Is it possible to have a 5 meter wire that has 10,000 amps or more current flowing? I assume this can be done with a step-down transformer and the voltage could be 2V and lower... I assume that most of these wires have low resistance? So that heat would be considered.
Yes, however these are usually referred to as "bus bars" and not wires. They are essentially bars or rods of copper (or another conductor if required) that carry the current.
You would use a step down transformer if your power source was higher voltage and lower current than what you need. Since you haven't specified where you're getting this power from we can't recommend anything.
Yes. The lower resistance the better. Often copper is used for this, but in some power installations they do use superconductors. That's pretty tricky due to the requirement for cooling.
You can use several bus bars in parallel to permit greater air flow, reducing your need for more metal to carry the same current, or the need for active cooling.
10kA is going to require a lot of copper, even over 5m, and copper is expensive so you'll want to carefully balance the use of the system as a whole against how much copper and cooling you may need.
For 10,000A it might be more "bar" than "wire", this baby handles megawatts at 50v:
Are you talking 10kA constant or a brief pulse? For a brief pulse you could get away with pretty small wires, as long as they don't vaporize.
Stop and think about this. 2 V x 10 kA = 20 kW. Are you prepared to source that much power? Are you prepared to deal with that much heat being produced by your 5 meter cable?
Instead of guessing or asking, do the math. You want a resistance of 2 V / 10 kA = 200 µΩ. The resistivity of copper at 20°C is 1.68 nΩm. (1.68n Ωm)(5 m)/(200µ Ω) = 42µ m2. That would be a square crossection 6.5 mm on a side, or a round one 7.3 mm in diameter.
However, that was at 20°C. Dumping 20 kW into a hunk of copper 5 m long and 7.3 mm in diameter is going to heat it quickly, so it won't stay near 20°C very long. This time you do the math. Calculate the total volume of copper, look up the density of copper to get the mass of copper, look up the specific heat, and then compute how fast the temperature will rise with 20 kW applied to that mass of copper. These are all very straight forward calculations, just like the ones above. You just need to look up the physical constants, just like I did above.
OK, I googled it for you. (42µ m2)(5 m) = 210µ m3 = 210 (cm)3. The density of copper is 8940 kg/m3. (8940 kg/m3)(210µ m3) = 1.88 kg. The specific heat of copper is 386 mJ/g°K. (386 mJ/g°K)(1877 g) = 725 J/°K, or 1°K(=°C) rise for each 725 J added to the cable. (1°C / 725 J)(20,000 J/s) = 28°C/s temperature rising rate. If you start at 20°C, for example, turn on the 10 kA, then in 2.9 second the cable will be hot enough to boil water. The melting point of copper is 1083°C. It would take 38.5 seconds to reach that point if not heat is lost. However, that's a long enough time to lose considerable heat power to the ambient air, especially with some deliberately moving air. In any case, you can run this for maybe 20 seconds at a time without anything bad happening if you allow it to cool down between runs.
If the copper is arranged as a sheet to get more surface area for the same crossectional area, then it will be able to lose heat to ambient air much more rapidly. Note that a circular crossection is the worst shape for this purpose. In that case the whole cable has a surface area of 178 in2, which requires 113 W/in2 transfer to stabalize just at the melting point. Of course at that point the temp difference is 1063°C, so maybe that's plausible with forced air. That comes out to a heat sink rating of 9.4°C/W.
Anyway, this is all just basic physics with constants that can be looked up on the internet. Not really much electronics here.
As other answers said, with a big enough cross-section you can carry 10 kA.
I want to focus on somethign else I think you might be misunderstanding.
It depends what your load needs.
If your load needs 10 kA at 2 V, you will want the voltage dropped in the bus bar to be much less than 2 V, or you will want the voltage at the supply end of the bus bar to be higher than 2 V to allow for drop over the bus bar. If you input 2 V to the bus bar and the voltage drop across the bus bar is 2 V, as suggested in Olin's answer, then there the voltage applied to the load would be 0, which is probably not what you wanted. (Not that Olin's answer is wrong, but I think he was interpreting to mean that the voltage is much higher than 2 V but you are allowed to drop 2 V in transmission)
But more efficient might be to use a higher voltage and lower current on the bus bar, and then use your step-down transformer to drop the voltage down to 2 V right at the load.
For example, you could send 100 A at 200 V over the bus bar, which would not require nearly such a big cross-section. Then step down at the load end to get the low voltage and high current that your load wants.
There's one limiting factor that no one else has mentioned: electromigration. As the current density in a conductor increases to extremely high levels the electrons start to move the metal atoms at a significant rate. Because a conductor is not ideal, its cross-section area will vary slightly. Points on the conductor with a smaller cross-section will have a higher current density, so the metal atoms tend to be removed from thinner spots and deposited in thicker spots. This just makes the thinner spots even thinner so the process eventually causes the conductor to fail as an open circuit.
When I worked in integrated circuits with aluminum wiring, the maximum allowed value of current density was around \$10^5\$ A/cm\$^2\$. Yes, 100 kA per square centimeter. These levels did occur fairly often. Of course the conductors would be extremely hot if they were not embedded in glass (amorphous SiO\$_2\$), which does a good job of removing the heat. As an aside, electromigration could also cause circuit failure when the metal atoms piled up at one end of a conductor and the bulge cracked the insulating layers.
Another thing to consider is the magnetic field generated. If I remember my A-level physics:
i.e. force applied to a conductor in a magnetic field (Newtons) is equal to the product of magnetic flux density (Teslas), current (Amps) and length (meters).
So in your case if you bought a loudspeaker magnet (~1T) close to lets say a 10cm section of the conductor, the resulting force would be 1 * 10000 * 0.1 = 1000N, or roughly the weight of a 100Kg mass. Not insignificant.
Another point of interest. The aluminium smelting process is a user of obscenely large currents. The Hall–Héroult process wikipedia page mentions 220kA. It links to this manufacturer's page claiming 350kA.