Skip to main content

Electromagnetic Wave Equation in Conducting Media (i.e. Lossy dielectric or Partially Conducting)

Maxwell's Equations: Maxwell's equation of the electromagnetic wave is a collection of four equations i.e. Gauss's law of electrostatic, Gauss's law of magnetism, Faraday's law of electromotive force, and Ampere's Circuital law. Maxwell converted the integral form of these equations into the differential form of the equations. The differential form of these equations is known as Maxwell's equations.

  1. $\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\rho}{\epsilon_{0}}$

  2. $\overrightarrow{\nabla}. \overrightarrow{B}=0$

  3. $\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$

  4. $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$

    Modified form:

    $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}+\mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t}$

For Conducting Media:

Current density $(\overrightarrow{J}) = \sigma \overrightarrow{E} $
Volume charge distribution $(\rho)=0$
Permittivity of Conducting Media= $\epsilon$
Permeability of Conducting Media=$\mu$

Now, Maxwell's equation for Conducting Media:

$\overrightarrow{\nabla}. \overrightarrow{E}=0 \qquad(1)$
$\overrightarrow{\nabla}. \overrightarrow{B}=0 \qquad(2)$
$\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} \qquad(3)$
$\overrightarrow{\nabla} \times \overrightarrow{H}= \overrightarrow{J}$

Modified form for Conducting Media:

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}+\mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t}$

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \sigma \overrightarrow{E}+ \mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t} \qquad(4)$

Now, On solving Maxwell's equation for conducting media i.e perfect dielectric and lossless media, gives the electromagnetic wave equation for conducting media. The electromagnetic wave equation has both an electric field vector and a magnetic field vector. So Maxwell's equation for conducting medium gives two equations for electromagnetic waves i.e. one is for electric field vector($\overrightarrow{E}$) and the second is for magnetic field vector ($\overrightarrow{H}$).

Electromagnetic wave equation for conducting media in terms of $\overrightarrow{E}$:

Now from equation $(3)$

$\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} $

Now take the curl on both sides of the above equation$

$\overrightarrow{\nabla} \times (\overrightarrow{\nabla} \times \overrightarrow{E})=-\overrightarrow{\nabla} \times \frac{\partial \overrightarrow{B}}{\partial t} $

$(\overrightarrow{\nabla}. \overrightarrow{E}).\overrightarrow{\nabla} - (\overrightarrow{\nabla}. \overrightarrow{\nabla}).\overrightarrow{E}=-\frac{\partial}{\partial t} (\overrightarrow{\nabla} \times \overrightarrow{B}) \qquad(5)$

We know that

$\overrightarrow{\nabla}. \overrightarrow{E}=0 $
$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$
$\overrightarrow{\nabla} \times \overrightarrow{B}=\sigma \mu \overrightarrow{E} + \mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t} $

Now substitute these values in equation $(5)$. So

$ -\nabla^{2}.\overrightarrow{E}=-\frac{\partial}{\partial t} \left(\sigma \mu \overrightarrow{E} + \mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t} \right)$

$ -\nabla^{2}.\overrightarrow{E}=-\mu \frac{\partial}{\partial t} \left(\sigma \overrightarrow{E} + \epsilon \frac{\partial \overrightarrow{E}}{\partial t} \right)$

$ \nabla^{2}.\overrightarrow{E}=\mu \epsilon \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} + \sigma \mu \frac{\partial \overrightarrow{E}}{\partial t} $

$ \nabla^{2}.\overrightarrow{E}-\mu \epsilon \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} - \sigma \mu \frac{\partial \overrightarrow{E}}{\partial t}=0 $

The value of $\frac{1}{\sqrt{\mu \epsilon}}= v$. Where $v$ is the speed of the electromagnetic wave in the conducting medium. So the above equation is often written as

$ \nabla^{2}.\overrightarrow{E}-\frac{1}{v^{2}}\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} - \sigma \mu \frac{\partial \overrightarrow{E}}{\partial t}=0 $

This is an electromagnetic wave equation for conducting media in terms of electric field vector ($\overrightarrow{E}$).

