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Lenz's Law and counter torque in relation to electrostatic induction

  1. May 20, 2008 #1
    All right, I was just wandering about something. First, when we have a electric generator, than the rotor in it will rotate easily until current is drawn from it, correct? In other words, counter torque won't be produced until current is drawn. If this isn't true, than just stop reading the rest of this post, because it won't make much sense (I read this about generators in my physics book, but i don't quite understand it. Anyone?)

    OK, so now let's attach a metal plate to the positive and negative end of the generator so that each plate is charged to a respective voltage. Take a large strip of conductive metal, and place the strip underneath the positive and negative ends so that a positive charge is electrostatically induced on one end of the strip and a negative charge on the other. Due to this, will not a large current form that goes from the positive to the negative end of the strip of metal (the negative end of the strip is connected to a chassis ground)?

    So here's the question. Assuming no current flows from the positive and negative induction plate, which is connected to the generator, will this not reduce significantly the counter torque of the generator rotor? You're still getting power, but no current is flowing from the generator. I remember reading in my physics book that only when current is drawn from the generator is a significant counter torque induced in the rotor.

    Any thoughts? I probably left out something that i didn't consider but that's why i'm asking! :yuck:Thanks in advance!
     
  2. jcsd
  3. May 21, 2008 #2
    I'm guessing not much of a current forms. Though a voltage may form. In a battery, the positive side is continuously supplied electrons and the negative side supplies them. In this case, this doesn't happen. So an external circuit with electrons being injected into the positive side would have to be created somehow?

    I don't know, any ideas?
     
  4. May 21, 2008 #3
    Displacement current exists. The plates are large so that the capacitance is substantial. The displacement current has an associated magnetic field just as would a conduction current. Displacement or conduction, current is current, and current is always accompanied by a magnetic field. The said field acts in a manner to produce a counter torque.
     
  5. May 21, 2008 #4
    Thanks cabraham for the response!
    Just for clarification, are you saying that current in the metal strip is not present, assuming the strip isn't connected to anything else?

    It's a good point that displacement current will exist and cause a counter torque. I haven't been exposed to this concept yet, so i looked it up. It appears that the magnetic field produced outside the capacitator between the magnetic strip and metal positive/negative plates is given by

    B = u*I/(2*pi*r). Where I = is the displacement current and u is a constant. The displacement current is given by

    I = e*(rate of change of electric flux) where e is a constant.

    So, based on the equations, it appears on first glance that while it would produce a counter torque via the magnetic field of the 'displacement current', this would be less than if 'conduction current' were actually going through the generator wires. The many coil loops in the actual generator would produce a significantly strong magnetic field and cause a greater counter torque?
    Thanks!
     
  6. May 21, 2008 #5
    The displacement current I'm referring to is in the generator's winding. This current has a magnetic field and a counter torque results. This current charges and discharges the capacitor formed by the large plates.

    But now that I've given it some thought, we should consider the phase difference between the generated output voltage and the disp current. The 90 degree phase would result in a different final result. For a resistive load, the Vout and Iout are in phase so that the counter torque exactly opposes the input torque. With disp current due to capacitive loading the counter torque is 90 deg from the input torque. So, the torques result in no work being done. Power is transferred to the capacitor, then back to the generator, back to the cap, etc.

    It makes sense. The net input power, torque times radian speed, is zero on the average for the ideal case. There are losses due to friction, air resistance, copper resistance, core loss, etc. A small net power exists.

    Does this answer your original question? BR.

    Claude
     
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