JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Magnetic Effects of Current

Add the field at the centre from the three pieces of the wire. Each straight arm is semi-infinite and tangent to the circle, so its perpendicular distance from the centre is $R$; a semi-infinite wire whose near end subtends $45^{\circ }$ gives $\frac{{\mu }_{0}I}{4\pi R}\left(1-\frac{1}{\sqrt{2}}\right)$ from the incoming arm and $\frac{{\mu }_{0}I}{4\pi R}\left(1+\frac{1}{\sqrt{2}}\right)$ from the outgoing arm, while the full circular turn contributes $\frac{{\mu }_{0}I}{2R}$. Taking the loop and the outgoing arm out of the page and the incoming arm into the page, the net field is $-\frac{{\mu }_{0}I}{4\pi R}\left(1-\frac{1}{\sqrt{2}}\right)+\frac{{\mu }_{0}I}{2R}+\frac{{\mu }_{0}I}{4\pi R}\left(1+\frac{1}{\sqrt{2}}\right)=\frac{{\mu }_{0}I}{4\pi R}\left(\frac{2}{\sqrt{2}}+2\pi \right)=\frac{{\mu }_{0}I}{2\pi R}\left(\pi +\frac{1}{\sqrt{2}}\right)$.