JEE Advanced 2022 Paper 2, Physics Section 3 Q4: Magnetic Field due to Electric Current

Compute the contribution from each segment and add as vectors along $\hat{k}$.

Semicircle of radius $L/2$ centered at $O$: $B_{\text{semi}}=\frac{{\mu }_{0}I}{4(L/2)}=\frac{{\mu }_{0}I}{2L}$.

Quarter circle of radius $L/4$ centered at $O$: $B_{\text{quarter}}=\frac{{\mu }_{0}I}{8(L/4)}=\frac{{\mu }_{0}I}{2L}$.

Long straight segment at perpendicular distance $L$ from $O$, contributing only over a finite arc that subtends a $45°$ angle at $O$ (so $\sin \theta =1/\sqrt{2}$): $B_{\text{straight}}=\frac{{\mu }_{0}I}{4\pi L}\cdot \frac{1}{\sqrt{2}}=\frac{{\mu }_{0}I}{4\sqrt{2}\pi L}$.

By right-hand rule the currents in all three pieces produce a field into the page ($-\hat{k}$).

Sum: $\vec{B}=-\left(\frac{{\mu }_{0}I}{2L}+\frac{{\mu }_{0}I}{2L}+\frac{{\mu }_{0}I}{4\sqrt{2}\pi L}\right)\hat{k}=-\frac{{\mu }_{0}I}{L}\left(1+\frac{1}{4\sqrt{2}\pi }\right)\hat{k}$. Option (C).