JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Magnetic Effects of Current

The field at $P$ is the sum of the contributions of the semicircular arc and the two straight pieces. The arc of radius $r$ carries current through a half turn, giving half the central field of a full loop, $B_{\text{arc}}=\frac{{\mu }_{0}i}{4r}$. The upper wire is semi-infinite and its near end sits at the foot of the perpendicular from $P$, so it contributes $B_{\text{wire}}=\frac{{\mu }_{0}i}{4\pi r}$. The lower wire runs along the line through $P$, so its current elements are directed along the line to $P$ and it adds nothing. The surviving contributions act in the same sense, so