Ampere's Circuital Law and its Modification

Ampere's Circuital Law Statement:

When the current flows in any infinite long straight conductor then the line integration of the magnetic field around the current-carrying conductor is always equal to the $\mu_{0}$ times of the current.

$\int \overrightarrow{B}. \overrightarrow{dl} = \mu_{\circ} i$


Let us consider, An infinite long straight conductor in which $i$ current is flowing, then the magnetic field at distance $a$ around the straight current carrying conductor
Ampere’s Circuital Law for Infinite Long Straight Current-Carrying Conductor
$B=\frac{\mu_{\circ}}{2 \pi} \frac{i}{a}$

Now the line integral of the magnetic field $B$ in a closed loop is

$\oint \overrightarrow{B}. \overrightarrow{dl} = \oint \frac{\mu_{\circ}}{2 \pi} \frac{i}{a} dl$

$\oint \overrightarrow{B}. \overrightarrow{dl} = \frac{\mu_{\circ}}{2 \pi} \frac{i}{a} \oint dl$

$\oint \overrightarrow{B}. \overrightarrow{dl} = \frac{\mu_{\circ}}{2 \pi} \frac{i}{a} \left(2 \pi a \right) \qquad \left( \because \oint dl= 2 \pi a \right)$

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

Modified Ampere's circuital law:

When the capacitor is placed in between the conductors then a current flows in the capacitor which is known as displacement current $\left(i_{d} \right)$. So modified Ampere's circuital law:

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