Simple (?) Theoretical Question - Current

Discuss how polywell fusion works; share theoretical questions and answers.

Moderators: tonybarry, MSimon

Is there a Current?

Poll ended at Fri Oct 01, 2010 3:24 pm

Yes
3
75%
No
1
25%
Maybe
0
No votes
 
Total votes: 4

KitemanSA
Posts: 6179
Joined: Sun Sep 28, 2008 3:05 pm
Location: OlyPen WA

Simple (?) Theoretical Question - Current

Post by KitemanSA »

I have been told that current is the flow of charge from an anode to a cathode (or vice versa).

If I have an electron source (head lamp filiment) and a positive target (ring) with a slight magnetic field to guide the electrons away (torus) and the electrons race toward the ring, miss, continue past, and climb back to a sink (flat plate) on the other side at the same potential as the source...

Is there a current?

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm

Post by chrismb »

There is a start-up current when the space begins to charge up with electrons in transit, and once there is a given space-charge of electrons, then there is no net current, though there is a power consumption due to the electrons being raised by their work-function to escape the thermionic emission. In that regards, the sink is then at the work-function below the source, so you can then see a current will flow between the space-charge around the filament which is at a different potential to the sink.

So the question is also 'at what point in time' do you want the answer. At start up, or steady state?

KitemanSA
Posts: 6179
Joined: Sun Sep 28, 2008 3:05 pm
Location: OlyPen WA

Post by KitemanSA »

If the electrons could be metered one by one so that there was one and only one electron at a time zipping past the ring, current?

D Tibbets
Posts: 2775
Joined: Thu Jun 26, 2008 6:52 am

Post by D Tibbets »

As mentioned, you need to consider the energy used to release the electrons from the cathode. But other than that, with the electrons flowing down a potential well and back up, and recycling (like in a perfect Polywell). There would be some initial current from the cathode, but then none as the electrons are recirculated, so there would be no net current and no net potential IE: no power expended.
If the electrons ground at the top of the potential well after giving up (almost) all of their kinetic energy to the potential well, there is current but no potential (net voltage) used. Again power would be zero. Power = volts * Amps. IE: no energy would used in this steady state system. In the real world, you would never reach this state, but you might come close. A super conductor fits this model. There is a lot of current, but no potential drop, so no power, heating, etc. This may seem strange since in a copper wire the heating of the wire (work) is determined by the current, not the voltage. But, remember, you have to have some voltage to drive the current through the resistive wire. The more resistance, the greater the volts needs to be (or perhaps I should say- the greater the voltage drop needs to be). But at the end, the voltage drops to zero at the end (ground). Again, ignoring the cost of extracting the electrons from the cathode, if you take the example of one battery discharging through a wire to another battery, you could consider that no energy was expended in this (superconducting) wire. There is energy expended in the inefficient chemical reactions at each battery though.
In plasmas, where there are often currents mentioned, I am sometimes confused whether they are referring to the net neutral mixing of charged particles (limited by quasineutrality-no net current flow), or if they are referring to charged particle flows to a sink like a grounded metal shell.
Then there are local concentrations of charged particles (beyond quasineutrality limits?) within a plasma- waves, shock fronts, oscillations, etc. Are these considered 'local' currents?

At least that is my take.

iDan Tibbets
To error is human... and I'm very human.

BenTC
Posts: 410
Joined: Tue Jun 09, 2009 4:54 am

Post by BenTC »

Just thinking out loud...

How is the hole left in the source by emission filled ? Otherwise a positive charge will build up at the source and any emmitted electron will be drawn back to the source.

How is the plate on the other side maintained at the same potential as the source? They must be electrically connected, otherwise electrons that accumulate on the plate will change the potential wrt to the source.

So electrons flow out of the filament, through the ring, into the plate and then cycle back the filament Hence there a current exists.

However I would not call individually metered electrons "a current".
In theory there is no difference between theory and practice, but in practice there is.

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm

Post by chrismb »

KitemanSA wrote:If the electrons could be metered one by one so that there was one and only one electron at a time zipping past the ring, current?
As BenTC's response. You can't just take a moving electron and say 'is this a current'. Well, you could if you like and if so then 'yes' is the answer to your question.

Basically, in the case of a thermionic emitter, I think the answer is yes merely because, as Dan says, the electrons flow down some gradient. When they get to the sink then 'holes' must travel through the conductor to the sink to neutralise the electrons as more electrons get emitted. So you have two halves of a circuit with both holes and electrons moving to the sink. The 'circuit' has no net current flowing 'around' it. Pick an electron and call it a current if it moves, then, yeah, you've got current.

Does this help, or confuse? Can you be clearer on the objective of asking your question?

TallDave
Posts: 3140
Joined: Wed Jul 25, 2007 7:12 pm
Contact:

Post by TallDave »

climb back to a sink (flat plate) on the other side at the same potential as the source...Is there a current?
No.

Assuming NO electrons reach your anode and NO electrons upscatter to carry energy to the plate, no energy is being transferred, and you would quickly arrive at an equilibrium in which no more electrons were accumulating due to space charge limits.

