JEE Main 2025 Jan 22 Shift 2, Physics Q22: Magnetic Field due to Electric Current

Point P is on the line through both wires, on the far side of Y from X. Distance from X to P: $r_X=6+4=10\text{cm}=0.10\text{m}$. Distance from Y to P: $r_Y=4\text{cm}=0.04\text{m}$.

Since the currents are opposite, at point P (outside the pair, on Y's side) the fields produced by X and Y point in opposite directions. Take the field due to Y as positive.

Magnitude from each wire: $B=\frac{{\mu }_{0}I}{2\pi r}$.

Net field at P:

$B_{\text{net}}=\frac{{\mu }_0(4)}{2\pi (0.04)}-\frac{{\mu }_0(5)}{2\pi (0.10)}=\frac{{\mu }_0}{2\pi }\left(\frac{4}{0.04}-\frac{5}{0.10}\right)$

$=\frac{{\mu }_0}{2\pi }(100-50)=\frac{{\mu }_0}{2\pi }(50)$

$=\frac{4\pi \times 10^{-7}}{2\pi }(50)=2\times 10^{-7}\times 50=1\times 10^{-5}\text{T}$.

$\therefore x=1$.