Faraday's paradox: homopolar generator on a train

  • Thread starter carrz
  • Start date
  • #26
atyy
Science Advisor
14,565
2,914
We are talking about the case where the magnet is fixed to the disk and rotates together with it. Surely magnetic field of the magnet will not separate from its source but will always be at the same location relative to the magnet and at the same location relative to every point on the disk.

As you said, the magnetic field is a vector field, an arrow at every point in space. Treat the magnet as the source of the field, as you say, and rotate the disk by a a fixed angle about its axis. Will the arrows in space change, given the symmetry of the magnetic source about the axis of rotation?
 
  • #27
vanhees71
Science Advisor
Insights Author
Gold Member
17,974
8,941
For this topic, I recommend to carefully read the Feynman Lectures vol. II. As I've written earlier (I thought even in this thread, perhaps it's one of the deleted postings, although I don't know, why it should be off-topic). So here it is again:

What's basic are the local Maxwell equations, among the Faraday's Law (using Heaviside-Lorentz units)
[tex]\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.[/tex]
The problem in many textbooks is that this fundamental law, which is always correct, is put into integral form not carefully enough, when it comes to moving surfaces and their boundaries. The correct derivation is pretty simple. The first step is to integrate the equation over an arbitrary surface [itex]F[/itex] with boundary [itex]\partial F[/itex] and using Stokes's Theorem, leading to
[tex]\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \int_F \mathrm{d}^2 \vec{F} \cdot \partial_t \vec{B}.[/tex]
The tricky point is now to put the time derivative on the right side of the equation out of the integral, and there many textbooks omit the change of the surface with time if it is moving. Doing everything correctly, you can easily show (using also Gauss's Law for the magnetic field, i.e., [itex]\vec{\nabla} \cdot \vec{B}=0[/itex] that
[tex]\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}-\frac{1}{c} \int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{B}).[/tex]
Here, [itex]\vec{v}=\vec{v}(t,\vec{r})[/itex] is the velocity of the boundary curve as a function of time and the location of the point on the surface. All integrals have to be taken at one fixed time, [itex]t[/itex], of course. Finally we combine the line integrals on both sides of the equation to get the correct form of Faraday's Law in integral form
[tex]\int_{\partial F} \mathrm{d} \vec{r} \cdot \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right )=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}.[/tex]

The homopolar generator, however, is easier studied using a more microscopic picture of the situation, than just using Faraday's Law. Suppose we have a long cylindrical bar magnet and assume homogeneous magnetization in its interior (for simplicity). Then also [itex]\vec{B}[/itex] is homogeneous in the interior of the magnet,
[tex]\vec{B}=B \vec{e}_z.[/tex]
Now rotate the magnet around the [itex]z[/itex] axis and wait a little until a stationary condition is established. Then the total force on the conduction electrons in the magnet must be 0, i.e.,
[tex]-e n \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)=0,[/tex]
i.e.,
[tex]\vec{E}=-\frac{\vec{v}}{c} \times \vec{B}.[/tex]
Thus there is an electric field due to the Hall effect: The electrons drift within the magnet and build a surface-charge density on its surface such that the electric and magnetic forces are compensated as in our ansatz.

Now we have
[tex]\vec{v} = \vec{\omega} \times \vec{r} \; \Rightarrow \; \vec{E}=-\frac{B \omega}{c} (\vec{e}_z \times \vec{r}) \times \vec{e}_z = -\frac{B \omega \rho}{c} \vec{e}_{\rho}.[/tex]
Here [itex]\rho[/itex] is the usual radial cylinder coordinate.

Thus a voltmeter connecting the center of the cylinder with it's rim (using, e.g., coal brushes) reads
[tex]U=\frac{B \omega R^2}{2c}.[/tex]
Of course, there is no contradiction with Faraday's Law, because the magnetic flux is constant in time in this situation and the same is true for the line integral, because [itex]\vec{E}+\vec{v}/c \times \vec{B} =0[/itex].
 
