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General => General Chat => Topic started by: Circlotron on September 23, 2016, 05:37:30 am

Title: Faraday homopolar disc problem
Post by: Circlotron on September 23, 2016, 05:37:30 am
So we get a C shaped magnet with closely spaced pole pieces and between (but not touching) those pole pieces we put a copper disc with its centre aligned with the magnet pole piece centre. Then we pass a direct current from the centre to the edge of the disc via a slip ring and brush on the edge. If the disc is free, it will rotate around it's centre. Nothing new here. All dates back to the 1830's.

However...

If we grab hold of that disc to prevent it from rotating, we would presume that an equal but opposite torque is exerted on the magnet. If it were not so, we would have set the physics world on its ear with our new reactionless motor.

The only alternative is that that reaction torque sets the magnet rotating. Now the big question is - does this rotating magnet induce a counter EMF in the disc? Present day physics says "no". The trouble is, this would mean that we have produced an electric motor that operates and produces mechanical power when supplied with a current but has no voltage drop across it. (Ignoring resistive losses)

A normal DC motor, when rotating, generates a counter EMF that opposes the supply voltage and therefore allows a predictable current to pass though it, the current being (supply voltage - counter EMF) / motor resistance.

Our homopolar motor with a stationary disc either does NOT produce a counter torque on the magnet structure, or if it does, the rotating magnet does NOT induce a counter EMF in the disc and therefore does not draw any electrical power while producing mechanical work because of zero voltage drop across it.

Neither situation seems satisfactory. Choose your poison.
Title: Re: Faraday homopolar disc problem
Post by: NiHaoMike on September 23, 2016, 05:57:17 am
Real conductors have resistance that need voltage to overcome. If you try to use a superconductor, it will interact with the magnetic field in such a way as to completely change the situation.

It's a little like the classical "connect a charged capacitor to an uncharged one" problem where calculations show that half the energy disappears. The question now is where the energy goes in an ideal circuit. There would be no resistance to absorb it. One answer is that it gets radiated, but an ideal circuit wouldn't do that either. The real answer is that it's physically impossible to build such an ideal circuit since there will always be inductance. If you could prevent it from radiating, it will just act as a LC tank circuit, never settling to steady state.
Title: Re: Faraday homopolar disc problem
Post by: Circlotron on September 23, 2016, 08:26:59 am
Mike, the subject is electromagnetism and whether or not an axially rotating magnetic pole piece induces an EMF across the radius of the disc. I understand what you are saying about resistance and the real world, but it doesn't materially alter the situation I have presented here.
Title: Re: Faraday homopolar disc problem
Post by: John Coloccia on September 23, 2016, 10:35:46 am
Our homopolar motor with a stationary disc either does NOT produce a counter torque on the magnet structure, or if it does, the rotating magnet does NOT induce a counter EMF in the disc and therefore does not draw any electrical power while producing mechanical work because of zero voltage drop across it.

How did you arrive at this conclusion?
Title: Re: Faraday homopolar disc problem
Post by: Circlotron on September 23, 2016, 11:09:35 am
Which part?
Title: Re: Faraday homopolar disc problem
Post by: helius on September 23, 2016, 11:26:45 am
I think both parts actually derive from the same premiss, so they are not really a contradiction, but simply begging the question. If you assume that a magnetic field has an axial symmetry, then rotating it around its axis doesn't change anything, so it can neither lead to a back-EMF nor can it be the cause of a reaction torque.
Title: Re: Faraday homopolar disc problem
Post by: amyk on September 23, 2016, 11:29:32 am
Use the (Lorentz) Force... (https://en.wikipedia.org/wiki/Faraday_paradox#Using_the_Lorentz_force) :D
Title: Re: Faraday homopolar disc problem
Post by: Circlotron on September 23, 2016, 11:34:48 am
so it can neither lead to a back-EMF nor can it be the cause of a reaction torque.
It 100% most definitely produces a torque on the disc to cause it to try and rotate, as I fact I have seen with my own experiments some years ago. The magnet assembly is effectively the stator of this motor, and to say that this stator does not try and twist the opposite direction to the disc opens up a big can of worms. My precise reason for starting this thread.
Title: Re: Faraday homopolar disc problem
Post by: Circlotron on September 23, 2016, 11:42:46 am
Use the (Lorentz) Force... (https://en.wikipedia.org/wiki/Faraday_paradox#Using_the_Lorentz_force) :D
That level of mathematics is beyond me, but if there is an obvious flaw in the reasoning in my first post then please point it out!
Title: Re: Faraday homopolar disc problem
Post by: alsetalokin4017 on September 23, 2016, 01:47:39 pm
https://www.youtube.com/watch?v=wFMq1Cvtg1s (https://www.youtube.com/watch?v=wFMq1Cvtg1s)

