Does Lenz' law apply to a conventional homopolar generator?

[jadaro2600]

That's whats puzzling me.  It is obvious that in a conventional generator where there are a plurality of poles, flux is constantly changing.  But an HPG is acyclical.  I seems to me that the flux is uniform and unchanging, yet there is "juice".  So, either I am wrong about the flux or Mr. Lenz is wrong about the drag. ;-)

Lee

You fail to realize the idea of changing flux.  You seem to think that it has to be increasing or decreasing..  All you have to do is move through it for current to form.  The field strength doesn't appear to be changing, but the gradient of strength is still there.

Of course, I'm always wrong about these things.

You need to realize that just because the flux doesn't appear to be changing, doesn't mean that it isn't.  Once you set it into motion, things are changing with respect to the reference frame.

The IS back torque on these devices, though you may calculate it to be zero, it will appear as abnormally large friction.

[leeanderthal]

You fail to realize the idea of changing flux.  You seem to think that it has to be increasing or decreasing..  All you have to do is move through it for current to form.  The field strength doesn't appear to be changing, but the gradient of strength is still there.

It's hard for me to imagine.  A uniform field strength, constant rotational speed and a constant load and the flux is changing?  I've always attributed the drag in an HPG to huge eddy currents and brushes. 
To quote a famous E. engineer and inventor, Charles F. Kettering "a problem well stated is a problem half solved".  It seems as though determining what the problem is ...is the problem.  If what you say is true, we are all wasting our time on HPG's.  If I am correct, all we have to do is get rid of the eddys and the brushes and OU happens.  Simple as that.

Lee

[jadaro2600]

The problem here is that, in order to take off the current, you have to create a path between the axis of rotation and the rim of the disk.  This cannot rotate with the disk, or no current is produced.  It is therefore logical , that it is the radial path which is where the current is being generated.

In such a radius, the rotational speed ( in surface feet per minute ) is least near the center and greatest near the periphery.  THIS is where the change is occurring; the path's inner atoms are spinning slower than the path of the outer atoms.

And yes, eddy currents play a part in this; but it is the relativity which creates the apparition of a changing magnetic flux.  Even though the field strength is the same, it is as if to say that there is one end of a wire moving faster than the other in an even field; therefore a there is an affective difference experienced in the wire.

What would generate more current?  A wire of a given distance moving at a given speed, or the that same wire of a given distance moving at three times that speed?

It's really all about what the individual atoms are experiencing and contributing as a whole.

[leeanderthal]

The problem here is that, in order to take off the current, you have to create a path between the axis of rotation and the rim of the disk.  This cannot rotate with the disk, or no current is produced.  It is therefore logical , that it is the radial path which is where the current is being generated.

In such a radius, the rotational speed ( in surface feet per minute ) is least near the center and greatest near the periphery.  THIS is where the change is occurring; the path's inner atoms are spinning slower than the path of the outer atoms.

And yes, eddy currents play a part in this; but it is the relativity which creates the apparition of a changing magnetic flux.  Even though the field strength is the same, it is as if to say that there is one end of a wire moving faster than the other in an even field; therefore a there is an affective difference experienced in the wire.

What would generate more current?  A wire of a given distance moving at a given speed, or the same wire a given distance moving at three times that speed?

The current flow wouldn't necessarily have to be a radial line, if the entire rim was immersed in a liquid conductor. Nevertheless, I understand you point.  This difference in relative speed is somewhat offset by the fact that the lines of flux will tend to be more concentrated near the center (slower portion) of the disk. If the magnet is disk shaped and as large as the conductor disk.
We also can't overlook the drag cause by induced current in the stationary exterior circuity.


Lee

[jadaro2600]

The current flow wouldn't necessarily have to be a radial line, if the entire rim was immersed in a liquid conductor. Nevertheless, I understand you point.  This difference in relative speed is somewhat offset by the fact that the lines of flux will tend to be more concentrated near the center (slower portion) of the disk. If the magnet is disk shaped and as large as the conductor disk.

Typically, there will be a radial line of current regardless of whether or not it's a liquid brush, such a mercury or liquid sodium, or gallium, even if the brush extends around the entire disk, it will want to take the shortest path.

We also can't overlook the drag cause by induced current in the stationary exterior circuity.

Lee

This path is supposedly where the rotation pushes off of.. "the reference frame"...or so I've been told.

I thought it might be interesting to use a laser to power one, but I can't figure out how this would work without that reference frame.