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Copper penny failling between poles of magnet

  1. Mar 24, 2008 #1
    1. The problem statement, all variables and given/known data
    A copper penny falls on a path that takes it between the poles of a magnet. Does it hit the ground going faster, slower, or at the same speed.


    3. The attempt at a solution

    Clearly the penny undergoes a change in magnetic flux. Since it receives a change in magnetic flux we expect that a current either begins within the penny, becomes stronger or reverses. Although there is a slight shift with the current it does not affect the mass of the penny or the effect of its gravitational constant. Because of this I expect there to be no change in velocity of the penny relative to the ground but a shift in velocity of current, I, within the penny. Is this correct?
     
    Last edited: Mar 24, 2008
  2. jcsd
  3. Mar 24, 2008 #2

    G01

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    Not quite. You are correct that there are eddy currents induced in the penny due to the change in flux, but remember that currents loops (think of the eddy currents as forming tiny current loops all over the penny) act as tiny magnetic dipoles. So, will these current loops experience a force from the B field? If so, how does this force affect the current loops, and thus the penny?

    HINT: Try thinking in terms of Lenz's Law.
     
  4. Mar 25, 2008 #3
    Hows about this: By Lenz's law I expect that the penny will do work against the magnet. Since there is no change in potential energy this change has to be through kinetic energy. Since there is also no change in the mass of the penny, it must be velocity that is expected to change. This change will be a negative change which indicates that the velocity of the penny relative to the ground will slow down. The kinetic energy that is lost is converted into heat.
     
  5. Mar 25, 2008 #4

    G01

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    You overall end result is correct, but I think you may be a little confused conceptually.

    In particular, the penny does not do work on the magnet, but the magnetic field from the magnet does work on the penny against the motion of the penny. Other than that, everything sounds good.:smile:
     
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