Biot-Savart Law & Circular Coils question from JEE Main 2023 (6 Apr, Shift 1)

Answer: 628

1. Each loop: $B_0 = \frac{\mu_0 I}{2R}$; the loops are in perpendicular planes, so $B = \sqrt{2}\,B_0 = \frac{\sqrt{2}\,\mu_0 I}{2R}$.
2. With $I = \sqrt{2}\text{ A}$: $B = \frac{2\mu_0}{2 \times 0.2} = 5\mu_0$.
3. $B = 5 \times 4 \times 3.14 \times 10^{-7} = 62.8 \times 10^{-7} = 628 \times 10^{-8}\text{ T}$.
4. So the answer is $628$ (in units of $10^{-8}\text{ T}$).