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E.m.f across the ends of a rod rotating in a magnetic field

  1. May 22, 2008 #1
    I am trying to derive the formula for E.m.f across the ends of a rod rotating in a magnetic field when the field is perpendicular to the plane of rotation:

    [tex]
    |\epsilon| = B \omega r^2
    [/tex]

    using the Lorentz Formula only. Suppose, there is a rod of length 'R'. Let, 'n' be the no. of electrons per unit length. Let, at a distance 'r' from the center of rotation, there be a small length element dr. The no. of electrons in this length element is [itex]ndr[/itex]. Also, the velocity, [itex]v = r\omega[/itex] is perpendicular to the field always as it lies in the plane of rotation of the rod. Hence, the Lorentz force here is given as:

    [tex]
    dF = en r\omega B dr
    [/tex]

    What do i do after this? Do integrate it? This integral only talks about a 'net force' on all the electrons. How do I use it to compute the e.m.f across the two ends? Do I first compute work using [itex]\int W = \int F.dr[/itex] and the differentiate w.r.t charge. But, what charge is the Work a function of?
     
  2. jcsd
  3. May 29, 2008 #2

    mukundpa

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    Homework Helper

    I think the EMF will be [tex]\frac{1}{2}[/tex] B [tex]\omega[/tex] r[tex]^{2}[/tex]
     
  4. May 29, 2008 #3

    mukundpa

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    Homework Helper

    The magnetic force on every electron is balanced by the force due to electric field. Thus

    eE = Bev = Be [tex]\omega[/tex] r

    and d[tex]\xi[/tex] = Edr
     
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