Two long parallel wires carry currents of 1.19 A 1.19 A and 4.27 A 4.27 A. The magnitude of the force per unit length acting on each wire is 7.73 * 10 - 5 N / m 7.73*10-5 N/m. Find the separation distance of the wires expressed in millimeters.

Question

Damon Paul

Physics

21 April, 11:16

Two long parallel wires carry currents of 1.19 A 1.19 A and 4.27 A 4.27 A. The magnitude of the force per unit length acting on each wire is 7.73 * 10 - 5 N / m 7.73*10-5 N/m. Find the separation distance of the wires expressed in millimeters.

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Answers (1)

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Camila Clay · Answer 1

Answer: r = 1.33 * 10^-5 mm

Explanation: The force per unit length of 2 current carrying conductor is given by the formulae below.

F/L = (u*I1*I2) / (2πr)

Where F/L = force per unit length = 7.73*10^-5mm = 0.0775 mm.

I1 = current on the first conductor = 4.27A

I2 = current on the second conductor = 1.19A

r = distance between both conductors.

u = permeability of free space = 1.25*10^-6

By substituting parameters, we have that

0.0775 = 1.25*10^-6 * 4.27 * 1.19 / 2πr

By cross multiplication

0.0775 * 2πr = 1.25*10^-6 * 4.27 * 1.19

r = 1.25*10^-6 * 4.27 * 1.19 / 0.0775 * 2π

r = 0.00000635162 / 0.0775 * 2π

r = 1.33 * 10^-5 mm