Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Current flowing in wire A, Ia = 8.0 A

Current flowing in wire B, Ib = 5.0 A

Distance between the two wires, r = 4.0 cm = 0.04 m

Length of a section of wire A, L = 10 cm = 0.1 m

Force exerted on length L due to the magnetic field is given as:

F = μ0 Ia Ib L/2 πr

Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹

F  = (4π x 10⁻⁷  x 8 x 5 x 0.10 )/2 π x 0.04  = 2 x 10⁻⁵

The magnitude of force is 2 x 10^-5 N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.