Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A 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×10⁻⁵ T. The value of x is . Take permeability of f…

Question

Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A 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×10⁻⁵ T. The value of x is . Take permeability of free space as μ₀=4π×10⁻⁷ SI units.

Accepted Answer

1. P is outside the pair, 4 cm from Y and 10 cm from X. B_X=(2×10⁻⁷×5)/(0.10)=1×10⁻⁵ T, B_Y=(2×10⁻⁷×4)/(0.04)=2×10⁻⁵ T. Opposite currents give opposite fields at an outside point: B=2-1=1×10⁻⁵ T → 1

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