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Moving magnetic dipole!

  1. Jan 22, 2014 #1

    ShayanJ

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    Gold Member

    Does anyone know how to calculate the magnetic field or flux of a uniformly moving magnetic dipole with magnetic dipole moment [itex] \vec{M} [/itex] pointing along the direction of motion?
    Thanks
     
  2. jcsd
  3. Jan 22, 2014 #2

    mfb

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    Staff: Mentor

    If the contribution from its motion is relevant, I think I would calculate it in its rest frame, and apply a Lorentz transformation on this field.
     
  4. Jan 22, 2014 #3

    ShayanJ

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    Here's the formula for the magnetic field of a magnetic dipole:
    [itex]
    \vec{B}=\frac{\mu_0}{4\pi} [\frac{ 3\hat{r}(\hat{r}\cdot\vec{M})-\vec{M} }{ r^3 }]
    [/itex]
    And the Lorentz transformation of the magnetic field(in the absence of an electric field) is:
    [itex]
    \vec{B}'=\gamma\vec{B}-\frac{\gamma^2}{\gamma+1}\vec{\beta}(\vec{\beta}\cdot\vec{B})
    [/itex]
    Now putting [itex] \vec{M}=M\hat{x} [/itex] and [itex] \vec{\beta}=\beta\hat{x} [/itex],I got the following:
    [itex]
    \vec{B}'=\frac{\mu_0 m \gamma}{4\pi r^4}[3x\hat{r}-r\hat{x}+\frac{\gamma \beta \hat{x}}{\gamma+1}(\frac{3x^2}{r}-r)]
    [/itex]
    Is everything OK with my calculations?
    And one other thing...I'm going to show this result to some people who are not aware of the Lorentz transformation of the electromagnetic field.Also the question is from somewhere which makes me almost sure that there should be another way other than using the Lorentz transformation for the field.But I can remember no other way!!!
    So maybe the motion doesn't matter!
    The question is asking that if one places a conducting circular loop in front of such a moving magnetic dipole,perpendicular to its direction of motion,what will be the force opposing the motion of the dipole?
     
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