Solution of electromagnetic wave equations in conducting media

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

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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. $\overrightarrow{\nabla} \times \overrightarrow{H}= \overrightarrow{J}$
   
   
   Modified form:
   
   
   $\overrightarrow{\nabla} \times \overrightarrow{H}= \overrightarrow{J}+ \frac{\partial \overrightarrow{D}}{\partial t}$

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{H}= \overrightarrow{J}+ \frac{\partial \overrightarrow{D}}{\partial t} $

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

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} $

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

But for conducting medium:

$\overrightarrow{\nabla}. \overrightarrow{E}=0 \qquad \left( From \: equation (1) \right)$

$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$

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

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 $\overrightarrow{H}$:

Now from equation $(4)$

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

Now take the curl on both sides of the above equation

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

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

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

But for conducting media:

$\overrightarrow{\nabla}. \overrightarrow{H}=0 \qquad \left( from \: equation (2) \right)$

$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$

$\overrightarrow{\nabla} \times \overrightarrow{E}= -\frac{\partial \overrightarrow{B}}{\partial t} \qquad \left( from \: equation (3) \right)$

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

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

$-\nabla^{2}.\overrightarrow{H}=- \sigma \mu \frac{\partial \overrightarrow{H}}{\partial t} - \mu \epsilon \frac{\partial^{2} H}{\partial t^{2}} \qquad (\because \overrightarrow{B}=\mu \overrightarrow{H})$

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

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

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

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Displacement Current

Description of Displacement Current: The concept of displacement current was first introduced by Maxwell purely on the theoretical ground.

Maxwell postulates that "It is not only current in a conductor that produces a magnetic field but a changing electric field (or time varying electric field) in vacuum or in dielectric also produces the magnetic field. It means that a changing electric field is equivalent to a current which flows as long as the electric field is changing. This equivalent current in a vacuum or dielectric produces the same magnetic effect as an ordinary or conductor current in a conductor. This equivalent current is known as displacement current".

According to the Maxwell modified ampere's law.

$\oint \overrightarrow{B}. \overrightarrow{dl}= \mu_{\circ}i+\mu_{\circ}i_{d}$

Where $i_{d}$ = Displacement Current

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Energy density in electromagnetic waves in free space

What is the energy density in the electromagnetic wave in free space?

The total energy stored in electromagnetic waves per unit volume due to the electric field and the magnetic field is called energy density in the electromagnetic wave in free space.

$U=\epsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}$

Derivation of Energy density in electromagnetic waves in free space:

The energy per unit volume due to the electric field is

$U_{E}= \frac{1}{2} \overrightarrow{E}.\overrightarrow{D} \qquad(1)$

The energy per unit volume due to the magnetic field is

$U_{B}= \frac{1}{2} \overrightarrow{B}.\overrightarrow{H} \qquad(2)$

The total energy density of electromagnetic waves is

$U=U_{E}+U_{B} \qquad(3)$

Now substitute the value of $U_{E}$ and $U_{B}$ in equation$(3)$ then we get

$U=\frac{1}{2} \left( \overrightarrow{E}.\overrightarrow{D}+\overrightarrow{B}.\overrightarrow{H} \right)$ $U=\frac{1}{2} \left( \overrightarrow{E}.\epsilon_{0}\overrightarrow{E}+\overrightarrow{B}.\frac{1}{\mu_{0}}\overrightarrow{B} \right) \qquad ( \because \overrightarrow{B}= \mu_{0} \overrightarrow{H} \:OR \: \overrightarrow{D}= \epsilon_{0}\overrightarrow{E} )$

$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\frac{B^{2}}{\mu_{0}} \right) \qquad ( \because \overrightarrow{E}\overrightarrow{E}= E^{2} \:OR \: \overrightarrow{B}.\overrightarrow{B}=B^{2})$

$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\frac{E^{2}}{c^{2} \mu_{0}} \right) \qquad ( \because B=\frac{E}{c})$

$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\frac{E^{2}}{c^{2} \mu_{0}} \right) $

$U=\frac{1}{2} \left( \epsilon_{0} E^{2}+\epsilon_{0} E^{2} \right) \qquad ( \because c=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}})$

$U=\frac{1}{2} \left(2 \epsilon_{0} E^{2} \right) $

$U= \epsilon_{0} E^{2} $

Similarly, the energy density of electromagnetic waves in free space in terms of the magnetic field $B$ can be written as:

$U= \frac{B^{2}}{\mu_{0}} $

The average value of energy density in the electromagnetic waves in free space:

Now we will find the average value of energy density in the electromagnetic wave in free space from the above equation $U= \epsilon_{0} E^{2} $. So we get

$\left< U \right> = \epsilon_{0} \left< E^{2} \right>$

$\left< U \right> = \epsilon_{0} \frac{E_{0}^{2}}{2} \qquad \left (\because \left< E^{2} \right>=\frac{E_{0}^{2}}{2} \right)$

$\left< U \right> = \epsilon_{0} E_{rms}^{2} \qquad \left (\because E_{rms}^{2}=\frac{E_{0}^{2}}{2} \right) \qquad (4)$

We know that

$\left< \overrightarrow{S} \right> = \frac{E_{rms}^{2}}{Z_{0}} .\hat{n} \qquad (5)$

Now divide the equation $(5)$ by equation$(4)$

$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\frac{E_{rms}^{2}}{Z_{0}} .\hat{n}}{\epsilon_{0} E_{rms}^{2}}$

$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\hat{n}}{\epsilon_{0} Z_{0}}$

$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\frac{\hat{n}}{\sqrt{\epsilon_{0} \mu_{0}}} \qquad(\because z_{0}= \sqrt{\frac{\mu_{0}}{\epsilon_{0}}})$

$\frac{\left< \overrightarrow{S} \right>}{\left< U \right>}=\hat{n} c \qquad(\because c= \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}})$

$ \left< \overrightarrow{S} \right>=\hat{n} c \left< U \right> $

The energy flow per unit area per unit time in an electromagnetic wave is the product of energy density, speed of light, and the direction of propagation.

