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Finding the strobilization of the magnetization at sphrerical coordinates.

  1. Mar 7, 2009 #1
    1. The problem statement, all variables and given/known data

    If the magnetization of a sphere is:
    [tex]\vec{M} = \hat{\phi }\frac {r}{R}sin\theta M_{o}[/tex]

    How much are the captive streams:
    [tex]J_{b}=\vec{\bigtriangledown }\times \vec{M}[/tex]
    [tex]K_{b}=\vec{M}\times \hat{n}[/tex]

    3. The attempt at a solution
    I find that the first is:
    [tex]J_{b} = \frac {2M_{o}}{R}(cos\theta \hat{r} - sin\theta \hat{\theta })[/tex]
    but i don't know if that is correct and the point that i really want help is how to work the K.

    Thank you a lot.
     
  2. jcsd
  3. Mar 7, 2009 #2

    gabbagabbahey

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    I've never heard the term "captive streams" before, in English [itex]\vec{J}_b[/itex] and [itex]\vec{K}_b[/itex] are usually called "bound currents"

    Your solution for the bound volume current [itex]\vec{J}_b[/itex] is correct :smile:....As for the bound surface current [itex]\vec{K}_b[/itex], what is the outward normal direction [itex]\hat{n}[/itex] for a spherical surface?
     
  4. Mar 8, 2009 #3
    Hello gabbagabbahey,
    You are right about the term "bound currents".
    For the surface current [itex] \vec{K}_b [/itex] , the normal direction [itex] \hat{n} [/itex] is always vertical to the spherical surface.But how are the coordinates of [itex] \hat{n} [/itex] to put them to the type of strobilization..??
    Thanks a lot for your interest.
     
  5. Mar 8, 2009 #4

    gabbagabbahey

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    I'm not sure what "strobilization" means.... are you translating this from another language?

    Also, the surface normal is always perpendicular to the surface, not "vertical":wink:....In spherical coordinates, [tex]\hat{n}=\hat{r}[/tex]....do you see why?
     
  6. Mar 8, 2009 #5
    With the word strobilization I mean the [tex]\vec{M}\times \hat{n}[/tex].
    Sorry but my English are not so good with the physics terms..

    Oh yes, i see. So the surface current bound must be:
    [tex]K_{b}=\frac{r}{R}sin\theta M_{o}\hat{\theta }[/tex] ???
     
  7. Mar 8, 2009 #6

    gabbagabbahey

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    looks good to me:approve:

    I guess by "strobilization", you maybe mean "vector cross product"?
     
  8. Mar 8, 2009 #7
    gabbagabbahey first of all, thanks a lot.
    And the next problem is how is the A(x) with these current bounds??
     
  9. Mar 8, 2009 #8

    gabbagabbahey

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    vchris, first of all, your welcome!:smile:

    And next, use the integral equations for the vector potential A:

    [tex]\vec{A}(\vec{r})=\frac{\mu_0}{4\pi} \int_{\mathcal{V}} \frac{\vec{J}_b}{|\vec{r}-\vec{r}'|}dV'+\frac{\mu_0}{4\pi} \oint_{\mathcal{S}} \frac{\vec{K}_b}{|\vec{r}-\vec{r}'|}dA'[/tex]
     
  10. Mar 8, 2009 #9
    I know this formula. I will try to find a solution and i will upload it to be able to compare...!!!!
    Thanks again!!
     
    Last edited: Mar 8, 2009
  11. Mar 9, 2009 #10
    I found that the surface current bound gives:
    [tex]\frac{\mu_oMr}3sin\theta \hat{e_\varphi}[/tex] for r<R
    [tex]\frac{\mu_oMR^3}3\frac{1}{r^2}sin\theta \hat{e_\varphi }[/tex] for r>R
    But i can't find a solution for the other current bound
     
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