Faraday's disk and "absolute" magnetic fields

In summary, the magnetic field of a magnet is absolute in the same sense that the forces felt by an object rotating or linearly accelerating are absolute. However, the field can be left out of the equation entirely and the relationship between the electrons in a spinning disk and an absolutely non rotating frame is what causes the current flow.
  • #36
Buckethead said:
I suspected you were going to call me on that. Replace the magnet with the Helmholtz apparatus such that the wire is always in the homogeneous area of the field.

Okay. Well, as far as I know, if the field is homogeneous, it's not changing as you move the coil and you have no current flow in the wire. Or, in terms of flux, the rate of change of magnetic flux through the wire is zero and no voltage or current is produced, just as predicted by Faraday's law of induction.

Buckethead said:
Sorry, that was a bit vague. The difficulty I'm having is the fact that if I move an electron across field lines or field lines across an electron, the electron will feel the Lorentz force, so the field lines seem to have a property of being able to translate through space. Does that make them real?

I don't think so. Almost every time you imagine a situation where field lines are moving, what you are actually having is a change in the magnetic field. It is this change that causes the force on the charged particles. This isn't Lorentz's law, it's Faraday's law.

Ask yourself what a field line represents. To quote from wiki:

A vector field defines a direction at all points in space; a field line for that vector field may be constructed by tracing a topographic path in the direction of the vector field. More precisely, the tangent line to the path at each point is required to be parallel to the vector field at that point.

So what we're really talking about with field lines are imaginary lines that are drawn parallel to the vectors of a vector field at every point. It's not the lines that are important, it's the vector field.
 
Physics news on Phys.org
  • #37
Buckethead said:
What you did not take into account is that when you rotate the stator and leave the disk stationary, the stator now becomes the disk and the disk the stator, so the disk rotates in both cases.
That's too much sleight of hand for me to deal with. :smile:
 
  • #38
Drakkith said:
Okay. Well, as far as I know, if the field is homogeneous, it's not changing as you move the coil and you have no current flow in the wire. Or, in terms of flux, the rate of change of magnetic flux through the wire is zero and no voltage or current is produced, just as predicted by Faraday's law of induction.

I think there is current in the wire, but I don't know this for sure. But the analogous situation is indeed the Faraday disk where the disk is spinning through a cylindricaly symmetrical field, so any given (non moving) electron is moving through a homogeneous field. One important difference is that as the electron moves it does move thought a changing flux, but at the time it was first motivated to move, it was in a homogeneous field.
Drakkith said:
So what we're really talking about with field lines are imaginary lines that are drawn parallel to the vectors of a vector field at every point. It's not the lines that are important, it's the vector field.
And as a vector field it's only the field strength at any given point that matters. OK, but there is still the problem of the homogeneous field that an electron in a disk experiences during rotation.
 
  • #39
Dale said:
If the disk is moving then I don't see how the flux is not changing.
I changed the magnet to a Helmholtz apparatus to guarantee a homogeneous field.
Dale said:
The fields do not have the property of translation either.
Assuming a current in the wire located in the Helmholtz field (and that assumption seems to be under debate), being caused by the apparatus transcribing a small circle as in my description, then why would this not indicate translation of the field?
 
  • #40
Buckethead said:
I think there is current in the wire, but I don't know this for sure.

You can look at Faraday's law and do the math. If the change of flux is zero, then there cannot be a voltage in the wire.

Buckethead said:
But the analogous situation is indeed the Faraday disk where the disk is spinning through a cylindricaly symmetrical field, so any given (non moving) electron is moving through a homogeneous field. One important difference is that as the electron moves it does move thought a changing flux, but at the time it was first motivated to move, it was in a homogeneous field.

I already explained this in post #9. Lorentz's law doesn't require an inhomogeneous field. A charged particle moving through a magnetic field in any direction other than parallel with the field vectors will experience a force. It's right there in the Lorentz force formula.

Buckethead said:
And as a vector field it's only the field strength at any given point that matters.

