Two long parallel wires that are 0.30 m apart carry currents of 5.0 A and 8.0 A in the opposite direction. Find the magnitude of the force (for 1 meter of length) that each wire exerts on the other wire and indicate if the force is attractive or repulsive.

Answer: 2.66*10^-5 N/m

Explanation: If two current carrying conductor are separated by a distance (r) from each other, they exert a force per unit length on each other (either attractive or repulsive) which generates a magnetic field, the force per unit length is defined by the formulae below.

F/L = (u*I1 * I2) / (2πr)

Where F/L = force per unit length

u = permeability of free space = 1.256*10^-6 mkgs^-2A^-2

I1 = current on first conductor = 5A

I2 = current on second conductor 8A

r = distance between conductors = 0.30m

F/L = (1.256*10^-6 * 5 * 8) / 2 * 3.142 * 0.30

F/L = 0.00005024/1.8849

F/L = 0.0000266 = 2.66*10^-5 N/m