Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2011 (AIEEE)

Answer: (d)

1. Split the half ring into thin strips of angle $d\theta$, each a long wire with current $dI = \frac{I}{\pi} d\theta$ at distance $R$.
2. $dB = \frac{\mu_0 dI}{2\pi R}$; the components along the symmetry line cancel, the perpendicular ones add: $B = \int_0^{\pi} \frac{\mu_0 I}{2\pi^2 R} \sin\theta \, d\theta$.
3. $B = \frac{\mu_0 I}{2\pi^2 R} \times 2 = \frac{\mu_0 I}{\pi^2 R}$
4. So the answer is (d) $\frac{\mu_0 I}{\pi^2 R}$.