Answer:
The magnitude of the force per unit length is 2.145 x 10⁻⁵ N/m and the direction of the force is outward or repulsive since the current in the two parallel wires are flowing in opposite direction.
Explanation:
Given;
distance between the parallel wires, r = 5.0 cm = 0.05 m
current in the first wire, I₁ = 1.65 A
current in the second wire, I₂ = 3.25 A
The magnitude of the force per unit length between the two wires is calculated as follows;
[tex]\frac{F}{l} =\frac{\mu_0 I_1 I_2}{2\pi r} \\\\\frac{F}{l} =\frac{4\pi \times 10^{-7} \times 1.65 \times 3.25}{2\pi \times 0.05} \\\\\frac{F}{l} = 2.145 \times 10^{-5} \ N/m[/tex]
Therefore, the magnitude of the force per unit length is 2.145 x 10⁻⁵ N/m and the direction of the force is outward or repulsive since the current in the two parallel wires are flowing in opposite direction.
