An infinitely long wire lying along z-axis carries a current I, flowing in positives z-direction. There is no other current. Consider a circle in x-y plane with centre at (2,0,0)m and radius 1m. Divide the circle in small segments and let dl denote the length of a small segment in anticlockwise direction, as shown.

Consider a closed amperian loop ABCOA. No current is passing throught the area of the loop

$\therefore \text{∮}\overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=0\text{ or, }{\int }_{\mathrm{ABC}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}+{\int }_{\mathrm{C}}^{\mathrm{A}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=0$

or, ${\int }_{\mathrm{ABC}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=-{\int }_{\mathrm{C}}^{\mathrm{A}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}$

but the magnetic field is perpendicular to the x axis at all point on it.

$\therefore {\int }_{\mathrm{C}}^{\mathrm{A}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=0;\therefore {\int }_{\mathrm{ABC}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=0$

Now, consider a square shaped Amperian loop PQRS. O is the centre of the given circle.

${\int }_{\text{RSPQR }} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}={\mathrm{\mu }}_0\mathrm{I}$ or ${\int }_{\mathrm{R}}^{\mathrm{S}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}+{\int }_{\mathrm{S}}^{\mathrm{P}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}+{\int }_{\mathrm{P}}^Q \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}+{\int }_Q^{\mathrm{R}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}={\mathrm{\mu }}_0\mathrm{I}$

All 4 integrals must be equal.

$\therefore {\int }_{\mathrm{S}}^{\mathrm{P}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{4};\therefore {\int }_O^{\mathrm{P}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=\frac{1}{2}\frac{{\mathrm{\mu }}_0\mathrm{I}}{4}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{8}$

For the given circle

${\text{∮}}_{\text{OTP }}\overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=0 \Rightarrow {\int }_{\mathrm{O}}^{\mathrm{T}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}+{\int }_{\mathrm{T}}^{\mathrm{P}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}+{\int }_{\mathrm{P}}^{\mathrm{O}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=0$

or, $0+{\int }_{\mathrm{T}}^{\mathrm{P}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=-{\int }_{\mathrm{P}}^{\mathrm{O}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}={\int }_{\mathrm{O}}^{\mathrm{P}} \overrightarrow{\mathrm{B}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{8}$