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

`i_2 = 978750 A`

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

Step-by-step explanation:

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

`F = (\mu_0 i_1 i_2)/(2 \pi d)`

here we know that

`F = 7.83 * 10 N/m`

`d = 7.0 cm = 0.07 m`

`i_1 = 28 A`

now we will have

`F = (4\pi * 10^(-7) (28.0)(i_2))/(2\pi (0.07))`

`7.83 * 10 = (2* 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