Magnetic Field of Straight Wires question from JEE Main 2025 (8 Apr, Shift 2)

Answer: (b)

1. Current divides inversely as resistance: $i_1 = \frac{2I}{3}$ in ABC, $i_2 = \frac{I}{3}$ in ADC.
2. One side at distance $\frac{a}{2}$: $B_s = \frac{\mu_0 i}{4\pi (a/2)}(2\sin 45^\circ) = \frac{\sqrt{2}\mu_0 i}{2\pi a}$.
3. ABC (clockwise) and ADC (anticlockwise) give opposite fields; the outer straight wires pass through the centre line and give zero.
4. $B = 2\cdot\frac{\sqrt{2}\mu_0}{2\pi a}(i_1 - i_2) = \frac{\sqrt{2}\mu_0}{\pi a}\cdot\frac{I}{3} = \frac{\sqrt{2}\mu_0 I}{3\pi a}$
5. So the answer is (b).