What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its pl

Question

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.)

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niczorrrr 3 years 2021-07-27T12:44:25+00:00 1 Answers 267 views 0

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  1. Sigrid
    0
    2021-07-27T12:46:14+00:00
    Reply

    Complete question:

    What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

    Answer:

    The peak emf generated by the coil is 15.721 kV

    Explanation:

    Given;

    Radius of coil, r = 0.250 m

    Number of turns, N = 500-turn

    time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

    magnetic field strength, B = 0.425 T

    Induced peak emf = NABω

    where;

    A is the area of the coil

    A = πr²

    ω is angular velocity

    ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s =  60 rev/s

    Induced peak emf = NABω

                                   = 500 x (π x 0.25²) x 0.425 x 376.738

                                   = 15721.16 V

                                   = 15.721 kV

    Therefore, the peak emf generated by the coil is 15.721 kV

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