Why Does a Moving Charge's Magnetic Field Follow an Inverse Square Law?

Ahmes
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The magnetic field of a moving charge is:
\boldsymbol{B} = \frac{\mu_0}{4\pi} \frac{q \boldsymbol{v}\times \boldsymbol{\hat{r}}}{r^2}
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 \boldsymbol{m} = \iiint \boldsymbol{x} \times \boldsymbol{J}(\boldsymbol{x}) d^3 x it appears a moving charge, \boldsymbol{J}=q \boldsymbol{v} \delta^3 (x) has a zero dipole moment.

So how could this be explained?
Thank you.
 
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A moving charge is a non stationary current distribution, so the last two formulae are no longer valid to describe its magnetic field.
 
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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