Physical Significance of Maxwell's Equations

Physical Significance:

The physical significance of Maxwell's equations obtained from integral form are given below:

Maxwell's First Equation:

1. The total electric displacement through the surface enclosing a volume is equal to the total charge within the volume.

2. It represents Gauss Law.

3. This law is independent of time. Charge acts as source or sink for the lines of electric force.

Maxwell's Second Equation:

1. The total magnetic flux emitting through any closed surface is zero. An isolated magnet do not exist monopoles.

2. There is no source or sink for lines of magnetic force.

3. This is time independent equation.

Maxwell's Third Equation:

1. The electromotive force around the closed path is equal to the time derivative of the magnetic displacement through any surface bounded by the path.

2. This gives relation between electric field $E$ and magnetic induction $B$.

3. This expression is time varying i.e. $E$ is generated by time variation of $B$.

4.This gives relation in variation of E with the time variation of $B$ or $H$.

5. This is a mathematical form of Faraday's law of electromagnetic induction or Lenz's Law.

Maxwell's Fourth Equation:

1. The magneto-motive force around the closed path is equal to the conduction current plus displacement current through any surface bounded by the path.

2. This is time dependent wave equation.

3 This is a mathematical form of Ampere circuital law.

4. Magnetic induction $B$ can be generated from $J$ and time variation of magnetic displacement $D$.

5. This relates the space variation of $B$ with time variation of $D$.

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Characteristics of Electromagnetic Wave

Electromagnetic Wave:

An electromagnetic wave is the combined effect of an electric field and magnetic field which carry energy from one place to another.

When an electric field and the magnetic field are applied perpendicular to each other then a wave propagates perpendicular to both the electric field and the magnetic field. This wave is called the electromagnetic wave.

Characteristics of Electromagnetic Wave:

1.) Electric and Magnetic Fields: Electromagnetic waves are produced through the mutually perpendicular interaction of electric and magnetic fields. The propagation of the wave is also perpendicular to both the electric field and the magnetic field.

2.) Wave Nature of electromagnetic waves: Electromagnetic waves are characterized by their wave-like behavior, so they exhibit the properties such as wavelength, frequency, amplitude, and velocity. This wave-like behavior of electromagnetic waves can undergo phenomena like interference, diffraction, and polarization.

3.) The spectrum of an electromagnetic wave: In the spectrum of electromagnetic waves, All the wavelengths and frequencies of the waves are included such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays etc.

4.) Speed of electromagnetic wave: The speed of electromagnetic waves in free space or vacuum is equal to the speed of light in free space or vacuum, which is approximately 299,792 $Km/s$.

5.) Transverse waves of an electromagnetic wave: Electromagnetic waves are transverse waves which mean that the oscillations of the electric and magnetic fields occur perpendicular to the direction of wave propagation.

6.) Dual nature of electromagnetic wave: Electromagnetic waves have both wave-like and particle-like behavior. They can be described as a stream of particles called photons, each photon carrying a specific amount of energy (quantum). This duality is described by the wave-particle duality principle in quantum mechanics.

7.) Energy transfer in electromagnetic waves: Electromagnetic waves transport energy through space. The amount of energy carried by each wave depends on its frequency. Higher frequency waves, such as gamma rays and X-rays, carry more energy than lower frequency waves like radio waves.

8.) Absorption, Reflection, and Transmission of Electromagnetic waves: Electromagnetic waves can be absorbed by certain materials, reflected off surfaces, or transmitted through transparent substances. The behavior of waves at boundaries depends on factors such as the angle of incidence, the nature of the material, and the frequency of the wave.

9.) Electromagnetic Induction of electromagnetic wave: When electromagnetic waves interact with conductive materials or circuits, they can induce electric currents or voltages. This principle is the basis for technologies like antennas, wireless communication, and electromagnetic sensors.

10.) Electromagnetic Interactions of electromagnetic waves: Electromagnetic waves can interact with matter in various ways, including absorption, scattering, and emission. These interactions are utilized in fields such as optics, spectroscopy, medical imaging, and telecommunications.

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Electric and magnetic field vector are mutually perpendicular to each other in electromagnetic wave

In electromagnetic waves, the electric field vector and magnetic field vector are mutually perpendicular to each other (Proof)

The general solution of the wave equation for the electric field vector and magnetic field vector are respectively given below

$\overrightarrow{E}= E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(1)$

$\overrightarrow{B}= B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(2)$

Here $E_{\circ}$ and $B_{\circ}$ are the complex amplitude of electric field vector $\overrightarrow{E}$ and magnetic field vector $\overrightarrow{B}$ respectively and $\overrightarrow{k}$ is the propagation constant.

