Two long parallel wires X and Y, separated by a distance of 6 cm… | JEE Main 2025 (22 Jan, Shift 2)

Q19JEE Main 2025 · 22 JAN · SHIFT 2Medium

Magnetic Field of Straight Wires

Two long parallel wires $X$ and $Y$ , separated by a distance of 6 cm, carry currents of 5 A and 4A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is $x \times 10^{-5}\text{ T}$ . The value of $x$ is _________. Take permeability of free space as $\mu_0 = 4\pi \times 10^{-7}$ SI units.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

P is outside both wires: $10\text{ cm}$ from X and $4\text{ cm}$ from Y. For opposite currents the fields at an outside point are opposite.
$B_X = \frac{2 \times 10^{-7} \times 5}{0.10} = 1 \times 10^{-5}\text{ T}$ , $B_Y = \frac{2 \times 10^{-7} \times 4}{0.04} = 2 \times 10^{-5}\text{ T}$
$B = B_Y - B_X = 1 \times 10^{-5}\text{ T}$ , so $x = 1$

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