Of course, it's pretty hard to keep the electrons off your anode and electrons all at the source energy; in reality the electrons would maxwellianize and some would end up at the bottom of the well while others carried energy to the plate, while still others carried lots of energy across the B field to the anode. But for purposes of your hypothetical, and considering only the elements in the scenario, I think the answer is no.
Last edited by TallDave on Fri Sep 24, 2010 5:49 pm, edited 1 time in total.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm

Post by chrismb »

TallDave wrote:Assuming NO electrons reach your anode and NO electrons upscatter to carry energy to the plate, you would quickly arrive at an equilibrium in which no more electrons were accumulating due to space charge limits. And since your electrons have no reason to go to the plate (it doesn't attract them), I don't see a flow happening there either.
I think that's right, but not for the right reasons. I think the question being asked is assuming that energy is collected by the electrons as it is accelerated by the field towards the target. It is the filament-target field that gives the electrons energy, and the inertia, together with the magnetic field, means that the electron doesn't actually get there.

The thing is, there is also a field between this sink and the target. As the electron, under inertia in the mag field, follows a path to the sink, it is actually going back up that potential field and, thus, looses the energy again. One might naively say that there is no energy expended, because the electron goes down then back up the 'same field', but that the electrons origin is actually a work-function-worth of potential less than the energy it gives back up in dropping back to the sink. So the power consumed is merely the total emitted charge x work function, plus the other EM emissions from the filament.

This all comes under 'parallel plate/magnetron condition' physics, and there is a limiting potential, or mag field, that means particles can't get across the gap before being turned back around on themselves.

TallDave
Posts: 3140
Joined: Wed Jul 25, 2007 7:12 pm
Contact:

Post by TallDave »

Yes, I suppose it depends on where the plate is. Put the anode exactly equidistant between plate and emitter, and the answer is no: they leave with the same energy. Put the plate closer to the anode than the emitter, and the electrons are giving up kinetic energy when they hit it, so the answer is yes. Put it farther, and they never get there, so the answer is no.

Again, I'm ignoring all the physical impossibilities here.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

D Tibbets
Posts: 2775
Joined: Thu Jun 26, 2008 6:52 am

Post by D Tibbets »

Keep in mind the net effects involved. In a conductive wire, the outer shell electrons normally drift from one atom to another in a random manner. So long as there is no input of new electrons there is no net effect so there is no net current, though there are a lot of semi free electrons moving around (equivalent to electrons moving back and forth in a potential well, sort of). It requires a electromotive force to push the electrons (or charged particles out one end. Certainly movement of electrons within a conductor- even if they stay in that local area may have useful or harmful effects when then allowed to escape that locality. Think of capaciters , resonate circuits.
Holes are places where electrons leave, and before the next electron replaces it. Easy to appreciate. But, then you start talking about the holes moving, like in semiconductors, or oppositely charged particles moving in a plasma and it becomes increasingly confusing. Then throw in superconductors , Debye shielding, quantum mechanics, relativity,etc. and the descriptions become even more obtuse.

Debye lengths are straight forward in the conditions where they are defined- enough time for an equalibrium to be established, and with infanitly small currents. I once asked how the debye lengths / shielding was effected by more than very tiny currents, or by non Maxwellion charged particle distributions, but I didn't get any feedback.

Dan Tibbets
To error is human... and I'm very human.

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm

Post by chrismb »

Before answering, I still think it is necessary to understand whether it is a question of the 'steady state' scenario, or other, and whether a current is considered flowing even if there is no net charge passing around a circuit, or whether it is simply defined as charges in motion. These things need to be defined in the question to answer it, and it would help if the objective of the question was stated.

WizWom
Posts: 371
Joined: Fri May 07, 2010 1:00 pm
Location: St Joseph, MO
Contact:

Post by WizWom »

As long as some electrons reach to magnetically protected grid, there is a current. You will never be able to run a polywell without the electron guns being on.
Wandering Kernel of Happiness

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm

Post by chrismb »

WizWom wrote:You will never be able to run a polywell without the electron guns being on.
Is that true? In my device, there is no need for any ion or electron sources because the lost energy from non-fusing scattering losses are mostly losses due to ionisation of the neutrals. So there is a ready source of ions, 'cos almost no collisions end up in fusion [statistically speaking].

hanelyp
Posts: 2261
Joined: Fri Oct 26, 2007 8:50 pm

Post by hanelyp »

In so far as the magnetic shielding on the grid is imperfect, there will be a current between the cathodes and the grid. Assuming a steady state polywell past the startup transient, there will be an almost equal number of the electrons circulating in and out of cusps, far exceeding the number reaching the grid.

As far as needing to keep the cathodes running, I recall speculation that with fusion products escaping and leaving electrons behind, if fuel is injected as neutral molecules that might in some cases be enough electrons to replaces losses. Some means of heating electrons would then be needed.

D Tibbets
Posts: 2775
Joined: Thu Jun 26, 2008 6:52 am

Post by D Tibbets »

Ah, but is the recirculating electrons a current? They are moving back and forth is a complex potential well with no net flow. at least in a perfect system. Up scattered electrons can overcome this potential well and reach the walls, That would definitely be a current. So three (?) electron currents. Those transported through the magnetic fields, those up scattered electrons reaching the wall, and those electrons (up scattered or otherwise) reaching unshielded magrid structures and supports. The up scattered current is needed to carry away the too hot electrons, the other currents need to be minimized.

Dan Tibbets
To error is human... and I'm very human.

Post Reply