Five long parallel wires in an xy plane are separated by distance d = 40.0 cm. The currents into the page are i1 = 2.00 A, i3 = 0.15 A, i4 = 3.00 A and i5 = 1.50 A; the current out of the page is i2 = 5.00 A. What is the magnitude of the net force per unit length acting on wire 3 due to the currents in the other wires?

The answer is = 5.81*10-7 N/m

Explanation:

From the above question, the first step to take is to find the magnitude of the net force per unit length acting on wire 3 due to the currents in the other wires

Solution

Given

The magnitude of the net force per unit length moving on wire 3 due to the currents in the other wires  is

= I₃* B

= I₃*[-μ₀I₁/(2π*2d) + μ₋₀I₂/(2πd) + μ₀I₄/(2πd) + μ₀I₅/(2π*2d)] (+y direction)

Thus,

= μ₀I₃/(2πd) * (-I₁/2 + I₁ + I₄ + I₅/2)

Therefore, the magnitude of the net force per unit length acting on wire 3 due to the currents in the other wires is = 5.81*10-7 N/m

answer; (d) spontaneous processes are characterized by the overall conversion of order to disorder;

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!