B Faraday's disk and "absolute" magnetic fields

[Buckethead]

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 demo starts at time 0:30

Thanks for the video. Case #5 was interesting. However this does not lead to the conclusion that there is no meaning to the absolute rotation of the disk. Case #5 generates a voltage because the upper part of the stator is moving through the magnetic field and the strength of this field is less at this point than it is near the path between the electrodes on the disk.

 

[TSny]

Thanks for the video. Case #5 was interesting. However this does not lead to the conclusion that there is no meaning to the absolute rotation of the disk. Case #5 generates a voltage because the upper part of the stator is moving through the magnetic field and the strength of this field is less at this point than it is near the path between the electrodes on the disk.

The following two cases produce the same emf:
(1) Rotate the disk at 10 rad/s with respect to the lab while keeping the stator and external parts of the circuit at rest relative to the lab.
(2) Keep the disk at rest with respect to the lab while rotating the stator and external parts of the circuit at 10 rad/s in the opposite direction relative to the lab.

If you rotate both the disk and the external circuit in the same direction at the same angular speed, no current is induced.

So, it's not absolute rotation of the disk that matters. What matters is the relative motion of the disk with respect to the external circuit. At least that's how I see it.

 

[sweet springs]

Hi. Tensor equation of Lorentz Force

Assures that frames of reference does not matter with physics. Best.

 

[Buckethead]

The following two cases produce the same emf:
(1) Rotate the disk at 10 rad/s with respect to the lab while keeping the stator and external parts of the circuit at rest relative to the lab.
(2) Keep the disk at rest with respect to the lab while rotating the stator and external parts of the circuit at 10 rad/s in the opposite direction relative to the lab.

If you rotate both the disk and the external circuit in the same direction at the same angular speed, no current is induced.

So, it's not absolute rotation of the disk that matters. What matters is the relative motion of the disk with respect to the external circuit. At least that's how I see it.

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.

Also I want to add that the two situations are not symmetrical. When you rotate the disk you are rotating one wire (the disk) through the field, when you rotate the stator you are moving two wires through the field and the two wires oppose each other. Normally this would result in a null current, however because the two wires are in different parts of the field they will see different flux strengths so the net result is you get current.

 

[Dale]

I'm thinking that a magnetic field does have the property of rotation but that the angular velocity is always 0.

I don't see any value in this distinction, but I suspect it probably is harmless.

Imagine a large disk magnet not rotating, but rather moving in a circle such that the radius of the circle is much less than the radius of the magnet. A small piece of wire enveloped by the field but not experiencing any flux strength change

If the disk is moving then I don't see how the flux is not changing.

I originally liked the idea of rotation of a field being undefined, but it seems inconsistent with the idea of field lines being able to have the property of translation.

The fields do not have the property of translation either.

 

[Drakkith]

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.

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.

 

[TSny]

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.

 

[Buckethead]

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.

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.

 

[Buckethead]

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.

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?

 

[Drakkith]

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.

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.

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?

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?

 

[Buckethead]

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.

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?

 

[Dale]

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?

 

[Buckethead]

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

 

[Buckethead]

That's too much sleight of hand for me to deal with.

I hope you were kidding because actually it's not sleight of hand.

 

[TSny]

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?

 

[sweet springs]

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.

 

[Paul Colby]

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?

 

[Drakkith]

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?

 

[sweet springs]

Re posts #46 and #47 　

Sharing a figure of faraday's disk, https://upload.wikimedia.org/wikipe...Faraday_disc.PNG/180px-Solid_Faraday_disc.PNG , as for the case that the brush is fixed to the disk and whole the circuit rotates, Lorentz forces working at the disk and the wire cancel so there occurs no emf. whatever rotation of the magnet is. Best.

 

[vanhees71]

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

 

[sweet springs]

So you say post #46 was wrong? I would appreciate your correction for my study.

 

[jartsa]

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.

 

[Buckethead]

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.

 

[Buckethead]

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.

 

[TSny]

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.

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?

 

[Buckethead]

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.

 

[Drakkith]

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.

 

[Buckethead]

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.

Oh, that's too bad. I've enjoyed your input and insight. It's been a helpful conversation. Thanks for sharing your thoughts.

 

[Drakkith]

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.

 

[sweet springs]

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.