  • #28
126
0
Suppose we have a long cylindrical bar magnet... Now rotate the magnet around the [itex]z[/itex] axis and wait a little until a stationary condition is established.

The magnet in a hompolar generator does not rotate around any of its axis but around an axis that is outside of the magnet. When rotated the magnet's path describes a circle just like it would if it was stationary on a train which goes around the Earth. There is also no any waiting in operation of homopolar generator.


Then the total force on the conduction electrons in the magnet must be 0, i.e.,
[tex]\vec{E}=-\frac{\vec{v}}{c} \times \vec{B}.[/tex]

Force between "conduction electrons" in the magnet and what other thing? What is that velocity of and what is it relative to?


Thus there is an electric field due to the Hall effect: The electrons drift within the magnet and build a surface-charge density on its surface such that the electric and magnetic forces are compensated as in our ansatz.

I see, only that should work even when the magnet and the disk are stationary in every respect, but it doesn't, there is no any current induced unless the disk as a whole is not actually moving. Also, voltage produced by Hall effect would be independent of the disk motion, and in our case it is proportional to its speed.
 
  • #29
Dale
Mentor
Insights Author
2020 Award
31,516
8,272
The tricky part of a homopolar generator is understanding the current density. It is not obvious.

As was mentioned by vanhees71 back in post 6, there is no possibility of a paradox here. This is simply an example of a hidden complexity which makes rigorous analysis complicated.
 
Last edited:
  • #30
vanhees71
Science Advisor
Insights Author
Gold Member
17,974
8,941
Ad #28: Then I misunderstood your question in #1. Perhaps you have a drawing of the setup somewhere?

I thought, you mean the usual setup, where a wheel is rotating in an external magnetic field or by rotating the magnet itself. The latter was the one I discussed.

The current in this case is due to the rotation of the magnet not by a current due to an external battery. Perhaps I shouldn't have called it "Hall effect". It's just the rearrangement of charges due to the force on the electrons from the electromagnetic field. In the stationary state this force must be 0.
 
  • #31
126
0
The tricky part of a homopolar generator is understanding the current density. It is not obvious

Unfortunately that paper doesn't seem to address this particular setup we are talking about.


As was mentioned by vanhees71 back in post 6, there is no possibility of a paradox here. This is simply an example of a hidden complexity which makes rigorous analysis complicated.

I'm not sure how to call it, but as far as Wikipedia article goes it certainly lacks explanation and it says there is unresolved debate on how to even model "rotating magnet" setup to begin with. I don't mind if an explanation is hidden and complex, I just don't see there is any explanation to be considered so far.
 
Last edited by a moderator:
  • #32
126
0
Ad #28: Then I misunderstood your question in #1. Perhaps you have a drawing of the setup somewhere?

I thought, you mean the usual setup, where a wheel is rotating in an external magnetic field or by rotating the magnet itself.

popup_3.jpg


There are three combinations and two paradoxes:

1.) disk rotates, magnet stationary -> induced current
2.) magnet rotates, disk stationary -> no current (paradox 1)
3.) disk and magnet rotate together -> induced current (paradox 2)


We are talking about the 3rd case where the magnet is fixed to the disk and rotates together with it. Imagine slapping a magnet on the rim of your bicycle wheel, that's all there is to it.
 
  • #33
1,498
123
There is a homopolar generator on a train moving at 215 km/h. The magnet of the generator is attached to the disk so they would rotate together, but they are stationary now, except that they are moving along with the train. Is there any current generated by the generator?

I expect the answer will be "no", so next I ask what is the difference between the disk and the magnet rotating together, and them moving together along in a train?


There is no voltage when a homopolar generator and a voltmeter are traveling on a train, and the voltmeter is measuring the voltage of the generator that is not being cranked.


If you move the voltmeter relative to the generator, then there is a voltage. The voltmeter prongs must move too, they must slide on the generator disk. If the velocity of the voltmeter happens to be 0, then the train is doing all the work required to push charges trough the voltmeter.
 
  • #34
6,054
391
Magnetic field is three-dimensional potential vector field, you can visualize it as a doughnut.

