Question

Two 1-meter long vertical, parallel wires 12.5 cm apart each carry a current. The first wire carries 31 Amps upward. If it exerts an attractive force of 5.1 x 10- 4 N on the second wire, what is the magnitude and direction of the current in the second wire?

Answer 1

The force per unit length between two parallel wires separated by a distance of d is expressed as follows.

F/L = uo*I1*I2 / 2pi*d

I2 = (F/L)(2pi*d/uo*I1)

Given :-

F/L = 5.1 x 10^-4 N

I1 = 31 Amps

d = 0.125 m

uo = 4pi x 10^-7

I2 = (5.1 x 10^-4)[(2pi x 0.125) / (4pi x 10^-7 x 31)]

I2 = 10.28 Amps

The magnetic force between two parallel wires is attractive if the current flow in the two wires is same. The magnetic

force between two parallel wires is repulsive if the current flow in the two wires is opposite to each other.