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
    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


    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.


    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
    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


    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.

    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