mutual inductance

Discussion in 'Homework Help' started by lemon, Feb 13, 2010.

  1. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    Hi:
    a) The flux linking a coil of 2000 turns fall from 0.25Wb to zero in 12.0s. What is the magnitude of the induced emf in the coil?

    emf=d(Nδ)/dt=(2000x0.25)/12=41.67v

    b) The current in one coil of a pair of coils changes at 0.85A/s. Calculate the induced emf in the other coil if this coil is open circuit and the mutual inductance of the pair is 400mH.

    Vs=-M x dip/dt=-400x10^-3 x 0.85 = -0.34V

    Would somebody kindly check this please?:)
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    You are correct. :)
     
  3. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    yeah!!
    I think that is the only time this semester
     
  4. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    Thanks for checking mik3
    But should the answer to b) -0.34v be a positive number?
    Or is it ok that voltage is negative as this is just showing the direction of the induced emf?
     
  5. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    It is ok to be negative since it has to cause a current (if possible) opposite to the current creating the magnetic field as to cancel the magnetic field which induced the emf (Lenz's law).
     
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