Question

Two long current-carrying wires run parallel to each other and are separated by a distance of 5.00 cm. If the current in one wire is 1.65 A and the current in the other wire is 3.25 A running in the opposite direction, determine the magnitude and direction of the force per unit length the wires exert on each other.

Answer 1

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;

$\frac{F}{l} =\frac{\mu_0 I_1 I_2}{2\pi r} \\\\\frac{F}{l} =\frac{4\pi \times 10^{-7} \times 1.65 \times 3.25}{2\pi \times 0.05} \\\\\frac{F}{l} = 2.145 \times 10^{-5} \ N/m$

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.