Now

$\overrightarrow{\nabla} \times \overrightarrow{E}= \left( \hat{i} \frac{\partial}{\partial x} + \hat{i} \frac{\partial}{\partial x} +\hat{i} \frac{\partial}{\partial x} \right). \left( \hat{i}E_{x} + \hat{j}E_{y} + \hat{k}E_{z} \right) $

$\overrightarrow{\nabla} \times \overrightarrow{E} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ E_{x} & E_{y} & E_{z} \\ \end{vmatrix}$

$\overrightarrow{\nabla} \times \overrightarrow{E} = \hat{i} \left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] - \hat{j} \left[ \frac{\partial E_{z}}{\partial x} -\frac{\partial E_{x}}{\partial z} \right] + \hat{k} \left[ \frac{\partial E_{y}}{\partial x} -\frac{\partial E_{x}}{\partial y } \right] \qquad(3)$

Here $E_{x}$, $E_{y}$ and $E_{z}$ are the component of electric field vector $\overrightarrow{E}$ in $x$, $y$, and $z$ direction. So the component form of the equation $(1)$

$\overrightarrow{E_{x}}= E_{\circ x} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

$\overrightarrow{E_{y}}= E_{\circ y} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

$\overrightarrow{E_{z}}= E_{\circ z} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

We know that:

$\overrightarrow{k}.\overrightarrow{r}=\left( \hat{i}k_{x} + \hat{j}k_{y} + \hat{k}k_{z} \right). \left( \hat{i}x + \hat{j}y +\hat{k}z \right)$

$\overrightarrow{k}.\overrightarrow{r}=x k_{x} + yk_{y} + zk_{z} $

So above equation can be written as:

$\overrightarrow{E_{x}}= E_{\circ x} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(4)$

$\overrightarrow{E_{y}}= E_{\circ y} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(5)$

$\overrightarrow{E_{z}}= E_{\circ z} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(6)$

Now find that derivative from the equation $(4)$, equation $(5)$, and equation $(6)$ then substitute these values in equation $(3)$, So we get

$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = \frac{\partial E_{z}}{\partial y} - \frac{\partial E_{y}}{\partial z} $

$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = \frac{\partial}{\partial y} \left( E_{\circ z} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \right) -\frac{\partial}{\partial z} \left( E_{\circ y} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \right) $

$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = \left(i k_{y} E_{z} - i k_{z} E_{y} \right) $

$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = i \left( k_{y} E_{z} - k_{z} E_{y} \right) \qquad(7)$

Similarly

$\left[ \frac{\partial E_{z}}{\partial x} -\frac{\partial E_{x}}{\partial z} \right] = i \left( k_{x} E_{z} - k_{z} E_{x} \right) \qquad(8)$

$\left[ \frac{\partial E_{y}}{\partial x} -\frac{\partial E_{x}}{\partial y} \right] = i \left( k_{x} E_{y} - k_{y} E_{x} \right) \qquad(9)$

Now substitute the value of equation $(7)$, equation $(8)$, and equation $(9)$ in equation $(3)$

$\overrightarrow{\nabla} \times \overrightarrow{E} = i \left[\hat{i} \left( k_{y} E_{z} - k_{z} E_{y} \right) - \hat{j} \left( k_{x} E_{z} - k_{z} E_{x} \right) _ \hat{k} \left( k_{x} E_{y} - k_{y} E_{x} \right) \right]$

$\overrightarrow{\nabla} \times \overrightarrow{E} = i \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ k_{x} & k_{y} & k_{z} \\ E_{x} & E_{y} & E_{z} \\ \end{vmatrix}$

$\overrightarrow{\nabla} \times \overrightarrow{E} = i \left( \overrightarrow{k} \times \overrightarrow{E} \right) \qquad(10)$

According to Maxwell's third equation

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

$\overrightarrow{\nabla} \times \overrightarrow{E}= - \frac{\partial}{\partial t} \left( B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \right) \quad \left\{From \: equation \: (2)\right\}$

$\overrightarrow{\nabla} \times \overrightarrow{E}= i \omega \overrightarrow{B} \qquad(11)$

From equation $(10)$ and equation $(11)$

$i \left( \overrightarrow{k} \times \overrightarrow{E} \right) = i \omega \overrightarrow{B}$

$ \left( \overrightarrow{k} \times \overrightarrow{E} \right) = \omega \overrightarrow{B} \qquad(12)$

$\therefore$ Magnetic field vector $(\overrightarrow{B})$ is perpendicular to both electric field vector $(\overrightarrow{E})$ and propagation of wave vector $(\overrightarrow{k})$.

