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Electromagnetic induction

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi! I have a problem with this exercise:
    A squared loop of length L and mass M can rotate around one of its vertices and it's on a vertical plane. In the region below the suspension point there is a uniform magnetic field B perpendicular to the plane of the loop.
    If the loop rotates at a constant angular velocity w, find the electromagnetically induced EMF.

    Here's a picture:
    http://www.allfreeportal.com/imghost/thumbs/756421Untitled1.png [Broken]

    3. The attempt at a solution
    [tex]\phi(B)=\frac{L^2B}{2}tan(\omega t)[/tex]
    [tex]EMF=-\dot{\phi}(B)[/tex]

    Now the question: why does the text tell me the mass of the loop?:confused:
    Has it something to do with the fact that the angular velocity is constant?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 2, 2009 #2

    Hao

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    I don't think M is necessary. If there is a second part to the question, it may be relevant there.

    However, you should take a second look at your expression for φ.

    Based on the units you are using, φ is the magnetic flux.
    [tex]\phi = \int B dA[/tex]
    For a uniform field and an area A inside that field,
    [tex]\phi = B \int dA = B A_{inside\hspace B}[/tex]

    According to your φ, it goes to infinity every half period, which would imply that either L or B goes to infinity, which is not the case.

    To find φ properly, it is necessary to express the area of the loop that is in the magnetic field as a function of angle, and hence a function of time. As the loop is a square, the expression may be a little complicated.

    Next, Farady's Law of Induction tells us:
    [tex]EMF = -\frac{d \phi}{d t} \neq -\phi[/tex]

    Note that when the entire loop is completely above or below the suspension point (a whole π radians of its motion per cycle), φ is constant, and hence EMF = 0.
     
    Last edited: Jun 2, 2009
  4. Jun 2, 2009 #3
    Yes, there is a second part: calculate the energy dissipated in one oscillation by the induced current.
     
  5. Jun 2, 2009 #4

    Hao

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    I can only imagine energy being dissipated if the loop has a finite resistance.

    I hope the question gives more information on what the loop is made of.
     
    Last edited: Jun 2, 2009
  6. Jun 2, 2009 #5
    oh sorry.. the loop has a resistance R.
     
  7. Jun 2, 2009 #6

    Hao

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    Once you have found the EMF as a function of time, you will be able to find power, and hence, by integration over one cycle, the energy dissipated.
     
  8. Jun 5, 2009 #7
    Thank you!
     
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