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#### John Heath

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##### Re: Faraday homopolar disc problem
« Reply #25 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.

#### John Coloccia

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##### Re: Faraday homopolar disc problem
« Reply #26 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.

#### T3sl4co1l

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##### Re: Faraday homopolar disc problem
« Reply #27 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...

Tim
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#### John Heath

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##### Re: Faraday homopolar disc problem
« Reply #28 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

#### Circlotron

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##### Re: Faraday homopolar disc problem
« Reply #29 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.

#### John Coloccia

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##### Re: Faraday homopolar disc problem
« Reply #30 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

Did you try what I suggested, without the disk?

#### T3sl4co1l

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##### Re: Faraday homopolar disc problem
« Reply #31 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.   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
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#### John Heath

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##### Re: Faraday homopolar disc problem
« Reply #32 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

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.

#### Circlotron

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##### Re: Faraday homopolar disc problem
« Reply #33 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?

#### helius

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##### Re: Faraday homopolar disc problem
« Reply #34 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.

#### T3sl4co1l

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##### Re: Faraday homopolar disc problem
« Reply #35 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 Aharonovâ€“Bohm 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
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#### John Heath

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##### Re: Faraday homopolar disc problem
« Reply #36 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.

#### Circlotron

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##### Re: Faraday homopolar disc problem
« Reply #37 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.

#### T3sl4co1l

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##### Re: Faraday homopolar disc problem
« Reply #38 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
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#### John Heath

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##### Re: Faraday homopolar disc problem
« Reply #39 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.

#### tatus1969

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##### Re: Faraday homopolar disc problem
« Reply #40 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.
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