Electrodynamic tether on the Moon/Mars?

[jumpjack]

Moon and Mars have no magnetic field, but that they are immersed into solar magnetic field.
If I place a long wire along north pole, perpendicular to Sun, while Moon/Mars moves along its orbit, the wire cuts magnetic field lines; this should produce a difference of potential between fare edges of the wire... or not?
If I connect a second wire to first one, but I completely insulate it (how?) from solar magnetic field, would I get an electric generator?
Of course the DDP would change during day and year, until it reverses, but a diode could prevent current from flowing in wrong direction.

 

[Baluncore]

The current through the diode would charge one end sphere (planet), then stop flowing once the voltage had built up.

A voltage difference is part of the generator, but you can only extract energy while a current flows. Power is the product of voltage and current. You actually need a circuit with a return path, or a loop. There is no need for Mars, the loop can start and end at Earth. There is no insulation required on the wire, space is good enough.

There will be a force on the generator wire that will push it sideways. The coils in an electric generator must be held firmly on the armature or field. You must fix the wire loop in space, so it does not move relative to the Earth.

Unfortunately the Moon does not rotate relative to the Earth, or we could wind a wire once around the armature Moon and use the Earth's magnetic field to generate power on the Moon, until we extracted all the energy and stopped the moon rotating.

 

[jumpjack]

I think you completely misunderstood my post.
Forget Mars.
Just replace "Moon/Mars" by "Moon" in the post above.

Look at the picture:

Will electrons in the wire be moved along it by Lorentz force?
If I connect a second wire parallel to the first one, but completely isolated shielded from solar magnetic field, will I have a closed circuit containing a generator?

 

[sophiecentaur]

The current through the diode would charge one end sphere (planet), then stop flowing once the voltage had built up.

If a coaxial system were used with an iron cylinder in between the conductors, I have a feeling there could be a constant current flowing along inner and outer conductors due to the different fields and hence different emf's generated in inner and outer conductors. But Electromagnetism is not intuitive so I could be just plain wrong. ??

 

[jumpjack]

I cannot find precise values for solar EM on Earth, only a couple of clues:
The magnetic field at an average place on the Sun is around 1 Gauss, about twice as strong as the average field on the surface of Earth (around 0.5 Gauss).
https://www.windows2universe.org/sun/sun_magnetic_field.html

The Sun’s magnetic field
During solar minimum, the magnetic field of the Sun looks similar to Earth’s magnetic field. It looks a bit like an ordinary bar magnet with closed lines close to the equator and open field lines near the poles. Scientist call those areas a dipole. The dipole field of the Sun is about as strong as a magnet on a refrigerator (around 50 gauss). The magnetic field of the Earth is about 100 times weaker.
https://www.spaceweatherlive.com/en/help/the-interplanetary-magnetic-field-imf

Which one is right?!?

Is that true that Earth EM is 100 times weaker than Sun EM?
Would this mean that to generate same amount of DDP generated in Tethered Experiment using Sun EM would require a wire 1/100 as long? (it was 20 km long, hence 0.2 km).

 

[jumpjack]

I eventually found some equations to work with:
https://ocw.mit.edu/courses/aeronau...2015/lecture-notes/MIT16_522S15_Lecture25.pdf

http://articles.adsabs.harvard.edu/...SASP.476...61I&defaultprint=YES&filetype=.pdf

If I understand correctly, the Difference Of Potential in the tether is given by:
Voc = v * B * L
v = perpendicular tether speed across EM [m/s]
B = EM field [Tesla]
L = tether length [m]

And the current in the tether, disregarding the "ionosphere items", as we are on the Moon:
I = Voc / Rt
Rt = tether resistance

Hence:
I = v*B*L/Rt

But
Rt = rho * L / A
rho = resistivity
L = length
A = section area

hence:

I = v * B * A / rho (no more "L")

being:
A= 3.14 * r^2 [m^2]
rho = 1.68E-08 [Ohm*m]

I = 3.14/rho * v * B * r^2 = 1.86E08 * v * B * r^2

Orbit speed of Moon around Sun can be assumed as the same of Earth, 108000 km/h, i.e. ~30000 m/s:
v = 3E+04 [m/s]

Sun EM on Moon should be around 1E-09 [T] (confirmations?)
B = 1E-09 [T]

So:
I = 1.86E08 * 3E+04 * 1E-09 * r^2

I = 5.58E+03 * r^2

This is a weird result, as the final current would depend just on tether section and not on its length...
Is this result correct?

Assuming a 10mm radius tether (10^-2 m), we would get:

I = 5.58E+03 * 1E-4

I = 0.558 Ampere (regardless of length)

If it is true that tether length does not matter (and I can't see how it could), if I'd get a block of copper 1m x 1m x 1m, would it generate 5580A?!?

 

[anorlunda]

v = perpendicular tether speed across EM [m/s]

Be careful here. That is the velocity relative to the field. In the case of the Earth-moon system, the field moves with the objects, so if the tether was attached to the Earth-Moon, it would rotate with the field and the velocity relative to the field is zero.

 

[jumpjack]

Be careful here. That is the velocity relative to the field.

To the SUN field, I am talking about Sun EMF.

 

[jumpjack]

nobody can help?