The ratio of the energy densities of the electric field and magnetic field:

So from above equation $U_{E}=\epsilon_{0} E^{2}$ and equation $U_{B}=\frac{B^{2}}{\mu_{0}}$, we can find the ratio between them i.e.

$\frac{U_{E}}{U_{B}}=\frac{\epsilon_{0} E^{2}}{\frac{B^{2}}{\mu_{0}}}$

$\frac{U_{E}}{U_{B}}=\frac{\epsilon_{0} \mu_{0} E^{2}}{B^{2}}$

$\frac{U_{E}}{U_{B}}=\frac{c^{2}}{c^{2}}$

$\frac{U_{E}}{U_{B}}=1$

$U_{E}=U_{B}$

So the energy density of the electric field is the same as the energy density of the magnetic field.

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Poynting Vector and Poynting Theorem

Poynting Vector:

The rate of flow of energy per unit area in plane electromagnetic wave is known as Poynting vector. It is represented by $\overrightarrow{S}$. It is a vector quantity.

$\overrightarrow{S}=\overrightarrow{E} \times \overrightarrow{H}$

$\overrightarrow{S}=\frac{1} {\mu_{0}} (\overrightarrow{E} \times \overrightarrow{B})$

Poynting Theorem (Work energy theorem):

The most important aspect of electrodynamics is:

• Energy density stored with an electromagnetic wave
• Energy Flux associated with an electromagnetic wave

To derive the energy density and energy flux. We consider the conservation of energy in small volume elements in space. The work done per unit volume by an electromagnetic wave:

$W=\overrightarrow{J}.\overrightarrow{E} \qquad(1)$

This work done also consider as energy dissipation per unit volume. This energy dissipation must be connected with the net decrease in energy density and energy flow out of the volume.

$W=\overrightarrow{E}.\left( \overrightarrow{\nabla} \times \overrightarrow{H} - \frac{\partial \overrightarrow{D} }{\partial t}\right) \qquad \left( \because \overrightarrow{H} = \overrightarrow{J} + \frac{\partial \overrightarrow{D} }{\partial t}\right) \qquad (2)$

Now we employ the vector identity

$\overrightarrow{\nabla}. (\overrightarrow{E} \times \overrightarrow{H})= \overrightarrow{H} (\overrightarrow{\nabla} \times \overrightarrow{E})-\overrightarrow{E}.(\overrightarrow{\nabla} \times \overrightarrow{H}) \qquad (3)$

From equation $(2)$ and equation $(3)$

$ W= \overrightarrow{H} (\overrightarrow{\nabla} \times \overrightarrow{E})-\overrightarrow{\nabla}.(\overrightarrow{E} \times \overrightarrow{H}) - \overrightarrow{E} \left( \frac{\partial \overrightarrow{D} }{\partial t}\right) $

$ W= \overrightarrow{H} \left( \frac{-\partial \overrightarrow{B} }{\partial t}\right)-\overrightarrow{\nabla}.(\overrightarrow{E} \times \overrightarrow{H}) - \overrightarrow{E} \left( \frac{\partial \overrightarrow{D} }{\partial t}\right) $

$ JE= -\overrightarrow{\nabla}.(\overrightarrow{E} \times \overrightarrow{H}) - \frac{\partial}{\partial t} \left( \frac{\overrightarrow{H}.\overrightarrow{B}+\overrightarrow{E}.\overrightarrow{D} }{2} \right) $

$ -JE= \overrightarrow{\nabla}.\overrightarrow{S} + \frac{\partial}{\partial t} \left( \frac{\overrightarrow{H}.\overrightarrow{B}+\overrightarrow{E}.\overrightarrow{D} }{2} \right) \qquad (\because \overrightarrow{S}= \overrightarrow{E} \times \overrightarrow{H})$

$ -JE= \overrightarrow{\nabla}.\overrightarrow{S} + \frac{\partial U}{\partial t} \qquad \left( \because U = \frac{\overrightarrow{H}.\overrightarrow{B}+\overrightarrow{E}.\overrightarrow{D} }{2} \right) $

$ \overrightarrow{\nabla}.\overrightarrow{S} + \frac{\partial U}{\partial t} =-JE $

This equation represents the conservation of energy principle. It is also known as Poynting theorem. Here Negative signs of work done to represent that electromagnetic flow with energy flux as continuity energy density. So this equation is also known as the continuity equation.

