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How to derive the curl of E equation in the frame of the conductor?

  1. Jun 12, 2014 #1
    According to wikipedia, "The moving magnet and conductor problem", I stopped at the equation shown in the attachment.
    It said that the curl of the E` ( electric field in the frame of the conductor) is equal to minus of the dot product of the velocity of the conductor and the del multiplied by the magnetic field.

    How to derive this formula?
     

    Attached Files:

  2. jcsd
  3. Jun 13, 2014 #2

    UltrafastPED

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  4. Jun 14, 2014 #3
    But i have 3 concerns:
    1) The above equation should equal to ∇B(x') ∂vt/∂t not to the composite ∇B(x') ° ∂vt/∂t
    2) How (v°∇)B is driven from ∇B(x') ° v?
    3) In wikipedia, it is a dot product (v.∇) B not a composite function (v°∇)B
     
    Last edited: Jun 14, 2014
  5. Jun 14, 2014 #4
    ∂B`(x`)/∂t = ∂B(x`+vt)/∂t

    ∂B/∂t= (∂B/∂x).(∂x/∂t) given that x=x`+vt

    but ∂x/∂t= v

    so ∂B/∂t= (∂B/∂x). v
     
  6. Jun 14, 2014 #5

    UltrafastPED

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    Yeah, a dot not a composition. Couldn't find the dot in the menu. Gotta learn LaTex someday I keep saying to myself!
     
  7. Jun 14, 2014 #6
    Fine, so again how (v.∇) B is reached? it should be v (∇.B). In other words, the only operator that is acting on B should be ∇ not (v.∇).
     
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