Request about experiments on the linear-motion Faraday paradox

[olgerm]

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.

This has made me curious about the linear version of the Faraday paradox.
A conductor placed atop a magnet, both at rest in one scenario. In another, both moving together in uniform linear motion.

 

[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]

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.

 

[greswd]

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.

Then what occurs during the transition? Thanks

 

[olgerm]

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.

Only the angular speed is important, whether the body is moving lineary at same time is not important.

 

[greswd]

Only the angular speed is important, whether the body is moving lineary at same time is not important.

So what's the mathematics? Cos the last time, you said:

This was about linear motion scenario not about disc breaking scenario.

 

[olgerm]

So what's the mathematics? Cos the last time, you said:"This was about linear motion scenario not about disc breaking scenario."

You said that you mean by transition to go from linear motion to a slight curve to full rotational motion, not disc breaking scenario.

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

My answer was about transition in that sense.
Do you want more explanation about disc breaking?

 

[artis]

@greswd , Who said the linear generator doesn't work?

One more time, there is no difference between the linear metal sheet or when the sheet is rolled up in a drum and the flat pancake disc they are all the same from the viewpoint of laws of physics, they all work on the idea of electrons feeling the Lorentz force as the metal cuts a homogeneous magnetic field at 90 degrees.
The only difference is that some of those geometries are much harder to make while the disc is easier so most people go wit the disc.

Why do you care about the transition and what do you even mean by that ? There is no transition as far as I know, the only thing that changes is the geometry , physics doesn't change. I think you ask about these transitions because there is a wrong perception that somehow the "linear generator doesn't work" but I already said it does.
Try it out, take a thin copper or aluminum metal sheet, salvage or get a lot of magnets , align them all so that the same poles face upwards, make stationary brushes and then move the sheet back and forth over the magnets while the sides of the sheet touch some stationary brushes or sliding contacts, voltage will be very very small but with a sensitive voltmeter you might pick up some mV.

 

[greswd]

You said that you mean by transition to go from linear motion to a slight curve to full rotational motion, not disc breaking scenario.

My answer was about transition in that sense.

Only the angular speed is important, whether the body is moving lineary at same time is not important.

ok, why is only the angular velocity important when the Lorentz force is dependent on the linear velocity?

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.

Also, how does this quote relate to the quote above?

 

[greswd]

@greswd , Who said the linear generator doesn't work?

One more time, there is no difference between the linear metal sheet or when the sheet is rolled up in a drum and the flat pancake disc they are all the same from the viewpoint of laws of physics, they all work on the idea of electrons feeling the Lorentz force as the metal cuts a homogeneous magnetic field at 90 degrees.

You see, the Lorentz force formula uses the relative velocity seen in your rest frame. Not exactly following the principle of relativity.

 

[olgerm]

why is only the angular velocity important when the Lorentz force is dependent on the linear velocity?

because if you change frame of reference all $\vec{E}$, $\vec{B}$ and $\vec{v}$ change in manner that frameinvariant quantities remain the same.

 

[olgerm]

@greswd , Who said the linear generator doesn't work?

Depends what you mean by linear generator. I say that linear generator in meaning that it is a generator, which all parts are moving together lineary, does not work.

 

[greswd]

because if you change frame of reference all $\vec{E}$, $\vec{B}$ and $\vec{v}$ change in manner that frameinvariant quantities remain the same.

Sorry, I don't understand what you're saying

 

[artis]

Depends what you mean by linear generator. I say that linear generator in meaning that it is a generator, which all parts are moving together lineary, does not work.

Of course it doesn't but then the Faraday disc also doesn't work when you rotate the brushes together with the disc, the same rules apply

 

[olgerm]

Of course it doesn't but then the Faraday disc also doesn't work when you rotate the brushes together with the disc, the same rules apply

Rotational and linear case may not give the same result.
I am not 100% sure. But I think Faraday disc would generate electricity if you rotated its brushes together with its disc.

 

[artis]

No it would not, think about it, because the brushes and the connecting wire between them would rotate with the same angular velocity and direction that the disc rotates which means that the the current in the brushes would be in the same direction as that in the disc , so the result is two currents running oppositely to one another which sums to zero current in the loop.

