Faraday Disc

Discussion in 'Math' started by BCELECTRIC, Jul 9, 2014.

  1. BCELECTRIC

    Thread Starter New Member

    Jun 18, 2014
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    I have a copper disc that is 10" in diameter and 1" thick if I spin this at 2500 rpm's what kind of voltage and current can I get off this disc.

    Thanks BC
     
  2. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    That depends on the local strength of terrestrial magnetism in your area, the orientation of your disc AND the number and distribution of your current pick off brushes. The Earths magnetic field is an extremely weak source for exciting induction in a generator.
     
  3. BCELECTRIC

    Thread Starter New Member

    Jun 18, 2014
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    I am open to any suggestions you might have, I am looking for the brushes that would work best this project. Can I add electronics to help with the excitation?
     
  4. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    Faraday used copper/brass leaf brushes. Something along those lines would be fine.

    The formula is also quite simple.

    Assuming you are aligned with the north south horizontal component of the earths field and supposing the earths field to be .2 cgs units of magnetic flux.

    H(flux) * n(r.p.sec. of disc) * pi * radius(squared)(must be in cm.)
    divided by 100,000,000

    equals volts generated.

    This will be about .005 or less volts.

    You REALLY need a stronger magnetic field to generate any kind of measurable voltage.
     
  5. LordKelvin001

    New Member

    Jul 11, 2014
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    Hello I am trying to pin down the voltage/current generated With a 10 inch (25.4 cm) Faraday copper disk. Between two magnets with a 2600 Gauss each. It will be moving on the slow end of 2500 RPM; and the fast end would be 5000 RPM. I could really use a complete brake down of the the math processes.Thank you
     
  6. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    In the old formula I gave here, I used Gaussian (cgs) units.

    It is really simple to figure.

    Revs per second of disc. RPM's divided by 60.
    We all know pi...
    radius squared(in centimeters)
    Flux density which you have given in Gauss(cgs units)

    Divide by 100,000,000

    Ans. is in volts.

    CURRENT will be another matter. Counter EMF will distort your field changing the calculations of your output. The higher the current drawn the stronger the distortion(and the work needed to maintain the rotational velocity). Homopolar generators can generate enormous amounts of current, and to counter the distortion caused by those currents, working generators of days past would purposefully drive the field magnets VERY deeply into magnetic saturation. This prevented the counter EMF fields from having as much affect on the stator field and proved to be beneficial to this type of generator.

    The current possible in your case will be sufficient to stall your driving source unless you have it hooked up to a V8 motor or something equally ridiculous.
    Mainly, the resistance of your load will determine how much current you can draw. That and how many horsepower you are willing to hook up to it. :)
     
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