How's that for a can of worms....   
Title: Re: Faraday homopolar disc problem
Post by: SeanB on September 23, 2016, 05:49:48 pm
Try making a ball bearing motor....
Title: Re: Faraday homopolar disc problem
Post by: T3sl4co1l on September 23, 2016, 06:06:56 pm
http://mysite.du.edu/~jcalvert/tech/faraday.htm (http://mysite.du.edu/~jcalvert/tech/faraday.htm)

;)

Browse the rest of the site BTW -- great education!

The resolution to the riddle is that the magnetic field is a product of rotational symmetry.  Rotating the magnet necessarily, cannot have any effect, because it's already spinning!

Using a horseshoe magnet adds confusion, but the fields remain the same.

Where is the counter-torque generated?  On the loop shorting across the disc: between the axis and the slip ring contact.

It still seems unbelievable, because one can mentally construct a situation where B field is penetrating the disc, and exists nowhere else (or at least, nowhere asymmetric and nonrotating).  But this is a faulty construction!  The laws of electromagnetism do not allow sharp cutoffs to fields, nor is a field allowed to decay over any distance, without also fringing away.

Thus, whether by distance or by shielding, the magnetic field must always be whole (it loops back on itself, because \$\nabla \cdot \vec{B} = 0\$ ).  And anywhere within that field that you might place a return wire, it must necessarily be exposed to the same enclosed flux.  You can equally well ask, what should happen if the disk is nonrotating and the contact is revolved around it?  And again, the answer is the same, a voltage is created.

Tim
Title: Re: Faraday homopolar disc problem
Post by: John Coloccia on September 23, 2016, 06:33:32 pm
Ah...I understand the question how. The Faraday "paradox". What a crappy way of teaching this or talking about it, to say that there's some paradox when it's really quite simple. When the magnet spins, there's no change in flux. It's a mistake to think of the field as being "attached" to points on the magnet. It's just like if you had a spherical light source and the source was rotating.
Title: Re: Faraday homopolar disc problem
Post by: alsetalokin4017 on September 23, 2016, 06:58:02 pm
OK... now hold that thought in your mind while you consider this:

https://www.youtube.com/watch?v=JRby1Wilv-Q (https://www.youtube.com/watch?v=JRby1Wilv-Q)
Title: Re: Faraday homopolar disc problem
Post by: IanB on September 23, 2016, 06:58:28 pm
But there is a torque on a rotating symmetrical magnet, as seen by my experiment here:

https://youtu.be/0iVOG2eMAGY

There is a force on the wire carrying the current, and since the wire is fixed in space the magnet experiences a reaction force and spins. The magnet poles are oriented up and down, so there is rotational symmetry in this system.
Title: Re: Faraday homopolar disc problem
Post by: helius on September 23, 2016, 08:53:13 pm
How can you say the torque isn't on the conductive plane (with the painted sectors) between the axis and the perimeter?
Title: Re: Faraday homopolar disc problem
Post by: CatalinaWOW on September 23, 2016, 10:17:20 pm
OK... now hold that thought in your mind while you consider this:

https://www.youtube.com/watch?v=JRby1Wilv-Q (https://www.youtube.com/watch?v=JRby1Wilv-Q)

Were I to look for an explanation of this I would explore the interaction of the boiling nitrogen with the magnet.  It appears that the magnet is slightly heavy on one side, giving it a mildly preferred orientation wrt gravity.  It is effectively a very low friction pendulum, with a natural frequency of a little over a second.  Which seems to be roughly the bubble rate of the nitrogen.  So now you are driving a high Q mechanical resonator near its resonant frequency.  Minor shape variances on the magnet would cause a preference for one direction over the other. 

