greswd
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artis said: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
artis said:I believe the linear moving sheet also becomes polarized just like the disc becomes
Well, we ought to investigate the transition cases.olgerm said:This was about linear motion scenario not about disc breaking scenario.
Depends what will be between the pieces, what is capacity of pieces, what shape the pieces are and etc.greswd said:Well, we ought to investigate the transition cases.
yeah, you can set your own parameters.olgerm said:Depends what will be between the pieces, what is capacity of pieces, what shape the pieces are and etc.
How could it violate principle of relativity? Set your own parameters and explain how it does violate principle of relativity.greswd said:yeah, you can set your own parameters.The issue is how to not violate the principle of relativity.
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.artis said: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.
greswd said: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.
olgerm said:How could it violate principle of relativity? Set your own parameters and explain how it does violate principle of relativity.
I do not understand your question. You where interested about transitional cases so I suppose you know yourself what you meant by transition.greswd said:So what's the mathematics of the "transition"?
The rotational Faraday homopolar generator works, the linear one doesn't.olgerm said:I do not understand your question. You where interested about transitional cases so I suppose you know yourself what you meant by transition.
artis said: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.
greswd said: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.
olgerm said:Only the angular speed is important, whether the body is moving lineary at same time is not important.
olgerm said:olgerm said:Only the angular speed is important, whether the body is moving lineary at same time is not important.
olgerm 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.greswd said:So what's the mathematics? Cos the last time, you said:"This was about linear motion scenario not about disc breaking scenario."
My answer was about transition in that sense.greswd said:What sort of mathematics describes the EMF during the transition between both cases? From linear motion to a slight curve to full rotational motion.
olgerm said: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.
ok, why is only the angular velocity important when the Lorentz force is dependent on the linear velocity?olgerm said:Only the angular speed is important, whether the body is moving lineary at same time is not important.
Also, how does this quote relate to the quote above?olgerm said: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.
artis said:@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.
because if you change frame of reference all ##\vec{E}##, ##\vec{B}## and ##\vec{v}## change in manner that frameinvariant quantities remain the same.greswd said:why is only the angular velocity important when the Lorentz force is dependent on the linear velocity?
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.artis said:@greswd , Who said the linear generator doesn't work?
Sorry, I don't understand what you're sayingolgerm said: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 said: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.
Rotational and linear case may not give the same result.artis said: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
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 said: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.
greswd said:Sorry, I don't understand what you're sayingolgerm said:because if you change frame of reference all ##\vec{E}##, ##\vec{B}## and ##\vec{v}## change in manner that frameinvariant quantities remain the same.
If there are no brushes, the disc still gets polarized, with the rim or the center having a higher electron density. @olgermartis said: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
@artis @jartsagreswd said:You see, the Lorentz force formula uses the relative velocity seen in your rest frame. Not exactly following the principle of relativity.
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).artis said: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 said: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 said:If there are no brushes, the disc still gets polarized, with the rim or the center having a higher electron density. @olgerm