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

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