Electromagnetic wave equation for conducting media in terms of $\overrightarrow{B}$:

Now from MAxwell's equation $(4)$

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \sigma \overrightarrow{E}+ \mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t}$

Now take the curl on both sides of the above equation

$\overrightarrow{\nabla} \times (\overrightarrow{\nabla} \times \overrightarrow{B})=\overrightarrow{\nabla} \times \left( \mu \sigma \overrightarrow{E}+ \mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t} \right) $

$(\overrightarrow{\nabla}. \overrightarrow{B}).\overrightarrow{\nabla} - (\overrightarrow{\nabla}. \overrightarrow{\nabla}).\overrightarrow{B} \\ =\overrightarrow{\nabla} \times \sigma \mu \overrightarrow{E}+ \mu \epsilon \left( \overrightarrow{\nabla} \times \frac{\partial \overrightarrow{E}}{\partial t} \right)$

$(\overrightarrow{\nabla}. \overrightarrow{B}).\overrightarrow{\nabla} - (\overrightarrow{\nabla}. \overrightarrow{\nabla}).\overrightarrow{B} \\ =\sigma \mu \left( \overrightarrow{\nabla} \times \overrightarrow{E} \right)+ \mu \epsilon \frac{\partial }{\partial t}\left( \overrightarrow{\nabla} \times \overrightarrow{E} \right) \qquad(6)$

We know that

$\overrightarrow{\nabla}. \overrightarrow{B}=0$
$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$
$\overrightarrow{\nabla} \times \overrightarrow{E}= -\frac{\partial \overrightarrow{B}}{\partial t}$

Now substitute these values in equation $(6)$. So

$-\nabla^{2}.\overrightarrow{B}=- \sigma \mu \frac{\partial \overrightarrow{B}}{\partial t} - \mu \epsilon \frac{\partial^{2} B}{\partial t^{2}}$

$\nabla^{2}.\overrightarrow{B}= \sigma \mu \frac{\partial \overrightarrow{B}}{\partial t} + \mu \epsilon \frac{\partial^{2} B}{\partial t^{2}} $

$\nabla^{2}.\overrightarrow{B}-\sigma \mu \frac{\partial \overrightarrow{B}}{\partial t} - \mu \epsilon \frac{\partial^{2} B}{\partial t^{2}}=0 $

The value of $\frac{1}{\sqrt{\mu \epsilon}}= v$. Where $v$ is the speed of the electromagnetic wave in the conducting medium. So the above equation is often written as

$\nabla^{2}.\overrightarrow{B} - \frac{1}{v^{2}} \frac{\partial^{2} B}{\partial t^{2}}-\sigma \mu \frac{\partial \overrightarrow{B}}{\partial t}=0 $

This is an electromagnetic wave equation for conducting media in terms of electric field vector ($\overrightarrow{B}$).

Comments

Popular Posts

Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig

Fraunhofer diffraction due to a single slit

Let $S$ be a point monochromatic source of light of wavelength $\lambda$ placed at the focus of collimating lens $L_{1}$. The light beam is incident normally from $S$ on a narrow slit $AB$ of width $e$ and is diffracted from it. The diffracted beam is focused at the screen $XY$ by another converging lens $L_{2}$. The diffraction pattern having a central bright band followed by an alternative dark and bright band of decreasing intensity on both sides is obtained. Analytical Explanation: The light from the source $S$ is incident as a plane wavefront on the slit $AB$. According to Huygens's wave theory, every point in $AB$ sends out secondary waves in all directions. The undeviated ray from $AB$ is focused at $C$ on the screen by the lens $L_{2}$ while the rays diffracted through an angle $\theta$ are focussed at point $p$ on the screen. The rays from the ends $A$ and $B$ reach $C$ in the same phase and hence the intensity is maximum. Fraunhofer diffraction due to

Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ $\frac{d^{2} \psi(x)}{d x