Magnetic Field of Straight Wires question from JEE Main 2024 (1 Feb, Shift 1)

Answer: 72

1. Side $a = \frac{4\pi}{6} = \frac{2\pi}{3}$ m; distance of centre from a side $d = \frac{\sqrt{3}a}{2} = \frac{\pi}{\sqrt{3}}$ m.
2. Each side: $\frac{\mu_0 I}{4\pi d}(2\sin 30^\circ) = \frac{\mu_0 I}{4\pi d}$; six sides: $B = \frac{6 \times 10^{-7} I}{d}$.
3. $B = \frac{6 \times 10^{-7} \times 4\pi\sqrt{3}}{\pi/\sqrt{3}} = 6 \times 10^{-7} \times 12 = 72 \times 10^{-7}$ T
4. So $x = 72$.