With a little money or a well equipped lab you could check this by conductively coupling a colder liquid to the nitrogen bath.  You could slow or eliminate the boiling.  Lots of choices for the colder liquid.  Nitrogen under vacuum.  Neon.  Or go beast mode and use hydrogen.

Or do it like an engineer and use a piece of cardboard to shield the suspended magnet from the gas boiling off and from any spray from the bursting bubbles.
Title: Re: Faraday homopolar disc problem
Post by: alsetalokin4017 on September 23, 2016, 11:57:36 pm
I don't have a video of it but the experiments have been done. The magnet rotates even when shielded from the turbulence of the boiling LN2.
Title: Re: Faraday homopolar disc problem
Post by: T3sl4co1l on September 24, 2016, 12:04:42 am
But there is a torque on a rotating symmetrical magnet, as seen by my experiment here:

<embed>

There is a force on the wire carrying the current, and since the wire is fixed in space the magnet experiences a reaction force and spins. The magnet poles are oriented up and down, so there is rotational symmetry in this system.

Obviously, it doesn't matter if:
1. The magnet is in free space, projecting a field through the conductive disk, or
2. the magnet is attached to the disc, above or below or within it, or
3. the magnet IS the disc.

What you've created is only one thing:

A mechanism for the generation of verbal ambiguity.

The torque is acting on the magnet (the whole physical object), because of the magnet (the invisible magnetic field, permeating the physical object and the space around it), and the current flowing through/over the object (the plating and material bulk are conductive).

There is no ambiguity in the physical phenomenon. ;)

Tim
Title: Re: Faraday homopolar disc problem
Post by: John Coloccia on September 24, 2016, 12:15:54 am
I don't have a video of it but the experiments have been done. The magnet rotates even when shielded from the turbulence of the boiling LN2.

The most obvious conclusion would be that the assumed symmetry of one of the magnets is actually not symmetrical for some reason. It's hard to imagine what else could account for it. The next question would be WHY isn't it symmetrical, and I wouldn't even have a guess. Obviously something is changing, though.

I know strange things happen to the flux lines in superconductors, more than just the Meissner effect depending how it's setup, so maybe it's something to do with the setup?
Title: Re: Faraday homopolar disc problem
Post by: IanB on September 24, 2016, 12:37:01 am
Responding to my video above, both helius and T3sl4co1l have pointed out that current is flowing radially from the edge of the spinning disk magnet to the center, and that there are magnetic field lines running vertically through the magnet between top and bottom faces. The current is therefore intersecting the field lines at right angles and this is generating a force that causes the magnet to spin (the paper disk is just for show). The "mystery" is not a mystery.

It is curious perhaps that the cylindrical magnet as a conductive metal disk, and the cylindrical magnet as a source of magnetic field lines, are two independent components of the system. It is not the "magnet" that spins, but the "conductive metal disk".
Title: Re: Faraday homopolar disc problem
Post by: T3sl4co1l on September 24, 2016, 03:06:24 pm
So, since the magnet is already spinning (in the sense that the electron spins, within, are producing a coherent magnetic field), riddle this:

If a magnet is suspended in free space, then its temperature raised to the Curie point, say by heating it with random-polarized light), does it begin spinning?

Or in other words, does the magnetization contain (microscopic) angular momentum, which is released as (macroscopic) rotation of the object, when the spins are decoupled (by raising the temperature)?

Tim
Title: Re: Faraday homopolar disc problem
Post by: John Coloccia on September 24, 2016, 03:33:42 pm
So, since the magnet is already spinning (in the sense that the electron spins, within, are producing a coherent magnetic field), riddle this:

If a magnet is suspended in free space, then its temperature raised to the Curie point, say by heating it with random-polarized light), does it begin spinning?

Or in other words, does the magnetization contain (microscopic) angular momentum, which is released as (macroscopic) rotation of the object, when the spins are decoupled (by raising the temperature)?

