Faraday's paradox: homopolar generator on a train

[carrz]

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?

 

[voko]

Consider the motion of electrons relatively to the magnetic field in both cases.

 

[carrz]

Consider the motion of electrons relatively to the magnetic field in both cases.

When the magnet is attached to the disk and they are rotated together there is actually current induced, that's the paradox. And if that works, then I don't see why wouldn't stationary generator on a train work, but I don't think it actually does, hence the questions. What's your answer?

 

[voko]

Are you familiar with the Lorentz force?

 

[carrz]

Are you familiar with the Lorentz force?

Yes.

 

[vanhees71]

Of course, there is no paradox. The only paradox is that people use the Faraday Law in integral form in a wrong or incomplete way in forgetting the "magnetic partic" of the electromotive force. It's pretty well explained on the Wikipedia

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

Of course, there is no difference whether you do the experiment in a moving train or at rest (as long as you can consider both reference frames as inertial frames). You only have to use the Lorentz transformation and the covariance of Maxwell's equations under Lorentz transformations to argue about that. There's not even a need to do an explicit calculation to prove it by this argument.

 

[carrz]

Of course, there is no difference whether you do the experiment in a moving train or at rest (as long as you can consider both reference frames as inertial frames). You only have to use the Lorentz transformation and the covariance of Maxwell's equations under Lorentz transformations to argue about that. There's not even a need to do an explicit calculation to prove it by this argument.

Can you explain what is the difference between the disk and the magnet rotating together, and them moving together along in a train? In other words, in both cases they are moving without any relative velocity, so why would rotating them together induce current, and why transporting them together on a train would not induce current?

 

[atyy]

I'm not sure, but why not just make the rule that the disk must be rotating? The disk is not rotating on the train so why would there be a current?

If you move the magnet and disk with constant speed in a straight line on the ground, there won't be a current. The disk has to rotate.

 

[Nugatory]

next I ask what is the difference between the disk and the magnet rotating together, and them moving together along in a train?

I am missing something here. In a homopolar generator the disk and the magnet do not rotate together; the disk rotates in the static magnetic field produced by fixed magnet, right?

 

[atyy]

I am missing something here. In a homopolar generator the disk and the magnet do not rotate together; the disk rotates in the static magnetic field produced by fixed magnet, right?

Yes, I was puzzled too. I thought maybe he means if the disk and magnet rotate together, there is still a current. If this is true, then I think it is because at a slow speed, the magnet can just be treated as producing a rotating magnetic field, but because of the rotational symmetry, the magnetic field seen by the disk is the same as if the magnet were not rotating. So all that matters if that the disk rotates relative to an inertial frame (about the axis of symmetry of the magnet)

 

[carrz]

I'm not sure, but why not just make the rule that the disk must be rotating?

Because the point is to compare their rotational and translational movement, and why would one induce current and not the other when in both cases their relative velocity is zero.

The disk is not rotating on the train so why would there be a current?

I agree with that question but I also ask, why not?

If you move the magnet and disk with constant speed in a straight line on the ground, there won't be a current. The disk has to rotate.

Now we're talking. But why? In both cases there is no any relative velocity between them, and in both cases they are moving, let's say relative to Earth. So why not? What is different when they are rotating and when they are moving in a straight line?

 

[carrz]

I am missing something here. In a homopolar generator the disk and the magnet do not rotate together; the disk rotates in the static magnetic field produced by fixed magnet, right?

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)

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

 

[voko]

In both cases there is no any relative velocity between them

It is not important that the magnet and disk have no relative velocity. It is important that charges in the disk and the magnetic field have a relative velocity that is not completely aligned with the magnetic field.

EDIT: The correct statement is: what is important is that the velocity of charges and direction of the magnetic field are not completely aligned, then there is non-zero Lorentz force acting on the charges.

 

[Nugatory]

What is different when they are rotating and when they are moving in a straight line?

I'm still confused. Have we established that there is a difference?

There will be a current if the conductor is moving relative to the magnet, whether rotating or straight line. If the magnet and the conductor are at rest relative to one another, there will be no current if they are moving inertially. Is it established that there will be a current if the magnet and the conductor are both undergoing uniform circular motion together?

 

[carrz]

I'm still confused. Have we established that there is a difference?

I gather everyone so far thinks there is a difference, so that non rotating generator on a moving train will not induce current, and stationary rotating generator will induce current even if the disk and the magnet are rotating together without any relative velocity.

Is it established that there will be a current if the magnet and the conductor are both undergoing uniform circular motion together?

Yes, that's why it's called "Faraday's paradox", see here:

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

 

[carrz]

It is not important that the magnet and disk have no relative velocity. It is important that charges in the disk and the magnetic field have a relative velocity that is not completely aligned with the magnetic field.

How is it that charges in the disk and the magnetic field have a relative velocity if they are rotating together?

 

[voko]

How is it that charges in the disk and the magnetic field have a relative velocity if they are rotating together?

I actually mi-stated the condition: what is important is that the velocity of charges and direction of the magnetic field are not completely aligned, then there is non-zero Lorentz force acting on the charges.

 

[carrz]

I actually mi-stated the condition: what is important is that the velocity of charges and direction of the magnetic field are not completely aligned, then there is non-zero Lorentz force acting on the charges.

