Use J Not I

From,

$B=-i\cfrac{dE}{dx}$

$B=-i\cfrac{dE}{dt}\cfrac{dt}{dx}$

$B=-i\cfrac{dE}{dt}\cfrac{1}{v}$

If we are dealing with a wire carrying moving charges, we know that the effects of the moving charges cancels before and after the point where the normal drop from the point $B$ is to be calculated.  The only effect on $B$ is confined to a disc directly under, at the base of the normal.  By conservation of flux from a fixed source, in this case a fixed current density,

$E=\cfrac{q}{2\pi\varepsilon_o r}$    as oppose to    $E=\cfrac{q}{4\pi\varepsilon_o r^2}$

a disc of perimeter $2\pi r$ not a sphere of area $4\pi r^2$.

So,

$B=-i\cfrac{d}{dt}\left\{{\cfrac{q}{2\pi\varepsilon_o r}}\right\}\cfrac{1}{v}$

where $v$ is the average speed of the moving charges and yet $\varepsilon_o$ is unchanged.

$B=-i\cfrac{1}{2\pi\varepsilon_o r}\cfrac{dq}{dt}\cfrac{1}{v}$

$B=-i\cfrac{1}{2\pi\varepsilon_o r}\cfrac{I}{v}$

$B=-i\cfrac{J}{2\pi\varepsilon_o r}$

$|B|=\cfrac{J}{2\pi\varepsilon_o r}$

Which is not the often published formula of $B=\cfrac{\mu_o I}{2\pi r}$. $-i$ signifies that $B$ is perpendicular to $J$ in the right hand sense.  The scaling factor of $\cfrac{1}{\varepsilon_o}$ as opposed to $\mu_o$ is easily forgiven except for energy formulae, in which case, another constant is applied to get rid of this artificial $\mu_o$.  For example, dividing by the wave impedence defined as $Z=\sqrt{\cfrac{\mu_o}{\varepsilon_o}}$.

The real concern is $J$, current density instead of $I$, current.  If we compare this with the E field due to a line charge,

$E=\cfrac{\lambda}{2\pi\varepsilon_o r}$

where $\lambda$ is the line charge density.  We see a good analogy between E and B field if we drop $\mu_o$ and its unit all together.