Similarly, from $\overrightarrow{\nabla} \times \overrightarrow{B}$, we get

$ \left( \overrightarrow{k} \times \overrightarrow{B} \right) = -\frac{\omega}{c} \overrightarrow{E} \qquad(13)$

Thus, In an electromagnetic wave, the electric field and magnetic field vector are perpendicular to each other and also to the direction of propagation of the wave.

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Transverse Nature of Electromagnetic Wave

Electromagnetic waves are transverse in nature: (Proof)

The general solution of the wave equation for the electric field and magnetic field are respectively given below

$\overrightarrow{E}= E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(1)$

$\overrightarrow{B}= B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(2)$

Here $E_{\circ}$ and $B_{\circ}$ are the complex amplitude of electric field vector $\overrightarrow{E}$ and magnetic field vector $\overrightarrow{B}$ respectively and $\overrightarrow{k}$ is the propagation constant.

Now

$\overrightarrow{\nabla}. \overrightarrow{E}= \left( \hat{i} \frac{\partial}{\partial x} + \hat{i} \frac{\partial}{\partial x} +\hat{i} \frac{\partial}{\partial x} \right). \left( \hat{i}E_{x} + \hat{j}E_{y} + \hat{k}E_{z} \right) $

$\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\partial}{\partial x} \left(E_{x} \right)+ \frac{\partial}{\partial y} \left(E_{y} \right) + \frac{\partial}{\partial z} \left(E_{z} \right) \qquad(3)$

Here $E_{x}$, $E_{y}$ and $E_{z}$ are the component of electric field $\overrightarrow{E}$ in $x$, $y$, and $z$ direction. So the component form of the equation $(1)$

$\overrightarrow{E_{x}}= E_{\circ x} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

$\overrightarrow{E_{y}}= E_{\circ y} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

$\overrightarrow{E_{z}}= E_{\circ z} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

Here

$\overrightarrow{k}.\overrightarrow{r}=\left( \hat{i}k_{x} + \hat{j}k_{y} + \hat{k}k_{z} \right). \left( \hat{i}x + \hat{j}y +\hat{k}z \right)$

$\overrightarrow{k}.\overrightarrow{r}=x k_{x} + yk_{y} + zk_{z} $

So above equation can be written as:

$\overrightarrow{E_{x}}= E_{\circ x} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(4)$

$\overrightarrow{E_{y}}= E_{\circ y} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(5)$

$\overrightarrow{E_{z}}= E_{\circ z} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(6)$

Now find that derivative of the equation $(4)$ along the direction of $x$ then

$\frac{\partial E_{x}}{\partial x} = i k_{x} E_{\circ x} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} $

$\frac{\partial E_{x}}{\partial x} = i k_{x} E_{x} $

Similarly, the derivative of the equation $(5)$, and equation $(6)$ along the direction of $y$ and $z$ then

$\frac{\partial E_{y}}{\partial y} = i k_{y} E_{y} $

$\frac{\partial E_{z}}{\partial z} = i k_{z} E_{z} $

Now substitute the value of $\frac{\partial E_{x}}{\partial x}$, $\frac{\partial E_{y}}{\partial y}$, and $\frac{\partial E_{z}}{\partial z}$ in equation $(3)$

$\overrightarrow{\nabla}. \overrightarrow{E}= i k_{x} E_{x} + i k_{y} E_{y} + i k_{z} E_{z}$

$\overrightarrow{\nabla}. \overrightarrow{E}= i \left( k_{x} E_{x} + k_{y} E_{y} + k_{z} E_{z} \right)$

$\overrightarrow{\nabla}. \overrightarrow{E}= i \left( \hat{i}k_{x} + \hat{j}k_{y} + \hat{k}k_{z} \right). \left( \hat{i}E_{x} + \hat{j}E_{y} + \hat{k} E_{z} \right)$

$\overrightarrow{\nabla}. \overrightarrow{E}= i \left( \overrightarrow {k} . \overrightarrow {E} \right) \qquad(7)$

From Maxwell's first equation in free space:

