How do I convert the units of a moving magnetic field to volts?
You want the 'weber' unit for magnetic flux, and the
significant equation is that the EMF in the external
circuit formed by the windings and load  is

$$EMF = \int{{\bf E} \centerdot d{\bf L}} =  {{d\Phi}\over {dt}}$$
$$ Flux == {\Phi} = \int{\bf B} \centerdot d{\bf A}$$
where the first integral is taken over the whole helical coil path, and
Phi, the flux, is cumulative over all the
turns whose enclosed area has a magnetic field, B.  This is the Maxwell-Faraday law, and it IS practical, but
not simple.  The development of motors and
generators depending on this phenomenon took some decades.

A 'magnet' has some external B-field, and that field, both in magnitude and
direction, matters.   You cannot complete the calculation without the field
B known in all of the region.  

That EMF is a voltage equivalent value, and is applied to
the whole circuit.   If the external circuit is a short, current
flows to completely eliminate the flux change (though of course
there's always some resistance in common wire).   This is because
the current in the wire CAUSES flux that opposes the external field
which is changing.

The equation holds whether the flux change is the cause of the EMF, or if
a voltage is applied to the coil to cause a flux change.


  what's my voltmeter and ammeter going to say ?


The voltmeter, the ammeter, and the wires to connect them, will determine
that.   An ideal ammeter is an electric short circuit, and an ideal voltmeter
is an electrical open circuit; connecting both to the coil terminals would
be impractical (would show zero volts). There would be current, but only
to the meter, not to an external load...