Question

(a) Calculate the (time-averaged) energy density of an electromagnetic plane wave in a conducting medium (Eq. 9.138). Show that the magnetic contribution always dominates.

(b) Show that the intensity is$\mathbf{\left(}\frac{\mathbf{k}}{\mathbf{2}\mathbf{\mu }\mathbf{\omega }}\mathbf{\right)}{\mathit{E}}_{\mathbf{0}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{z}}$

Short answer

(a) The energy density of an electromagnetic plane wave in a conductive medium is$u=\frac{k^2}{2\mu {\omega }^2}{E}_0^2{e}^{-2kz}$ .

(b) It is proved that the intensity is $\left(\frac{k}{2\mu \omega }\right)E_0^2{e}^{-2xz}$.

Step-by-step solution

Expression for the energy density of an electromagnetic plane wave:

Write the expression for the energy density of an electromagnetic plane wave.

$u=\frac{1}{2}(\epsilon E^2+\frac{1}{\mu }{B}^2)$ …… (1)

Here,$\epsilon$ is the permittivity of free space, E is the electric field, $\mu$Is the permeability of free space, and B is the magnetic field.

Determine the energy density of an electromagnetic plane wave in a conductive medium:

(a)

Write the expression for the electric field.

$\tilde{E}(z,t)=E_0{e}^{-kz}\cos (kz-\omega t+{\delta }_E)\widehat{x}$

Squaring on both sides.

$E^2=E_0^2{e}^{-2kz}{\cos }^2(kz-\omega t+{\delta }_E)$

Write the expression for the magnetic field.

$\tilde{B}(z,t)=B_0{e}^{-kz}\cos (kz-\omega t+{\delta }_E+\varphi )\widehat{y}$

Squaring on both sides.

$B^2=B_0^2{e}^{-2kz}{\cos }^2(kz-\omega t+{\delta }_E+\varphi )$

Substitute$E^2=E_0^2{e}^{-2kz}{\cos }^2(kz-\omega t+{\delta }_E)$ and$B^2=B_0^2{e}^{-2kz}{\cos }^2(kz-\omega t+{\delta }_E+\varphi )$ in equation (1).

$\begin{array}{c}u=\frac{1}{2}(\epsilon E_0^2{e}^{-2kz}{\cos }^2(kz-\omega t+{\delta }_E)+\frac{1}{\mu }{B}_0^2{e}^{-2kz}{\cos }^2(kz-\omega t+{\delta }_E+\varphi ))\\ u=\frac{1}{2}{e}^{-2kz}[\epsilon E_0^2{\cos }^2(kz-\omega t+{\delta }_E)+\frac{B_0^2}{\mu }{\cos }^2(kz-\omega t+{\delta }_E+\varphi )]\\ u=\frac{1}{2}{e}^{-2kz}[\frac{\epsilon }{2}{E}_0^2+\frac{1}{2\mu }{B}_0^2]\end{array}$

Using the relation between $B_0$and$E_0$ , it is known that:

$B_0=E_0\sqrt{\epsilon \mu \sqrt{1+{\left(\frac{\sigma }{\epsilon \omega }\right)}^2}}$

Calculate the energy density of an electromagnetic plane wave in a conductive medium.

$\begin{array}{c}u=\frac{1}{2}{e}^{-2kz}[\frac{\epsilon }{2}{E}_0^2+\frac{1}{2\mu }{\left(E_0\sqrt{\epsilon \mu \sqrt{1+{\left(\frac{\sigma }{\epsilon \omega }\right)}^2}}\right)}^2]\\ u=\frac{1}{4}{e}^{-2kz}\epsilon E_0^2\frac{2}{\epsilon \mu }\frac{k^2}{{\omega }^2}\\ u=\frac{k^2}{2\mu {\omega }^2}{E}_0^2{e}^{-2kz}\end{array}$

Show that the magnetic contribution always dominates:

Write the expression for the magnetic energy.

$u_{\text{mag}}=\left(\frac{B_0^2}{2\mu }\right)\frac{1}{2}{e}^{-2kz}$

Write the expression for the density electrostatic energy density.

$u_{\text{elec}}=\frac{1}{2}{e}^{-2kz}\frac{\epsilon }{2}{E}_0^2$

Take the ratio of magnetic energy density and electrostatic energy density.

$\begin{array}{c}\frac{u_{\text{mag}}}{u_{\text{elec}}}=\frac{\left(\frac{B_0^2}{2\mu }\right)\frac{1}{2}{e}^{-2kz}}{\frac{1}{2}{e}^{-2kz}\frac{\epsilon }{2}{E}_0^2}\\ \frac{u_{\text{mag}}}{u_{\text{elec}}}=\frac{\left(\frac{B_0^2}{\mu }\right)}{\epsilon E_0^2}\\ \frac{u_{\text{mag}}}{u_{\text{elec}}}=\frac{1}{\epsilon \mu }\epsilon \mu \sqrt{1+{\left(\frac{\sigma }{\epsilon \omega }\right)}^2}\end{array}$

On further solving,

$\begin{array}{c}\frac{u_{\text{mag}}}{u_{\text{elec}}}=\sqrt{1+{\left(\frac{\sigma }{\epsilon \omega }\right)}^2}>1\\ u_{\text{mag}}>u_{\text{elec}}\end{array}$

Hence, the magnetic contribution is always the dominator.

Therefore, the energy density of an electromagnetic plane wave in a conductive medium is$u=\frac{k^2}{2\mu {\omega }^2}{E}_0^2{e}^{-2kz}$ .

Show that the intensity is (k2μω)E02e-2xz(k2μω)E02e-2xz:

(b)

Write the expression for intensity.

$I=cu$

Substitute$u=\frac{k^2}{2\mu {\omega }^2}{E}_0^2{e}^{-2kz}$ and $c=\frac{\omega }{k}$in the above expression.

$\begin{array}{c}I=\frac{k^2}{2\mu {\omega }^2}{E}_0^2{e}^{-2kz}\frac{\omega }{k}\\ I=\left(\frac{\omega }{k}\right)\frac{k^2}{2\mu {\omega }^2}{E}_0^2{e}^{-2kz}\\ I=\frac{k}{2\mu \omega }{E}_0^2{e}^{-2kz}\end{array}$

Therefore, it is proved that $I=\frac{k}{2\mu \omega }{E}_0^2{e}^{-2kz}$.