Question

A vertical straight wire carrying an upward 28-A current exerts an attractive force per unit length of 7.83 X 10 N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?

Answer 1

Answer:

$i_2 = 978750 A$

Since the force between wires is attraction type of force so current must be flowing in upward direction

Explanation:

Force per unit length between two current carrying wires is given by the formula

$F = \frac{\mu_0 i_1 i_2}{2 \pi d}$

here we know that

$F = 7.83 \times 10 N/m$

$d = 7.0 cm = 0.07 m$

$i_1 = 28 A$

now we will have

$F = \frac{4\pi \times 10^{-7} (28.0)(i_2)}{2\pi (0.07)}$

$7.83 \times 10 = \frac{2\times 10^{-7} (28 A)(i_2)}{0.07}$

$i_2 = 978750 A$

Since the force between wires is attraction type of force so current must be flowing in upward direction