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