Where$ \overrightarrow{\nabla}.\overrightarrow{S}$ - The flow of energy$U$ - Energy density of electromagnetic filed$S$ - Energy flux or Poyting vector

If Current density $\overrightarrow{J}=0$ Then

$ \overrightarrow{\nabla}.\overrightarrow{S} + \frac{\partial U}{\partial t} = 0 $

$ \overrightarrow{\nabla}.\overrightarrow{S} = \frac{\partial U}{\partial t} $

$ \overrightarrow{\nabla}.\overrightarrow{S} = \frac{\partial }{\partial t} (Storage \: energy) $

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Characteristic impedance of electromagnetic wave

We know that the electromagnetic wave propagates perpendicular to both electric field and magnetic field which can describe as

$\overrightarrow{k} \times \overrightarrow{E}= \omega \overrightarrow{B} \qquad(1)$

If $\hat{n}$ is a unit vector in the direction of the propagation then

$\overrightarrow{k}=k \hat{n}$

Substitute these values in equation$(1)$ then we get

$k(\hat{n} \times \overrightarrow{E})= \omega \overrightarrow{B}$

$\overrightarrow{B}= \frac{k}{\omega}(\hat{n} \times \overrightarrow{E}) \qquad(2)$

But the value of $k$ and $\omega$ is

$k=\frac{2\pi}{\lambda}$

$\omega=2 \pi \nu$

Then value of $\frac{k}{\omega}=\frac{1}{c}$

Now substitute the value of $\frac{k}{\omega}$ in equation$(2)$ then we get

$\overrightarrow{B}= \frac{1}{c}(\hat{n} \times \overrightarrow{E})$

The magnitude form of the above equation can be written as

$B=\frac{E}{c}$

$\frac{E}{B}=c$

$\frac{E}{\mu_{0}H}=c \qquad (\because B=\mu_{0} H)$

$\frac{E}{H}=\mu_{0}c$

$\frac{E}{H}=\frac{\mu_{0}}{\sqrt{\mu_{0} \epsilon_{0}}} \qquad(\because c=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}})$

$\frac{E}{H}= \sqrt{\frac{\mu_{0}}{\epsilon_{0}}}$

The term $\frac{E}{H}$ has dimensions of the impedance and is known as characteristic impedance or intrinsic impedance of free space. It is represented by $(Z_{0})$.

$Z_{0}=\frac{E}{H}=\sqrt{\frac{\mu_{0}}{\epsilon_{0}}}$

Now substitute the value of $\mu_{0}$ and $\epsilon_{0}$ i.e.

$\mu_{0}=4\pi \times 10^{-7}$

$\epsilon_{0}=8.854 \times 10^{-12}$

$Z_{0}=\sqrt{\frac{4\pi \times 10^{-7}}{8.854 \times 10^{-12}}}$

$Z_{0}=376.73\: \Omega $

$Z_{0}=120\pi \: \Omega $

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Radiation pressure of electromagnetic wave

When an electromagnetic wave strikes a surface then its momentum changes. the rate of change of momentum is equal to the applied force. this force acting on the unit area of the surface exerts a pressure called radiation pressure$(P_{rad})$.

Let us consider a plane electromagnetic wave incident normally on a perfectly absorbing surface of area $A$ for a time $t$. If energy $U$ is absorbed during this time then momentum $P$ delivered to the surface is given according to Maxwell's prediction by

$P=\frac{U}{C} \qquad(1)$

If $S$ is the energy flow per unit area per unit time i.e. Poynting vector then the energy density

$U=SAt \qquad(2)$

From equation $(1)$ and equation $(2)$

$P=\frac{SAt}{c}$

$P=UAt \qquad (\because U=\frac{S}{c})$

$\frac{P}{t}=UA \qquad (3)$

If average force $(F)$ acting on the surface, is equal to the average rate of change of momentum $(P)$, is delivered to the surface then

$F=\frac{P}{t} \qquad(4)$

Now from equation$(3)$ and equation$(4)$ we get

$F=UA \qquad(5)$

The radiation pressure $(P_{rad})$ exerted on the surface is

$P_{rad}=\frac{F}{A} \qquad(6)$

Now substitute the value of $F$ from equation$(5)$ in equation$(6)$ then we get

$P_{rad}=\frac{UA}{A}$

$P_{rad}=U$

Hence, the radiation pressure exerted by a normally incident play electromagnetic wave on a perfect absorber is equal to the energy density of the wave.

For a perfect reflector or for a perfect reflecting surface, the radiation after reflection has momentum equal in magnitude but opposite in direction to the incident radiation. Then the momentum imparted to the surface will therefore be twice as on perfect absorber i.e.

$P_{rad}=2U$

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Energy flow in the electromagnetic wave in free space

Derivation of energy flow in the electromagnetic wave in free space:

The Poynting vector is given by

$\overrightarrow{S}=\overrightarrow{E} \times \overrightarrow{H} \qquad(1)$

$\overrightarrow{S}=\frac{1}{\mu_{0}} ( \overrightarrow{E} \times \overrightarrow{B} ) \qquad(2) \qquad (\because \overrightarrow{B}= \mu_{0} \overrightarrow{H})$

We know that the characteristic impedance equation i.e.

$\overrightarrow{B}=\frac{1}{\mu_{0}c}(\hat{n} \times \overrightarrow{E}) \qquad(3)$

Now substitute the value of $\overrightarrow{B}$ in equation$(2)$

$\overrightarrow{S}=\frac{1}{\mu_{0}c} [\overrightarrow{E} \times (\hat{n} \times \overrightarrow{E})]$

$\overrightarrow{S}=\frac{1}{\mu_{0}c} [(\overrightarrow{E}.\overrightarrow{E}) \hat{n}- (\overrightarrow{E}.\hat{n}) \overrightarrow{E})] \qquad(4)$

As $\overrightarrow{E}$ is perpendicular to $\hat{n}$ so $\overrightarrow{E} . \hat{n}=0$ then we get for above equation$(4)$

$\overrightarrow{S}=\frac{1}{\mu_{0}c} E^{2} \hat{n}$

From the above equation, we can conclude $\overrightarrow{S}$ has the same direction as $\hat{n}$ which is the direction of wave propagation.

