Respuesta :
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
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.
