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13 August, 20:44

The coil in a 60-Hz ac generator has 125 turns, each having an area of 3.0 * 10-2 m2 and is rotated in a uniform 0.12-T magnetic field. What is the peak output voltage of this generator?

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Answers (2)
  1. 13 August, 21:31
    0
    Given that,.

    Frequency

    f = 60Hz

    Number of turns

    N = 125turns

    Surface area of coil

    A = 3 * 10^-2 m²

    Magnetic field

    B = 0.12T

    Voltage peak to peak? I. e the EMF

    EMF is given as

    ε = - dΦ/dt

    Where Φ is magnetic flux and it is given as

    Φ = NBA Cosθ

    Where N is number of turns

    B is magnetic field

    A is the cross sectional area

    And θ is the resulting angle from the dot product of area and magnetic field

    Where θ = ωt and ω = 2πf

    Then, θ = 2πft

    So, your magnetic flux becomes

    Φ = NBA Cos (2πft)

    Now, dΦ / dt = - NBA•2πf Sin (2πft)

    dΦ / dt = - 2πf • NBA Sin (2πft)

    So, ε = - dΦ/dt

    Then,

    ε = 2πf • NBA Sin (2πft)

    So, the maximum peak to peak emf will occur when the sine function is 1

    I. e Sin (2πft) = 1

    So, the required peak to peak emf is

    ε = 2πf • NBA

    Substituting all the given parameters

    ε = 2π * 60 * 125 * 0.12 * 3 * 10^-2

    ε = 169.65 Volts

    The peak to peak voltage is 169.65 V
  2. 13 August, 21:49
    0
    Answer: 169.67 V

    Explanation:

    From Faraday's Law, we know that

    emf = - d (magnetic flux) / d t

    magnetic flux = N B A cos (θ), where

    N = number of turns

    B = magnetic field

    A = area of coils

    cos (θ) = results from dot product of magnetic field vector and area vector.

    θ = a function of time as the coil is rotating.

    emf = N B A (2πf) sin (2π f t)

    f = frequency

    sin 2πft = 1

    The peak output voltage is then

    emf = (125) * (0.12) * (3*10^-2) * (2 * 3.142 * 60)

    emf = 0.45 * 377.04

    emf = 169.67 V
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