JEE Main 2025 Apr 8 Shift 2, Physics Q14: Magnetic Field due to Electric Current

Current entering at A splits between the two parallel paths ABC (resistance $r$) and ADC (resistance $2r$) in inverse proportion to their resistances.

So $i_1=\frac{2I}{3}$ through ABC and $i_2=\frac{I}{3}$ through ADC.

For a finite straight wire, the magnetic field at perpendicular distance $d$ from its midpoint is $B=\frac{{\mu }_{0}i}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$; here each side of the square subtends $45^{\circ }$ at the centre, and the perpendicular distance from the centre to each side is $a/2$.

Field magnitude from one side carrying current $i$: $B=\frac{{\mu }_{0}i}{4\pi (a/2)}(2\sin 45^{\circ })=\frac{{\mu }_{0}i\sqrt{2}}{2\pi a}$.

By the right-hand rule, both sides of the ABC branch produce fields pointing in the same direction (say $+\hat{k}$), and both sides of the ADC branch produce fields in the opposite direction ($-\hat{k}$), because the two currents circulate opposite ways around the centre.

Net field: $B=\frac{{\mu }_0\sqrt{2}}{2\pi a}\left[2\cdot \frac{2I}{3}-2\cdot \frac{I}{3}\right]=\frac{{\mu }_0\sqrt{2}}{2\pi a}\cdot \frac{2I}{3}=\frac{\sqrt{2}{\mu }_{0}I}{3\pi a}$, matching option (3).