Can a magnetic fields/forces do work on a current carrying wire?

[andrien]

well,I don't know what one is trying to prove.magnetic field can not do work on charges in motion because of the well known lorentz law.but there is a mechanical energy associated with a magnetic dipole when put into magnetic field and the loop which is in the magnetic field is a magnetic dipole in the figure.

 

[vanhees71]

You have to distinguish between work and force! Of course there are magnetic forces which cause acceleration, torques and all that, but they do no work on, single charges and magnetic dipoles (like single electrons, atoms, etc.) charge and magnetization distributions (like in macroscopic electrodynamics).

 

[andrien]

well then what is -m.B when a magnetic dipole is placed into magnetic field.

 

[Miyz]

You have to distinguish between work and force! Of course there are magnetic forces which cause acceleration, torques and all that, but they do no work on, single charges and magnetic dipoles (like single electrons, atoms, etc.) charge and magnetization distributions (like in macroscopic electrodynamics).

They do work on magnetic dipoles...
So far I don't understand why you don't agree.

 

[harrylin]

[..]then the B field cannot do all of the work or energy would not be conserved. [..]

Perhaps there was a misunderstanding about the original question? I think that the question was not if the B field does all of the work, but if the B field can do no work at all.

 

[Dadface]

Noticed that this is not a really popular topic around the forums huh?

Also noticed its a very very misunderstood FIELD... Seriously! I looked around here and there... No simple and common answer about this. I can say its not understood properly and strangely everyone is confused from other sites and countless sources.

Hi Miyz.This is from your post (number five).I highlighted the word "FIELD".Was the pun intended?
I think that some of the main areas of disagreement here are possibly trivial and to do with peoples definitions of "fields".

 

[vanhees71]

well then what is -m.B when a magnetic dipole is placed into magnetic field.

That's potential energy of the dipole in the magnetic field.

 

[andrien]

That's potential energy of the dipole in the magnetic field.

of course,that is and one can use the principal of virtual work to find the force.
F=∇(m.B)
so why are you saying that magnetic field does no work on magnetic dipole.

 

[cabraham]

No, I strictly disagree with the statement that magnetic fields do work on particles or macroscopic bodies. I think Maxwell's theory and particularly its quantized version (Quantum Electrodynamics) is a correct description of electromagnetism!

So do we. We all accept Maxwell's theory. It's just a matter of how it applies to a specific case. Maxwell's equations are the basis for my position. I've shown in detail the forces/torques & directions involved. You have not shown anything as to how it is the E force that spins the rotor. Until you do so, you haven't made a case at all. Draw the vector diagrams. What force/torque acts on the rotor? What is the origin of said force/torque?

Nobody prevails by default. Not accepting my position does not make you right. Proving me wrong only proves me wrong, it doesn't prove you right. To prove your position you must illustrate that the torque on the rotor times the angle, which is work, is done by an E force. You have not thus far. Don't take this as an attack, you have a good understanding of this subject. Even great minds need to scratch their head & think through a problem as involved as this one. BR.

Claude

 

[vanhees71]

Of course the origin of the force/torque is the magnetic field. I've never denied this. I don't deny that, perhaps, I'm wrong, but then please explain to me where the term of the "magnetic field doing work" is missing in Poynting's theorem. This is not for a specific example but a very general statement for any system of charges, currents and the electromagnetic field.

Let's see whether it goes through for the magnetic moment of a point particle. Its magnetization density is given by
$$\vec{M}(\vec{x})=\vec{m} \delta^{(3)}(\vec{x}),$$
where I've assumed that it's a particle sitting at the origin at rest. The power according to Poynting's theorem is given by
$$P=c \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; \vec{E} \cdot (\vec{\nabla} \times \vec{M})=c \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; (\vec{E} \times \vec{m}) \cdot \vec{\nabla} \delta^{(3)}(\vec{x}).$$
Integration by parts and integrating out the $\delta$ distribution yields
$$P=-c \vec{\nabla} \cdot (\vec{E} \times \vec{m})=-c (\vec{\nabla} \times \vec{E}) \cdot \vec{m}=(\partial_t \vec{B}) \cdot \vec{m}.$$
In the last step I've used Faraday's Law.

