Equal current i flows in two segments of a circular loop in the direction shown in figure. Radius of the loop is r. The magnitude of magnetic field induction at the centre of the loop is

Magnetic field induction at O due to current through ACB is ${\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0\mathrm{i\theta }}{4\mathrm{\pi r}}$

It is acting perpendicular to the paper downwards. Magnetic field induction at O due to current through ADB is

${\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{i}(2\mathrm{\pi }-\mathrm{\theta })}{\mathrm{r}}$

It is acting perpendicular to paper upwards.

$\therefore$ Total magnetic field at O due to current loop is

  $\begin{array}{c}\mathrm{B}={\mathrm{B}}_2-{\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{i}}{\mathrm{r}}(2\mathrm{\pi }-\mathrm{\theta })-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{i}}{\mathrm{r}}\mathrm{\theta }\\ =\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi }}\frac{\mathrm{i}}{\mathrm{r}}(\mathrm{\pi }-\mathrm{\theta })\end{array}$