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The Axial and the Transverse Emf

  1. Sep 2, 2010 #1
    Let us consider a rectangular plate moving horizontally in a vertical magnetic field with its length perpendicular to its direction of motion.This may be closely approximated by the wing of a flying aircraft.The axial emf is given by the traditional formula, E=BLV.
    Now when the "axial emf" is being developed a transient current flows from one end to the other along the length of the conductor.The external field should deflect this current and the edges should get charged.Therefore apart from the axial emf(=BLV) we must have a "Transverse Emf" between the edges.
    Let us now connect the tips(ie the ends) of the conductor by a wire that is magnetically shielded.Current now continuously flow along the length of the conductor and the connecting wire due to the action of the axial emf=BLV. This current gets continually deflected by the vertical magnetic field.The situation simply represents the Hall Effect,of course in a moving conductor[assume the velocity to be constant]. So we should have a typical Hall Voltage between the edges.
    I have tried to deduce an alternative expression for the Hall voltage in relation to moving conductors and you will find it in the uploaded presentation.The presentation is based on the article "On Motional Emf".
    Link:http://www.eurojournals.com/ejsr_39_1_11.pdf [Broken]
    There is an interesting issue connected with the traditional axial emf.If observation is made from the moving conductor itself the value of the Lorentz force =0[since v=0].So how does the electron move from one end to the other of the conductor.It is of course due to the effect of relativistic transformations. The moving conductor experiences an electric force which is axial and this drives the electron to the other end of the conductor. This matter has been discussed in the article as well as in the uploaded presentation.
     

    Attached Files:

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 2, 2010 #2
    This appears to be the same principle as the Faraday disk dynamo, described by Faraday ~1831 (Note: this is NOT Faraday's law of induction). The largest homopolar generator unit ever built is the 500 megajoule homopolar generator built by Oliphant at the Australian National University in Canbarra Australia to power a synchrotron in the 1950's and early 1960's. See

    http://en.wikipedia.org/wiki/Homopolar_generator

    It could produce ~4 megawatts for ~ 100 seconds. See construction details and photos at

    http://www.google.com/url?sa=t&sour...4tnCCw&usg=AFQjCNEk9LcgKHhM3GuNsV3jej15Y3Y49w

    Bob S
     
  4. Sep 3, 2010 #3
    In the disc generator the electrons move either from the periphery of the rotating disc towards the center or from the center towards the border. Now as soon as the motion starts it should again get deflected by the magnetic field. The resulting motion again interacts with the magnetic field.But due to the rotation of the disc the cross-radial motion of the electron is always there producing the radial emf due to the action of the Lorentz Force. One could also think of the radial motion,generating a cross radial emf(by the action of the Lorentz force). But this wont be very much useful in the context of the circular configuration of the disc[perhaps]. [But in the case of a conductor having a linear motion we can think of two distinct emfs --the axial and the transverse one].Again one could consider the whole situation from the perspective of the disc itself, that is by attaching the reference frame to the disc .How does the Lorentz Force generate here,considering the zero velocity of the electron?Of course the electric forces are generated by the field transformations[Relativistic] and once the motion starts the magnetic force becomes operative.The situation is more complex and interesting from the theoretical point of view.I have discussed this issue in relation to linear motion[translation] with the help of differential equations[in the article].
    I have also tried to deduce an expression for Transverse Emf[= Hall Voltage]in the steady state condition.Though the expression is different in form from the expression for the typical Hall Voltage it is equivalent to the traditional expression.[Slides 17,18 and 19]

    Transverse Emf=B^2eVl/k

    B: Magnetic Field
    V: Speed of the conductor
    l:width of the conductor
    k:Constant of proportionality[Resistive force proportional to velocity of the electron]
    The above expression may be shown equivalent to the Hall voltage [slides mentioned above].
     
    Last edited: Sep 3, 2010
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