Question

Two long current-carrying wires run parallel to each other and are separated by a distance of 4.00 cm. If the current in one wire is 1.50 A and the current in the other wire is 3.15 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 force per unit length the wires exert on each other is 2.36 × 10^-5 N/m.

Since the wires carry currents running in the opposite direction, the force is therefore repulsive.

Explanation: Please see the attachments below

Answer 2

Answer:

The magnitude of the force F = $- 2.363*10^{-5} \ \ N$

The direction of the force is in opposite direction.

Explanation:

The expression for force per unit length between two parallel wires carrying current $I_1$ and $I_2$ can be written as:

$F= \frac{\mu_oI_1I_2}{2 \pi a}$

$\mu_o =$ permeability constant = $4 \pi * 10 ^{-7} \ H/m$

a = distance between the wires = 4.00 cm = 0.04 cm

$I_1 = 1.50 \ A$

$I_2 = 3.15 \ A$

replacing our values into the above equation; we have:

$F= \frac{4 \pi*10^{-7}*1.5*3.15}{2 \pi*0.04}$

$F = 2.363*10^{-5} \ \ N$

Hence, If the current flows in the same direction, then , the force is said to be attractive (+ve) . However, if the direction of the current flow is opposite; then the force is said to be repulsive (-ve).

From the question given, the current flows in opposite direction in the wires, thus the force is said to be repulsive.

Thus ; F = $- 2.363*10^{-5} \ \ N$