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

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explanation:

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