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Magnetic field of a moving charge

  1. Oct 7, 2006 #1
    The magnetic field of a moving charge is:
    [tex]\boldsymbol{B} = \frac{\mu_0}{4\pi} \frac{q \boldsymbol{v}\times \boldsymbol{\hat{r}}}{r^2}[/tex]
    This is an inverse square law.

    But also we know that every localized current distribution (and a moving particle is most obviously a localized current distribution) appears from very far away as a dipole moment - which field is an inverse cube law.

    Also using [itex]\boldsymbol{m} = \iiint \boldsymbol{x} \times \boldsymbol{J}(\boldsymbol{x}) d^3 x[/itex] it appears a moving charge, [itex]\boldsymbol{J}=q \boldsymbol{v} \delta^3 (x)[/itex] has a zero dipole moment.

    So how could this be explained?
    Thank you.
    Last edited: Oct 7, 2006
  2. jcsd
  3. Oct 7, 2006 #2
    A moving charge is a non stationary current distribution, so the last two formulae are no longer valid to describe its magnetic field.
  4. Oct 7, 2006 #3
    When Jackson develops these formulae he doesn't demand the current distribution to be stationary, although I can see why it is not.
    OK, thank you.
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