-"Elektrizität für Anfänger"!

[nova_robotics]

Cheat: Assuming a 12 V & 3 W LED bulb which draws 0.25 A,
and a 1.5 sq mm/ AWG 16 lead (having an inherent voltage drop of 6 V /km),
the applied battery voltage would only have to be 3 600 012 V [or 3.600012 MV].

Surprisingly, the light will illuminate at much less voltage. There are capacitive and inductive effects at work. Think of the wires as two plates of a parallel plate capacitor, or the loop as two tightly coupled dipole antennas. Even if you cut the wires (at the farthest points) the bulb would still illuminate at perhaps a few hundred, or a few thousand volts. I suspect it would also oscillate because that's essentially a giant RLC circuit with a 0 ohm termination at each end.