Some things to consider:
The power in watts (which is all heat) equals the current squared times the resistance. This means if you keep the same winding, but reduce the current (you can only do this by reducing the voltage), the heat will be reduced by the square of the current.
So for my coil:
at 6 volts I had 6 A and 1 ohm:
Power = (current squared) x resistance = 6x6x1 = 36 watts
If I reduce the voltage to 3 volts, coil is still 1 ohm:
Power = (current squared) x resistance = 3x3x1 = 9 watts
That is 300% less hot. Still too hot? maybe.
Chuck, where do you get your current carrying data from? Do you have a link? #20 wire is good for 11 amps in free air, but I have no idea what it's good for in a coil.
E