Question

In the figure below, five long parallel wires in an xy plane are separated by distance d = 30.0 cm. The currents into the page are i_1 = 1.90 A, i_3 = 0.500 A, i_4 = 5.00 A, and i_5 = 1.90 A; the current out of the page is i_2 = 3.80 A. What is the magnitude of the net force per unit length acting on wire 3 due to the currents in the other wires?

Answer 1

Parallel wires attract each other if their currents are in the same direction, and repel each other if they are in opposite directions.

The force is = F = μ(I_a)(I_b)L/2π(D) where The distance b/w the wire 'a' and wire 'b' = D =d(a - b)

Also I_{1 =} I₅ = 1.9 A, I₂ = 3.8 A , I₄ = 5.0 A

F₃ = F₃₁ + F₃₂ + F₃₄ + F₃₅ = F₃₂ + F₃₄ (F₃₁ = -F₃₅)

= μI₃L/2*pi*d[I₂ + I₄] = (10^-7 *0.5*L)*8.8/0.3 = 1.46*L*10^-6 N

Put L and find Force on wire 3.