You don't need much to heat up wire. I think maybe 10W is a good estimate for your application.
If you take about 1 foot of AWG 30 which has a resistance of about 0.10320 ohms then that means you have to run 1A through it to get that.
The issue is that your appling 13V to it and so it will actually draw about
125A. Now if you had a 1V source then you could do it.
(Although I'm neglecting that the resistance will increase when it is heated... maybe by a factor of 10(happens atleast with a lightbulb))
So ultimate what you have to have is some type of conductor that has a large resistance.
if you can get something that has about 10ohm then that means you can draw about 1.3A through it and dissipate 17W.
Before you worry about making it variable you might want to actually get something to work. Adding in a series resistor isn't going to cut it. If you find a 5 ohm pot using the 10 ohm wire above then you have 10 to 15 ohms... this gives you a dissipation from 7.5W to 16.9W but the pot will need to dissipate about 5W(3.76 to be exact).
A light bulb filament is about 10 ohms cold and 100 ohms hot. The above does not take into account this change.
Normal coper wire isn't going to work because your voltage is to high producing extreme currents. So either you need to reduce the voltage(A 1VDC
20A source would work nice) or increase the resistance.
A better way would be to use AC and an SCR. Its very easy to control the power into the load and vary it.
Your best bet is to find some high resistance conducting wire that is made for that sort of stuff. Normal wire isn't going to work. I think what is commonly used for this sorta stuff is nichrome:
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Once you find some thing to use, if you haven't already, and are still confused then post what you have.
For some high power pots,
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So if you had a nominal Rw ohm hot wire and Rp ohm pot then
When Rp is all the way up,
I = V/(Rw + Rp) Pw = I^2*Rw Pp = I^2*Rp
and when Rp is off,
I = V/Rw Pw = I^2*Rw Pp = 0
You can use these formulas to find out the max current drawn(V/Rw) and the max dissipations in the devices(I^2*Rw for the hot wire and I^2*Rp for the pot). These all should be less than the max ratings. You can also use them to solve for Rw and Rp given whatever power you you want to dissipate in the wire.
Note that when Rw goes up because it gets hot it will decrease all the values.
Incase you are confused then what I would do is get the nichrome AWG 31 which has a resistance of about 8.523 and one of those 10ohm 12.5W pots.
This means you have approximately what I gave above. If you want to have more power then all you have to do is use a larger wire size like AWG 29... but this will increase the current and might burn up the pot(shouldn't as its about a factor of 2 which I think gives a pot dissipation of about 10W).
In any case that one link seems to give some examples. I'm kinda rambling now ;/
Jon