Calculating Current Draw in Electromagnetic Coil

[frankenstein]

My situation is this. I have 6,560 ft (4 lb) of polyurethane-nylon coated #27 gauge round wire wound in a coil with an air core of 1 1/8 in. wide x 4 in. long x 4 in deep. I don't want to destroy my precious electromagnet so I'm afraid to give it some current. What would happen if I attached a 12 volt car battery? The wire at 6560 ft has a resistance of 337.6 \Omega. Using ohms law that voltage across any pure resistance is equal to the strength of the current in the resistance times the strength of the resistance... we get the equation: V=IR. Rearranging it: I=V/R

With a 12 volt battery and that given resistance I should be drawing .0355 Amps?
But I want 1200vdc - 6,000vdc or higher.
With 1200vdc following the equation I should be drawing 3.55 Amps?

Won't I burn the insulation through and short? How can I achieve a very high DC voltage while maintaining an amperage under the wire's conservative amp rating of .268 Amps (based on 750 circulare mils per Amp) or the wire's Max Amp rating of .403 Amps (based on 500 circular mils per Amp)?

My basic question: How can I operate this specific electromagnet with a 360vdc - 6,000vdc power supply at an amperage that will not hinder the performance of the electromagnet?

It may be helpful to know that I am switching this electromagnet off and on many times per second, reversing polarities every time it's turned on. I'm using a high voltage magnetic reed switch to pulse the negative attaching a full wave bridge rectifier BEFORE the switch to rid myself of all the sparking and to charge another battery. :) see any problems with my circuit other than it being a normal HV electric chair? lol

 

[Redbelly98]

A better approach is to first figure out what strength magnet you need.

My basic question: How can I operate this specific electromagnet with a 360vdc - 6,000vdc power supply at an amperage that will not hinder the performance of the electromagnet?

If we go with the 0.268 A figure, you can apply at most

V = I R = 0.268 x 338 V = 90V

(And, it would be good to measure the resistance with an ohmeter, rather than relying on a calculation.)

Also, your wire insulation is probably rated for at most a few hundred volts, possibly less. Don't even think about applying kilovolts unless you know the voltage rating of the wire+insulation you have.

edit:
p.s. Welcome to PF!