Tim

Angular momentum is conserved. It's either there or it's not, so I'm not sure how it could stored up and then released.
Title: Re: Faraday homopolar disc problem
Post by: amyk on September 24, 2016, 04:51:55 pm
The homopolar generator problem becomes even more fun when relativity (https://en.wikipedia.org/wiki/Moving_magnet_and_conductor_problem) is involved. :)
Title: Re: Faraday homopolar disc problem
Post by: TerraHertz on September 25, 2016, 01:08:52 am
I don't have a video of it but the experiments have been done. The magnet rotates even when shielded from the turbulence of the boiling LN2.

Next experiment: suspend the magnet in a vacuum vessel above the superconductor.
My guess is the pendulum torque is due to small differences in surface temp of the magnet acquired from the surrounding temp-gradient gas, and differences in the statistical vector of average force due to gas molecules rebounding from the magnet surface.
Prediction: in a vacuum, it won't rotate.
Title: Re: Faraday homopolar disc problem
Post by: John Heath on October 23, 2016, 09:54:01 am
Interesting discussion. The problem can be dissected by separating variables such as magnet , copper disk and contacts to copper disk. Rotating magnet will not not generate voltage in the copper disk. Rotating the copper disk only will generate a voltage. Rotating the contacts to the copper disk only will also generate a voltage ?? Bigger surprise. If you rotate the copper disk and contacts to the copper disk together it will not generate a voltage. To add clarity a volt meter tie-wrapped to the copper disk with one lead solder to the outside and other lead to the center of the copper disk will not show a voltage when the disk is spinning.  It seems the difference in contacts to disk rotation speed is the main point of generating a voltage. Spinning the contacts or spinning the copper disk is equivalent. This begs the question how could stationary contacts measure a volt on the disk while a meter spinning with the copper disk says there is no voltage??  Is it the scraping of a contact along a copper disk that is generating the voltage ? Is it a relativistic effect that the spinning meter would share with the spinning disk leading to a null result  More questions than answers. Have a video that demonstrates how rotating the contacts only will generate a voltage.

https://www.youtube.com/watch?v=gduYoT9sMaE (https://www.youtube.com/watch?v=gduYoT9sMaE)
Title: Re: Faraday homopolar disc problem
Post by: John Coloccia on October 23, 2016, 10:19:51 am
This begs the question how could stationary contacts measure a volt on the disk while a meter spinning with the copper disk says there is no voltage??  Is it the scraping of a contact along a copper disk that is generating the voltage ? Is it a relativistic effect that the spinning meter would share with the spinning disk leading to a null result  More questions than answers. Have a video that demonstrates how rotating the contacts only will generate a voltage.

Remove the disc, short the contacts and repeat the experiment. I suspect waving a wire/metal above the magnet is enough to generate a significant voltage, especially looking into a high impedance like a scope.
Title: Re: Faraday homopolar disc problem
Post by: T3sl4co1l on October 23, 2016, 12:07:23 pm
Using brushes or spokes is an interesting non-sequitur.  Suppose you make them rotationally symmetric:
- Instead of just one disk, use two, stacked.  (The contacts have brushes for both.  Now the front and back discs can rotate independently, as can the brushes.)
- Instead of brushes, on arms, at one angular position, use many.  The voltage generated is independent of angle, so the brushes can be moved to any location, or turned into a complete surface of revolution.
- Finally, use cylindrical walls with brush faces on either end.

Now the top and bottom faces (disks), and inside and outside walls (cylinders), can rotate independently of each other and the magnet.  (Okay, so you'll have to do some mechanical finagling to get all four truly independent!)

In all cases, rotation of the magnet has no effect.  You will find rotation of the far disc has little effect (because, obviously, it's furthest from the magnet!), and rotation of the near disc, or the cylinders, has the most effect.

If the inner and outer cylinders rotate in the same direction, is there a voltage?  If they rotate in opposite directions, is there a voltage?

