# I Request about experiments on the linear-motion Faraday paradox

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1. Feb 24, 2019

### greswd The Faraday paradox is a very curious example in the topic of relative motion.

An experiment demonstrating the curious results is shown in the video below:

A conductor placed atop a magnet, both at rest in one scenario. In another, both moving together in uniform linear motion.

The linear scenario is expected to produce very different results from the rotational scenario.

Therefore, I'm also interested in the transition from linear to rotational motion and vice versa.
Small segments of a circular arc approximate straight lines, small segments of rotational motion approximate linear motion.

Experiments of the transitional scenarios should ostensibly yield explanations as to why the two scenarios produce different results.

Thus, so far, what results have such experiments yielded?

2. 3. Feb 25, 2019

### BvU Not clear to me if you watched part 2 as well ?

4. Feb 25, 2019

### greswd I have, it doesn't mention any linear or transition experiments afaia.

5. Feb 25, 2019

### BvU My bad, I focused on the video. Linear motion is even easier (I assume you don't intend to go to relativistic speeds and Lorentz transforms).
Same Lorentz force expression.

6. Feb 25, 2019

### olgerm Where is the voltage measured? aka where would wires to voltmeter be connected?

You should consider, that $\vec{\epsilon}=\frac{E}{q}=\int(dl \frac{F}{q})=\frac{\int(dl (E+v\times B))}{q}$ and maxwell equations.

In 1. case E=0 and v=0. So $\epsilon$ must be 0.

Last edited: Feb 25, 2019
7. Feb 25, 2019

### greswd No worries. In rotational motion, the conductor and magnet moving together generates EMF. In linear motion, I guess it is zero EMF.
I'm curious about the experimental results in transitional cases.

8. Feb 25, 2019

### greswd Could be one wire at each end of the conductor plate, and the setup being moved perpendicular to the connective line.

That means a velocity vector perpendicular to an imaginary line connecting both wires across the plate.

I don't know if the wire ends being connected to the moving plate or scraping along the surface of the moving plate makes a difference.

9. Feb 25, 2019

### BvU transitional results. 10. Feb 25, 2019

### greswd Lol. But in all seriousness, someone ought to have investigated this before.

Last edited: Feb 25, 2019
11. Feb 25, 2019

### olgerm it should be that
$\epsilon=\frac{E}{q}=\oint(dl \frac{F}{q})=\oint(dl (\vec{E}+\vec{v}\times \vec{B}))$

12. Feb 25, 2019

### greswd ah ok, thanks
by the way, how do you think the transition occurs?

13. Feb 26, 2019

### olgerm What do you mean by transition?

14. Feb 26, 2019

### olgerm To measure voltage or generate eletricity you need a closed circut.
$U=\oint(dl*(\vec{E}+\vec{v}\times \vec{B}))=\oint(dl*\vec{E})+\oint(dl*(\vec{v}\times \vec{B}))=\frac{\partial \oint (dS*\vec{B})}{\partial t*c}+\oint(dl*(\vec{v}\times \vec{B}))$. I do not see how yet, but I think that if you simplify it more you may get that U=0. U is voltage.

15. Feb 28, 2019

### greswd The linear case is expected to give different results from the rotational case. But I wonder about the transition, in between those two cases.

16. Feb 28, 2019

### jartsa If we zoom into a small area on an average sized Faraday disk, we will see just some co-moving particles, so we can say that it's not the magnet straight below the conductor that causes a Lorentz-force on the charges.

So we can guess it's the other parts of the magnet. Those parts of the magnet that are in motion relative to the part of the conductor that we are zooming into.

17. Feb 28, 2019

### olgerm Should be that:
$U=\oint(dl*(\vec{E}+\vec{v}\times \vec{B}))=\oint(dl*\vec{E})+\oint(dl*(\vec{v}\times \vec{B}))=\oint(dl*(\vec{v}\times \vec{B}))-\frac{\partial \oint (dS*\vec{B})}{\partial t}$

U is not 0 in all closed circuts only if $rot(\vec{v}\times \vec{B})-\frac{\partial \vec{B}}{\partial t}=\vec{v}(\nabla \cdot \vec{B}) - \vec{B}(\nabla \cdot \vec{v}) + (\vec{B}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{B}-\frac{\partial \vec{B}}{\partial t}$

according to Maxwells II equation div(B)=0 the expression simplifies to:
$rot(\vec{v}\times \vec{B})-\frac{\partial \vec{B}}{\partial t}=\vec{B}(\nabla \cdot \vec{v}) + (\vec{B}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{B}-\frac{\partial \vec{B}}{\partial t}$

since the idea based on a rigid body $div(\vec{v})=0$ the expression simplifies to:
$rot(\vec{v}\times \vec{B})-\frac{\partial \vec{B}}{\partial t}=\vec{B}(\nabla \cdot \vec{v})+(\vec{v}\cdot \nabla)\vec{B}-\frac{\partial \vec{B}}{\partial t}$

Since 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 $(\vec{v}\cdot \nabla)\vec{B}=\frac{\partial \vec{B}}{\partial t}$ the expression simplifies to:
$rot(\vec{v}\times \vec{B})-\frac{\partial \vec{B}}{\partial t}=\vec{B}(\nabla \cdot \vec{v})=\vec{B}\cdot \omega$

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.

Last edited: Feb 28, 2019
18. Feb 28, 2019

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

19. Mar 2, 2019

### jartsa 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.

Last edited: Mar 2, 2019
20. Mar 3, 2019

### artis The video is a bit misleading I think because keeping the magnet and disc stationary while simply moving the brushes should not generate current due to lorentz force , at least not from the disc as the electrons in the disc don't experience any force in such a situation , so I think he is picking up "noise" from the wires wiggling in a surrounding B field.

Also I have understood that the Faraday disc is very interesting with regards that it involves special relativity because there is only EMF and current when the disc is closed with a return path that rotates with a different speed or is stationary with respect to the disc , in other words there always must be relative motion between the two sides. In fact I think you can have a linear homopolar generator/Faraday disc but you must provide electrical contacts to the metal sheet sliding atop a stationary or moving magnet and the contacts and the wire connecting them must be stationary or move with a different speed than the sheet.

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.
I think this is what happens..
Also as the OP said , a large enough disc resembles a linear metal wire or sheet moving in a B field if small portions/slices are considered of the disc and again they too would need a closing loop rotating at different speed to have any current in the loop or force exerted by the slice to move. 21. Mar 3, 2019

### jartsa Return path moves relative to a magnet, so voltage is generated in the return path. Disk does not move relative to a magnet, so there is no voltage generated, so there is no voltage that could cancel that one voltage, so there is a voltage in the circuit.

I have a question: Does a spinning disk shaped magnet generate a Lorentz-force on a metal rod above it?

Or do the many microscopic Lorentz-forces that the many microscopic magnets inside the disk-magnet exert on the charges inside the rod cancel out each other?

If they cancel out, then I made an error earlier in post #13.