Hi,
You have 16 times the capacitance, so you need 1/16 times the resistance.
You've quartered the length of the wire, so that quarters the resistance, and you've doubled the wires working area so that quarters the resistance again which means the resistance is then 1/16 of the original.
The 2ms fusing current of #29 copper wire is over 400 amps. The 2ms fusing current of #23 copper wire is over 1600 amps. The thing is, if the copper got up to 1082 degrees C and didnt melt yet, the insulation would be in trouble. So i dont think the fusing current is going to matter that much, but could help comparatively. Since we do see about 4 times the fusing current for a 2ms pulse, it should be ok.
Then we have the heat buildup to consider. The fusing current (and the wire insulation degradation) is based on an ambient temperature of 30 degrees C, but if the coil heats up during operation then that looks like an ambient temperature rise which brings the fusing current down. It would help then to estimate the coils temperature rise.
If the max current is 48 amps and the max voltage is 40 volts that's a peak power of nearly 2000 watts, but it's at a duty cycle of only one third of one percent (0.0033 percent) which brings us down to about 6 watts average power dissipation, at worst case. The full power wont be present for the full 2ms so we'll estimate at one half of the peak, which brings us down to about 3 watts average power dissipation (we could double check this).
The remaining question then is what size is the coil. We need to know the diameter and the total length of the coil turns end to end. We can then figure out if the coil will get too hot or not.
The coil sounds like it is about 0.8 inches diameter by about 0.6 inches long but that's just a guess.
So then, what are the dimensions of the coil?
Also, what was the temperature rise of the old coil under the said operating conditions?