JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Magnetic Effects of Current

Wire X carries 5 A and wire Y carries 4 A, in opposite directions, separated by 6 cm. Point P lies 4 cm from Y on the side away from X, so P is $6+4=10$ cm from X.

The magnetic field of a long straight wire at perpendicular distance r is $B=\frac{{\mu }_{0}I}{2\pi r}$.

Field due to X at P (10 cm = 0.10 m):
$B_X=\frac{{\mu }_0(5)}{2\pi (0.10)}=\frac{25{\mu }_0}{\pi }$

Field due to Y at P (4 cm = 0.04 m):
$B_Y=\frac{{\mu }_0(4)}{2\pi (0.04)}=\frac{50{\mu }_0}{\pi }$

Since the currents in X and Y are anti-parallel and P lies beyond Y (outside both wires), by the right-hand rule the two fields at P point in opposite senses, so the resultant is their difference:
$B=B_Y-B_X=\frac{50{\mu }_0}{\pi }-\frac{25{\mu }_0}{\pi }=\frac{25{\mu }_0}{\pi }$

Substituting ${\mu }_0=4\pi \times 10^{-7}$:
$B=\frac{25\times 4\pi \times 10^{-7}}{\pi }=100\times 10^{-7}=1\times 10^{-5}\text{ T}$

So x = 1.