Question

Prove ampere's circuital law for a long straight wire carrying a current I

Answer 1

As per the Ampere's circuital law,

The integral of magnetic field along a closed curve equals the sum of the electric current passing thorough the cross-section of the closed curve times permeability.

$\oint \overrightarrow{B} \overrightarrow{dl}=\mu_oI$

Let I current flowing in a current carrying conductor which have the radius r

Current enclosed in a cross section area $\pi r^2$ is $I_{enclosed}=\dfrac{\pi r^2}{\pi a^2}I=\dfrac{r^2}{a^2}I$

Taking magnetic field integral along the loop,

$\oint \overrightarrow{B}\overrightarrow{dl}=\overrightarrow{B}.2\pi\overrightarrow{ r}$

So, $\overrightarrow{B}2\pi r=\mu_o\dfrac{r^2}{a^2}I$

$\overrightarrow{B}=\mu_o\dfrac{rI}{2\pi a^2}$