JEE Main · 2011 · 10 Apr 2011 · Offline · Magnetic Effects of Current

Treat the semicircular sheet as a bundle of fine parallel wires. A strip at angle $\theta$ carries $di=\frac{I}{\pi R}Rd\theta =\frac{I}{\pi }d\theta$, and its field at a point on the axis, $dB=\frac{{\mu }_{0}di}{2\pi R}$, points perpendicular to the line joining the strip to the axis. By symmetry the components along the diameter cancel and only those perpendicular to it survive. Integrating $\frac{{\mu }_{0}I}{2{\pi }^{2}R}\sin \theta$ over $\theta$ from $0$ to $\pi$ gives a factor $2$, so the net field is $B=\frac{{\mu }_{0}I}{{\pi }^{2}R}$.