I wonder how much heat is generated in a KM long spool of wire with that much juice flowing through it? Wouldn’t it become a huge electromagnet?
Forget the cord, go microwave
en.m.wikipedia.org
Good question and good caution.
I have been lazy at times when doing site "jumper" wires -- like maybe needing 200 feet and a having 500 roll of single wire -- so I just roll out some, leave most on the reel and jump from the ends -- hooking that in a circuit -- and YEAH! it makes a Choke / Induction Coil -- hums, gets warm, creates a voltage drop.
BUT -- that is not what is happening with a Cord. With a cord, BOTH legs (or sides, or hot + return) -- however you want to say this -- are coiled together with each other inside a cord -- so the field(s) created by the wires being in a coil shape pretty much cancel each other out. If you think about with signal wires you may played with -- they are sometimes even twisted together (called "twisted pair") -- that is make the overall loop(s) really tiny, and cancel out induction.
But as far as the overall Voltage Drop (the heat part of your question) is there whether stretched out straight or all rolled up. It is partly from the resistance of the wire, but much more so from the Amps (or Current or "I") going through the wire. Since we are trying to get power down range (Volts X Amps = Watts) -- we crank the Voltage UP, so we can keep the Amps DOWN. The heat loss equation is also in Watts (Amps ^2 X Resistance = Watts -- due to Heat Loss). Since Amps is squared in this case -- it is a HUGE factor.
That is why John Deere is running theirs at 2400 Volts. When Volts go UP, Amps can come DOWN.
Not sure if this one means 2400V Phase-to-ground, but if so, that means about 4000 Volts phase-to-phase. (always need to check that stuff when Europe is involved) And if that is 3 phase, we use SQRT(3) to average it out, 4000 V X 1.73 = 6900 effective volts. So going back to the JD claim of 200,000 watts at the tractor end (about 200 HP) -- means they are sending less than 30 amps down the cord (Watts = Volts X Amps) --> (200,000 = 4000 X 1.73 X Amps) . . . Amps = 28.9. So the low amps mean low heat loss in the cord, overall.