Question

A straight wire carries a 12-A current eastward and a second straight wire carries a 14-A current westward. The wires are separated by a distance of 42 cm. The force on a 6.4 m length of one of the wires is ________

Answer 1

`5.12\cdot 10^(-4) N`

Step-by-step explanation:

The force exerted between two current-carrying wires is given by

`F=(\mu_0 I_1 I_2 L)/(2 \pi r)`

where

`\mu_0`is the vacuum permeability

I1 and I2 are the two currents

L is the length of the segment of wire on which we want to calculate the force

r is the distance between the wires

In this problem we have:

`I_1 = 12 A\\I_2 = 14 A\\r = 42 cm = 0.42 m\\L = 6.4 m`

Substituting into the formula, we find:

`F=((4\pi \cdot 10^(-7))(12)(14)(6.4))/(2 \pi (0.42))=5.12\cdot 10^(-4) N`

And since the direction of the two currents is opposite, the force between the wires is repulsive.