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

[Ahmes]

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.

 

[pseudovector]

A moving charge is a non stationary current distribution, so the last two formulae are no longer valid to describe its magnetic field.

 

[Ahmes]

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.