Both the direction and magnitude of the field vectors matter. That's why there's a cross product in the Lorentz force formula. Are you familiar with what a cross product is?

Buckethead said:
Assuming a current in the wire located in the Helmholtz field (and that assumption seems to be under debate), being caused by the apparatus transcribing a small circle as in my description, then why would this not indicate translation of the field?

Given the definition and use of a field in physics, how can you translate a field?
 
  • #41
Drakkith said:
You can look at Faraday's law and do the math. If the change of flux is zero, then there cannot be a voltage in the wire.
Drakkith said:
I already explained this in post #9. Lorentz's law doesn't require an inhomogeneous field. A charged particle moving through a magnetic field in any direction other than parallel with the field vectors will experience a force. It's right there in the Lorentz force formula.
OK I'm confused. First you say a moving wire in the Helmholtz apparatus would not experience current because of the homogeneous field (change in flux is 0) and then you say it would. Where am I getting confused?
 
  • #42
Buckethead said:
I changed the magnet to a Helmholtz apparatus to guarantee a homogeneous field.
So you have a Helmholtz coil with a uniform field in the middle. And there is a wire in the homogenous region. And then you wiggle the coils so that the wire stays in the homogenous region.

How is the wire oriented with respect to the field and what direction is the coil wiggled?
 
Last edited:
  • #43
Dale said:
So you have a Helmholtz coil with a uniform field in the middle. And there is a wire in the homogenous region. And then you wiggle the coils so that the wire stays in the homogenous region.

How is the wire oriented with respect to the field and what direction is the coil wiggled?
The wire is perpendicular to the field and the circle scribed by the wiggle is in the plane that is also perpendicular to the field
 
  • #44
TSny said:
That's too much sleight of hand for me to deal with. :smile:
I hope you were kidding because actually it's not sleight of hand.
 
  • #45
Buckethead said:
I hope you were kidding because actually it's not sleight of hand.
No, I wasn't kidding.

But I think I must be misinterpreting what you mean by "absolute rotation of the disk". Does that mean rotation "relative to the fixed stars" as in Newton's rotating bucket experiment?
 
  • #46
Hi. For our discussion I would write down below the table of rotation experiment results.

(d,c,m)=
----------
(s,s,s)=0
(r,s,s)=1
(s,r,s)=1
(r,r,s)=0
(s,s,r)=0
(r,s,r)=1
(s,r,r)=1
(r,r,r)=0
----------
where
d : the disk
c : the circuit other than the disk
m : the magnet

s : still
r : rotate in an amount

(d,c,m) = 0 no emf, 1 emf.

Relative rotation of the disk and other part of the circuit matters. Rotation of magnet has nothing to do with it in non relativistic theory at least. Best.
 
Last edited:
  • #47
So, there is a case of interest to me in which I'd like opinions on. If the brushes are replaced by fixed connections, say a screw and a length of wire attached to the disk. If the disk is reciprocated sinusoidally in angle and the magnetic is fixed, is there a voltage generated? Also, very important to me, is there a corresponding tongue associated with a sinusoidal applied current?
 
  • #48
Buckethead said:
OK I'm confused. First you say a moving wire in the Helmholtz apparatus would not experience current because of the homogeneous field (change in flux is 0) and then you say it would. Where am I getting confused?

That's not what I said. I said that moving the coil would not produce a voltage in the wire because the change in the magnetic flux is zero (Faraday's law). I did not say that moving the wire in the field would not produce a voltage (Lorentz's law).

Perhaps you're getting confused because there are two different laws here?
 
  • #50
You make very many confusing words! Look at my really simple calculation for a special case. It should be very clear, how in addition to the magnetic field also an electric field is created when rotating a magnet. You only have to calculate everything relativistically, which is not a surprise since the homopolar generator is an example, where "relativistic effects" are large! There's no Farada-disk paradox left!

http://th.physik.uni-frankfurt.de/~hees/pf-faq/homopolar.pdf
 
  • Like
Likes Paul Colby
  • #51
So you say post #46 was wrong? I would appreciate your correction for my study.
 