Ooh, a brand new riddle... have fun... ;D ;D

Tim
Title: Re: Faraday homopolar disc problem
Post by: John Heath on October 30, 2016, 09:57:08 pm
There are hills on the ocean. It is caused by under water mountains. the mountain is denser than water so there is more gravity causing a slight hill on the ocean surface. The interesting thing about these ocean gravity hills is you can not water ski down them as from gravity's point of view it is a straight line not a curved hill. With the gravity ocean hill in mind I would like to return to the Faraday paradox. Turning the copper disk only will generate a voltage on the contacts. Turning the contacts only will generate a voltage.  ? Turning the copper disk or turning the contacts is equivalent. Either one turned will generate a voltage. Turning the disk clockwise = positive voltage . Turning the contacts clockwise is a negative voltage. One more step into the abyss. If disk is positive and contacts are negative for clockwise turning then turning both contacts and disk clockwise turning clockwise equals 0 volts. This is somewhat like the ocean gravity hills where the hill is there but you can not water ski down the hill much like the meter and contacts that are rotating with the disk can not measure the voltage as the same voltage is being generated in the meter leads to make the measurement therefore no current therefore no voltage.

The implication of this is one meter spinning with the disk will read 0 voltage while an identical meter stationary with sliding brushes on the disk will read a voltage. If I could take a picture at the exact time the sliding brush is in the same place as the spinning meter probe on the outside of the spinning disk one meter would say 1 volt while the other meter says 0 volts measuring the same location at the same time  :scared: 
Title: Re: Faraday homopolar disc problem
Post by: Circlotron on October 31, 2016, 11:25:04 am
In the original post I mentioned that my imaginary setup has a magnetic pole piece that does not allow any external field to escape and induce an EMF into the brushes or measuring leads.
Title: Re: Faraday homopolar disc problem
Post by: John Coloccia on October 31, 2016, 01:29:33 pm
There are hills on the ocean. It is caused by under water mountains. the mountain is denser than water so there is more gravity causing a slight hill on the ocean surface. The interesting thing about these ocean gravity hills is you can not water ski down them as from gravity's point of view it is a straight line not a curved hill. With the gravity ocean hill in mind I would like to return to the Faraday paradox. Turning the copper disk only will generate a voltage on the contacts. Turning the contacts only will generate a voltage.  ? Turning the copper disk or turning the contacts is equivalent. Either one turned will generate a voltage. Turning the disk clockwise = positive voltage . Turning the contacts clockwise is a negative voltage. One more step into the abyss. If disk is positive and contacts are negative for clockwise turning then turning both contacts and disk clockwise turning clockwise equals 0 volts. This is somewhat like the ocean gravity hills where the hill is there but you can not water ski down the hill much like the meter and contacts that are rotating with the disk can not measure the voltage as the same voltage is being generated in the meter leads to make the measurement therefore no current therefore no voltage.

The implication of this is one meter spinning with the disk will read 0 voltage while an identical meter stationary with sliding brushes on the disk will read a voltage. If I could take a picture at the exact time the sliding brush is in the same place as the spinning meter probe on the outside of the spinning disk one meter would say 1 volt while the other meter says 0 volts measuring the same location at the same time  :scared:

Did you try what I suggested, without the disk?
Title: Re: Faraday homopolar disc problem
Post by: T3sl4co1l on October 31, 2016, 01:41:31 pm
In the original post I mentioned that my imaginary setup has a magnetic pole piece that does not allow any external field to escape and induce an EMF into the brushes or measuring leads.

I think you will find, such a setup remains an object of pure imagination. ;D  It's very easy to do partial textbook problems like "a charged particle enters and flies through a region of uniform \$\vec{B}\$ directed out of the page", but the real world does not admit fields that can simply be willed into existence!  There is a necessary gradient along the way, and since it's a vector field, the gradient carries a direction as well, and divergence and curl (according to the rules of E&M: namely, \$\nabla \cdot \vec{B} = 0\$ and \$\nabla \times \vec{B} =\$  [contained currents] ).  For the field to remain 'vertical' but reduce in magnitude, the gradient has to be stretched out infinitely long; because if it were finite, the vectors would 'bulge out' (fringing field), and no longer be vertical.  the field diverges, locally, around the edges of the pole pieces or whatever (which can be seen as local monopoles, thus seemingly violating the first law; of course, the poles themselves are double-ended, so the whole system is fine).