How/why are they not aligned when rotating together? And why are they aligned when moving together in a straight line?

 

[atyy]

Now we're talking. But why? In both cases there is no any relative velocity between them, and in both cases they are moving, let's say relative to Earth. So why not? What is different when they are rotating and when they are moving in a straight line?

Is it established that there will be a current if the magnet and the conductor are both undergoing uniform circular motion together?

Wikipedia's explanation http://en.wikipedia.org/wiki/Faraday_paradox seems reasonable to me. This is the explanation that voko has been giving in the thread.

"This mechanism agrees with the observations: an EMF is generated whenever the disc moves relative to the magnetic field, regardless of how that field is generated.

The use of the Lorentz equation to explain the Faraday Paradox has led to a debate in the literature as to whether or not a magnetic field rotates with a magnet. Since the force on charges expressed by the Lorentz equation depends upon the relative motion of the magnetic field to the conductor where the EMF is located it was speculated that in the case when the magnet rotates with the disk but a voltage still develops, that the magnetic field must therefore not rotate with the magnetic material as it turns with no relative motion with respect to the conductive disk."

Bolding above is mine. It looks like the only tricky part is whether the magnetic field moves with the magnet. Classically, a magnetic field would be produced by a current loop, and if the current loop executes circular motion, its motion is non-inertial, and the charges in the loop will accelerate producing radiation, ie. the magnetic field will move. However, at very slow speeds, this should be a small effect and the magnetic will not move.

 

[carrz]

Wikipedia's explanation http://en.wikipedia.org/wiki/Faraday_paradox seems reasonable to me. This is the explanation that voko has been giving in the thread.

I'd say it's lacking and it leaves many questions open.

The use of the Lorentz equation to explain the Faraday Paradox has led to a debate in the literature as to whether or not a magnetic field rotates with a magnet.

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."

Since the force on charges expressed by the Lorentz equation depends upon the relative motion of the magnetic field to the conductor where the EMF is located it was speculated that in the case when the magnet rotates with the disk but a voltage still develops, that the magnetic field must therefore not rotate with the magnetic material as it turns with no relative motion with respect to the conductive disk."

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

Classically, a magnetic field would be produced by a current loop, and if the current loop executes circular motion, its motion is non-inertial, and the charges in the loop will accelerate producing radiation, ie. the magnetic field will move. However, at very slow speeds, this should be a small effect and the magnetic will not move.

I don't see what radiation has to do with any of this. It's just Lorentz force that is supposed to explain it all.

 

[voko]

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

Magnetic field is a vector. You can visualize it as a pencil, with its tip pointing in the direction of the magnetic field. If you rotate the pencil along its long axis, will the tip point in any different direction?

 

[atyy]

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

My thinking is the same as voko's in post 21. In the Faraday paradox http://en.wikipedia.org/wiki/Faraday_paradox, the magnet is cylindrical. Since the magnet is cylindrical, if you rotate it by a particular angle about its cylindrical axis, and leave it at the new angle, the magnetic field remains the same.

 

[Dale]

I agree. EM fields themselves don't have any velocity (although changes in the fields can certainly have a velocity).

 

[Drakkith]

Multiple off-topic posts have been deleted. Please stay on topic.

 

[carrz]

Magnetic field is a vector. You can visualize it as a pencil, with its tip pointing in the direction of the magnetic field. If you rotate the pencil along its long axis, will the tip point in any different direction?

Magnetic field is three-dimensional potential vector field, you can visualize it as a doughnut. But in any case pencil rotation and direction of its axis does not compare to homopolar generator.

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.

 

[atyy]

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?

 

[vanhees71]

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)
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$
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 $F$ with boundary $\partial F$ and using Stokes's Theorem, leading to
$$\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \int_F \mathrm{d}^2 \vec{F} \cdot \partial_t \vec{B}.$$
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., $\vec{\nabla} \cdot \vec{B}=0$ that
$$\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}-\frac{1}{c} \int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{B}).$$
Here, $\vec{v}=\vec{v}(t,\vec{r})$ 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, $t$, 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
$$\int_{\partial F} \mathrm{d} \vec{r} \cdot \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right )=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}.$$

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 $\vec{B}$ is homogeneous in the interior of the magnet,
$$\vec{B}=B \vec{e}_z.$$
Now rotate the magnet around the $z$ 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.,
$$-e n \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)=0,$$
i.e.,
$$\vec{E}=-\frac{\vec{v}}{c} \times \vec{B}.$$
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
$$\vec{v} = \vec{\omega} \times \vec{r} \; \Rightarrow \; \vec{E}=-\frac{B \omega}{c} (\vec{e}_z \times \vec{r}) \times \vec{e}_z = -\frac{B \omega \rho}{c} \vec{e}_{\rho}.$$
Here $\rho$ 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
$$U=\frac{B \omega R^2}{2c}.$$
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 $\vec{E}+\vec{v}/c \times \vec{B} =0$.

 

[carrz]

Suppose we have a long cylindrical bar magnet... Now rotate the magnet around the $z$ 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.,
$$\vec{E}=-\frac{\vec{v}}{c} \times \vec{B}.$$

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.

 

[Dale]

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.

 

[vanhees71]

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.

 

[carrz]

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.

 

[carrz]

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.

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.

 

[jartsa]

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.

 

[voko]

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.

 

[carrz]

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.