You are confusing yourself by overcomplicating things. Consider first that the magnet is such that its magnetic field is uniform and perpendicular to the disk (this is a good approximation of the field of any magnet not too close to its edges). The pencil approximation works here and is enough to explain the "paradox".

Then you can consider the more general case, where it is helpful to examine separately the components of the field parallel and perpendicular to the disk.
 
  • #35
126
0
Consider first that the magnet is such that its magnetic field is uniform and perpendicular to the disk (this is a good approximation of the field of any magnet not too close to its edges).

And?


The pencil approximation works here and is enough to explain the "paradox".

The pencil analogy does not compare, it does not "work", and you never even attempted to explain how it is supposed to work.


Then you can consider the more general case, where it is helpful to examine separately the components of the field parallel and perpendicular to the disk.

Consider you have said nothing about rotation of the disk, without which there is no any current induced.
 
  • #36
126
0
...and the voltmeter is measuring the voltage of the generator that is not being cranked.

How's that?


If you move the voltmeter relative to the generator, then there is a voltage. The voltmeter prongs must move too, they must slide on the generator disk. If the velocity of the voltmeter happens to be 0, then the train is doing all the work required to push charges trough the voltmeter.

Voltmeter is not moving in none of the three setups.

1.) disk rotates, magnet stationary -> induced current
2.) magnet rotates, disk stationary -> no current (paradox 1)
3.) disk and magnet rotate together -> induced current (paradox 2)

http://en.wikipedia.org/wiki/Faraday's_Paradox
 
  • #37
6,054
391
carrz, you are being confrontational and that's not going to get you anywhere.
 
  • #38
1,176
84
How's that?




Voltmeter is not moving in none of the three setups.

1.) disk rotates, magnet stationary -> induced current
2.) magnet rotates, disk stationary -> no current (paradox 1)
3.) disk and magnet rotate together -> induced current (paradox 2)

http://en.wikipedia.org/wiki/Faraday's_Paradox

https://www.physicsforums.com/showthread.php?t=737583

The above thread is where I attached a paper by Dr. Munley of Univ of Va. He explains the disk but not very completely. Understanding how the charges move from center to rim of disk is the key. I will comment if needed. No paradox at all here, just geometry and following the Lorentz force on the electrons in the disk, looking at the path.

1) Consensus, we indeed have induced current as Faraday predicts.
2) No induced current, as is expected. Magnet moving does not change the field in the disk. Disk is stationary and electrons have no velocity wrt B field, so no Lorentz force acts on them. No paradox at all.
3) Disk & magnet rotating together is identical to case 1). Rotating the magnet does not change the B field encountered by the disk. No paradox at all.

Comments/questions/feedback are welcome, best regards.

Claude
 
  • #39
1,498
123
How's that?

I just meant the voltmeter measures some voltage, voltage is zero volts in that spesific case.

Voltmeter is not moving in none of the three setups.

1.) disk rotates, magnet stationary -> induced current
2.) magnet rotates, disk stationary -> no current (paradox 1)
3.) disk and magnet rotate together -> induced current (paradox 2)


Ok.

Let me ask a question: Bob is cranking a homopolar generator. Bob is applying a torque on the disk and on the magnet, but the disk and the magnet are rotating at constant rate. What is applying an opposite torque on the disk and the magnet? (Disk and magnet rotate together in this generator)
 
  • #40
Drakkith
Staff Emeritus
Science Advisor
21,242
5,060
I'd say it's lacking and it leaves many questions open.

I disagree. I read through it and the explanation looked fairly thorough.


Also this:
- "Several experiments have been proposed using electrostatic measurements or electron beams to resolve the issue, but apparently none have been successfully performed to date."

As the wiki says, in case 2 the magnet is rotated and no current is observed. Therefore either the magnetic field is not rotating with the magnet, or this type of rotating magnetic field does not induce a current. Either one agrees with observations.

That part doesn't make sense. How can possibly magnetic field not rotate with the magnetic material?