$\overrightarrow{\nabla}. \overrightarrow{E}= 0 \qquad(8)$

From equation $(7)$ and equation $(8)$

$i \left( \overrightarrow {k} . \overrightarrow {E} \right)=0$

$ \overrightarrow {k} . \overrightarrow {E} = 0$

From the above equation, we can conclude that the electric field is perpendicular to the direction of propagation of the wave i.e. $\overrightarrow {E}\perp \overrightarrow {k}$

Similarly, The same result is obtained from $\overrightarrow{\nabla}. \overrightarrow {B}$ i.e. $ \overrightarrow {k} . \overrightarrow {B} = 0$, So we can conclude that the magnetic field is also perpendicular to the direction of propagation i.e. $\overrightarrow {B}\perp \overrightarrow {k}$

Thus, "The electromagnetic waves are transverse in nature"

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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:

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

The wave equation of electric field vector:

$\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$

The wave equation of magnetic field vector:

$\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$

Now the solution of electromagnetic wave for electric field vector.

Differentiate with respect to $t$ of equation $(3)$

$\frac{\partial \overrightarrow{E}}{\partial t}=i \omega E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

Again differentiate with respect to $t$ of the above equation:

$\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}=i^{2} \omega^{2} E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

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

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.

$\left.\begin{matrix} \frac{\partial^{2} B_{x}}{\partial z^{2}}=- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) B_{x} \\ \frac{\partial^{2}B_{y}}{\partial z^{2}} =- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) B_{y} \end{matrix}\right\} \quad(7)$

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:

$\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{-\beta \overrightarrow{r}}e^i{(\alpha \overrightarrow{r} - \omega t)} \qquad(14)$

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

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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. According to Modified Maxwell's Forth equation:

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

$\overrightarrow{J} = \overrightarrow{\nabla} \times \overrightarrow{H} - \frac{\partial \overrightarrow{D} }{\partial t} \qquad (2)$

Now subtitute the value of $\overrightarrow{J}$ in equation $(1)$. Therefore we get

$W=\overrightarrow{E}.\left( \overrightarrow{\nabla} \times \overrightarrow{H} - \frac{\partial \overrightarrow{D} }{\partial t}\right) \qquad(3) $

Now we employ the vector identity

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

$\overrightarrow{E}.(\overrightarrow{\nabla} \times \overrightarrow{H}) = \overrightarrow{H} (\overrightarrow{\nabla} \times \overrightarrow{E})-\overrightarrow{\nabla}. (\overrightarrow{E} \times \overrightarrow{H})\qquad (4)$

From equation $(3)$ and equation $(4)$

$ 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) $

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

$ W= -\overrightarrow{\nabla}.(\overrightarrow{E} \times \overrightarrow{H}) -\frac{1}{2}\left [2 \overrightarrow{H} \left( \frac{-\partial \overrightarrow{B} }{\partial t}\right) + 2 \overrightarrow{E} \left( \frac{\partial \overrightarrow{D} }{\partial t}\right) \right] $

$ W= -\overrightarrow{\nabla}.(\overrightarrow{E} \times \overrightarrow{H}) -\frac{1}{2}\left [ \overrightarrow{H} \left( \frac{\partial \overrightarrow{B} }{\partial t}\right)+\overrightarrow{H} \left( \frac{\partial \overrightarrow{B} }{\partial t}\right) + \overrightarrow{E} \left( \frac{\partial \overrightarrow{D} }{\partial t}\right)+ \overrightarrow{E} \left( \frac{\partial \overrightarrow{D} }{\partial t}\right) \right] $

Put $B=\mu H$ and $D=\epsilon E$ in the above equation

$ W= -\overrightarrow{\nabla}.(\overrightarrow{E} \times \overrightarrow{H}) -\frac{1}{2}\left [ \overrightarrow{H} \left( \frac{\partial \overrightarrow{B} }{\partial t}\right)+\frac{\overrightarrow{B}}{\mu} \left( \frac{\partial \left(\mu \overrightarrow{H}\right) }{\partial t}\right) + \overrightarrow{E} \left( \frac{\partial \overrightarrow{D} }{\partial t}\right)+ \frac{\overrightarrow{D}}{\epsilon} \left( \frac{\partial \left( \epsilon \overrightarrow{E}\right) }{\partial t}\right) \right] $

