Re: induced emf again

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

On Tue, 30 Apr 2002, kowalskil wrote:

> Suppose I remove the rigid frame over which the rod was
> sliding. The same rod is moved in a vacuum, still
> perpendiculary to the magnetic field B. Are there going to be
> static charges of opposite sign at the ends of the rod or not?

I failed to specifically address this question in the response
that I posted earlier at

 <http://www.csupomona.edu/~ajm/special/rodnrail.pdf>

When the rod moves (without the rigid frame) through the magnetic
field as shown below left, the vxB force is responsible for the
accumulation of static charges at either end of the rod. In this
frame we would say that the charge polarization increases until
the polarization electric field balances the vxB "force."


         |+|                                |+|
         |+|                           ^    |+|
    x    | |-->                  x     |    | |
  B_ext  |-|                   B_ext   |    |-|
         |-|                         E_ext  |-|

     Lab frame                      Rod frame

But, in the frame of the rod itself, the polarization is the
result of the existence of an *external* electric field as shown
above right. This electric field arises as a result of the Lorentz
transformation and has a magnitude given to a good aproximation by
E_ext = vB_ext as long as v << c. In this frame we would say that
the charge polarization increases until the polarization electric
field balances the external electric field.

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm