Question

In standard household wiring, parallel wires about 1 cm apart carry currents of about 15 A.

Answer 1

Answer:

0.00225 N/m

Explanation:

Parameters given:

Current in first wire, I(1) = 15A

Current in second wire, I(2) = 15A

Distance between two wires, R = 1cm = 0.01m

The force per unit length between two current carrying wires is:

F/L = μ₀I(1)I(2)/2πR

μ₀ = 4π * 10^(-7) Tm/A

F/L = [4π * 10^(-7) * 15 * 15] / (2π * 0.01)

F/L = 2.25 * 10^(-3) N/m or 0.00225 * 10^(-3) N/m

Answer 2

Given Information:

distance between wires = d = 1 cm = 0.01 m

Current in wire_1 = I₁ = 15 A

Current in wire_2 = I₂ = 15 A

Required Information:

Force per unit length = F/L = ?

Answer:

F/L = 0.0045 N/m

Explanation:

When two parallel wires carrying a current I₁ and I₂ are separated by a distance d then the force between these wires is given by

F/L = μ₀I₁ I₂/2πd

Where μ₀= 4πx10⁻⁷ N/A² is the permeability of free space

This force will be attractive when the current in the two wires are in the same direction otherwise, the force will be repulsive.

F/L = 4πx10⁻⁷*15*15/2π*0.01

F/L = 0.0045 N/m

Therefore, the force per unit length between these two wires is 0.0045 N/m

If we know the length of the wires then we will get the force in Newtons.