  • #52
Paul Colby said:
So, there is a case of interest to me in which I'd like opinions on. If the brushes are replaced by fixed connections, say a screw and a length of wire attached to the disk. If the disk is reciprocated sinusoidally in angle and the magnetic is fixed, is there a voltage generated? Also, very important to me, is there a corresponding tongue associated with a sinusoidal applied current?

My opinion is:

1: There is a voltage generated, between the center of the disk and the rim of the disk. Because there is a Lorentz force exerted on the electrons on the disk ... if the electrons are forced to follow the rotation of the disk.

2: There is a torque generated, the magnet applies a torque on the disk, the disk applies a reaction torque on the magnet. To be more specific the current carrying electrons of the disk feel a Lorentz force, and those electrons transmit that force to the rest of the disk somehow, I guess the force is transmitted by friction, in other words by resistance.

I mean when voltage is applied between the center of the disk and the rim of the disk, then there exists those torques.
 
  • Like
Likes Paul Colby
  • #53
Drakkith said:
That's not what I said. I said that moving the coil would not produce a voltage in the wire because the change in the magnetic flux is zero (Faraday's law). I did not say that moving the wire in the field would not produce a voltage (Lorentz's law).

Perhaps you're getting confused because there are two different laws here?
It is my understanding that moving the coil will produce a voltage. We are looking at a relative situation here, moving the wire relative to the coil or moving the coil relative to the wire. I understand the formulas used in each case is different but the result has to be the same, this is just relativity. I must be misunderstanding you.
 
  • #54
TSny said:
No, I wasn't kidding.

But I think I must be misinterpreting what you mean by "absolute rotation of the disk". Does that mean rotation "relative to the fixed stars" as in Newton's rotating bucket experiment?
More accurately, an object is considered rotating if it experiences rotational forces (proper acceleration due to rotation). If the disk experiences these forces than I call this absolute rotation but it is probably more proper to say a disk is rotating if its angular velocity is > 0.
 
  • #55
Buckethead said:
More accurately, an object is considered rotating if it experiences rotational forces (proper acceleration due to rotation).
But we know that a current can be generated when the disk is not experiencing rotational forces if we rotate the external wires of the circuit instead of rotating the disk.
upload_2017-7-2_13-26-9.png

But, if I understand some of your previous remarks, you are not claiming that the circular disk must have absolute rotation. You are claiming that some part of the apparatus must have absolute rotation. So, in the picture on the left, it would be the disk that has absolute rotation while in the picture on the right it is the wires that have absolute rotation. But how could you prove that it's absolute rotation that matters, rather than relative motion of the disk and wires?
 
  • #56
TSny said:
But we know that a current can be generated when the disk is not experiencing rotational forces if we rotate the external wires of the circuit instead of rotating the disk.
View attachment 206432
But, if I understand some of your previous remarks, you are not claiming that the circular disk must have absolute rotation. You are claiming that some part of the apparatus must have absolute rotation. So, in the picture on the left, it would be the disk that has absolute rotation while in the picture on the right it is the wires that have absolute rotation. But how could you prove that it's absolute rotation that matters, rather than relative motion of the disk and wires?

Because the disk has no effect on the wires. Move two parallel wires past each other in a magnetic field and each wire experiences forces only because of each wire and the field, not because you have two wires going past each other.
 
  • #57
Buckethead said:
It is my understanding that moving the coil will produce a voltage. We are looking at a relative situation here, moving the wire relative to the coil or moving the coil relative to the wire. I understand the formulas used in each case is different but the result has to be the same, this is just relativity.

I can't disagree with that and I think we're a bit beyond my knowledge level here. I think I'll step out of this conversation before I put my foot in my mouth. :-p
 
  • #58
Drakkith said:
I can't disagree with that and I think we're a bit beyond my knowledge level here. I think I'll step out of this conversation before I put my foot in my mouth. :-p
Oh, that's too bad. I've enjoyed your input and insight. It's been a helpful conversation. Thanks for sharing your thoughts.
 
