First of all: Don’t do this at home! You will ruin my copyright and it is dangerous, too. Heavy current is nothing to be dealt with by non-engineers.
now I am able to describe the experiment in detail. We need an car battery with 40Ah and 12V. This battery will be connected to a switching power supply that delivers 70A DC. The switching power supply connects once again to a heavy current cable. Therefore we need at least 50 meters of heavy current cable. That cable will again be connected to the power supply so we have a circuit.
Then it is only a question of arrangement and direction. We take a plate of wood of one squaremeter. We begin to fix the cable one meter straight. Then we go in a half circle back to the beginning of the plate and attach another one meter straight but ten cm above the first. We repeat it unless the whole plate has eleven straight one meter cables all divided by ten cm.
Then we need some kind of launching pad. We take another two plates of wood, which have one squaremter either too. We let there be a distance of about 10 cm between and fasten them together so we have a kind of bin. This bin needs to be fastened too so it won’t leave the ground with zero gravity.
Then the experiment will start. We will put on 70A on the cable, put the launching pad about five cm away from the plate. The plate will be set so the charge is going from west to east exactly. All eleven cable pieces shoud tend to east. And then we take cork and throw it in the launching pad. The prediction is that it will return and will fly out the launching pad against the sun 🙂
The equations used, are the following:
R = rho * l / Querschnitt A
R = rho(Alu) * 50m / 50mm²
R = rho(Alu) m / mm²
R = 2.65 * 10^-2 Ohm
U = R * I = 2.65 * 10^-2 Ohm * 70 A = 1.8 V
So the cable has an cross-section A of 50mm². and the cable will be 50 meters long. The specific resistance of aluminum used is 2.65 * 10^-2 Ohm * mm² / m.
So we got to modify the power supply to change from 12V to 2V. That will be the task for Dami or Doc Greg because I can not modify a power supply. But then we will have 70A with 2V.
The magnetic field established will be:
my(0) = 4 * Pi * 10^-7
B = my(0) * I / 2Pi * d = 60 * 10^-6 T
d = 20cm
I = (60 * 10^-6 T / 4 * Pi * 10^ -7) * 2 * Pi * 2 * 10^-1 m
I = (600 / 4 Pi) * 2 * Pi * 2 * 10^-1 m
I = 600 * 10^-1
I = 60A
This means we need at least 60A to make a magnetic field in a distance of 20cm that is as big as the magnetic filed of the earth. So simply we would compensate the magnetic and electrical filed of the earth.
Yeah, I hope you wish us the best and again don’t try this at home, 70 Ampere are too much.