artis

Personally I don't think any force is exerted on the rod above the magnet apart from if it is from a ferromagnetic material is will feel attraction to the magnet.
My theoretical knowledge is not perfect so I can't say but from what I have read from experiments, spinning a magnet around it's axis doesn't change the field , if the magnet is symmetrical then the field is homogeneous and symmetrical and physically rotating the magnet material doesn't change that. Treat the rod above the magnet as part of a disc so there is no difference and as far as we know if the disc does not spin itself then there is no current generated. To the best of my knowledge.

In other words the B field is static because it's strength doesn't change , it is symmetrical and homogeneous, for the electrons in a stationary conductor to feel a force they need to either move themselves through a B field or have a B field that changes strength or polarity with time so a time varying field , but here we have a stationary field and a stationary conductor.

olgerm

Gold Member
If we were to reverse a film about a fast spinning Faraday disk breaking into pieces, then we would have a film about many small linear Faraday generators becoming one large Faraday generator.
At the beginning of this film the generators would not produce anything (no Lorentz-force is produced). At the end of the film the generator works like a Faraday generator is supposed to work (there is a Lorentz-force).
So between the beginning and the end there must be a transition occurring. The system is in an interesting transitional state.
the generator is producing electricity if charge-carrying material(disc) between stator(brush) and center of disc is moving crosswise to vector from stator(brush) to center of the disc.
If disc break into pieces and there is non-conductive material between pieces the generator does not produce electricity because there is no connection between stator(brush) and center of the disc.

greswd

the generator is producing electricity if charge-carrying material(disc) between stator(brush) and center of disc is moving crosswise to vector from stator(brush) to center of the disc.
If disc break into pieces and there is non-conductive material between pieces the generator does not produce electricity because there is no connection between stator(brush) and center of the disc.
but there could be an EMF within a broken piece, causing the piece to become polarized.

greswd

If we were to reverse a film about a fast spinning Faraday disk breaking into pieces, then we would have a film about many small linear Faraday generators becoming one large Faraday generator.

At the beginning of this film the generators would not produce anything (no Lorentz-force is produced). At the end of the film the generator works like a Faraday generator is supposed to work (there is a Lorentz-force).

So between the beginning and the end there must be a transition occurring. The system is in an interesting transitional state.
yes! you got it!

greswd

I have a question: Does a spinning disk shaped magnet generate a Lorentz-force on a metal rod above it?
According to the video, spinning the magnet does nothing. @artis

greswd

if the body is moving lineary(not spinning) $\omega=0$ the expression simplifies to:
$rot(\vec{v}\times \vec{B})-\frac{\partial \vec{B}}{\partial t}=0$

Therefore the linear generator does not work. Maxwell equations+ and loretz force confirm that in both frames of reference.
Sorry, why does this mean that the linear generator does not work?

To me it seems to only state that $rot(\vec{v}\times \vec{B})=\frac{\partial \vec{B}}{\partial t}$

olgerm

Gold Member
Sorry, why does this mean that the linear generator does not work?
To me it seems to only state that $rot(\vec{v}\times \vec{B})=\frac{\partial \vec{B}}{\partial t}$
because to produce electricity you need a closed circuit in which electromotive force(U) is not 0. U can be expressed as $U=\oint(dl*(\vec{E}+\vec{v}\times \vec{B}))=\oint(dl*\vec{E})+\oint(dl*(\vec{v}\times \vec{B}))=\oint(dl*(\vec{v}\times \vec{B}))-\frac{\partial \oint (dS*\vec{B})}{\partial t}$.
If you take smaller circuits and connect these together then EMF of the new circuit is sum of the smaller circuits. EMF infinitesimaly small circuit is $rot(\vec{v}\times \vec{B})-\frac{\partial \vec{B}}{\partial t}$ and I showed that all these have 0 EMF. All closed circuits can be composed of infinitesimaly small circuits. Therefore all closed circuits must have 0 EMF.

