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26 August, 02:38

An electric generator contains a coil of 99 turns of wire, each forming a rectangular loop 73.9 cm by 34.9 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 2.96 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1200 rev/min about an axis perpendicular to the magnetic field?

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Answers (2)
  1. 26 August, 05:00
    0
    12078.46 V

    Explanation:

    Applying,

    E₀ = BANω ... Equation 1

    Where: E₀ = maximum emf, B = magnetic Field, A = Area of the coil, N = Number of turns of the coil. ω = angular velocity

    Given: N = 99 turns, B = 2.96 T, ω = 1200 rev/min = (1200+0.10472) = 125.664 rad/s

    A = L*W, where L = Length = 93.9 cm = 0.939 m, W = width = 34.9 cm = 0.349 m

    A = (0.939*0.349) = 0.328 m²

    Substitute into equation 1

    E₀ = 99 (2.96) (0.328) (125.664)

    E₀ = 12078.46 V

    Hence the maximum value of the emf produced = 12078.46 V
  2. 26 August, 05:01
    0
    The maximum emf induced in the loop is 9498.268 V

    Explanation:

    Given;

    number of turns of coil, N = 99 turns

    area of the rectangular loop, A = 0.739 m x 0.349 m = 0.2579 m²

    magnetic field strength, B = 2.96 T

    angular speed of the loop, ω = 1200 rev/min

    angular speed of the loop, ω (rad/s) = (2π x 1200) / 60 = 125.68 rad/s

    The maximum value of the emf produced is calculated using the formula below;

    ξ = NABω

    Substitute the given values and calculate the maximum emf induced;

    ξ = (99) (0.2579) (2.96) (125.68)

    ξ = 9498.268 volts

    Therefore, the maximum emf induced in the loop is 9498.268 V
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