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Magnetic Field and Work

  1. Apr 3, 2008 #1
    How does a magnetic field do work on the electrons in a wire?

    If F = q v x B, the magnetic field is always perpendicular to the velocity of the free charge. After a small time interval, dx = v dt, so that v and x are in the same direction.

    For the magnetic field to do work on the charge, dW = F dot dx. But F and dx are perpendicular, aren't they?, so no work seems to be done.
  2. jcsd
  3. Apr 3, 2008 #2


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    That's right. Magnetic forces never do any work.
  4. Apr 3, 2008 #3


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    A time varying magnetic field will lead to an electric field that does work.
    An example is Faraday's law.
  5. Apr 3, 2008 #4
    An electric generator does work because the field changes?
  6. Apr 4, 2008 #5


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    A magnetic field acting on a permanent or electric magnetic can do work.
  7. Apr 4, 2008 #6


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    That is how an induction motor works.
    Other motors use a permanent or electric magnet as an armature,
    and a magnetic field does do work on the armature.
  8. Apr 5, 2008 #7
    Pam, thank you for answering.

    You threw me so far. I am now sure you are applying Faraday's law in integral form,
    [tex]\int _{_\partial S} \overline{E} ds = -\int _{S} \partial_{t}\overline{B}dxdy[/tex] ,
    it must surely be.

    This makes some very good sense.

    I could not possibly see how to apply Faraday's law in differential form,

    [tex]\nabla\times \overline{E} = -\partial_{t}\overline{B}[/tex].

    Unfortunately, I still cannot.

    In the vicinity of the wire B is unchanging, we should presume. In such a manner [tex]\partial_{t}\overline{B}[/tex] should be zero as well, in this locally manner. But should this not make [tex]\nabla \times \overline{E}[/tex] zero as well?

    Something I am missing.
  9. Apr 5, 2008 #8
    Can this be understood in terms of a field that doesn't act at a distance?

    Does a locally acting 4-vector potential, A and Dirac's equation of A acting on the phase of the free electrons in a conductor make some sense of it?
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