Two straight, parallel wires a and b carry currents in opposite directions, and are separated by a distance d. The magnitude of the force ex

Question

Two straight, parallel wires a and b carry currents in opposite directions, and are separated by a distance d. The magnitude of the force exerted by each wire on a segment of length L of the other wire is F. The vector sum of these forces is

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Latifah 4 years 2021-08-21T16:52:32+00:00 1 Answers 6 views 0

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  1. thaiduong
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    2021-08-21T16:54:01+00:00
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    Answer:

    Zero

    Explanation:

    The magnetic field due to the first wire on the second wire with current i₁ in the first wire at a distance d from the second wire is B₁ = μ₀i₁/2πd.

    The magnetic force due to this field on the second wire of length segment, L and current i₂ is F₁ = Bi₂L = (μ₀i₁/2πd)i₂L = μ₀i₁i₂L/2πd = F

    The magnetic field due to the second wire on the first wire with current i₂ in the first wire at a distance d from the second wire is B₂ = μ₀i₂/2πd.

    The magnetic force due to this field on the first wire of length segment, L and current i₁ is F₂ = Bi₁L = (μ₀i₂/2πd)i₁L = μ₀i₁i₂L/2πd = F

    Since their magnetic fields are in opposite directions, according to the right hand rule, their forces would also be in opposite directions.

    So F₁ = F and F₂ = -F

    So their vector sum F₁ + F₂ = F +(-F) = F – F = 0

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