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.