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

Answer:

$5.12\cdot 10^{-4} N$

Explanation:

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

$F=\frac{\mu_0 I_1 I_2 L}{2 \pi r}$

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:

$F=\frac{(4\pi \cdot 10^{-7})(12)(14)(6.4)}{2 \pi (0.42)}=5.12\cdot 10^{-4} N$

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