The thickness/material of the conductor determines current capacity while insulation thickness/material determines voltage rating. Do not multiply the ratings to get some sort of "power rating" -- that is not how cables work.
In other words, you need an appropriate gauge for 50A regardless of the voltage. You then also need the appropriate insulation rating for safety with respect to the voltage.
The reason, since you asked in a comment, is that current is a quantity of electrons over time (1 A = 1 coulomb/second). Voltage is the difference of potential between the voltage source and ground. Thus, you can have a variable voltage at a given current, and by definition the same number of charges pass through the conductor per unit time.
Analogies are never 100%, but water in a pipe is sometimes used to help conceptualize this. Voltage would be akin to the water pressure, while current would be analogous to the flow rate. You can have the same flow rate at different pressures. Higher pressure would require thicker walls while higher flow rates would require a larger diameter pipe.
Additional remarks, based on comments:
Voltage Drop & Conductor Cooling
Wires aren't superconductors[citation needed] so you have to think of them as low value resistors. When you do this, it should become obvious that the wire will, like a resistor, dissipate some power and impart a voltage drop. A voltage drop is simply the difference in voltage across a component or wire. If you're familiar with series resistor circuits, then you know each resistor drops voltage proportional to its resistance. The wire does, too.
For a given length, a wire with a larger diameter will have less resistance, and thus a lower voltage drop.
The power law (\$P=I^2R\$) shows that the cable will dissipate some amount of power as heat proportional to the current through it.
A 12 gauge wire has 5.2 mΩ of resistance per meter. A 6 gauge wire has 1.3 mΩ per meter (source). The power (heat) dissipated by a meter of wire carrying 50 A will be 13 watts for the 12 gauge, and 3.25 watts for the 6 gauge.
Heat is generated by the wire has to go somewhere. If it is in a bundle and not air-cooled, the heat will increase the temperature of the wire/bundle.
The temperature rise above ambient, for 50 A through 12 gauge wire in free air, would be about 100 degrees. Comparatively, for 6 gauge wire, it is negligible.
Source
If you had some sort of cooling system to remove the heat from smaller-gauge wire, you could, in fact, use it. But now you've got a whole cooling system to install, maintain, and power!
A final thought, if you need 50 amperes for your experiment at some distance from the power source, you might consider using AC and stepping it up and down in voltage with transformers, just as power transmission lines do. For example if your load is 250 watts, you could supply 50 volts AC at 5 amperes, send that over 22 AWG wire, then step it down in voltage using a 10:1 transformer rated for 50 A on the secondary winding. If you need DC, use a bridge rectifier to convert from AC to DC. These additional items add some cost and complexity, so you'll have to weigh this against the cost and hassle of purchasing and using 6 AWG wire.
