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jbond:
Two conductors spaced by an insulator form a capacitor. Two wires in a cable have a spacing that is not too great, and between them is insulating material. This material will have a dielectric constant greater than 1. Even if the dielectric constant was 1 because the wires were in a vacuum there is capacitance.
As an example 2 conductor #12 Romex has about 20 pfd capacitance per foot, based on a short sample of some old wire. Assuming your cable had the same capacitance, then 500 ft is 10,000 pfd or 0.01 mfd. Your cable might have a greater coupling capacitance.
At 60 Hz 0.01 mfd has a capacitive reactance of 265,000 ohms. The maximum current from this at 120 V would be about 0.5 MA. If the LED in the pilot lamp is one that has high brightness, means little current to produce useable light, then you are likely to see some residual light.
If the input wire to the pilot lamp is connected to nothing (supposed to be off), then you do not have a loop circuit for magnetic flux lines to cut and therefore no magnetic field induced current. There will be an induced voltage but no current. Even if you had a closed circuit I doubt you would induce 1 V in the loop and this would never light the LED with the amount of series resistance that is in the pilot lamp.
The use of the wording "induced voltage" would normally apply to a voltage in a conductor generated from a changing magnetic field. For example, connect a coil of wire to a sensitive meter and move a strong magnet near the coil. With a small 1200 turn coil about 1.5 x 1.5 inches outside and a strong magnetic I can get 5 to 10 MV deflection on a Simpson 270 on the 250 MV (50 microamp) range.
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