A current of 19.5 mA is maintained in a single circular loop with a circumference of 2.30 m. A magnetic field of 0.710 T is directed paralle

Question

A current of 19.5 mA is maintained in a single circular loop with a circumference of 2.30 m. A magnetic field of 0.710 T is directed parallel to the plane of the loop. What is the magnitude of the torque exerted by the magnetic field on the loop?

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Thanh Thu 3 years 2021-08-17T18:29:14+00:00 1 Answers 4 views 0

Answers ( )

  1. haidang
    0
    2021-08-17T18:30:24+00:00
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    Answer:

    0.0058 Nm

    Explanation:

    Parameters given:

    Number of turns, N = 1

    Current, I = 19.5 mA = 0.0195 A

    Circumference = 2.3 m

    Magnetic field, B = 0.710 T

    Angle, θ = 90°

    Magnetic torque is given as:

    τ = N * I * A * B * sinθ

    Where A is area

    Circumference is given as:

    C = 2 * pi * r

    => 2.3 = 2 * pi * r

    => r = 2.3/(2 * 3.142)

    r = 0.366 m

    Area, A can now be found:

    A = pi * r² = 3.142 * 0.366²

    A = 0.42 m²

    Therefore, torque is:

    τ = 1 * 0.0195 * 0.42 * 0.71 * sin90

    τ = 0.0058 Nm

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