Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of the force o

Question

Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of the force on a 3.0-m length of one of the wires is equal to 8.0 μN, what is the greater of the two currents?

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Thành Công 3 years 2021-08-05T07:12:02+00:00 1 Answers 17 views 0

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  1. thienhuong
    0
    2021-08-05T07:13:25+00:00
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    Answer:

    Explanation:

    Magnitude of force per unit length of wire on each of wires

    = μ₀ x 2 i₁ x i₂ / 4π r    where i₁ and i₂ are current in the two wires , r is distance between the two and  μ₀ is permeability .

    Putting the values ,

    force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )

    = .67 i² x 10⁻⁴

    force on 3 m length

    = 3 x .67 x 10⁻⁴ i²

    Given ,

    8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²

    i²  = 3.98 x 10⁻²

    i = 1.995 x 10⁻¹

    = .1995

    =  0.2 A approx .

    2 i = .4 A Ans .

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