induce voltage sensor

Sweet!

Not enough information - you can't do it based on the live wire mains potential. You could get the 6 Volt potential, but not the 0.5 Amp current. The capacitive reactance between two wires at two meters separation is too low for efficient coupling of power (V x I). You can get V into an open circuit, but if you put a load on the wire and try to draw current, the capacitive source impedance will drop the coupled potential to zero. You can prove this to yourself. Assume an unrealistically high 10 pF/meter of parallel run of 25 kV wiring and your pickup wire. Work out the total capacitance between live wire and pickup wire, based on parallel length. Calculate reactance of that capacitance at mains frequency (50/60 Hz): Z = 1/(6.28*C*f). That impedance is in series between 25 kV and your 12 Ohm load. Now calculate the voltage divider: Vload = 25 kV *(12 Ohms/Z). For a ten meter parallel run, you will get about 10 mV loaded potential at 60 Hz.

In order to get real power, you need a loop pickup. That calculation requires the mains current, not the mains potential. Stories are told that farmers used to place large loops under cross-country power transmission lines that crossed their property and get free power that way. Power companies frown on that, no surprise.

1/2 amp is a lot of LEDs. I think this can be done using a copper coil on a core to a lesser core . At six feet you will need enough copper in the primary winding that it would be cheaper to buy a small generator. You are not planning on stealing this power are you? The power companies play for keeps. About 1/2 mile of 24 gauge copper around the end of a 12 inch diameter pole will induce enough electricity to generate some power. The power company will be able to measure the loss.

No I did not do this but I knew a farmer who did this to light up a pump building under some power lines. He got caught.