An infinitely long straight conductor is bent into the shape as shown in the figure, Current in it is i and the radius of the circular loop is r. The magnetic field at its centre is

Magnetic field at O due to ABCD straight conductor $=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}}$
Magnetic field at O due to the BEC circular conductor $=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{r}}$
The fields are in opposite direction. Hence the resultant field at O is $\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{r}}-\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}}(\mathrm{\pi }-1)$