Due to the Bargmann-Telegdi equation for a particle with spin in an external electromagnetic field (for a nice derivation, see the introductory chapter about classical electromagnetics of point particles of Itzykson, Zuber, Quantum Field Theory), $$\mathrm{d}_t \vec{m} \perp \vec{B}$$ for a particle at rest, and thus we can write
$$P=\frac{\mathrm{d}}{\mathrm{d} t} (\vec{m} \cdot \vec{B}).$$ This is consistent with the fact that there is a potential-energy contribution $U=-\vec{B} \cdot \vec{m}$ from the dipole-magnetic-field interaction, and again Poynting's theorem has proven to be correct! The total energy of fields and point particle with a dipol is of course conserved, as it must be.

 

[harrylin]

Of course the origin of the force/torque is the magnetic field. I've never denied this. I don't deny that, perhaps, I'm wrong, but then please explain to me where the term of the "magnetic field doing work" is missing in Poynting's theorem. This is not for a specific example but a very general statement for any system of charges, currents and the electromagnetic field. [..]

It appears to me that he doesn't specify that at all in his theorem - neither for electric or magnetic fields. He started out instead with:

"A space containing electric currents may be regarded as a field where energy is transformed at certain points into the electric and magnetic kinds by means of batteries, dynamos, thermoelectric actions, and so on [..]".

And:

"The aim of this paper is to prove that there is a general law for the transfer of energy, according to which it moves at any point perpendicularly to the plane containing the lines of electric force and magnetic force, and that the amount crossing unit of area per second of this plane is equal to the product of the intensities of the two forces, multiplied by the sine of the angle between them, divided by 4pi, while the direction of flow of energy is that in which a right-handed screw would move if turned round from the positive direction of the electromotive to the positive direction of the magnetic intensity. "

"The change per second in the sum of the electric and magnetic energies within a surface together with the heat developed by currents is equal to a quantity to which each element of the surface contributes a share depending on the values of the electric and magnetic intensities at the element."

From Poynting's explanations I get that, while he discussed the flow of energy, it was not his purpose to discuss if it is either the magnetic field or the electric field that "is doing work", or even that only one of the two would be doing all work. I may have overlooked it, but he seems to make no such suggestion - quite the contrary.

https://en.wikisource.org/wiki/On_the_Transfer_of_Energy_in_the_Electromagnetic_Field

 

[cabraham]

Of course the origin of the force/torque is the magnetic field. I've never denied this. I don't deny that, perhaps, I'm wrong, but then please explain to me where the term of the "magnetic field doing work" is missing in Poynting's theorem. This is not for a specific example but a very general statement for any system of charges, currents and the electromagnetic field.

Let's see whether it goes through for the magnetic moment of a point particle. Its magnetization density is given by
$$\vec{M}(\vec{x})=\vec{m} \delta^{(3)}(\vec{x}),$$
where I've assumed that it's a particle sitting at the origin at rest. The power according to Poynting's theorem is given by
$$P=c \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; \vec{E} \cdot (\vec{\nabla} \times \vec{M})=c \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; (\vec{E} \times \vec{m}) \cdot \vec{\nabla} \delta^{(3)}(\vec{x}).$$
Integration by parts and integrating out the $\delta$ distribution yields
$$P=-c \vec{\nabla} \cdot (\vec{E} \times \vec{m})=-c (\vec{\nabla} \times \vec{E}) \cdot \vec{m}=(\partial_t \vec{B}) \cdot \vec{m}.$$
In the last step I've used Faraday's Law.

Due to the Bargmann-Telegdi equation for a particle with spin in an external electromagnetic field (for a nice derivation, see the introductory chapter about classical electromagnetics of point particles of Itzykson, Zuber, Quantum Field Theory), $$\mathrm{d}_t \vec{m} \perp \vec{B}$$ for a particle at rest, and thus we can write
$$P=\frac{\mathrm{d}}{\mathrm{d} t} (\vec{m} \cdot \vec{B}).$$ This is consistent with the fact that there is a potential-energy contribution $U=-\vec{B} \cdot \vec{m}$ from the dipole-magnetic-field interaction, and again Poynting's theorem has proven to be correct! The total energy of fields and point particle with a dipol is of course conserved, as it must be.