 

[olgerm]

No it would not, think about it, because the brushes and the connecting wire between them would rotate with the same angular velocity and direction that the disc rotates which means that the the current in the brushes would be in the same direction as that in the disc , so the result is two currents running oppositely to one another which sums to zero current in the loop.

You are probably right. I can't get intuition of the situation. Would it not depend of shape of the wire from brush to center of disk and shape of B-field?

 

[artis]

Well I have read in theory that if the brushes and return wire could get magnetically shielded then in theory the current in the loop should be the sum of the current generated in the disc portion but in real life this is not possible at least haven't heard anyone done that.
The best practical way to "shield" the return path from getting any canceling current generated is to keep it still and that is the reason for the sliding contacts

But I think you got it , it's a Lorentz force generator so it's very peculiar to the B field and parts of a single loop moving while others staying still unlike in a conventional generator you simply change the field strength through a loop and get induced current.

 

[greswd]

because if you change frame of reference all $\vec{E}$, $\vec{B}$ and $\vec{v}$ change in manner that frameinvariant quantities remain the same.

Sorry, I don't understand what you're saying

@olgerm Sorry, need your help.

 

[greswd]

Of course it doesn't but then the Faraday disc also doesn't work when you rotate the brushes together with the disc, the same rules apply

If there are no brushes, the disc still gets polarized, with the rim or the center having a higher electron density. @olgerm

 

[greswd]

You see, the Lorentz force formula uses the relative velocity seen in your rest frame. Not exactly following the principle of relativity.

@artis @jartsa
maybe @olgerm is right and it does depend on the angular velocity. and this would be investigating a rarely mentioned chapter of physics.

 

[olgerm]

No it would not, think about it, because the brushes and the connecting wire between them would rotate with the same angular velocity and direction that the disc rotates which means that the the current in the brushes would be in the same direction as that in the disc , so the result is two currents running oppositely to one another which sums to zero current in the loop.

You are right, it can't produce DC current. EMF summarised over one full rotation in loop that consist of wire disc and brush must be proportional to change in B-field flux compared to last time, that the wire was in that position beacuse of maxwells III equation. And magnetic flux throught the loop must be same every time it gets to that position, because B-field is constant(not changing in time).

 

[greswd]

You are right, it can't produce DC current. EMF summarised over one full rotation in loop that consist of wire disc and brush must be proportional to change in B-field flux compared to last time, that the wire was in that position beacuse of maxwells III equation. And magnetic flux throught the loop must be same every time it gets to that position, because B-field is constant(not changing in time).

If there are no brushes, the disc still gets polarized, with the rim or the center having a higher electron density. @olgerm

There is a distinct physical effect in the rotational case which cannot be ignored.

 

[olgerm]

There is a distinct physical effect in the rotational case which cannot be ignored.

This was not about disc breaking scenario, but about (rotational) faraday generator, when brushes rotate together with disc.

 

[olgerm]

How exactly depends on details (shape of pieces, properties of the material etc), but generally, EMF would smoothly go to zero if the disc breaks.

 

[greswd]

This was not about disc breaking scenario, but about (rotational) faraday generator, when brushes rotate together with disc.

I wasn't referring to the disc-breaking scenario. Also, I said that brushes or no brushes, the disc gets polarized.

 

[greswd]

How exactly depends on details (shape of pieces, properties of the material etc), but generally, EMF would smoothly go to zero if the disc breaks.

That's an interesting assumption. You see, as it breaks, its tangential velocity remains unchanged, and by the Lorentz force law, the force should not change.

 

[olgerm]

That's an interesting assumption. You see, as it breaks, its tangential velocity remains unchanged, and by the Lorentz force law, the force should not change.

• If pieces are very far EMF is not directed to center of disc, but crosswise to it.
• If distances get bigger polarizing effects get smaller.
• If magnet field is from a magnet behind disc, the B-field applied to pieces is very small, if pieces are far from the magnet.

 

[greswd]

• If magnet field is from a magnet behind disc, the B-field applied to pieces is very small, if pieces are far from the magnet.

Let's say its a huge magnet, with a wide magnetic field.

