Hello- I'm trying to make a tri-polar motor, and am using 12-28 machine screws as my poles. Obviously this is a small motor, but does how thin or thick the iron (steel) core is affect how strong the magnetic field is? Or is it JUST current and amount of copper wire? Thanks!

 

The TYPE of iron or steel affects it.

Soft, low carbon steel, being the easiest to magnetize. Harder steel and high carbon cast iron resist magnetization more than soft steel and require more current and ampere turns of copper to create the field.

 

Thanks for the reply.
Another question since you seem knowledgeable:
How about the gauge of copper wiring? The amount of wraps? Again talking about a pretty small .216 inch diameter, 1 inch pole, how many wraps of wiring and which gauge should I look at?
Does the way I wrap it matter?

EDIT: I'll be running this at 9v

 

You need to determine how many amps(or milliamps) you can run the magnet at without it overheating. This means experimenting some. If you are going to use a 9 volt battery then you will be very limited on current output. If you use a 12 V Lead acid then you could melt the device down if you don't put a fuse in.

Google up Wire gauge chart/ the chart will tell you what the ohm resistance of the different wire sizes are. They usually list it for 1000 ft of wire. Say,a 28 gauge wire might be as small as you want to go, but if the resistance of the wire is to low(not long enough, or too large) excessive current will flow and overheat the magnet.

DUTY CYCLE: How many seconds on / Total time between activations Another good definition to understand with these type of devices. You can use large currents for short periods, but must then let the coils cool down for awhile.

For a nine volt battery I would go with some pretty small wire 28 gauge or maybe 26. Calculate the ohms(or use a meter) of the coil and use ohms law to figure current flow.

Keep the current flow figure smaller than the wire gauge chart ampacity of the wire.