The electromagnetic wave equations in conducting media:
For electric field vector:
$ \nabla^{2}.\overrightarrow{E}-\mu \epsilon\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} - \sigma \mu \frac{\partial \overrightarrow{E}}{\partial t}=0 \qquad(1)$
For magnetic field vector:
$\nabla^{2}.\overrightarrow{H} - \mu \epsilon \frac{\partial^{2} H}{\partial t^{2}}-\sigma \mu \frac{\partial \overrightarrow{H}}{\partial t}=0 \qquad(2)$
The wave equation of electric field vector:
$\overrightarrow{E}(\overrightarrow{r},t)=E_{0} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$
The wave equation of magnetic field vector:
$\overrightarrow{H}(\overrightarrow{r},t)=H_{0} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$
Now the solution of electromagnetic wave for electric field vector.
Differentiate with respect to $t$ of equation $(3)$
$\frac{\partial \overrightarrow{E}}{\partial t}=i \omega E_{0} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$
Again differentiate with respect to $t$ of the above equation:
$\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}=i^{2} \omega^{2} E_{0} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$
$\frac{\partial^{2} \overrightarrow{E}}{\partial^{2} t}=- \omega^{2} \overrightarrow{E}(\overrightarrow{r},t)$
Now substitute the value of the above equation in equation$(1)$
$\nabla^{2} \overrightarrow{E}=-\omega^{2} \mu \epsilon \overrightarrow{E} - i \omega \mu \sigma \overrightarrow{E}$
$\nabla^{2} \overrightarrow{E}=- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) \overrightarrow{E}$
This is the solution of the electromagnetic wave equation in conducting media for the electric field vector.
Now component form of the above equation:
$(\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} +\frac{\partial^{2}}{\partial z^{2}})(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) =- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right)(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) \qquad(5)$
If the wave is propagating along $z$ direction. Then for uniform-plane electromagnetic waves-
$\frac{\partial}{\partial x}=\frac{\partial}{\partial y}=0$
$\frac{\partial^{2}}{\partial x^{2}}=\frac{\partial^{2}}{\partial y^{2}}=0$
$E_{z}=0$
Now the equation $(5)$ can be written as:
$\frac{\partial^{2}}{\partial x^{2}} (\hat{i}E_{x}+\hat{j}E_{y})=- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right)(\hat{i}E_{x}+\hat{j}E_{y})$
Now separate the above equation in $x$ and $y$ components so
$\left.\begin{matrix}
\frac{\partial^{2} E_{x}}{\partial z^{2}}=- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) E_{x}
\\
\frac{\partial^{2}E_{y}}{\partial z^{2}} =- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) E_{y}
\end{matrix}\right\} \quad(6)$
The solution of electromagnetic wave for magnetic field vector can find out by following the above method.
Therefore $x$ and $y$ components of the solution of the electromagnetic wave equation for magnetic field vector can be written as. i.e.
$\left.\begin{matrix}
\frac{\partial^{2} H_{x}}{\partial z^{2}}=- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) H_{x}
\\
\frac{\partial^{2}H_{y}}{\partial z^{2}} =- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) H_{y}
\end{matrix}\right\} \quad(7)$
In the solution of electromagnetic wave equation $(6)$ and equation $(7)$. The term $\left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right)$ is equal to $k_{z}^{2}$. It is known as propagation constant $k_{z}$. Then
$k_{z}^{2}=\left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) \qquad(8)$
The propagation constant is the complex quantity so
$k_{z}=\alpha+i \beta \qquad(9)$
Now from equation $(8)$ and equation $(9)$
$\left(\alpha+i \beta \right)^{2}=\left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right)$
$\alpha^{2} - \beta^{2} +2 i \alpha \beta =\left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right)$
Now separate the real and imaginary terms:
$Real \: Term \rightarrow \alpha^{2} - \beta^{2} = \omega^{2} \mu \epsilon \quad (10)$
$Imaginary \: Term \rightarrow 2 \alpha \beta = \omega \mu \sigma \quad (11)$
On solving the equation $(10)$ and equation $(11)$
$\alpha= \omega \sqrt{\frac{\mu \epsilon}{2}} \left[ 1 + \left\{ 1+ \left( \frac{\sigma}{\epsilon \omega} \right)^{2} \right\}^{1/2} \right]^{1/2} \quad(12)$
$\beta= \omega \sqrt{\frac{\mu \epsilon}{2}} \left[ \left\{ 1+ \left( \frac{\sigma}{\epsilon \omega} \right)^{2} \right\}^{1/2} -1 \right]^{1/2} \quad(13)$
The wave equation $(3)$ of the electric field vector also can be written as:
$\overrightarrow{E}(\overrightarrow{r},t)=E_{0} e^{-\beta \overrightarrow{r}}e^i{(\alpha \overrightarrow{r} - \omega t)} \qquad(14)$
The above equation has an additional term $e^{-\beta \overrightarrow{r}}$ compared to the purely harmonic solution.
Where
$\alpha \rightarrow$ Attenuation Constant
$\beta \rightarrow$ Absorption Coefficient and Phase Constant
