Re the inductance: as per definition the current and voltage in a "perfect" inductance is 90 degrees apart, so the real power (which is V*I*cos(phi) should in theory be zero.
I have another theory why your PA might get so hot; in case you have a bridge-coupled PA you may have an offset output voltage, which in that case sees only the very low-resistance of the copper winding itself. This may cause a DC current of an appreciable size.
Try to connect a large capacitor in series with the winding and you get rid of the DC-component mentioned. If using electrolytic capacitors you may connect them in series back-to-back (+ to + sides) and in so doing get the equivalent of a bipolar capacitor of half the capacitance (provided the two capacitors are equal otherwise). Or use a bipolar type, such that are intended for passive filters in loudspeakers. Be sure though to use a sufficiently large capacitance, otherwise you will get some voltage drop over it. And watch out for resonance phenomenons (perhaps you want it, perhaps not).
/Pim