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greswd

because to produce electricity you need a closed circuit in which electromotive force(U) is not 0. U can be expressed as $U=\oint(dl*(\vec{E}+\vec{v}\times \vec{B}))=\oint(dl*\vec{E})+\oint(dl*(\vec{v}\times \vec{B}))=\oint(dl*(\vec{v}\times \vec{B}))-\frac{\partial \oint (dS*\vec{B})}{\partial t}$.
If you take smaller circuits and connect these together then EMF of the new circuit is sum of the smaller circuits. EMF infinitesimaly small circuit is $rot(\vec{v}\times \vec{B})-\frac{\partial \vec{B}}{\partial t}=\vec{v}(\nabla \cdot \vec{B}) - \vec{B}(\nabla \cdot \vec{v}) + (\vec{B}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{B}-\frac{\partial \vec{B}}{\partial t}$ and I showed that all these have 0 EMF. All closed circuits can be composed of infinitesimaly small circuits. Therefore all closed circuits must have 0 EMF.
I refer to my earlier post:
but there could be an EMF within a broken piece, causing the piece to become polarized.
All the linear-motion pieces could be expected to be polarized.

olgerm

Gold Member
All the linear-motion pieces could be expected to be polarized.

artis

I believe the linear moving sheet also becomes polarized just like the disc becomes, in both cases any current can be measured or generated only when the circuit is closed with a conducting path that is either stationary with respect to the moving part of the circuit or moving with a different speed, the generated voltage will then be equal to the field strength and the difference in speed between the two parts of the circuit, the larger the difference in speed the higher the voltage.

The reason I say that the Faraday generator can be also made linear is because there exists a Faraday disc/homopolar generator whose geometry is so called "drum" type instead of axial disc type. Basically a conductive tube or cylinder is rotating around it's vertical axis, the field must then be either from inside out or outside in and the cylinder cuts the field at 90 degrees just like the disc does , again we get generated current if brushes are applied at both ends of the cylinder. Theoretically one can imagine this cylinder being cut open along it's axis and rolled out flat becoming a sheet of metal and the same current generating principles would apply, as long as there are brushes and two parts of a circuit that move with different speed.

greswd

I believe the linear moving sheet also becomes polarized just like the disc becomes
the problem is, that might be a violation of the principle of relativity

greswd

Well, we ought to investigate the transition cases.

olgerm

Gold Member
Well, we ought to investigate the transition cases.
Depends what will be between the pieces, what is capacity of pieces, what shape the pieces are and etc.

greswd

Depends what will be between the pieces, what is capacity of pieces, what shape the pieces are and etc.
yeah, you can set your own parameters.

The issue is how to not violate the principle of relativity.

olgerm

Gold Member
yeah, you can set your own parameters.The issue is how to not violate the principle of relativity.
How could it violate principle of relativity? Set your own parameters and explain how it does violate principle of relativity.

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olgerm

Gold Member
Look at this simple "railgun" which is basically a linear homopolar motor/Faraday disc, if one applies current to the rails the third shunting rod moves along the rails, but if one puts a voltmeter across the rails, puts a magnet under or above the rails and moves the third rod by hand the voltmeter should read DC voltage output because imagine the rails , voltmeter and moving rod form a rectangular loop that is electrically closed , as you move the rod you change the cross-sectional area of the loop in other words you change the amount of B field lines through the loop which results in generated current in the loop.
This is not what OP meant by linear version of the Faraday paradox in his original post, because conductor and magnet are not moving together in your setup.
A conductor placed atop a magnet, both at rest in one scenario. In another, both moving together in uniform linear motion.

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greswd

How could it violate principle of relativity? Set your own parameters and explain how it does violate principle of relativity.
So what's the mathematics of the "transition"?

artis

Both the disc and linear Faraday "sheet" or whatever you call it both agree with relativity in fact I think they can only be explained with the help of Special relativity and Lorentz forces on electrons etc, that is the reason this was called the Faraday paradox in the beginning until the early 20th century.

I suggest @greswd really think about the "drum" type homopolar generator as that is essentially a rolled up version of a flat conducting sheet being moved across a B field at 90 degree angle. Same rules apply as with a disc rotating in a B field.

olgerm

Gold Member
So what's the mathematics of the "transition"?
I do not understand your question. You where interested about transitional cases so I suppose you know yourself what you meant by transition.

greswd

I do not understand your question. You where interested about transitional cases so I suppose you know yourself what you meant by transition.
The rotational Faraday homopolar generator works, the linear one doesn't.

What sort of mathematics describes the EMF during the transition between both cases? From linear motion to a slight curve to full rotational motion.

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