Biot-Savart Law & Circular Coils question from JEE Main 2019 (9 Jan, Shift 1)

Answer: (d)

1. Radial parts PQ and RS give no field at O. Arcs: $r_1 = 3$ cm, $r_2 = 5$ cm, $\theta = \dfrac{\pi}{4}$, currents in opposite senses.
2. $B = \dfrac{\mu_0 i\theta}{4\pi}\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right) = 10^{-7}\times 10 \times \dfrac{\pi}{4}\left(\dfrac{1}{0.03}-\dfrac{1}{0.05}\right)$
3. $B = 10^{-6} \times 0.785 \times 13.3 \approx 1.0 \times 10^{-5}$ T
4. So the answer is (d) $1.0 \times 10^{-5}$ T.