Question

In the figure below, 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?

Answer 1

the magnitude of the net force per unit length acting on wire 3 due to the currents in the other wires
= I₃*B

= I₃*[-μ₀I₁/(2π*2d) + μ₀I₂/(2πd) + μ₀I₄/(2πd) + μ₀I₅/(2π*2d)] (+y direction)

= μ₀I₃/(2πd) * (-I₁/2 + I₂ + I₄ + I₅/2)

= 5.81*10^-7 N/m