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15 August, 05:04

Two long parallel wires carry currents of 3.35 A and 6.99 A. The magnitude of the force per unit length acting on each wire is 6.03 * 10 - 5 N / m. Find the separation distance d of the wires expressed in millimeters.

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Answers (2)
  1. 15 August, 05:26
    0
    244mm

    Explanation:

    I₁ = 3.35A

    I₂ = 6.99A

    μ₀ = 4π*10^-7

    force per unit length (F/L) = 6.03*10⁻⁵N/m

    B = (μ₀ I₁ I₂) / 2πr ... equation i

    B = F / L ... equation ii

    equating equation i & ii,

    F / L = (μ₀ I₁ I₂) / 2πr

    Note F/L = B = F

    F = (μ₀ I₁ I₂) / 2πr

    2πr*F = (μ₀ I₁ I₂)

    r = (μ₀ I₁ I₂) / 2πF

    r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

    r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

    r = 0.244m = 244mm

    The distance between the wires is 244m
  2. 15 August, 07:44
    0
    d = 77.5mm

    Explanation:

    The expression used for calculating the force between two parallel wires carrying currents separated by a distance is;

    F12 = µµol1I2L/2πd where

    I1 and I2 are the currents in both wires

    L is the length of the wires

    d is the separation distance

    F12 is the magnitude of the force between them.

    µo is the constant of proportionality

    The equation can be rewritten as;

    F12/L = µµol1I2/2πd

    F12/L is the force per unit length acting on each wire = 6.03*10^-5 N/m

    I1 = 3.35A

    I2 = 6.99A

    µ = 1

    µo = 4π*10^-7N/A²

    d = ?

    Substituting this datas in the equation above to get the separation distance d we have;

    6.03*10^-5 = 1 * 4π*10^-7 * 3.35 * 6.99/2πd

    6.03*10^-5 = 4*10^-7*23.42/2d

    6.03*10^-5 = 9.37*10^-6/2d

    2d = 9.37*10^-6/6.03*10^-5

    2d = 1.55 * 10^-1

    d = (1.55 * 10^-1) / 2

    d = 0.155/2

    d = 0.0775m

    Since 1m = 1000mm

    0.0775m = (0.0775*1000) mm

    d = 77.5mm
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