In the current carrying conductor (AOCDEFG) as shown, the magnetic induction at the point \(\mathrm{O}\) is \(\left(\mathrm{R}_1\right.\) and \(\mathrm{R}_2\) are radii of are \(C D\) \& \(E F\) respectively, \(I\) = current in the loop, \(\mu_0=\) permeability of free space)

Using biot savart's law, it can be concluded that the magnetic field along the line of a straight current carrying conductor is zero.
Hence, magnetic fields due to sections AO, OC, DE \& FG are zero at point \(\mathrm{O}\).
Magnetic field at the center of a current carrying loop is given by:
\(B=\frac{\mu_0 I}{2 R}\)
Hence, magnetic field due to section \(\mathrm{CD}\) is:
\(\mathrm{B}_{\mathrm{CD}}=\left(\frac{1}{4}\right) \frac{\mu_0 \mathrm{I}}{2 \mathrm{R}_1}\) inside the plane of paper (using right hand thumb rule)
Here, the factor \(\frac{1}{4}\) appears as only one fourth of the total circumference contributes to the magnetic field.
Similarly, magnetic field due to section EF is:
\(\mathrm{B}_{\mathrm{EF}}=\left(\frac{1}{4}\right) \frac{\mu_0 \mathrm{I}}{2 \mathrm{R}_2}\) inside the plane of paper (using right hand thumb rule)
Total magnetic field at \(\mathrm{O}\) is:
\(\mathrm{B}=\frac{\mu_0 \mathrm{I}}{8}\left(\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_1}\right)\) inside the plane of paper.