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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Calantha 5 months 2021-08-05T17:36:34+00:00 1 Answers 15 views 0

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    2021-08-05T17:38:21+00:00

    Answer:

    The greater of the two currents is 0.692 A

    Explanation:

    Given;

    distance between the two parallel wires; r = 6 mm = 6 x 10⁻³ m

    let the current in the first wire = I₁

    then, the current in the second wire = 2I₁

    length of the wires, L = 3.0 m

    magnitude of force on the wires, F = 8 μN = 8 x 10⁻⁶ N

    The magnitude of force on the two parallel wires is given by;

    F = \frac{\mu_o I_1(2I_1)}{2\pi r}\\\\F = \frac{\mu_o 2I_1^2}{2\pi r}\\\\I_1^2 = \frac{F*2\pi r}{2\mu_o} \\\\I_1^2 = \frac{8*10^{-6}*2\pi (6*10^{-3})}{2(4\pi*10^{-7})}\\\\I_1^2 = 0.12\\\\I_1 = \sqrt{0.12}\\\\ I_1 =0.346 \ A

    the current in the second wire = 2I₁ = 2 x 0.346 A = 0.692 A

    Therefore, the greater of the two currents is 0.692 A

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