In the current carrying conductor (AOCDEFG) as shown, the magnetic induction at the point $\mathrm{O}$ is…
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.
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