JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Magnetic Effects of Current

Each straight part of the conductor lies along the line passing through the centre $O$, and a straight wire produces no magnetic field at points on its own line, so only the semicircular arc contributes at $O$. A full circular loop gives $\frac{{\mu }_{0}I}{2R}$ at its centre, and a semicircle, being half of it, gives $B=\frac{{\mu }_{0}I}{4R}$. Substituting $I=3$ A and $R=\frac{\pi }{10}$ m, $B=\frac{(4\pi \times 10^{-7})(3)}{4\times \frac{\pi }{10}}=\frac{4\pi \times 10^{-7}\times 3\times 10}{4\pi }=3\times 10^{-6}$ T. The magnetic field at the centre is $3\mu$T.