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Magnetic Field of Magnetic Dipole Moment

  1. Oct 11, 2008 #1
    I'm seeking some clarification on a topic found in Jackson 'Classical Electrodynamics' Chapter 5. This deals with the magnetic field of a localised current distribution, where the magnetic vector potential is given by
    [tex]\vec{A}(\vec{x}) = \frac{\mu_0}{4\pi} \frac{\vec{m}\times\vec{x}}{\left|\vec{x}\right|^3}[/tex]

    and the magnetic induction [tex]\vec{B}[/tex] is the curl of this [tex]\vec{B} = \nabla \times \vec{A}[/tex].
    I know that the end result should be
    [tex]\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi} \left[\frac{3\vec{n}\left(\vec{n}\cdot\vec{m} - \vec{m}}{\left|\vec{x}\right|^3}\right][/tex]
    Where [tex]\vec{n}[/tex] is the unit vector in the direction of [tex]\vec{x}[/tex]
    But I'm struggling to do this calculation for myself.

    I know that
    [tex]\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi} \left(\nabla \times \left(\frac{\vec{m}\times\vec{x}}{\left|\vec{x}\right|^3}\right)\right) \\
    = \frac{\mu_0}{4\pi}\left(\vec{m}\left(\nabla\cdot\frac{\vec{x}}{\left|\vec{x}\right|^3}\right) - \frac{\vec{x}}{\left|\vec{x}\right|^3}\left(\nabla\cdot \vec{m}\right)\right)[/tex]

    How does Jackson end up at the result quoted?
     
  2. jcsd
  3. Oct 11, 2008 #2

    clem

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    Your second term is wrong. It should be
    [tex]-({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right][/tex].
     
  4. Apr 4, 2009 #3
    Hi

    I'm new to this forum! Googleing for the questioned problem I found this forum. Having the same problem as the author of this thread, I want to ask if someone has a solution on how Jackson found the formula for [tex]\vec{B}[/tex]?

    I tried my luck with the product rule [tex]\nabla\times(\vec{A}\times\vec{B})=(\vec{B}\cdot\nabla)\vec{A}-(\vec{A}\cdot\nabla)\vec{B}+\vec{A}(\nabla\cdot\vec{B})-\vec{B}(\nabla\cdot\vec{A})[/tex] but came to no conclusion.

    kind regards,
    derivator

    (sorry for my english, it's not my mother tongue)
     
    Last edited: Apr 4, 2009
  5. Apr 4, 2009 #4

    clem

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    As I said in my post, the curl of the cross product should be
    [tex]\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi} \left(\nabla \times \left(\frac{\vec{m}\times\vec{x}}{\left|\vec{x}\right|^3}\right)\right) \\
    = \frac{\mu_0}{4\pi}\left(\vec{m}\left(\nabla\cdot\frac{\vec{x}}{\left|\vec{x}\right|^3}\right) - ({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right][/tex].
    This follows because m is a constant so only two terms of your four enter.
    Given this result, the first term is a delta function is not needed if r>0.
    The second term gives the result in the first post if the delta function singularity at the origin is again ignored.
    Jackson is a bit obscure on this.
     
  6. Apr 4, 2009 #5
    Hi,
    thanks for your answer.

    Could you get a bit more detailed on how to show that

    [tex]- ({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right]=\left[\frac{3\vec{n}\left(\vec{n}\cdot\vec{m}\right) - \vec{m}}{\left|\vec{x}\right|^3}\right][/tex]

    is correct?
     
  7. Apr 5, 2009 #6

    clem

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    You really need a better textbook.
    Sorry I can't seem to find my latex error below. If hyou can read or correct it, it should show you what to do.
    - ({\vec m}\cdot\nabla)\left[\frac{\vec x}{|{\vec x}|^3}\right]=
    -{\vec x} ({\vec m}\cdot\nabla)\frac{1}{|{\vec x}|^3}
    -\frac{1}{|{\vec x}|^3} ({\vec m}\cdot\nabla){\vec x}
     
    Last edited: Apr 5, 2009
  8. Apr 5, 2009 #7

    clem

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    The point is that the m.del acts on only one function at a time.
     
  9. Mar 22, 2011 #8
    [tex]
    - (\vec{m}\cdot\nabla)\left[\frac{\vec{x}}{|\vec {x}|^3}\right]=
    -{\vec x} ({\vec m}\cdot\nabla)\frac{1}{|{\vec x}|^3}
    -\frac{1}{|{\vec x}|^3} ({\vec m}\cdot\nabla){\vec x}
    [/tex]
     
  10. Mar 22, 2011 #9
    [tex]\nabla\frac{\vec{r}}{r^{3}}=\left(\nabla\frac{1}{r^{3}}\right)\vec{r}+\frac{\nabla\vec{r}}{r^{3}}=-\frac{3\vec{r}\vec{r}}{r^{5}}+\frac{\mathbf{I}}{r^{3}}[/tex]

    [tex]\vec{m}\cdot\nabla\frac{\vec{r}}{r^{3}}=\vec{m}\cdot\left(-\frac{3\vec{r}\vec{r}}{r^{5}}+\frac{\mathbf{I}}{r^{3}}\right)=-\frac{3(\vec{m}\cdot\vec{r})\vec{r}}{r^{5}}+\frac{\vec{m}}{r^{3}}[/tex]

    I is the unit tensor.
     
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