Amplitude of the induced voltage across the coil

Discussion in 'Homework Help' started by lee.perrin, Apr 1, 2010.

  1. lee.perrin

    Thread Starter New Member

    Mar 30, 2010
    1
    0
    If anyone can advise that would be great.

    1. The plane of a 5 turn coil of 5mm² cross sectional area is rotating a 1200 r.p.m in a magnetic field of 10mT.

    Q. Calculate the amplitude of the induced voltage across the coil

    Data:

    5 turns
    c = 1200 rpm
    A = 5cm²
    B = 10 mT

    2. φ = N.B.A φ = 5×10×10^-3 ×5 × 10^-2
    φ = 2.5 ×10^-3

    V = dφ/dt or N.A. dB/dt

    The equation thought to have been used is the following but the Length is not vissible to me.

    V/L = c x B c being the speed
    B field
    L lenth of wire

    3. I have tried two avenues please advise

    a: V(t) = -N•d(B•A)/dt = -N•B•dA/dt = N•B•ω•A•sin(ωt)
    V(t) =(5)(10x10^-3)(125.66370599999999)(5)sin(ωt) = 31.415 sin(ωt) volts.

    Or

    b: And using the following, saying t = 60 sec

    Then φ = N.B.A φ = 5×10×10^-3 ×5 × 10^-2
    φ = 2.5 ×10^-3

    And V = dφ/dt or N.A. dB/dt = 4.17x10^-3 V

    My concern here is that 1200rpm was not used.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    For simplicity I would assume a rectangular cross section for the coil.

    Suppose the rectangular side length orthogonally cutting the field is L and the top & bottom length is D. The top & bottom sides are assumed not to be cutting the field and produce no emf.

    The emf producing side of length L sweeps through a circular path length of ∏*D at 1200 rpm or 20 rotations per second. The side L then cuts the field at a maximum velocity of

    v=∏*D*20 m/sec

    The emf per side is then of the form

    e=Emax*sin(ωt)

    AND the total emf = 2*e

    You know the relationship between L, D and the given coil cross sectional area [CSA].

    You should be able to find Emax & ω from the derived rotational velocity v and the other information supplied. Emax will be in terms of L & D which you would then relate to the coil CSA.

    Remember:
    1. The general relationship for induced emf is E=B*L*v for a conductor of length L cutting a field of B Tesla at constant velocity v meters/sec.

    2. In this case the output is a sinusoid since either side of length L is cutting the field at a sinusoidally varying rate throughout the total cycle. But you do know the maximum rate (or velocity) at which the field is being cut.

    NB - There are 5 turns per side - not just one!
     
    Last edited: Apr 2, 2010
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