A current carrying loop consists of $3$ identical quarter circles of radius $\mathrm{R}$, lying in the positive quadrants of the $\mathrm{xy}$ , $\mathrm{yz}$ and $\mathrm{zx}$ planes with their centres at the origin, joined together. Find the direction and magnitude of $\mathrm{B}$ at the origin.

I current carrying arc having a radius $\mathrm{R}$ subtend angle $\theta$ at centre point then, magnetic field is given by $\mathrm{B}=\frac{\mu_{0} \mathrm{I} \theta}{4 \pi \mathrm{R}}$.

Magnetic field at $O$ from current carrying loop at $x y$-plane, $\overrightarrow{\mathrm{R}_{\mathrm{n}}}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{R}}\left(\frac{\pi}{2}\right) \hat{k}$

$\overrightarrow{\mathrm{R}_{\mathrm{n}}}=\frac{\mu_{0} \mathrm{I}}{4(2 \mathrm{R})} \hat{k}$

Magnetic field at $\mathrm{O}$ from current carrying loop at $y z$-plane, $\therefore \overrightarrow{\mathrm{R}_{\mathrm{n}}}=\frac{\mu_{0} \mathrm{I}}{4(2 \mathrm{R})} \hat{j}$

Magnetic field at $\mathrm{O}$ from current carrying at $z x$-plane,

$\therefore\overrightarrow{\mathrm{R}_{\mathrm{n}}}=\frac{\mu_{0} \mathrm{I}}{4(2 \mathrm{R})} \hat{j}$ $\therefore \overrightarrow{\mathrm{R}_{\mathrm{n}}} =\overrightarrow{\mathrm{R}_{1}}+\overrightarrow{\mathrm{R}_{2}}+\overrightarrow{\mathrm{R}_{3}}$ $=\frac{\mu_{0} \mathrm{I}}{4(2 \mathrm{R})}(\hat{i}+\hat{j}+\hat{k})$ $\overrightarrow{\mathrm{R}_{\mathrm{n}}}=\frac{\mu_{0} \mathrm{I}}{8 \mathrm{R}}(\hat{i}+\hat{j}+\hat{k})$