Answer:
The force per unit length (N/m) on the top wire is 16.842 N/m
Explanation:
Given;
distance between the two parallel wire, d = 38 cm = 0.38 m
current in the first wire, I₁ = 4.0 kA
current in the second wire, I₂ = 8.0 kA
Force per unit length, between two parallel wires is given as;
[tex]\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d }[/tex]
where;
μ₀ is constant = 4π x 10⁻⁷ T.m/A
Substitute the given values in the above equation and calculate the force per unit length
[tex]\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d } = \frac{4\pi *10^{-7}*4000*8000 }{2\pi *0.38} = 16.842 \ N/m[/tex]
Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m