91Ó°ÊÓ

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

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

Here,$\mathrm{Î\mu }$ is the permittivity of free space, E is the electric field, $\mathrm{μ}$Is the permeability of free space, and B is the magnetic field.

(a)

Write the expression for the electric field.

$\tilde{E}(z,t)=E_0{e}^{−kz}\cos (kz−\mathrm{Ó\neg }t+{\mathrm{δ}}_E)\widehat{x}$

Squaring on both sides.

$E^2=E_0^2{e}^{−2kz}{\cos }^2(kz−\mathrm{Ó\neg }t+{\mathrm{δ}}_E)$

Write the expression for the magnetic field.

$\tilde{B}(z,t)=B_0{e}^{−kz}\cos (kz−\mathrm{Ó\neg }t+{\mathrm{δ}}_E+\mathrm{Ï•})\widehat{y}$

Squaring on both sides.

$B^2=B_0^2{e}^{−2kz}{\cos }^2(kz−\mathrm{Ó\neg }t+{\mathrm{δ}}_E+\mathrm{Ï•})$

Substitute$E^2=E_0^2{e}^{−2kz}{\cos }^2(kz−\mathrm{Ó\neg }t+{\mathrm{δ}}_E)$ and$B^2=B_0^2{e}^{−2kz}{\cos }^2(kz−\mathrm{Ó\neg }t+{\mathrm{δ}}_E+\mathrm{Ï•})$ in equation (1).

$\begin{array}{c}u=\frac{1}{2}(\mathrm{Î\mu }{E}_0^2{e}^{−2kz}{\cos }^2(kz−\mathrm{Ó\neg }t+{\mathrm{δ}}_E)+\frac{1}{\mathrm{μ}}{B}_0^2{e}^{−2kz}{\cos }^2(kz−\mathrm{Ó\neg }t+{\mathrm{δ}}_E+\mathrm{Ï•}))\\ u=\frac{1}{2}{e}^{−2kz}[\mathrm{Î\mu }{E}_0^2{\cos }^2(kz−\mathrm{Ó\neg }t+{\mathrm{δ}}_E)+\frac{B_0^2}{\mathrm{μ}}{\cos }^2(kz−\mathrm{Ó\neg }t+{\mathrm{δ}}_E+\mathrm{Ï•})]\\ u=\frac{1}{2}{e}^{−2kz}[\frac{\mathrm{Î\mu }}{2}{E}_0^2+\frac{1}{2\mathrm{μ}}{B}_0^2]\end{array}$

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

$B_0=E_0\sqrt{\mathrm{Î\mu }\mathrm{μ}\sqrt{1+{\left(\frac{\mathrm{σ}}{\mathrm{Î\mu }\mathrm{Ó\neg }}\right)}^2}}$

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

$\begin{array}{c}u=\frac{1}{2}{e}^{−2kz}[\frac{\mathrm{Î\mu }}{2}{E}_0^2+\frac{1}{2\mathrm{μ}}{\left(E_0\sqrt{\mathrm{Î\mu }\mathrm{μ}\sqrt{1+{\left(\frac{\mathrm{σ}}{\mathrm{Î\mu }\mathrm{Ó\neg }}\right)}^2}}\right)}^2]\\ u=\frac{1}{4}{e}^{−2kz}\mathrm{Î\mu }{E}_0^2\frac{2}{\mathrm{Î\mu }\mathrm{μ}}\frac{k^2}{{\mathrm{Ó\neg }}^2}\\ u=\frac{k^2}{2\mathrm{μ}{\mathrm{Ó\neg }}^2}{E}_0^2{e}^{−2kz}\end{array}$

Write the expression for the magnetic energy.

$u_{\text{mag}}=\left(\frac{B_0^2}{2\mathrm{μ}}\right)\frac{1}{2}{e}^{−2kz}$

Write the expression for the density electrostatic energy density.

$u_{\text{elec}}=\frac{1}{2}{e}^{−2kz}\frac{\mathrm{Î\mu }}{2}{E}_0^2$

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

$\begin{array}{c}\frac{u_{\text{mag}}}{u_{\text{elec}}}=\frac{\left(\frac{B_0^2}{2\mathrm{μ}}\right)\frac{1}{2}{e}^{−2kz}}{\frac{1}{2}{e}^{−2kz}\frac{\mathrm{Î\mu }}{2}{E}_0^2}\\ \frac{u_{\text{mag}}}{u_{\text{elec}}}=\frac{\left(\frac{B_0^2}{\mathrm{μ}}\right)}{\mathrm{Î\mu }{E}_0^2}\\ \frac{u_{\text{mag}}}{u_{\text{elec}}}=\frac{1}{\mathrm{Î\mu }\mathrm{μ}}\mathrm{Î\mu }\mathrm{μ}\sqrt{1+{\left(\frac{\mathrm{σ}}{\mathrm{Î\mu }\mathrm{Ó\neg }}\right)}^2}\end{array}$

On further solving,

$\begin{array}{c}\frac{u_{\text{mag}}}{u_{\text{elec}}}=\sqrt{1+{\left(\frac{\mathrm{σ}}{\mathrm{Î\mu }\mathrm{Ó\neg }}\right)}^2}>1\\ u_{\text{mag}}>u_{\text{elec}}\end{array}$

Hence, the magnetic contribution is always the dominator.

Therefore, the energy density of an electromagnetic plane wave in a conductive medium is$u=\frac{k^2}{2\mathrm{μ}{\mathrm{Ó\neg }}^2}{E}_0^2{e}^{−2kz}$ .

(b)

Write the expression for intensity.

$I=cu$

Substitute$u=\frac{k^2}{2\mathrm{μ}{\mathrm{Ó\neg }}^2}{E}_0^2{e}^{−2kz}$ and $c=\frac{\mathrm{Ó\neg }}{k}$in the above expression.

$\begin{array}{c}I=\frac{k^2}{2\mathrm{μ}{\mathrm{Ó\neg }}^2}{E}_0^2{e}^{−2kz}\frac{\mathrm{Ó\neg }}{k}\\ I=\left(\frac{\mathrm{Ó\neg }}{k}\right)\frac{k^2}{2\mathrm{μ}{\mathrm{Ó\neg }}^2}{E}_0^2{e}^{−2kz}\\ I=\frac{k}{2\mathrm{μ}\mathrm{Ó\neg }}{E}_0^2{e}^{−2kz}\end{array}$

Therefore, it is proved that $I=\frac{k}{2\mathrm{μ}\mathrm{Ó\neg }}{E}_0^2{e}^{−2kz}$.