Question

A wire with a weight per unit length of 0.082 N/m is suspended directly above a second wire. The top wire carries a current of 30.6 A and the bottom wire carries a current of 60.4 A. Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. mm

Answer 1

d = 4.5079mm

Step-by-step explanation:

from the question we are given that

`(F)/(L)`= 0.082N/m

I₁= 30.6 A , I₂ = 60.4 A , and μ₀ = 4 π × 10⁻⁷ T · m / A

In order for the system to be in equilibrium, the repulsive magnetic force

per unit length on the top wire must equal the weight per unit length of the wire. Then we have

F/L = μ₀I₁I₂ / 2πd , making d as the subject of formular

we have that d = Lμ₀I₁I₂ / 2πF

= 4 π × 10⁻⁷ × 30.6 ×60.4 / 2π × 0.082

=45079.02× 10⁻⁷

in mm .. 45079.02× 10⁻⁷ × 10³

45079.02 ×10⁻⁴

d = 4.5079mm