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I Electric Generators and resistance

  1. Jul 15, 2016 #1
    I am curious as to how much energy is lost by resistance that exists between the magnet and the cooper coils.
    For instance, if a the work done to push a magnet through a copper coil tube produces a certain amount of power, how much of work is lost due to the magnet resistance as the magnet is slowed by the copper coils?

    upload_2016-7-15_15-8-22.png
     
  2. jcsd
  3. Jul 15, 2016 #2
    I was looking for a percentage number for the energy lost - thx
     
  4. Jul 16, 2016 #3

    CWatters

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    I think you will find its 100%.

    It seems to me you are asking.... If you do work against a force, what percentage of work are you doing against the force?

    The answer is 100% because you aren't doing work against anything else. There is no other load. No useful output.
     
  5. Jul 17, 2016 #4

    CWatters

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    You have probably seen this set up where a magnet falls through a copper tube...


    With a long tube it should be possible to achieve and measure the constant velocity with which the magnet falls. That would allow you to work out the rate at which potential energy is being dissipated in the copper due to eddy currents instead of it being converted to kinetic energy in the magnet.

    I think its pretty hard to calculate eddy currents and forces from first principles. Perhaps see some of the references in this thread..

    https://www.physicsforums.com/threa...tude-of-eddy-currents-retarding-force.822714/
     
  6. Jul 17, 2016 #5

    Merlin3189

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    While CW is correct and may have completely answered your question, I took you to mean something perhaps slightly different.
    I took to mean that this is some sort of dynamo/generator, with the copper being a solenoid rather than a solid tube. Then the resistance you feel and push against is caused by the current flowing in the coil, which is related to the electrical power produced. If no electrical load is attached, no current flows and there is no resistance*. All the work you do goes into the KE of the moving magnet and no further work is done when it moves with constant speed.
    When a load is connected, a current can flow, you will feel resistance and do work. The work you do will be equal to the total electrical energy dissipated in the circuit. Hopefully most of that will be usefully deployed in the load, but some will be dissipated in the resistance of the coil. The amount will depend on the details of coil and load and the speed at which you move the magnet.
    * You may do some work against air resistance and friction in the bearings which support the magnet shaft. And there will be some eddy current losses as CW mentions, but they will be small for a coil of fine wire wound as a solenoid, rather than a solid cylinder of copper. If these are negligible, then substantially all your work is converted to electrical energy, apart from a little needed to accelerate the magnet and remaining as KE when it leaves the solenoid.
     
  7. Jul 18, 2016 #6
    @Merlin3189 -- wow thank you for all the time to give that excellent explanation. I was thinking there was work being done and that work was in moving the magnet a certain distance forward in the coil/solenoid. I thought the resistance was not such a big factor as you mentioned for a small coil of fine wire wound up.

    I am probably wrong...but your sentence that said if the amount of copper is negligible ("coil of fine wire"), then all the KE to move the magnet would be converted to electrical energy .... are you saying that if the magnet size is large relative to the cooper...and work is done to move this magnet, then most of the KE to move the magnet could be converted to electrical energy? Doesn't the magnet strength/size induce a greater effect on the electrons in the copper? So once energy is used to accelerate the magnet forward, the momentum overcomes the resistance more easily?
     
  8. Jul 19, 2016 #7

    CWatters

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    If there is a load on the coil then yes there would be work done moving the magnet in the coil. If resistive and other losses are low then all of the work done by the applied force is delivered to the load. The applied force also puts some energy into the KE of the magnet (Ke = 0.5mV^2) but in theory you can get that back when the magnet slows down so that's usually ignored.

    No. Momentum doesn't make it easier to deliver power to the load. Momentum isn't a source of energy only a means of storing energy. If the mass of the magnet is increased the driving force has to deliver more energy to accelerate it to the same velocity but in theory more energy can be recovered when it slows down.

    Once it's going at a constant velocity all of the input power is delivered to the load (and losses if not negligible).
     
  9. Jul 20, 2016 #8

    Merlin3189

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    Again, I agree with CW's comments.
    Re. my own post,
    I talked about a fine wire coil versus a solid block of copper, because the eddy current losses are likely to be much smaller. You could still have the same mass of copper, but in a little larger space (allowing for insulation and gaps.)
    In a solid block, it is easy to have large currents circulating inside the copper. When the copper is in the form of a long wire, the induced emfs cause current to flow along the wire and through the load, but there are only short, small loops where current can circulate inside the wire. This is why the iron in transformers is made of multiple thin sheets rather than solid blocks. Eddy currents in the iron, which is a conductor, are smaller, so the power lost is lower.
     
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