$ W= -\overrightarrow{\nabla}.(\overrightarrow{E} \times \overrightarrow{H}) -\frac{1}{2}\left [ \overrightarrow{H} \left( \frac{\partial \overrightarrow{B} }{\partial t}\right)+\overrightarrow{B} \left( \frac{\partial \overrightarrow{H} }{\partial t}\right) + \overrightarrow{E} \left( \frac{\partial \overrightarrow{D} }{\partial t}\right)+ \overrightarrow{D} \left( \frac{\partial \overrightarrow{E} }{\partial t}\right) \right] $

$ W= -\overrightarrow{\nabla}.(\overrightarrow{E} \times \overrightarrow{H}) -\frac{1}{2}\left [ \frac{ \partial \left(\overrightarrow{B}.\overrightarrow{H}\right) }{\partial t} + \frac{ \partial \left( \overrightarrow{E}.\overrightarrow{D}\right) }{\partial t} \right] $

$ W= -\overrightarrow{\nabla}.(\overrightarrow{E} \times \overrightarrow{H}) - \frac{1}{2} \frac{\partial}{\partial t} \left( \overrightarrow{H}.\overrightarrow{B}+\overrightarrow{E}.\overrightarrow{D} \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) \qquad \left( \because W= \overrightarrow{J}.\overrightarrow{E} \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}$ $\rightarrow$ The flow of energy

$U$ $\rightarrow$ Energy density of electromagnetic filed

$S$ $\rightarrow$ 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:

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

The wave equation of electric field vector:

$\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$

The wave equation of magnetic field vector:

$\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$

Now the solution of electromagnetic wave for electric field vector.

Differentiate with respect to $t$ of equation $(3)$

$\frac{\partial \overrightarrow{E}}{\partial t}=i \omega E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

Again differentiate with respect to $t$ of the above equation:

$\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}=i^{2} \omega^{2} E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

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

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.

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

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

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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:

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

The wave equation of electric field vector:

$\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$

The wave equation of magnetic field vector:

$\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$

Now the solution of electromagnetic wave for electric field vector.

Differentiate with respect to $t$ of equation $(3)$

$\frac{\partial \overrightarrow{E}}{\partial t}=i \omega E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

Again differentiate with respect to $t$ of above equation:

$\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}=i^{2} \omega^{2} E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

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

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.

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

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

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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. $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$
   
   
   Modified Form:
   
   
   $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \left(\overrightarrow{J}+ \epsilon \frac{ \partial \overrightarrow{E}}{\partial t} \right)$

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:

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

Modified form for non-conducting media:

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu_{\circ} \epsilon_{\circ} \frac{ \partial \overrightarrow{E}}{\partial t} \qquad(4)$

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 ($\overrightarrow{B}$).

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

Now from Maxwell's 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

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

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

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

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

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 $\overrightarrow{B}$:

Now from Maxwell's equation $(4)$

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu_{\circ} \epsilon_{\circ} \frac{ \partial \overrightarrow{E}}{\partial t}$

Now take the curl on both sides of the above equation

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

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

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}=-\mu \epsilon \frac{\partial}{\partial t} (\frac{\partial \overrightarrow{B}}{\partial t})$

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

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

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

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

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

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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. $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$
   
   
   Modified Form:
   
   
   $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \left(\overrightarrow{J}+ \epsilon \frac{ \partial \overrightarrow{E}}{\partial t} \right)$

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:

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

Modified form for free space:

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu_{\circ} \epsilon_{\circ} \frac{ \partial \overrightarrow{E}}{\partial t} \qquad(4)$

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 ($\overrightarrow{B}$).

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

Now from Maxwell's equation $(3)$ in free space

$\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

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

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

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

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

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

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 $\overrightarrow{B}$:

Now from Maxwell's equation $(4)$

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu_{\circ} \epsilon_{\circ} \frac{ \partial \overrightarrow{E}}{\partial t}$

Now take the curl on both sides of the above equation

$\overrightarrow{\nabla} \times (\overrightarrow{\nabla} \times \overrightarrow{B})=\overrightarrow{\nabla} \times \mu_{\circ} \epsilon_{\circ} \frac{ \partial \overrightarrow{E}}{\partial t} $

$(\overrightarrow{\nabla}. \overrightarrow{B}).\overrightarrow{\nabla} - (\overrightarrow{\nabla}. \overrightarrow{\nabla}).\overrightarrow{B} \\ =\mu_{\circ} \epsilon_{\circ} \frac{\partial}{\partial t} (\overrightarrow{\nabla} \times \overrightarrow{E}) \qquad(6)$

But for free space:

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

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

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

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

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

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

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

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

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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 $\overrightarrow{S}$ 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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