Energy flow in an electromagnetic wave takes place in the direction of the propagation of the wave.

Here $\mu_{0} c=Z_{0}$ (The Characteristic impedance of free space)

$\overrightarrow{S}=\frac{1}{Z_{0}} E^{2} \hat{n}$

This is the equation of energy flow in the electromagnetic wave in free space.

The average energy flow over one period of the electromagnetic wave in free space:

Now the average energy flow of the above equation

$ \left< \overrightarrow{S} \right> =\frac{1}{Z_{0}} \left< E^{2} \right> \hat{n} \qquad(5)$

We know the electric field vector wave equation i.e. $\overrightarrow{E}=E_{0} e^{i(\overrightarrow{k}. \overrightarrow{r} -\omega t)} $

So the value of $\left< E^{2} \right>$ from above equation:

$\left< E^{2} \right>= \left< Re[E_{0} e^{i(\overrightarrow{k}. \overrightarrow{r} -\omega t)}]^{2} \right>$

$\left< E^{2} \right>= \left< E_{0}^{2}\: cos^{2}(\overrightarrow{k}. \overrightarrow{r} -\omega t) \right>$

For one period or cycle of electromagnetic wave the value of $cos^{2}(\overrightarrow{k}. \overrightarrow{r} -\omega t)=\frac{1}{2}$ then we get

$\left< E^{2} \right>= \frac{E_{0}^{2}}{2}$

$\left< E^{2} \right>= (\frac{E_{0}}{2})^{2}$

$\left< E^{2} \right>= E_{rms}^{2} \qquad \left (\because E_{rms} = \frac{E_{0}}{2}\right )$

Now substitute the value of $ \left< E^{2} \right>$ in equation $(5)$ then we get

$ \left< \overrightarrow{S} \right> =\frac{E_{rms}^{2}}{Z_{0}} \hat{n} $

This average energy flow equation over one period of the electromagnetic wave in free space.

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Momentum of electromagnetic wave

Derivation of momentum of electromagnetic wave:

Maxwell's had also predicted that electromagnetic waves transport linear momentum in the direction of propagation. Let a particle which has mass $m$ moving with velocity then the momentum of a particle,

$\overrightarrow{P}=m\overrightarrow{v} \qquad(1)$

According to mass-energy relation

$U=mc^{2}$

Here $U$ - Total energy of the particle

$m=\frac{U}{c^ {2}} \qquad(2)$

From equation $(1)$ and equation $(2)$

$\overrightarrow{P}=\frac{U}{c^{2}} \overrightarrow{v} \qquad(3)$

If the electromagnetic wave is propagating along the x-axis then

$\overrightarrow{v}=c \hat{i}$

Put this value in the above equation $(3)$

$\overrightarrow{P}=\frac{U}{c} \hat{i} \qquad(4)$

We know that the equation of energy flow in electromagnetic wave

$\overrightarrow{S}= \frac{1}{\mu_{0} c} E^{2} \hat{n}$

Here wave is propagating along x-axis i.e 

$\hat{n}=\hat{i}$

$\overrightarrow{S}= \frac{1}{\mu_{0} c} E^{2} \hat{i} \qquad(5)$

The energy density in plane electromagnetic wave in free space:

$U=\epsilon_{0} E^{2}$

Where $E$ - Magnitude of electric field

$E^{2}=\frac{U}{\epsilon_{0}} \qquad(6)$

Now substitute the value of $E^{2}$ in equation$(5)$

$\overrightarrow{S}= \frac{1}{\mu_{0} c} \frac{U}{\epsilon_{0}} \hat{i} $

$\overrightarrow{S}= \frac{c^{2}}{c} U \hat{i} \qquad (\because \frac{1}{\sqrt{ \mu_{0} \epsilon_{0}}}=c) $

$\overrightarrow{S}= c U \hat{i} $

$U \hat{i}=\frac{\overrightarrow{S}}{c} \qquad(7)$

Now substitute the value of $ U \hat{i} $ in equation $(4)$. Then

$\overrightarrow{P}=\frac{\overrightarrow{S}}{c}$

$\overrightarrow{P}=\frac{(\overrightarrow{E} \times \overrightarrow{B})}{ \mu_{0}c^{2}}$

$\overrightarrow{P}=\epsilon_{0}(\overrightarrow{E} \times \overrightarrow{B})$

This is the equation of "Momentum of electromagnetic wave"

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Solution of electromagnetic wave equations in free space

The electromagnetic wave equations in free space:

For electric field vector:

$\nabla^{2} \overrightarrow{E}=\frac{1}{c^{2}} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} \qquad(1)$

For magnetic field vector:

$\nabla^{2} \overrightarrow{H}=\frac{1}{c^{2}} \frac{\partial^{2} \overrightarrow{H}}{\partial t^{2}} \qquad(2)$

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$(1)$

$\nabla^{2} \overrightarrow{E}=\frac{-\omega^{2}}{c^{2}} \overrightarrow{E}(\overrightarrow{r},t)$

$\nabla^{2} \overrightarrow{E}=-(\frac{\omega}{c})^{2} \overrightarrow{E}(\overrightarrow{r},t)$

$\nabla^{2} \overrightarrow{E}=-k^{2} \overrightarrow{E}(\overrightarrow{r},t) \qquad (\because \frac{\omega}{c}=k )$

Where $k$ - Wave propagation Constant

$\nabla^{2} \overrightarrow{E} + k^{2} \overrightarrow{E}(\overrightarrow{r},t)=0 $

This is the solution of the electromagnetic wave equation in free space for the electric field vector.