First, realize that field lines/flux lines are not real objects. They are merely a way of representing the strength and direction of the field, just like contour lines on a map represent elevation. In reality the field strength is one smooth continuum with gradual transitions between field strengths of different areas. When the cylindrical magnet is rotated, the field remains exactly the same at all points and there is no change to induce a current in the disk.
 
  • #41
1,498
123
There is a homopolar generator on a train moving at 215 km/h. The magnet of the generator is attached to the disk so they would rotate together, but they are stationary now, except that they are moving along with the train. Is there any current generated by the generator?

I expect the answer will be "no", so next I ask what is the difference between the disk and the magnet rotating together, and them moving together along in a train?



I would like to answer the question: What is same between the disk and the magnet rotating together, and them moving together along in a train?

Answer: Electrons in the disk do not feel a lorentz force.

What happens if the brush of a homopolar generator accidentally becomes welded to the disk? Answer: The generator stops generating current, immediateletely when the motion between the disk and the wires disappears.

The disk and the wires are moving together in that damaged homopolar generator -> no current
The disk and the wires are moving together in that train -> no current
 
  • #42
126
0
https://www.physicsforums.com/showthread.php?t=737583

The above thread is where I attached a paper by Dr. Munley of Univ of Va. He explains the disk but not very completely. Understanding how the charges move from center to rim of disk is the key. I will comment if needed. No paradox at all here, just geometry and following the Lorentz force on the electrons in the disk, looking at the path.

I don't see that paper addresses the case where the magnet is rotating together with the disk.


2) No induced current, as is expected. Magnet moving does not change the field in the disk. Disk is stationary and electrons have no velocity wrt B field, so no Lorentz force acts on them. No paradox at all.

Electrons move around protons, and beside Biot-Savart magnetic fields they also have their spin magnetic moments. It works when we move a magnet relative to stationary coil, so conductor being stationary or not is not an issue, it must be something else. What's the difference?


3) Disk & magnet rotating together is identical to case 1). Rotating the magnet does not change the B field encountered by the disk. No paradox at all.

So why exactly would electrons move towards the rim of the wheel? And once they arrive there, why would they want go back to the center of the disk but only the other way around through the galvanometer and connecting wires?
 
  • #43
Drakkith
Staff Emeritus
Science Advisor
21,242
5,060
Electrons move around protons, and beside Biot-Savart magnetic fields they also have their spin magnetic moments. It works when we move a magnet relative to stationary coil, so conductor being stationary or not is not an issue, it must be something else. What's the difference?

The difference is that when you move a magnet towards a coil, the magnetic field is changing over time. When you rotate the magnet the field is not changing at all. And yes, it does matter if the conductor is stationary or not, as has been shown more than once.

So why exactly would electrons move towards the rim of the wheel? And once they arrive there, why would they want go back to the center of the disk but only the other way around through the galvanometer and connecting wires?

Because the electrons, being in motion through a magnetic field, experience a force that causes them to move. This means a voltage is developed in the circuit and so current flows. The galvanometer and wires provide a path for the current to flow
 
  • #44
126
0
When the cylindrical magnet is rotated, the field remains exactly the same at all points and there is no change to induce a current in the disk.

You are talking about this:
http://www.zamandayolculuk.com/cetinbal/AE/disk-dynamo1.gif [Broken]


I'm talking about this:
popup_3.jpg



For the second case I'm talking about you can not possibly say that magnetic field is not rotating with the magnet, it must be where its source is. There is no any theory or experiment that says otherwise.

So when the magnet rotates, in the 2nd setup I'm talking about, and the disk is stationary, the disk is actually "cutting" through different density flux lines just like stationary coil is cutting through the flux lines of a moving magnet, but there is no induction in the disk, only coil. And when the magnet rotates together with the disk, the disk is not "cutting" through any flux lines and yet the current is induced. It's exactly the opposite than Faraday's law of induction predicts, hence "paradox".
 
Last edited by a moderator:
  • #45
126
0
The difference is that when you move a magnet towards a coil, the magnetic field is changing over time. When you rotate the magnet the field is not changing at all.

