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 3
1 answer:
Answer:
d = 4.5079mm
Explanation:
from the question we are given that
= 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
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