JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Magnetic Effects of Current

The current entering at $A$ divides between the two halves in inverse proportion to their resistances, so the path $ABC$ of resistance $r$ carries $\frac{2I}{3}$ and the path $ADC$ of resistance $2r$ carries $\frac{I}{3}$. Each side of the square lies a perpendicular distance $\frac{a}{2}$ from the centre and subtends $45^{\circ }$ on either side, giving a field $\frac{{\mu }_{0}i}{4\pi (a/2)}(2\sin 45^{\circ })=\frac{\sqrt{2}{\mu }_{0}i}{2\pi a}$ per side. The two sides carrying $\frac{2I}{3}$ give $\frac{2\sqrt{2}{\mu }_{0}I}{3\pi a}$ in one sense, and the two carrying $\frac{I}{3}$ give $\frac{\sqrt{2}{\mu }_{0}I}{3\pi a}$ in the opposite sense. Their difference is the resultant field at the centre, $\frac{\sqrt{2}{\mu }_{0}I}{3\pi a}$.