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21 July, 00:03

What is the peak emf generated by rotating a a980-turn, 11cm diameter coil in the Earth's  5·10-5  T magnetic field, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 7 ms? g

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  1. 21 July, 03:18
    0
    The peak emf generated by the coil is 418.3 mV

    Explanation:

    Given;

    number of turns, N = 980 turns

    diameter, d = 11 cm = 0.11 m

    magnetic field, B = 5 x 10⁻⁵ T

    time, t = 7 ms = 7 x 10⁻³ s

    peak emf, V₀ = ?

    V₀ = NABω

    Where;

    N is the number of turns

    A is the area

    B is the magnetic field strength

    ω is the angular velocity

    V₀ = NABω and ω = 2πf = 2π/t

    V₀ = NAB2π/t

    A = πd²/4

    V₀ = N x (πd²/4) x B x (2π/t)

    V₀ = 980 x (π x 0.11²/4) x 5 x 10⁻⁵ x (2π/0.007)

    V₀ = 980 x 0.00951 x 5 x 10⁻⁵ x 897.71

    V₀ = 0.4183 V = 418.3 mV

    The peak emf generated by the coil is 418.3 mV
  2. 21 July, 03:48
    0
    E = 426mV

    Explanation:

    Given N = 980

    B = 5*10-⁵ T

    R = 11cm = 0.11m

    Δt = 7ms = 7*10-³s

    Area A = π*D²/4 = π*0.11²/4 = 0.0095m²

    E = - N*ΔBAωSinθ = NBA*2π/t

    The field and the Area are not changing but the angle between the area vector changes from 0 to 90° and Sinθ changes from 0 to 1. Therefore the flux changes from zero to maximum.

    So E = - 980*5.10*10-⁵*0.0095*2π / (7*10-³)

    E = 426*10-³V = 426mV
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