  • #59
Buckethead said:
Oh, that's too bad. I've enjoyed your input and insight. It's been a helpful conversation. Thanks for sharing your thoughts.

Thanks so much! One last thing, I think part of the problem in our final example is that when you're moving the Helmholtz coil in small circles, you're probably generating EM waves in the field, which we didn't take into account. I'm not sure how to account for those, so I'll have to let someone else step in.
 
  • #60
Thanks TSny for post #55 that suggests to improve my post #46 , with signature to mention the direction of emf, i.e.

(d,c,m)
----------
(s,s,s)=0
(r,s,s)=+1
(s,r,s)=-1
(r,r,s)=0
(s,s,r)=0
(r,s,r)=+1
(s,r,r)=-1
(r,r,r)=0
----------
Best.
 
  • #61
Buckethead said:
The wire is perpendicular to the field and the circle scribed by the wiggle is in the plane that is also perpendicular to the field
I haven't worked it out in detail, but I think there will be no EMF induced on such a wire from the static field. I.e. This is a physically different scenario from when the wire is wiggled in the homogenous region of the coil.

Buckethead said:
We are looking at a relative situation here, moving the wire relative to the coil or moving the coil relative to the wire.
It is not relative. Accelerometers attached to the wires or the coils will be different in the two different situations. These two situations are not symmetric, they are two physically different scenarios.
 
  • Like
Likes Buckethead
  • #62
A Faraday disk (even one that allows rotation of the magnet, the disk, and the return circuit independently) alone cannot demonstrate whether or not the magnetic field rotates with the magnet or is stationary or is somethi ng for which rotatio is not an apt descriptor.
A Faraday dish (with indipendently rotating components) shows that the requirement for current is relative rotation between the disk and the remainder of the circuit.
.
Perhaps with a setup where the return ircuit is significantly shielded and a magnetic flux return path is built away from the return circuit more certainty woukd be available.
 
  • #63
Dale said:
It is not relative. Accelerometers attached to the wires or the coils will be different in the two different situations. These two situations are not symmetric, they are two physically different scenarios.
This is a really good point and one that slipped by me entirely. If the wire is moving at a constant velocity linearly or the coil is doing so instead then I believe the situation is symmetric, but circular motion introduces acceleration and I can see how that would be different. I suppose it follows that even linear acceleration of the coil or wire is not symmetric. Is this true?
 
  • #64
Benbenben said:
A Faraday disk (even one that allows rotation of the magnet, the disk, and the return circuit independently) alone cannot demonstrate whether or not the magnetic field rotates with the magnet or is stationary or is somethi ng for which rotatio is not an apt descriptor.
A Faraday dish (with indipendently rotating components) shows that the requirement for current is relative rotation between the disk and the remainder of the circuit.
.
Perhaps with a setup where the return ircuit is significantly shielded and a magnetic flux return path is built away from the return circuit more certainty woukd be available.

The Faraday disk is a clumsy setup when trying to analyze the relationship between the disk and the field but this relationship is put to rest in an experiment done in the early 20th century (sorry I don't have a name for this or a reference). The setup consists of 2 concentric metal cylinders with a small gap between them and a single wire connecting the two cylinders located halfway down their length. A round magnet is located at one end of the cylinders around the axis of the cylinders.

The cylinders are rotated and the magnet is either held stationary or rotated relative to the cylinders (doesn't matter). After a while, remove the wire and stop the rotation and measure the voltage between the two cylinders (which is essentially a capacitor). In both cases there is a voltage. You can now discharge the cylinders and just rotate the magnet and no charge will show. This indicates the rotation of the cylinder relative to the field and not the magnet is what is causing the voltage.
 
  • #65
Buckethead said:
If the wire is moving at a constant velocity linearly or the coil is doing so instead then I believe the situation is symmetric,
Yes, I agree.

Buckethead said:
I suppose it follows that even linear acceleration of the coil or wire is not symmetric. Is this true?
I think that is correct also. If you were to write Maxwell's equations in an accelerating reference frame then they would look different than the usual form.
 