• If pieces are very far EMF is not directed to center of disc, but crosswise to it.
• If distances get bigger polarizing effects get smaller.

Then we assume near distances.

Its not about distance, its about linearity vs rotationality.

 

[olgerm]

we assume near distances.
Its not about distance, its about linearity vs rotationality.

after the disc breaks its pieces must get only füther and füther from each other as time passes. EMF is not directed to center of disc, but crosswise to it if time from breaking approaces infinity. polarizing effects lack to exist as time from breaking approaches infinity.

 

[olgerm]

why is only the angular velocity important when the Lorentz force is dependent on the linear velocity?

linear velocity is not important because if you change frame of reference all $\vec{E}$, $\vec{B}$ and $\vec{v}$ change in manner that frameinvariant quantities remain the same.

Sorry, I don't understand what you're saying

E,B,v are different in different frames of reference, but meaningful(frame invariant) claims same in all frames of reference. E,B,v are different in frames of reference, where linear generator is in rest and where it is moving, but whether it is generating power or not is same in both frames of reference.U is EMF.
In frame where linear generator is in rest:
$U=\oint(dl*(\vec{E}+\vec{v}\times \vec{B}))=\oint(dl*(\vec{0}+\vec{0}\times \vec{B}))=0$

In frame where linear generator is moving:
$U=\oint(dl*(\vec{E}+\vec{v}\times \vec{B}))=\oint(dl*\vec{E})+\oint(dl*(\vec{v}\times \vec{B}))=$
(because Maxwell's III equation)
$\oint(dl*(\vec{v}\times \vec{B}))-\frac{\partial \oint (dS*\vec{B})}{\partial t}$=
(beacause stokes theorem)
$\oint(dS*(rot(\vec{v}\times \vec{B})-\frac{\partial \vec{B}}{\partial t}))= \oint(dS*(\vec{v}(\nabla \cdot \vec{B}) - \vec{B}(\nabla \cdot \vec{v}) + (\vec{B}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{B}-\frac{\partial \vec{B}}{\partial t}))=$(because according to Maxwells II equation $div(B)=0$)
$\oint(dS*(\vec{B}(\nabla \cdot \vec{v}) + (\vec{B}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{B}-\frac{\partial \vec{B}}{\partial t}))=$
(because the idea based on a rigid body moving lineary with constant speed $\forall_i(\frac{\partial v}{\partial x_i}=0)$)
$\oint(dS*(-(\vec{v}\cdot \nabla)\vec{B}-\frac{\partial \vec{B}}{\partial t}))$
(beacuse we assume that all EM field is created by the magnet and magnetic field is soly determined by position of magnet, $B(t)=f(\vec{X_{magnet}}(t))$ time $\Delta t$ ago was $B(t-\Delta t)=f(\vec{X_{magnet}}(t-\Delta t))=f(\vec{X_{magnet}}(t)-\Delta t*v)$, it must be that $(\vec{v}\cdot \nabla)\vec{B}=\frac{\partial \vec{B}}{\partial t}$)
$-2*\oint(dS*(\vec{v}\cdot \nabla)\vec{B})$

It should be 0 in both frames of reference, but I probably made a sign error somewhere.

 

[artis]

if the disc breaks into pieces then each individual piece effectively becomes a small linear faraday generator because even a single electron rotating above a homogeneous B field experiences the Lorentz force and would be deflected sideways and the same thing happens in a conducting element being dragged through a B field at 90 degree angle which is essentially the Faraday disc.

Again the physics to the best of my knowledge does not change whether in the rotational or linear scenario.

 

[olgerm]

if the disc breaks into pieces then each individual piece effectively becomes a small linear faraday generator because even a single electron rotating above a homogeneous B field experiences the Lorentz force and would be deflected sideways and the same thing happens in a conducting element being dragged through a B field at 90 degree angle which is essentially the Faraday disc.
Again the physics to the best of my knowledge does not change whether in the rotational or linear scenario.

EMF goes smaller and smaller as time passes from the disc break because:

• If pieces are very far EMF is not directed to center of disc, but crosswise to it.
• If distances get bigger polarizing effects get smaller.
• If magnet field is from a magnet behind disc, the B-field applied to pieces is very small, if pieces are far from the magnet.