So, necessarily, you will have the brushes or leads passing through an opening in a pole-piece, where they subtend some enclosed flux.  As well there must always be. :)

Tim
Title: Re: Faraday homopolar disc problem
Post by: John Heath on November 01, 2016, 09:30:38 am
There are hills on the ocean. It is caused by under water mountains. the mountain is denser than water so there is more gravity causing a slight hill on the ocean surface. The interesting thing about these ocean gravity hills is you can not water ski down them as from gravity's point of view it is a straight line not a curved hill. With the gravity ocean hill in mind I would like to return to the Faraday paradox. Turning the copper disk only will generate a voltage on the contacts. Turning the contacts only will generate a voltage.  ? Turning the copper disk or turning the contacts is equivalent. Either one turned will generate a voltage. Turning the disk clockwise = positive voltage . Turning the contacts clockwise is a negative voltage. One more step into the abyss. If disk is positive and contacts are negative for clockwise turning then turning both contacts and disk clockwise turning clockwise equals 0 volts. This is somewhat like the ocean gravity hills where the hill is there but you can not water ski down the hill much like the meter and contacts that are rotating with the disk can not measure the voltage as the same voltage is being generated in the meter leads to make the measurement therefore no current therefore no voltage.

The implication of this is one meter spinning with the disk will read 0 voltage while an identical meter stationary with sliding brushes on the disk will read a voltage. If I could take a picture at the exact time the sliding brush is in the same place as the spinning meter probe on the outside of the spinning disk one meter would say 1 volt while the other meter says 0 volts measuring the same location at the same time  :scared:

Did you try what I suggested, without the disk?

The video I posted is not my experiment so I can not change it. However I know what you mean. Waving a wire in front of a magnet will produce a voltage. Suppose you are given the task of writing a program to simulate a meter  measuring the voltage across a wire in a changing magnetic field. You calculate the wire voltage. You now calculate the voltage across the meter leads as they are in the same changing magnetic field. If the wire and the meter leads follow an identical route it will always equal 0 volts as the two cancel out. On the other hand if the meter leads take a roundabout route outside the magnet field then a voltage across the wire can be made.

With this in mind let us go the the equator where we are moving 1000 miles per hour eastward through the earth's magnetic field. Under these conditions you can not measure a voltage across a wire as the earth's magnetic field is constant every where. No place to hide the meter leads from the magnetic field. The wire and meter lead voltages will always cancel out. In the above example yes as it is a small magnetic field that can be avoided by moving the meter leads away from the magnetic. This can not be done at the equator as the earth's magnetic field is just too large to work around it.
Title: Re: Faraday homopolar disc problem
Post by: Circlotron on November 01, 2016, 12:26:01 pm
Couldn't you lay 100 metres of wire in a north-south direction and have your long meter leads encased in steel tubing? Would the field be diverted around the meter leads, enabling you to measure any EMF in the wire?
Title: Re: Faraday homopolar disc problem
Post by: helius on November 01, 2016, 02:39:28 pm
I wouldn't say the Earth's magnetic field is too small / too flat to measure: magnetometers work. It can also have unintended effects such as a different deflection of the electron beam in a CRT display depending on where you are in the magnetic field. CRTs destined for the Northern hemiphere, Southern hemisphere, or the Equator need to be adjusted differently because of this.
Title: Re: Faraday homopolar disc problem
Post by: T3sl4co1l on November 02, 2016, 04:46:45 pm
Couldn't you lay 100 metres of wire in a north-south direction and have your long meter leads encased in steel tubing? Would the field be diverted around the meter leads, enabling you to measure any EMF in the wire?

You make a loop of wire with a toroidal shield around it; the flux enclosed by the inside part of the shield is the same flux that is enclosed by the loop, regardless.  (And that's assuming infinite mu for the shield, so that the B field at the wire can be ~zero.)

It's perfectly normal for fields to have an effect, even with B=0.  Take the Aharonov–Bohm effect (https://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect) for instance.

