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.

Read More

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.

Read More

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.

Read More

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 $(2)$

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

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

Read More

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-

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

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

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

$\oint_{v} \overrightarrow{\nabla}.\overrightarrow{E} dV- \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.

Read More

Conservative force and non conservative force

Conservative force: There are the following points that describe the conservative force-

1.) The conservative force depends only on the position of the particle and does not depend on the path of the particle.

2.) In a conservative force, the kinetic energy of the particle does not change between the positions.

Let us consider a particle is moving from position $A$ to position $B$ under the conservative force. If the kinetic energy of the particle at position $A$ and $B$ is $K_{i}$ ,$K_{f}$ respectively then for conservative force-

$K_{i}=K_{f}$

3.) In conservative force, the work done by the force in completing one round between any two positions is zero.

According to the work-energy theorem

$W=\Delta{K}$

$W=K_{f}-K_{i} \quad(1)$

For a conservative force, the kinetic energy of the particle does not change between the positions of the particle i.e., $ K_{f}=K_{i}$. So from equation $(1)$

$W=0$

Alternative Method:

The above statement can also be proven by the following method-

The work is done by the conservative force to move a particle from position $A$ to position $B$ = $W_{AB}$.

The work done by the conservative force to move a particle from position $B$ to position $A$ = $W_{BA}$

We know that the kinetic energy of the particle does not change between the positions, so the work done by the force between the positions will be equal and opposite. i.e

$W_{AB}=-W_{BA}$

$W_{AB}+W_{BA}=0$

So, from above, we can conclude that the net work done by force between the two positions in completing one round is zero.
4.) The central force is also known as the conservative force.
5.) The conservative force is always equal to the negative gradient of potential energy.

$\overrightarrow{F}=-\overrightarrow{\nabla}U$

$\overrightarrow{F}= -\frac{dU}{dr}$

Non-conservative force:

1.) The non-conservative force is path-dependent between the two positions and does not depend on the positions of the particle.

2.) The kinetic energy of the particle varies from one position to another due to friction.

3.) For a non-conservative force, the work done by the force in completing one round between two positions is not zero.

4.) This is not a central force.

5.) The non-conservative force does not equal the negative gradient of potential energy.

Read More