Question

State Ampere's circuital law and arrive at the expression for the magnetic field near a straight infinite current carrying wire.

Answer 1

The line integral of $\overrightarrow{B}.d\overrightarrow{l}$ around any closed path enclosing a surface is ${\mu }_0$ times the total steady current passing through the surface
Consider a straight conductor of infinite length in the plane of the paper
Let $I$ be the current through it from X to Y
Let $P$ be a point at a perpendicular distance $r$ from the straight wire.
Consider a circle of radius $r$ around the wire passing through $P$ as an amperian loop. The magnetic field $\overrightarrow{B}$ at $P$ due to the current in wire is tangential to the circle.
Hence the magneitc field $\overrightarrow{B}$ is parallel to the line element $d\overrightarrow{l}$ at $P$
$\oint \overrightarrow{B}.d\overrightarrow{l}=\oint Bdl\cos \theta =B\oint dl=B\left(2\pi r\right)$
According to Ampere's circuital law, $\oint \overrightarrow{B}.d\overrightarrow{l}={\mu }_{0}I$
$B\left(2\pi r\right)={\mu }_{0}I$
$B=\frac{{\mu }_{0}I}{2\pi r}$
$B=\frac{{\mu }_0}{4\pi }\frac{2I}{r}$