Question

In the figure below, five long parallel wires in an xy plane are separated by distance d = 55.0 cm. The currents into the page are i₁ = 2.70 A, i₃ = 0.450 A, i₄ = 4.80 A, and i₅ = 2.70 A; the current out of the page is i₂ = 5.40 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

Before going into the problem, let me state that

Parallel currents (those in the same direction )attract and anti-parallel (those in the opposite direction) currents repel.

Saying that, the wire 3 would experience attraction to wires 1, 4 and 5 and repulsion from wire 2.

The net force per unit length is just the sum of interactions with all other wires.

Force per unit length between two current carrying wires is given by F_ab = (μ_o / 2π ) (I_aI_b /d )

Therefore,

Force per unit length on wire 3:

F₃ = F₁₃ + F₂₃ + F₄₃ + F₅₃

F₁₃ is attractive force hence pulls wire 3 towards left, F₂₃ is repulsive, hence pushes towards right , F₄₃ is attractive , hence pulls wire 3 towards right and F₅₃ is attractive and pulls wire 3 towards right.

Taking right as +ve and left as -ve

F₃ =  (μ_o / 2π ) (I₁I₃ /2d) +   (μ_o / 2π ) (I₂I₃ /d ) +  (μ_o / 2π ) (I₄I₃ /d ) +  (μ_o / 2π ) (I₅I₃ /2d)

F₃ =  (μ_o / 2π ) ( (I₁I₃ /2d) + (I₂I₃ /d ) + (I₄I₃ /d ) + (I₅I₃ /2d) )

F₃ = 2 x 10^-7 ( - 2.7x0.45 /2d + 5.4 x 0.45 /d + 4.8 x 0.45 /d + 2.7 x0.45/ 2d)

F₃ = 2x 10^-7 (4.59/ d)

F₃ = 2x 10^-7 ( 4.59/ 0.55)

F₃ = 16.6909 x 10^-7  N