JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Magnetic Effects of Current

The field at $O$ comes from four long straight current portions, the two horizontal leads and the two perpendicular arms, each lying a perpendicular distance $a$ from $O$ and each running off to infinity from the foot of that perpendicular. A semi-infinite wire whose near end is at the foot of the perpendicular produces $B=\frac{{\mu }_{0}I}{4\pi a}\left(\sin 90^{\circ }+\sin 0^{\circ }\right)=\frac{{\mu }_{0}I}{4\pi a}$. The current sense makes all four contributions point the same way, so they add: $B_O=4\times \frac{{\mu }_{0}I}{4\pi a}=\frac{{\mu }_{0}I}{\pi a}$.