Cycloid motion of electrons
Date: Monday, December 20, 2004 @ 22:02:10 UTC
Topic: Science


Roland and Leo C. wrote (free-energy yahoo group): Hi all

Herewith a strategy to obtain electric power from electron emission in a vacuum.
This is not new, it has been covered by: http://www.geocities.com/CapeCanaveral/Lab/9222/magele.pdf and sundry others.

The principle is simple and well analysed. Charged particles in the presence of perpendicular magnetic (B) and electric fields (E) undergo cycloid motion, whereby they travel at a(n average) drift speed (E/B) perpendicular to both the electric and magnetic fields, while describing a cycloid motion.

A cycloid motion is essentially the path described by a point on one of the bicycle spokes, as the bicycle wheel moves forward. In general it's the path described by a point on a stick attached to the spoke outside the bicycle wheel. Imagine the stick just going through the ground, or imagine the wheel on a suspended track with the stick glued along a spoke but missing the track....

A perpendicular B and E field provide a ferry for isolated charged particles. The charged particles can move in only one (average) direction, regardless of their initial direction and speed.

So practically, what we need to achieve power from this simple setup is a vacuum tube, with electrodes each end ( flat plates as wide as the tube). Above and below the tube, we have charged flat plates applying an electric field. To the left and the right of the tube, we apply a magnetic field. Perhaps we put permanent magnets on both sides.

Then, we need a source of charged particles within the vacuum tube. We could use a standard hot filament, but my personal preference is a beta-decay radioactive source.

>From whatever charged particle source, the perpendicular E and B fields will provide an effective voltage across the vacuum tube.

Interestingly, regardless of the initial direction and velocity, the exit velocity of the particle is at least the initial velocity, and in many cases more. A particle in initially at rest, for example, has a non-zero average exit velocity.

So, what's the parameters to make this work?

The average exit momentum of the electron is crucial. The 'cyclotron radius' of the maximum electron momentum must be within the bounds of the tube. The cyclotron radius is (inversely) proportional to the B field strength.

So, all we have to do is provide a strong enough magnetic field to preclude the electron hitting the tube.

Caveat #1: Particles emitted with a major component in the direction of the B field will hit the side of tube. Cycloid movement is only in the plane perpendicular to the B field.

Solution: Target the charge (plus or minus). In my case, I'm targeting electrons, negative charge. I place a strong negatively charged plate both sides of the B field.

So what do we get?

In the case of the Edison filament bulb, we can recapture some of the heat loss.

More interestingly for me, we can capture B- radioactive decay completely as electric potential.

Roland


Stefan and Roland,

For publications describing other attempts based on this idea, see:

http://www.geocities.com/CapeCanaveral/Lab/9222/index.html
http://arxiv.org/ftp/physics/papers/0311/0311104.pdf
http://arxiv.org/PS_cache/cond-mat/pdf/0310/0310073.pdf

Fascinating stuff!

Regards,
Leo C.






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