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31 January, 19:02

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

What's the force per unit length between these wires?

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Answers (2)
  1. 31 January, 19:31
    0
    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.
  2. 31 January, 19:31
    0
    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
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