It's not the field at the wire that matters, so much as the loop integral around it; and the value of that integral is inseparable from the flux enclosed by the loop.  The fact that a loop integral (along a closed curve, with dimensions of length, but no dimension of thickness) equates directly with the integral over the enclosed surface (an open surface with dimensions of area), is a deep result of analytical calculus.  As E&M fields must be analytic, so too must this fact be true. :)

Tim
Title: Re: Faraday homopolar disc problem
Post by: John Heath on November 03, 2016, 03:38:59 am
Magnetic shielding for a changing magnetic field is possible as the shield will develop a counter EMF to null it out. However for a magnetic field that is not changing such as the earth's field it is not that easy. A local man made field that is opposite of the earth's field could cancel it out.  Another concern with making a measurement of a voltage on a wire from a magnetic field that is not changing in strength is energy conservation laws. If a magnet could produce a voltage on a wire without movement it would be free energy which is a big no no. No free lunch is physics so from energy conservation alone the measurement should be impossible to make. Somewhat like a condenser microphone that has a permanent voltage difference but you can never discharge it or an ocean water gravity hill but you can not water ski down this hill. In the case of the mono pole motor there is no voltage without movement. Movement requires energy so energy conservation laws are satisfied. This being said the nerd in me still thinks there has to be a way to make this voltage measurement and I suspect then key is to make sure no energy can come of it. 
Title: Re: Faraday homopolar disc problem
Post by: Circlotron on November 03, 2016, 11:17:40 am
Magnetic shielding for a changing magnetic field is possible as the shield will develop a counter EMF to null it out. However for a magnetic field that is not changing such as the earth's field it is not that easy.
If you pass a wire through a copper pipe, the pipe will shield the wire from a changing field just as you say. A steady field will pass straight through the pipe walls. With a a ferrous metal pipe e.g. steel the magnetism with enter the steel from one side of the pipe and take the low reluctance path around to the other side of the pipe and then exit and on its way rather than leave the steel and instead go through the internal air space and through the wire. That's why I would have thought you could shield the meter wires from the earth's mag field. But nothing ever seems so simple.

(http://i.stack.imgur.com/MUfr3.gif)
Title: Re: Faraday homopolar disc problem
Post by: T3sl4co1l on November 03, 2016, 03:20:40 pm
If you step-change the pipe-and-wire assembly in that field, at first there will be no induced voltage, and the pipe will have sheared the field lines in that direction.  Which is a graphical way of putting it: the pipe's outer wall has an inducted current, which you are seeing superimposed upon the ambient field.  This reflected field happens to be just the right magnitude to cancel out the field in the interior, so the induced voltage is zero.

Over time, the field lines seep through the wall.  Resistive materials act as a gooey trap for field lines.  (Type I superconductors have no resistance, and exclude field lines for all time.  Type II superconductors exhibit hysteresis loss -- called flux pinning -- so that, if the magnitude of influence is great enough, some flux will pass through anyway.  I don't know if this is a sudden thing or a gradual thing (consider Barkhausen noise in ferromagnetic materials!), but the effect is simply: nonzero AC resistance, and therefore some induced voltage in the wire.)

For the steel pipe, the field lines divert aroundthe wire, but sooner or later, the field lines trapped above and below the wire must move around the pipe, and the only way for that to happen is for the wire to be cut by lines.  In this case, there is an induced voltage, merely delayed by the losses of the steel.

For pipes of the same dimensions (dia, thickness), the induced voltage will be greater for steel, but not by a gross amount.  The time constant (i.e., how much the step change is slowed down by) is proportional to wall thickness (or rather, a power of it).

Tim
Title: Re: Faraday homopolar disc problem
Post by: John Heath on November 06, 2016, 05:10:25 am
There is a gradient in a magnetic field and there is the density of a magnetic field. The steel pipe will shield against a gradient in a magnetic field as there is a path of less resistance through the steel however it will not shield against the magnetic field itself. An analogy is a resistor that is 1000 volts at one end and 1001 volts at the other. A jumper across the resister will shield against a voltage gradient across the resistor but it will not shield against the average 1000 volt e field of the resister. A compass or a hall effect transistor can only measure the gradient of a magnetic field but not the magnetic field itself. If that sounds like word salad not to worry as I also find it a little confusing.
Title: Re: Faraday homopolar disc problem
Post by: tatus1969 on November 06, 2016, 07:06:03 am
I don't have a video of it but the experiments have been done. The magnet rotates even when shielded from the turbulence of the boiling LN2.

Next experiment: suspend the magnet in a vacuum vessel above the superconductor.
My guess is the pendulum torque is due to small differences in surface temp of the magnet acquired from the surrounding temp-gradient gas, and differences in the statistical vector of average force due to gas molecules rebounding from the magnet surface.
Prediction: in a vacuum, it won't rotate.
+1