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Aharonov Bohm effect

  1. Mar 22, 2005 #1
    When we derive the phase difference in the classical experiment of the solenoid as shown below


    Now what if we change the path from the sorce to screen so it follows a rectangular path like this


    the phase difference i.e equation 10 in the slides above will remain the same cause no matter what the path the the closed loop integral of the curl of A remains the same.

    NOw what if I change it a bit, and apply a parpendicular magnetic field throughout the region and my gauge is A=(-By/2, Bx/2,0)

    How will the phase differnce change of the rectangular path.
    and what about if A=(By,0,0) which should be the same as that of A=(-By/2, Bx/2,0)
    Last edited: Mar 22, 2005
  2. jcsd
  3. Mar 22, 2005 #2
    It seems to me that since, in general, the integration is taken over a different area, then the flux will be different, (eqn. 5); and thus will result in a different amount of phase difference, (eqn 10).

    As to your second question; I'm not exactly sure, but it seems to me if you are going to align the axis of the selonoid along the x axis (and the path remains in the x-y plane) then the vector potential should intersect each path with the same rotation --thus no phase difference. However, I'm assuming here the electrons are not spin polarized similarly; if such is the case then [itex]\delta\phi[/itex] may reappear in accordance with some quantum relation. :wink:

    Creator :cool:

    P. S. Nice explanation of AB effect.
    Last edited: Mar 22, 2005
  4. Mar 22, 2005 #3
    actually for the second case. the structure can be thought of inside the solenoid. cause there is a perpendicular magnetic field. IN the first case the path was outside the solenoid and it didn't had magnetic field only vector potential.

    sso i think the question is how does an actual magnetic field change its phase compared to just vector potential
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