Now the component form of the above equation:

$(\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} +\frac{\partial^{2}}{\partial z^{2}})(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) \\ =- k^{2}(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) \qquad(5)$

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:

$\frac{\partial^{2}}{\partial x^{2}}(\hat{i}E_{x}+\hat{j}E_{y})=- k^{2}(\hat{i}E_{x}+\hat{j}E_{y})$

Now separate the above equation in $x$ and $y$ components so

$\frac{\partial^{2} E_{x}}{\partial z^{2}}=- k^{2}E_{x}$

$\frac{\partial^{2}E_{y}}{\partial z^{2}} =- k^{2}E_{y}$

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.

$\frac{\partial^{2} H_{x}}{\partial z^{2}}=- k^{2}H_{x}$

$\frac{\partial^{2}H_{y}}{\partial z^{2}} =- k^{2}H_{y}$

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Solution of electromagnetic wave equations in non conducting media

The electromagnetic wave equations in non-conducting media :

For electric field vector:

$\nabla^{2} \overrightarrow{E}=\frac{1}{v^{2}} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} \qquad(1)$

For magnetic field vector:

$\nabla^{2} \overrightarrow{H}=\frac{1}{v^{2}} \frac{\partial^{2} \overrightarrow{H}}{\partial t^{2}} \qquad(2)$

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 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$(1)$

$\nabla^{2} \overrightarrow{E}=\frac{-\omega^{2}}{v^{2}} \overrightarrow{E}(\overrightarrow{r},t)$

$\nabla^{2} \overrightarrow{E}=-(\frac{\omega}{v})^{2} \overrightarrow{E}(\overrightarrow{r},t)$

$\nabla^{2} \overrightarrow{E}=-\alpha^{2} \overrightarrow{E}(\overrightarrow{r},t) \qquad (\because \alpha=\frac{\omega}{v} )$

Where $\alpha$ - Wave propagation Constant

$\nabla^{2} \overrightarrow{E} + \alpha^{2} \overrightarrow{E}(\overrightarrow{r},t)=0 $

This is the solution of the electromagnetic wave equation in non-conducting media for the electric field vector.

Now the component form of the above equation:

$(\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} +\frac{\partial^{2}}{\partial z^{2}})(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) =- \alpha^{2}(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) \qquad(5)$

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:

$\frac{\partial^{2}}{\partial x^{2}} (\hat{i}E_{x}+\hat{j}E_{y})=- \alpha^{2}(\hat{i}E_{x}+\hat{j}E_{y})$

Now separate the above equation in $x$ and $y$ components so

$\frac{\partial^{2} E_{x}}{\partial z^{2}}=- \alpha^{2}E_{x}$

$\frac{\partial^{2}E_{y}}{\partial z^{2}} =- \alpha^{2}E_{y}$

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.

$\frac{\partial^{2} H_{x}}{\partial z^{2}}=- \beta^{2}H_{x} \qquad \left( \because \beta=\frac{\omega}{v} \right)$

$\frac{\partial^{2}H_{y}}{\partial z^{2}} =- \beta^{2}H_{y}$

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Electromagnetic wave equation in non conducting media (i.e. Perfect dielectric or Lossless media)

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 for free space.

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. $\overrightarrow{\nabla} \times \overrightarrow{H}= \overrightarrow{J}$
   
   
   Modified Form:
   
   
   $\overrightarrow{\nabla} \times \overrightarrow{H}= \overrightarrow{J}+ \frac{\partial \overrightarrow{D}}{\partial t}$

For non-conducting media:

Current density $(\overrightarrow{J})=0$
Volume charge distribution $(\rho) \neq 0 $
Permittivity of non-conducting media= $\epsilon$
Permeability of non-conducting media=$\mu$

Now, Maxwell's equation for non-conducting media:

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

Modified form for non-conducting media:

$\overrightarrow{\nabla} \times \overrightarrow{H}= \frac{\partial \overrightarrow{D}}{\partial t} \qquad(4)$

Now, On solving Maxwell's equation for a non-conducting media i.e perfect dielectric and Lossless media, we get the electromagnetic wave equation for a non-conducting media. The electromagnetic wave equation has both an electric field vector and a magnetic field vector. So Maxwell's equations for non-conducting media give 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 non-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} $

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

But for non-conducting media:

$\overrightarrow{\nabla}. \overrightarrow{E}=\frac{\rho}{\epsilon} \qquad \left( From \: equation (1) \right)$

$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \frac{\partial \overrightarrow{D}}{\partial t} \qquad \left( From \: equation (4) \right)$

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

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

For non conducting media, If the volume charge distribution is uniform then the gradient of volume charge density is very small (or almost zero) so neglect the term gradient of volume charge density $(\nabla \rho)$ in above eqaution.

Or

The wave propagation does not contain the charges in most of the cases. Therefore we get

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

$ \nabla^{2}.\overrightarrow{E}=\mu \frac{\partial^{2}}{\partial t^{2}} (\epsilon \overrightarrow{E}) \qquad \left(\because \overrightarrow{D}=\epsilon \overrightarrow{E} \right)$

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

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

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

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

Electromagnetic wave equation for non-conducting media in terms of $\overrightarrow{H}$:

Now from equation $(4)$

$\overrightarrow{\nabla} \times \overrightarrow{H}= \frac{\partial \overrightarrow{D}}{\partial t}$

Now take the curl on both sides of the above equation

$\overrightarrow{\nabla} \times (\overrightarrow{\nabla} \times \overrightarrow{H})=\overrightarrow{\nabla} \times \frac{\partial \overrightarrow{D}}{\partial t} $

$(\overrightarrow{\nabla}. \overrightarrow{H}).\overrightarrow{\nabla} - (\overrightarrow{\nabla}. \overrightarrow{\nabla}).\overrightarrow{H}=\frac{\partial}{\partial t} (\overrightarrow{\nabla} \times \overrightarrow{D}) \qquad(6)$

But for non-conducting media:

$\overrightarrow{\nabla}. \overrightarrow{B}=0 \qquad \left( from \: equation (2) \right)$

$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$

$\overrightarrow{\nabla} \times \overrightarrow{D}= -\epsilon \frac{\partial \overrightarrow{B}}{\partial t} \qquad \left( from \: equation (3) \right)$

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

$ -\nabla^{2}.\overrightarrow{H}=-\frac{\partial}{\partial t} (\epsilon \frac{\partial \overrightarrow{B}}{\partial t})$

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

$ \nabla^{2}.\overrightarrow{H}=\epsilon \frac{\partial^{2}}{\partial t^{2}} (\mu \overrightarrow{H}) \qquad \left(\because \overrightarrow{B}=\mu \overrightarrow{H} \right)$

$\nabla^{2}.\overrightarrow{H}=\mu \epsilon \frac{\partial^{2} \overrightarrow{H}}{\partial t^{2}}$

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

$\nabla^{2}.\overrightarrow{H}=\frac{1}{v^{2}} \frac{\partial^{2} \overrightarrow{H}}{\partial t^{2}}$

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

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Electromagnetic wave equation in free space

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. $\overrightarrow{\nabla} \times \overrightarrow{H}= \overrightarrow{J}$
   
   
   Modified Form:
   
   
   $\overrightarrow{\nabla} \times \overrightarrow{H}= \overrightarrow{J}+ \frac{\partial \overrightarrow{D}}{\partial t}$

For free space:

Current density ($\overrightarrow{J}$) = 0
Volume charge distribution ($\rho$) = 0
Permittivity $\epsilon = \epsilon_{0}$
Permeability $\mu = \mu_{0}$

Now, Maxwell's equation for free space:

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

Modified form for free space:

$\overrightarrow{\nabla} \times \overrightarrow{H}= \frac{\partial \overrightarrow{D}}{\partial t} \qquad(4)$

Now, On solving Maxwell's equation for free space we get the electromagnetic wave equation for free space. The electromagnetic wave equation has both an electric field vector and a magnetic field vector. So Maxwell's equations for free space give two-equation for electromagnetic wave i.e. one is for electric field vector($\overrightarrow{E}$) and the second is for magnetic field vector ($\overrightarrow{H}$).

Electromagnetic wave equation for free space in term 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} $

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

But for free space:

$\overrightarrow{\nabla}. \overrightarrow{E}=0 \qquad$   {from equation (1)}

$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu_{0} \frac{\partial \overrightarrow{D}}{\partial t}\qquad $   {from equation (4)}

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

$ -\nabla^{2}.\overrightarrow{E}=-\frac{\partial}{\partial t} (\mu_{0} \frac{\partial \overrightarrow{D}}{\partial t})$

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

$ \nabla^{2}.\overrightarrow{E}=\mu_{0} \frac{\partial^{2}}{\partial t^{2}} (\epsilon_{0} \overrightarrow{E}) \qquad \left( \because \overrightarrow{D}=\epsilon_{0} \overrightarrow{E} \right)$

$\nabla^{2}.\overrightarrow{E}=\mu_{0} \epsilon_{0} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}$

The value of $\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}= c$. Where $c$ is the speed of the wave in free space. So the above equation is often written as

$\nabla^{2}.\overrightarrow{E}=\frac{1}{c^{2}} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}$

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

Electromagnetic wave equation for free space in term of $\overrightarrow{H}$:

Now from equation $(4)$

$\overrightarrow{\nabla} \times \overrightarrow{H}= \frac{\partial \overrightarrow{D}}{\partial t}$

Now take the curl on both sides of the above equation

$\overrightarrow{\nabla} \times (\overrightarrow{\nabla} \times \overrightarrow{H})=\overrightarrow{\nabla} \times \frac{\partial \overrightarrow{D}}{\partial t} $

$(\overrightarrow{\nabla}. \overrightarrow{H}).\overrightarrow{\nabla} - (\overrightarrow{\nabla}. \overrightarrow{\nabla}).\overrightarrow{H}=\frac{\partial}{\partial t} (\overrightarrow{\nabla} \times \overrightarrow{D}) \qquad(6)$

But for free space:

$\overrightarrow{\nabla}. \overrightarrow{B}=0 \qquad $  {from equation (2)}

$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$

$\overrightarrow{\nabla} \times \overrightarrow{D}= -\epsilon_{0} \frac{\partial \overrightarrow{B}}{\partial t} \qquad$   {from equation (3)}

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

$ -\nabla^{2}.\overrightarrow{H}=-\frac{\partial}{\partial t} (\epsilon_{0} \frac{\partial \overrightarrow{B}}{\partial t})$

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

$ \nabla^{2}.\overrightarrow{H}=\epsilon_{0} \frac{\partial^{2}}{\partial t^{2}} (\mu_{0} \overrightarrow{H}) \qquad \left(\because \overrightarrow{B}=\mu_{0} \overrightarrow{H} \right)$

$\nabla^{2}.\overrightarrow{H}=\mu_{0} \epsilon_{0} \frac{\partial^{2} \overrightarrow{H}}{\partial t^{2}}$

The value of $\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}= c$. Where $c$ is the speed of the wave in free space. So the above equation is often written as

$\nabla^{2}.\overrightarrow{H}=\frac{1}{c^{2}} \frac{\partial^{2} \overrightarrow{H}}{\partial t^{2}}$

This is the electromagnetic wave equation for free space in terms of electric field vector ($\overrightarrow{H}$).

