Question

Determine the magnitude of the force between two 42 m-long parallel wires separated by 0.03 m, both carrying 6.3 A in the same direction.

Answer 1

The magnitude of the force between the two parallel wires is 0.0111 N.

Step-by-step explanation:

Given;

length of the two parallel wires, L = 42 m

distance between the two wires, r = 0.03 m

current in both wires, I₁, I₂ = 6.3 A

Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

`F = (\mu_0 I_1I_2l)/(2\pi r)\\\\where;\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi *10^(-7) \ T.m/A \\\\F = ((4\pi *10^(-7))(6.3)^2(42))/(2\pi (0.03))\\\\F = 0.0111 \ N`

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.