(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, λ(z, t), and the current, I (z, t), on the inner conductor.

Question

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

[latex]\left.\begin{array}{l} E (s, \phi, z, t)=\frac{A \cos (k z-\omega t)}{s} \hat{ s } \B (s, \phi, z, t)=\frac{A \cos (k z-\omega t)}{c s} \hat{\phi}\end{array} ight\}[/latex] (9.197)

[latex]\left. \begin{matrix} ext { (i) } abla \cdot E =0 ext {, } & ext { (iii) } abla imes E =-\frac{\partial B }{\partial t}, \ ext { (ii) } abla \cdot B =0 ext {, } & ext { (iv) } abla imes B =\frac{1}{c^{2}} \frac{\partial E }{\partial t} ext {. } \end{matrix} ight\} [/latex] (9.177)

[latex]\left.\begin{array}{l} ext { (i) } \quad E ^{\|}= 0 \ ext { (ii) } B^{\perp}=0\end{array} ight\}[/latex] (9.175)
(b) Find the charge density, λ(z, t), and the current, I (z, t), on the inner conductor.

Accepted Answer

[latex] ext { (a) } abla \cdot E =\frac{1}{s} \frac{\partial}{\partial s}\left(s E_{s} ight)=0 ; abla \cdot B =\frac{1}{s} \frac{\partial}{\partial \phi}\left(B_{\phi} ight)=0 ; abla imes E =\frac{\partial E_{s}}{\partial z} \hat{ \phi }-\frac{1}{s} \frac{\partial E_{s}}{\partial \phi} \hat{ z }=-\frac{E_{0} k \sin (k z-\omega t)}{s} \hat{\phi} \stackrel{?}{=}[/latex]

[latex]-\frac{\partial B }{\partial t}=-\frac{E_{0} \omega}{c} \frac{\sin (k z-\omega t)}{s} \hat{ \phi } ( ext { since } k=\omega / c) ; abla imes B =-\frac{\partial B_{\phi}}{\partial z} \hat{ s }+\frac{1}{s} \frac{\partial}{\partial s}\left(s B_{\phi} ight) \hat{ z }=\frac{E_{0} k}{c} \frac{\sin (k z-\omega t)}{s} \hat{ s } \stackrel{?}{=}[/latex]

[latex]\frac{1}{c^{2}} \frac{\partial E }{\partial t}=\frac{E_{0} \omega}{c^{2}} \frac{\sin (k z-\omega t)}{s} \hat{ s } . ext { Boundary conditions: } E^{\|}=E_{z}=0 ; B^{\perp}=B_{s}=0[/latex].

(b) To determine λ, use Gauss’s law for a cylinder of radius s and length dz:

[latex]\oint E \cdot d a =E_{0} \frac{\cos (k z-\omega t)}{s}(2 \pi s) d z=\frac{1}{\epsilon_{0}} Q_{ enc }=\frac{1}{\epsilon_{0}} \lambda d z \Rightarrow \lambda=2 \pi \epsilon_{0} E_{0} \cos (k z-\omega t)[/latex]

To determine I, use [latex] ext { Ampére's }[/latex] law for a circle of radius s (note that the displacement current through this loop is zero, since E is in the [latex]\hat{ S }[/latex] direction): [latex]\oint B \cdot d l =\frac{E_{0}}{c} \frac{\cos (k z-\omega t)}{s}(2 \pi s)=\mu_{0} I_{ enc } \Rightarrow I=\frac{2 \pi E_{0}}{\mu_{0} c} \cos (k z-\omega t)[/latex] .

The charge and current on the outer conductor are precisely the opposite of these, since E = B = 0 inside the metal, and hence the total enclosed charge and current must be zero.

Question 9.32: (a) Show directly that Eqs. 9.197 satisfy Maxwell’s equation...

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