Two long parallel wires carrying currents 4 A and 3 A in opposite directions are placed at a distance of 5 cm from each other. A point P is at equidistance from both the wires such that the line joining the point P to the wires are perpendicular to each other. The magnitude of magnetic field at point P is ( ₀= permeability of free space =4 10⁻⁷ SI unit)

Step-by-step solution

Point P is equidistant from both wires, with vectors connecting it to the wires perpendicular. Setting the wires at (0,0) and (d,0) with d = 0.05 m , the equidistance condition x^2 + y^2 = (x-d)^2 + y^2 yields x = d/2 . Imposing perpendicularity of the vectors -x i - y j and (d-x) i - y j gives -xd + x^2 + y^2 = 0 , which becomes -d^2/2 + d^2/4 + y^2 = 0 after substituting x = d/2 , so y = d/2 . Choosing P = (0.025, 0.025) m , the distance from each wire is r = (0.025)^2 + (0.025)^2 = 0.025 2 = d/ 2 . The magnetic