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

The magnitude of the force F =
`- 2.363*10^(-5) \ \ N`

The direction of the force is in opposite direction.

Step-by-step explanation:

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

`F= (\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= (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`

Answer 2

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