Question

Two parallel wires are separated by 6.65 cm, each carrying 3.25 A of current in the same direction. (a) What is the magnitude of the force per unit length between the wires?

Answer 1

The magnitude of the force per unit of lenght between the wires are of F/L= 3.17 * 10⁻⁵ N/m.

Step-by-step explanation:

d=0.0665m

I1=I2= 3.25A

μo= 4π * 10⁻⁷ N/A²

F/L= (μo * I1 * I2) / (2π * d)

F/L= 3.17 * 10⁻⁵ N/m