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Equation of continuity for current density

We know the equation of continuity is

$\overrightarrow{\nabla}. \overrightarrow{J}+ \frac{\partial \rho}{\partial t}=0 \qquad(1)$

According to Maxwell's first differential equation

$\overrightarrow{\nabla}. \overrightarrow{D}=\rho \qquad(2)$

From equation $(1)$ and equation$(2)$

$\overrightarrow{\nabla}. \overrightarrow{J}+ \frac{\partial }{\partial t}(\overrightarrow{\nabla}. \overrightarrow{D})=0$

$\overrightarrow{\nabla}. (\overrightarrow{J}+ \frac{\partial \overrightarrow{D} }{\partial t}) =0 $

Where the term →
$(\overrightarrow{J}+ \frac{\partial \overrightarrow{D} }{\partial t})$ → solenoidal vector and it is also regarded as total current density for time varying electric field.
$D$ → The displacement vector
$\frac{\partial \overrightarrow{D} }{\partial t}$ → Displacement current density

The above equation is known as the "Equation of continuity for current density".

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Equation of continuity of electromagnetic wave

Definition:

The mathematical representation of the law of conservation of charge in differential form is called the "continuity equation".

Mathematical representation of Equation of continuity:

If $\overrightarrow{J}$ is the current density of a closed surface $S$ then the current through a closed surface is

$i=\oint_{S} \overrightarrow{J}. \overrightarrow{dS} \qquad(1)$

Let $V$ be the volume enclosed by the surface $S$. So the total charge in this volume-

$q=\oint_{V} \rho. dV \qquad(2)$

By the law of conservation of charge i.e. "Charge can neither be created nor destroyed". If some charge flows out from the volume per unit time giving rise to current density, the charge in the volume decreases at the same rate. So the current

$i=-\frac{\partial q}{\partial t}$

$i=-\frac{\partial}{\partial t} (\oint_{V} \rho. dV) \qquad (from \: equation(2) \: )$

$i=-\oint_{V} \frac{\partial \rho}{\partial t} dV \qquad(3)$

from equation $(1)$ and equation $(3)$

$\oint_{S} \overrightarrow{J}. \overrightarrow{dS}= -\oint_{V} \frac{\partial \rho}{\partial t} dV \qquad(4)$

According to Gauss's divergence theorem-

$\oint_{S} \overrightarrow{J}. \overrightarrow{dS}= \oint_{V} (\overrightarrow{\nabla}. \overrightarrow{J}) dV \qquad(5)$

From equation $(4)$ and equation $(5)$

$\oint_{V} (\overrightarrow{\nabla}. \overrightarrow{J}) dV = -\oint_{V} \frac{\partial \rho}{\partial t} dV$

$\oint_{V} (\overrightarrow{\nabla}. \overrightarrow{J}) dV + \oint_{V} \frac{\partial \rho}{\partial t} dV =0 $

$\oint_{V} [(\overrightarrow{\nabla}. \overrightarrow{J}) + \frac{\partial \rho}{\partial t}] dV =0 $

$\overrightarrow{\nabla}. \overrightarrow{J} + \frac{\partial \rho}{\partial t} =0 $

This equation is known as the equation of continuity and it is based on the conservation of charge.

For the study state $\frac{\partial{\rho}}{\partial{t}}=0$

$\overrightarrow{\nabla}. \overrightarrow{J}=0$

This means that in the steady state. There is no source or sink of current.

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Derivation of Maxwell's forth equation

Maxwell's fourth equation is the differential form of Ampere's circuital law.

$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \frac{\partial{\overrightarrow{D}}}{\partial{t}}$

Derivation:

According to Ampere's circuital law

$\oint_{l} \overrightarrow{B}. \overrightarrow{dl}=\mu_{0} i \qquad(1)$

According to Stroke's theorem-

$\oint_{l} \overrightarrow{B}.\overrightarrow{dl}=\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} \qquad(2)$

From equation$(1)$ and equation$(2)$

$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} = \mu_{0} i \qquad(3)$

where $i=\oint_{S} \overrightarrow{J}. \overrightarrow{dS} \qquad(4)$

So from equation$(3)$ and equation$(4)$

$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} = \mu_{0} \oint_{S} \overrightarrow{J}. \overrightarrow{dS}$

$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} - \mu_{0} \oint_{S} \overrightarrow{J}. \overrightarrow{dS}=0$

$\oint_{S} [(\overrightarrow{\nabla} \times \overrightarrow{B}) - \mu_{0} \overrightarrow{J}]. \overrightarrow{dS}=0$

$(\overrightarrow{\nabla} \times \overrightarrow{B}) - \mu_{0} \overrightarrow{J}=0$

$\overrightarrow{\nabla} \times \overrightarrow{B} = \mu_{0} \overrightarrow{J}$

We know that $\overrightarrow{B}=\mu_{0} \overrightarrow{H}$. So the above equation can be again written as-

$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J}$

Modified Maxwell's fourth equation:

The modified Maxwell's fourth equation is the differential form of the modified Ampere's circuital law.

