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.

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

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