A current I flow through a long straight conductor which is bent into a circular loop of radius R in the middle as shown in the figure.

The magnitude of the net magnetic field at point O will be

The magnetic field due to the straight conductor at point O is given by

$$\begin{gathered} {\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi R}}\otimes \\ \mathrm{and} \mathrm{that} \mathrm{due} \mathrm{to} \mathrm{circular} \mathrm{loop} \mathrm{of} \mathrm{radius} \mathrm{R} \mathrm{is} \\ {\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}} \\ \mathrm{As}, {\mathrm{B}}_1>{\mathrm{B}}_1 \therefore \mathrm{The} \mathrm{magnitude} \mathrm{of} \mathrm{net} \mathrm{magnetic} \mathrm{field} \mathrm{at} \mathrm{point} \mathrm{O} \mathrm{is} \\ {\mathrm{B}}_{\mathrm{net}}={\mathrm{B}}_2-{\mathrm{B}}_1=\frac{\mathrm{\mu }{1}_0\mathrm{I}}{2\mathrm{R}}-\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi R}} \\ =\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}}\left(1-\frac{1}{\mathrm{\pi }}\right) \end{gathered}$$