## 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:

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

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.