Magnetic field of those disk magnets is not changing when the magnets are stationary as well, so why is then current induced when they are stationary and the conducting disk is spinning?


And yes, it does matter if the conductor is stationary or not, as has been shown more than once.

It matters for a disk, not for a coil. The mystery relation is obviously geometrical, not just temporal.


Because the electrons, being in motion through a magnetic field, experience a force that causes them to move. This means a voltage is developed in the circuit and so current flows. The galvanometer and wires provide a path for the current to flow

But they have open pathway to go either way and potential difference is the same. If the disk is of less resistance than connecting wires wouldn't electrons rather want to go back the same way they came, when the magnet is on the opposite side of the disk for example?
 
  • #46
126
0
Speaking about coils...

faraday-law-1.jpg


1. coil moves, magnet stationary -> induced current
2. magnet moves, coil stationary -> induced current
3. coil and magnet move together -> current induced?
 
  • #47
vanhees71
Science Advisor
Insights Author
Gold Member
17,974
8,941
Ad #44: Finally you provided a clear picture. The upper one is the classical Faraday Disk setup. Of course, an EMF is induced. The calculation is precisely the same as the one I've given yesterday. It doesn't matter here, whether the magnet is rotating with the disk or not. The EMF is due to the drift of the electrons in the conductor which builds up an electric field.

The 2nd setup is not as clear to me, because it's not properly written what represents what. I guess the horse-shoe shaped thing is the magnet and you consider two cases: (a) only the disk is rotating and the magnet stays fixed and (b) the magnet is fixed at the disk and rotating with it. In both cases you measure an EMF. In case (a) it's qualitatively the same as with the first case: The magnetic field is time-independent and the electrons in the conductor drift due to the Lorentz force [itex]\vec{v}\times \vec{B}/c[/itex] building up an electric field. In case (b) the magnet is rotating and thus you have both a magnetic and an electric field. In principle you can calculate both by using Maxwell's equations, noting that the magnetization of the permanent magnet is equivalent to a current [itex]\vec{j}_{\text{mag}}=c \vec{\nabla} \times \vec{M}[/itex]. In any case the electrons in the conducting disk are drifting again due to the Lorentz force due to the electromagnetic field of the rotating magnet.
 
  • #48
vanhees71
Science Advisor
Insights Author
Gold Member
17,974
8,941
Ad #46: If the magnet and the coil move uniformly together, no EMF is induced. You can just Lorentz boost to the frame, where both are at rest, and thus you immediately see that there's no "voltage" shown by the galvanometer. This, of course, holds true in any reference frame, i.e., also in the frame where both magnet and coil move together.
 
  • #49
Drakkith
Staff Emeritus
Science Advisor
21,242
5,060
You are talking about this:

I'm talking about this:

In the 2nd picture, is the magnet the blue horseshoe bar on top? If so, that's a different setup than what's usually discussed. Typically the magnet is at least as big as the disk so that as the disk rotates, the field is static. If the magnet is much smaller than the disk, and placed at the edge, I expect there will be a different answer than normal.
 
  • #50
126
0
In the 2nd picture, is the magnet the blue horseshoe bar on top? If so, that's a different setup than what's usually discussed. Typically the magnet is at least as big as the disk so that as the disk rotates, the field is static. If the magnet is much smaller than the disk, and placed at the edge, I expect there will be a different answer than normal.

Yes, blue horseshoe thing is the magnet. That's the original design Faraday used, that's what he was talking about when he was puzzled with the paradox himself, so I don't think it produces different results than disk magnets, but we can analyze both and see how it fits.

220px-Faraday_disk_generator.jpg

http://en.wikipedia.org/wiki/Homopolar_generator
 

Related Threads on Faraday's paradox: homopolar generator on a train

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
21
Views
5K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
14
Views
4K
  • Last Post
2
Replies
46
Views
4K
  • Last Post
Replies
2
Views
2K
Replies
81
Views
20K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
6K
  • Last Post
Replies
5
Views
4K
Top