Answer:
The magnitude and direction of the force per unit length of one wire on the other is 2 x 10⁻⁵ N/m, attractive force.
Explanation:
Given;
distance between the two parallel wires, r = 10 cm = 0.1 m
current in the first wire, I₁ = 2A
current in the second wire, I₂ = 5 A
The force per unit length on each wire can be calculated as;
[tex]\frac{F}{L} = \frac{\mu_oI_1I_2}{2\pi r}[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]\frac{F}{L} = \frac{\mu_oI_1I_2}{2\pi r} \\\\\frac{F}{L} = \frac{4\pi*10^{-7}*2*5}{2\pi *0.1} \\\\\frac{F}{L} = 2 *10^{-5} \ N[/tex]/ m
The direction of the force between the two wires is attractive since the current in the two wires are in opposite direction.
Therefore, the magnitude and direction of the force per unit length of one wire on the other is 2 x 10⁻⁵ N/m, attractive force.
