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.