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Vector potential from magnetic field

  1. Nov 5, 2016 #1
    <<Mentor note: Moved from non-homework forum>>

    If a uniform magnetic field ##{\bf{B}}=B_{z}{\bf{\hat{z}}}## exists in a hollow cylinder (with the top and bottom open) with a radius ##R## and axis pointing in the ##z##-direction, then the vector potential

    $${\bf{A}}=\frac{BR^{2}}{2r}{\bf{\hat{\phi}}}?$$

    using Stokes's theorem.

    How can you prove this?
     
    Last edited by a moderator: Nov 6, 2016
  2. jcsd
  3. Nov 5, 2016 #2

    Simon Bridge

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    Start from the definition of the vector potential, in terms of the field, and apply stokes' theorem.
    What is it you are trying to prove exactly and why?
    If you want to prove the vector potential is consistent with the given field, then use the definition to get the feild from the potential.
     
  4. Nov 5, 2016 #3
    I am trying to prove that

    $${\bf{A}}=\frac{BR^{2}}{2r}{\bf{\hat{\phi}}}?$$

    using Stokes's theorem.

    This is something that has appeared in my reading of Sakurai.
     
  5. Nov 5, 2016 #4
    ##\displaystyle{{\bf B}=\nabla \times {\bf A}}##

    ##\displaystyle{\int {\bf B}\cdot{d {\bf l}}=\int (\nabla \times {\bf A})\cdot{d{\bf S}}}##

    ##\displaystyle{\int B_{r}(2\pi R)=\int (0,-\frac{BR^{2}}{2r^{2}},0)\cdot{d{\bf S}}}##

    ##\displaystyle{(0,0,0)=\int (0,-\frac{BR^{2}}{2r^{2}},0)\cdot{d{\bf S}}}##

    Does it look good so far?
     
  6. Nov 5, 2016 #5

    Simon Bridge

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    Third line does not make sense - please show your reasoning.
     
  7. Nov 6, 2016 #6

    Orodruin

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    The second line is not correct.

    Also note that the vector potential generally is not unique but subject to gauge transformations.
     
  8. Nov 6, 2016 #7

    Simon Bridge

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    Yah - that's where the reasoning likely went astray.
     
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