Physics - Magnetism Question with Solution | TestHub

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: $B=\frac{{\mu }_{0}I}{2R}$ 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.