We know the modified Ampere's circuital law-

$\oint_{l} \overrightarrow{B}. \overrightarrow{dl}=\mu_{0} i + i_{d}$

Where $i_{d}$ - Displacement current

So the derivation of the modified Maxwell equation is similar to the above derivation. Therefore the modified Maxwell's fourth equation can be written as-

$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \overrightarrow{J_{d}} \qquad(1)$

Where $\overrightarrow{J_{d}}$ - Displacement current density

And the value of $\overrightarrow{J_{d}}$ is

$\overrightarrow{J_{d}}=\epsilon_{0} \frac{\partial{\overrightarrow{E}}}{\partial{t}}$

$\overrightarrow{J_{d}}=\frac{\partial{\overrightarrow{D}}}{\partial{t}} \qquad(\because \overrightarrow{D}=\epsilon_{0} \overrightarrow{E})$

Now substitute the value of $\overrightarrow{J_{d}}$ in equation $(1)$, then

$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \frac{\partial{\overrightarrow{D}}}{\partial{t}}$

This is modified by Maxwell's fourth equation.

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Derivation of Maxwell's third equation

Maxwell's third equation is the differential form of Faraday's law induction.i.e

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

Derivation:

According to Faraday's Induced law-

$e=-\frac{\partial{\phi_{B}}}{\partial{t}} \qquad(1)$

According to Gauss's law of magnetism-

$\phi_{B}=\oint_{S} \overrightarrow{B}.\overrightarrow{dS} \qquad(2)$

Now substitute the value of $\phi_{B}$ in equation $(1)$

$e=-\frac{\partial}{\partial{t}} \oint_{S} \overrightarrow{B}.\overrightarrow{dS}$

$e=-\oint_{S} \frac{\partial{\overrightarrow{B}}}{\partial{t}}.\overrightarrow{dS} \qquad(3)$

The line integral of the electric field around a closed loop is called electromotive force. Thus

$e=\oint_{l} \overrightarrow{E}.\overrightarrow{dl} \qquad(4)$

from equation $(3)$ and $(4)$

$\oint_{l} \overrightarrow{E}.\overrightarrow{dl}=-\oint_{S} \frac{\partial{\overrightarrow{B}}}{\partial{t}}.\overrightarrow{dS} \qquad(5)$

According to Stroke's Theorem-

$\oint_{l} \overrightarrow{E}.\overrightarrow{dl}=\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{E}).\overrightarrow{dS} \qquad(6)$

from equation $(5)$ and equation $(6)$

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

$\oint_{S} [(\overrightarrow{\nabla} \times \overrightarrow{E})+ \frac{\partial{\overrightarrow{B}}}{\partial{t}}].\overrightarrow{dS}=0$

If the surface is arbitrary then-

$(\overrightarrow{\nabla} \times \overrightarrow{E})+ \frac{\partial{\overrightarrow{B}}}{\partial{t}}=0$

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

This is Maxwell's third equation.

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Derivation of Maxwell's second equation

Maxwell's second equation is the differential form of Gauss's law of magnetism.

As magnetic, monopoles do not exist in magnets and the magnetic field lines form closed loops. There is no source of the magnetic field from which the lines will either only diverge or only converge. Hence the divergence of the magnetic field is zero.

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

Derivation-

According to Gauss's law of magnetism

$\oint_{S} \overrightarrow{B}. \overrightarrow{dS}=0 \qquad(1)$

Now apply the Gauss's divergence theorem-

$\oint_{S} \overrightarrow{B}. \overrightarrow{dS}= \oint_{v} \overrightarrow{\nabla}.\overrightarrow{B}.dV \qquad (2)$

from equation $(1)$ equation $(2)$

$\oint_{v} (\overrightarrow{\nabla}.\overrightarrow{B}).dV =0$

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

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Derivation of Maxwell's first equation

Maxwell's first equation is the differential form of Gauss's law of electrostatics.i.e

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

Derivation:

According to Gauss's law for electrostatic-

$\oint_{s} \overrightarrow{E}.\overrightarrow{dS}=\frac{q}{\epsilon_{0}} \qquad(1)$

For continuous charge distribution inside the surface-

$q=\oint_{v}\rho.dV$

Where
$\rho$→Charge density
dV→Small volume

Now substitute the value of $q$ in equation $(1)$ then

$\oint_{s}\overrightarrow{E}.\overrightarrow{dS}=\frac{1}{\epsilon_{0}} \oint_{v}\rho.dV \qquad(2)$

Now according to Gauss's divergence theorem-

$\oint_{s} \overrightarrow{E}.\overrightarrow{dS}= \oint_{v} \overrightarrow{\nabla}.\overrightarrow{E} \qquad (3)$

From equation$(2)$ and equation$(3)$, we can write the above equation-

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

$\oint_{v} \overrightarrow{\nabla}.\overrightarrow{E}- \frac{1}{\epsilon_{0}} \oint_{v}\rho.dV=0 $

$\oint_{v} (\overrightarrow{\nabla}.\overrightarrow{E}- \frac{\rho}{\epsilon_{0}})dV=0 $

On solving the above equation-

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

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

This is Maxwell's first equation.

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