Biot-Savart Law & Circular Coils question from JEE Main 2026 (24 Jan, Shift 2)

Answer: (b)

1. Each loop is at axial distance $x = r$ from O: $B = \dfrac{\mu_0 I r^2}{2(2r^2)^{3/2}} = \dfrac{\mu_0 I}{4\sqrt{2}r}$ for current $I$.
2. $B_P = \dfrac{\mu_0 I}{4\sqrt{2}r}$, $B_Q = \dfrac{4\mu_0 I}{4\sqrt{2}r}$
3. Clockwise as seen from O means each field at O points away from O: $B_P$ towards P, $B_Q$ towards Q (opposite).
4. $B_{net} = B_Q - B_P = \dfrac{3\mu_0 I}{4\sqrt{2}r}$ towards Q
5. So the answer is (b).