}\frac{\partial E_s}{\partial \varphi }\hat{z}$

$$\begin{gathered} \nabla \times E=\frac{-E_{0}k\sin (kz-\omega t)}{s}\hat{\varphi } \\ \nabla \times E=-\frac{\partial B}{\partial t} \end{gathered}$$

Prove $\nabla \times B=\frac{1}{c^2}\frac{\partial E}{\partial t}$

$$\begin{gathered} \nabla \times B=\left(\frac{1}{s}\left(\frac{\partial B_z}{\partial \varphi }-\frac{\partial B_{\varphi }}{\partial z}\right)\hat{s}+\left(\frac{\partial B_s}{\partial z}-\frac{\partial B_z}{\partial s}\right)\hat{\varphi }\right)+\frac{1}{s}\left[\frac{\partial }{\partial s}\left(s{B}_{\varphi }\right)-\frac{\partial B_s}{\partial \varphi }\right]\hat{z} \\ \nabla \times B=-\frac{\partial B_{\varphi }}{\partial z}+\frac{1}{s}\frac{\partial }{\partial s}\left(s{B}_{\varphi }\right)\hat{z} \\ \nabla \times B=\frac{E_{0}k}{c}\frac{\sin (kz-\omega t)}{s}\hat{s} \\ \nabla \times B=\frac{1}{c^2}\frac{\partial E}{\partial t} \end{gathered}$$

Satisfy the boundary conditions.

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

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

$$\begin{gathered} \int E·da=\frac{E_0\cos (kz-\omega t)}{s}\left(2\pi s\right)dz \\ \frac{E_0\cos (lz-\omega t)}{s}\left(2\pi s\right)dz=\frac{Q_{\text{enclaged}}}{{\epsilon }_0}\mid \\ {E}_0\cos (kz-\omega t)\left(2\pi \right)dz=\frac{\lambda dz}{{\epsilon }_0} \end{gathered}$$

Hence, the charge density will be,

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

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

$$\begin{gathered} \int B·dl=\frac{E_0}{c}\frac{\cos (kz-\omega t)}{s}\left(2\pi s\right) \\ {\int }_{1}B·dl={\mu }_0{l}_{\text{er1dased}} \end{gathered}$$

Hence, the current density will be,

$I=\frac{2\pi {\epsilon }_0}{{\mu }_{0}C}\cos (kz-\omega t)$

Therefore, the charge density is $\lambda (z,t)=2\pi {\epsilon }_0{E}_0\cos (kz-\omega t)$, and the current density is $I=\frac{2\pi {\epsilon }_0}{{\mu }_{0}c}\cos (kz-\omega t)$.