  • #66
I may have missed it but: You can represent the EM field (in this case a magnet) as a 4by4 antisymmetric matrix, a good representation of a tensor. This tensor is cast into other reference frames by multiplying the tensor with a lorentz matrix/tensor. Both the rows and columns have to be transformed so multiplication has to be on both sides when viewing the situation as a matrix. Now the electrons have their own reference frame that determines the Lorentz force they feel.
Just because the magnetic field isn't changing in magnitude; it doesn't imply it isn't changing. The changing can always be interpreted as the generated electric field. Orthogonal to the pointwise magnetic field and relative velocity vectors.
If the electron were rotating with the field it shouldn't experience a Lorentz force.
Unfortunately, I don't recall if there are second order effects due to the acceleration the electron would be undergoing. One is certain Cherenkov radiation.
I pretty sure the above is a good representation of the mathematics/physics but am open to constructive criticism; I have done any research on this for ages.
 
  • #67
Buckethead said:
The Faraday disk is a clumsy setup when trying to analyze the relationship between the disk and the field but this relationship is put to rest in an experiment done in the early 20th century (sorry I don't have a name for this or a reference). The setup consists of 2 concentric metal cylinders with a small gap between them and a single wire connecting the two cylinders located halfway down their length. A round magnet is located at one end of the cylinders around the axis of the cylinders.

The cylinders are rotated and the magnet is either held stationary or rotated relative to the cylinders (doesn't matter). After a while, remove the wire and stop the rotation and measure the voltage between the two cylinders (which is essentially a capacitor). In both cases there is a voltage. You can now discharge the cylinders and just rotate the magnet and no charge will show. This indicates the rotation of the cylinder relative to the field and not the magnet is what is causing the voltage.

I'm not clear on the setup you are describing.

Are the cylinders locked to rotate together? If this is the case, I am surprised this is occurring without relative motion between the two.

...or are the cylinders rotating with respect one another? If that is the case, it is difficult to understand how you can distinguish between the net of relative motion between each of the cylinders and the magnet as opposed to relative motion just between the cylinders.
 
  • #68
Benbenben said:
I'm not clear on the setup you are describing.

Are the cylinders locked to rotate together? If this is the case, I am surprised this is occurring without relative motion between the two.

...or are the cylinders rotating with respect one another? If that is the case, it is difficult to understand how you can distinguish between the net of relative motion between each of the cylinders and the magnet as opposed to relative motion just between the cylinders.
I really need to have a reference for this experiment!
Benbenben said:
I'm not clear on the setup you are describing.

Are the cylinders locked to rotate together? If this is the case, I am surprised this is occurring without relative motion between the two.

...or are the cylinders rotating with respect one another? If that is the case, it is difficult to understand how you can distinguish between the net of relative motion between each of the cylinders and the magnet as opposed to relative motion just between the cylinders.
I really need to have a reference for this! I did have the Faraday disk in mind at one point and the cylinders seem to contradict what I remember (not hard but I would like explicit details).
 
  • #69
So, when a disk-shaped magnet spins, no electric field is detected. Why is that?

Let's say the disk is magnetic because there are microscopic current loops all over the disk. When the disk is spinning, we can say a microscopic current loop is approximately in linear motion. Now we know that a moving current loop is an electric dipole, there was a discussion about that some time ago here.

So, what kind of macroscopic electric field is caused by those microscopic dipoles? Well the rim of the disk becomes charged. Every point of such disk has the same potential. So there are no currents in static wires that are brushing the disk.
 
  • Like
Likes Dale
  • #70
TSny said:
There is no meaning to absolute rotation of the disk. As Paul Colby points out in post #20, the return path must also be considered.
Here is a video that shows various cases of parts of the setup moving relative to other parts.

The video link in your post is quite helpful. Relative motion of the disk and stator are clearly shown to be important. I think I converged to the wrong answer to the question I posed in post #47.
 
Back
Top