91影视

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

$u=\frac{1}{2}(蔚E^2+\frac{1}{渭}{B}^2)$ 鈥︹�� (1)

Here,$蔚$ is the permittivity of free space, E is the electric field, $渭$Is the permeability of free space, and B is the magnetic field.

(a)

Write the expression for the electric field.

$\tilde{E}(z,t)=E_0{e}^{鈭�kz}\cos (kz鈭�蝇t+{未}_E)\widehat{x}$

Squaring on both sides.

$E^2=E_0^2{e}^{鈭�2kz}{\cos }^2(kz鈭�蝇t+{未}_E)$

Write the expression for the magnetic field.

$\tilde{B}(z,t)=B_0{e}^{鈭�kz}\cos (kz鈭�蝇t+{未}_E+蠒)\widehat{y}$

Squaring on both sides.

$B^2=B_0^2{e}^{鈭�2kz}{\cos }^2(kz鈭�蝇t+{未}_E+蠒)$

Substitute$E^2=E_0^2{e}^{鈭�2kz}{\cos }^2(kz鈭�蝇t+{未}_E)$ and$B^2=B_0^2{e}^{鈭�2kz}{\cos }^2(kz鈭�蝇t+{未}_E+蠒)$ in equation (1).

$\begin{array}{c}u=\frac{1}{2}(蔚E_0^2{e}^{鈭�2kz}{\cos }^2(kz鈭�蝇t+{未}_E)+\frac{1}{渭}{B}_0^2{e}^{鈭�2kz}{\cos }^2(kz鈭�蝇t+{未}_E+蠒))\\ u=\frac{1}{2}{e}^{鈭�2kz}[蔚E_0^2{\cos }^2(kz鈭�蝇t+{未}_E)+\frac{B_0^2}{渭}{\cos }^2(kz鈭�蝇t+{未}_E+蠒)]\\ u=\frac{1}{2}{e}^{鈭�2kz}[\frac{蔚}{2}{E}_0^2+\frac{1}{2渭}{B}_0^2]\end{array}$

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

$B_0=E_0\sqrt{蔚渭\sqrt{1+{\left(\frac{蟽}{蔚蝇}\right)}^2}}$

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

$\begin{array}{c}u=\frac{1}{2}{e}^{鈭�2kz}[\frac{蔚}{2}{E}_0^2+\frac{1}{2渭}{\left(E_0\sqrt{蔚渭\sqrt{1+{\left(\frac{蟽}{蔚蝇}\right)}^2}}\right)}^2]\\ u=\frac{1}{4}{e}^{鈭�2kz}蔚E_0^2\frac{2}{蔚渭}\frac{k^2}{{蝇}^2}\\ u=\frac{k^2}{2渭{蝇}^2}{E}_0^2{e}^{鈭�2kz}\end{array}$

Write the expression for the magnetic energy.

$u_{\text{mag}}=\left(\frac{B_0^2}{2渭}\right)\frac{1}{2}{e}^{鈭�2kz}$

Write the expression for the density electrostatic energy density.

$u_{\text{elec}}=\frac{1}{2}{e}^{鈭�2kz}\frac{蔚}{2}{E}_0^2$

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

$\begin{array}{c}\frac{u_{\text{mag}}}{u_{\text{elec}}}=\frac{\left(\frac{B_0^2}{2渭}\right)\frac{1}{2}{e}^{鈭�2kz}}{\frac{1}{2}{e}^{鈭�2kz}\frac{蔚}{2}{E}_0^2}\\ \frac{u_{\text{mag}}}{u_{\text{elec}}}=\frac{\left(\frac{B_0^2}{渭}\right)}{蔚E_0^2}\\ \frac{u_{\text{mag}}}{u_{\text{elec}}}=\frac{1}{蔚渭}蔚渭\sqrt{1+{\left(\frac{蟽}{蔚蝇}\right)}^2}\end{array}$

On further solving,

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

Hence, the magnetic contribution is always the dominator.

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

(b)

Write the expression for intensity.

$I=cu$

Substitute$u=\frac{k^2}{2渭{蝇}^2}{E}_0^2{e}^{鈭�2kz}$ and $c=\frac{蝇}{k}$in the above expression.

$\begin{array}{c}I=\frac{k^2}{2渭{蝇}^2}{E}_0^2{e}^{鈭�2kz}\frac{蝇}{k}\\ I=\left(\frac{蝇}{k}\right)\frac{k^2}{2渭{蝇}^2}{E}_0^2{e}^{鈭�2kz}\\ I=\frac{k}{2渭蝇}{E}_0^2{e}^{鈭�2kz}\end{array}$

Therefore, it is proved that $I=\frac{k}{2渭蝇}{E}_0^2{e}^{鈭�2kz}$.