The classical electromagnetic effect is perfectly consistent with the lone
electrostatic effect but with special relativity taken into consideration.
The simplest hypothetical experiment would be two identical parallel
infinite lines of charge (with charge per unit length of $ \lambda \ $
and some non-zero mass per unit length of $\rho \ $ separated
by some distance $ R \ $. If the lineal mass density is small enough
that gravitational forces can be neglected in comparison to the electrostatic
forces, the static non-relativistic repulsive (outward) acceleration (at the instance
of time that the lines of charge are separated by distance $ R \ $)
for each infinite parallel line of charge would be:
$$ a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} $$
If the lines of charge are moving together past the observer at some
velocity, $ v \ $, the non-relativistic electrostatic force would appear to be
unchanged and that would be the acceleration an observer traveling along
with the lines of charge would observe.
Now, if special relativity is considered, the in-motion observer's clock
would be ticking at a relative rate (ticks per unit time or 1/time) of $ \sqrt{1 - v^2/c^2} $
from the point-of-view of the stationary observer because of time dilation. Since
acceleration is proportional to (1/time)2, the at-rest observer would observe
an acceleration scaled by the square of that rate, or by $ {1 - v^2/c^2} \ $,
compared to what the moving observer sees. Then the observed outward
acceleration of the two infinite lines as viewed by the stationary observer would be:
$$ a = \left(1 - v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} $$
or
$$ a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} - \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e - F_m }{\rho} $$
The first term in the numerator, $ F_e \ $, is the electrostatic force (per unit length) outward and is
reduced by the second term, $ F_m \ $, which with a little manipulation, can be shown
to be the classical magnetic force between two lines of charge (or conductors).
The electric current, $ i_0 \ $, in each conductor is
$$ i_0 = v \lambda \ $$
and $ \frac{1}{\epsilon_0 c^2} $ is the magnetic permeability
$$ \mu_0 = \frac{1}{\epsilon_0 c^2} $$
because $ c^2 = \frac{1}{ \mu_0 \epsilon_0 } $
so you get for the 2nd force term:
$$ F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R} $$
which is precisely what the classical E&M textbooks say is the magnetic force (per unit length)
between two parallel conductors, separated by $ R \ $, with identical current $ i_0 \ $.