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

(N/A) The magnetic field due to a current-carrying arc of radius $R$ subtending an angle $\theta$ at the center is given by $B = \frac{\mu_0 I \theta}{4 \pi R}$.
For a quarter circle,$\theta = \frac{\pi}{2}$,so the magnitude of the magnetic field is $B = \frac{\mu_0 I}{8 R}$.
$1$. For the quarter circle in the $xy$-plane,the magnetic field at the origin is $\vec{B}_1 = \frac{\mu_0 I}{8 R} \hat{k}$.
$2$. For the quarter circle in the $yz$-plane,the magnetic field at the origin is $\vec{B}_2 = \frac{\mu_0 I}{8 R} \hat{i}$.
$3$. For the quarter circle in the $zx$-plane,the magnetic field at the origin is $\vec{B}_3 = \frac{\mu_0 I}{8 R} \hat{j}$.
The total magnetic field at the origin is the vector sum: $\vec{B} = \vec{B}_1 + \vec{B}_2 + \vec{B}_3 = \frac{\mu_0 I}{8 R} (\hat{i} + \hat{j} + \hat{k})$.
The magnitude is $|\vec{B}| = \frac{\mu_0 I}{8 R} \sqrt{1^2 + 1^2 + 1^2} = \frac{\sqrt{3} \mu_0 I}{8 R}$.