1st bold: I'm glad you agree that B force produces torque. So we have consensus that B force, not E force, is, as you put it, "the origin of the torque". I agree with you completely. We will never argue over that. So what is the definition of "work". In linear motion (non-rotational) it is force times distance (dot product) times cosine of angle if both are constant. If one or both vary, work is the integral of force dot product w/ differential length over the path.

With rotational motion, work is the integral of torque dotted with angular displacement. A torque acting on the rotor multiplied by the radian angle measure the rotor has turned equals the work done on the rotor. If the torque was a constant 1.00 Newton-meter, & spins the rotor through an angle of 1.00 radian, then 1.00 joule of work has been done. So the magnetic force, B, exerts 1.00 N of force with a moment arm of 1.00 meter, for a torque of 1.00 N-m, spinning the rotor through 1.00 radian of angle, then B has done 1.00 joule of work on the rotor.

In reality, the torque changes as the rotor spins, so instead of simple multiplication, an integration must be done. Anyway, since we agree B produces torque, all we have to do is integrate said torque wrt angle, & we obtain the work done by B.

Poynting, Maxwell, Faraday, Ampere, etc. laws are all upheld here. Your equations & facts presented are certainly correct, but nowhere did you apply them to the question at hand. If you still disbelieve me, please draw a sketch indicating the quantities & show where the work is being done & by which force. I grow tired of asking but this is important. Otherwise, best regards.

Claude

 

[Dale]

Conservation of energy is upheld beyond a doubt. Even a CCS exciting the motor has to conserve energy.

Yes, definitely. I was trying to find a simplification that would make the calculations easier. My basic goal is to find the simplest calculation that represents a motor and look at the energy balances to determine what the work done on the rotor is equal to. Then we can look at how that changes and determine which terms in the EM energy equation represent the work on the rotor.

OK, so in general the following expression represents the conservation of energy for EM fields (equation 1033 here):

$$\nabla \cdot \left( E \times B \right) + \frac{\partial}{\partial t}\left( \frac{E^2}{2} + \frac{B^2}{2} \right) - E \cdot j = 0$$

The first term is the energy flux, the second term is the energy in the field, and the third term is usually interpreted to be the work done on matter. It is the meaning of this third term which is in dispute in this thread.

The total work done on the rotor in a motor can be given by
$$P=I^2 R+\tau \omega$$

The first term is the resistive losses which increases the thermal energy of the rotor and the second term is the mechanical work done by the rotor.

My claim is that E.j=P and therefore by energy conservation all of the work is done by the usual term.

If I understand it correctly, your claim is that E.j = I²R < P which implies that all of the mechanical work done by the rotor, τω, has come from the other terms in the energy conservation equation (whether the first or second is not important since both have B terms).

Is this, in your opinion, an accurate statement of the disagreement? If so, then we can in principle calculate P and I²R and compare them to E.j to determine the correct answer. Do you agree?

EDIT: I had to make some late corrections to how I had originally posted. The above equation for EM energy is the time derivative of the energy density, so we will have to do some integration to get energy or power. Also, I had accidentally written IR instead of I²/R.

 

[Per Oni]

Cabraham, remember, you were the first to accuse people of being rude!

In 232 you state: “Remember there is mutual inductance between rotor & stator. Each receive energy from the other. Although I is constant, LI2/2 still changes…….”

Then in 239 you say: “Nobody said a permag "gives energy" to the motor.”

If according to you each item receives energy of the other then surely I can make the conclusion that one of the transfers includes the stator giving energy LI2/2 to the rotor, and therefore to the motor? Furthermore, it is entirely up to me to imagine this stator to be a permanent magnet.

I still also refute most other (crucial for this thread) points you made in 239 but this will have to do for now.

 

[cabraham]

Cabraham, remember, you were the first to accuse people of being rude!

In 232 you state: “Remember there is mutual inductance between rotor & stator. Each receive energy from the other. Although I is constant, LI2/2 still changes…….”

Then in 239 you say: “Nobody said a permag "gives energy" to the motor.”

If according to you each item receives energy of the other then surely I can make the conclusion that one of the transfers includes the stator giving energy LI2/2 to the rotor, and therefore to the motor? Furthermore, it is entirely up to me to imagine this stator to be a permanent magnet.

I still also refute most other (crucial for this thread) points you made in 239 but this will have to do for now.

Maybe I should not have replied to your permag statement. You mentioned that if a permag did work it would lose its energy, to which I replied that it does not give up its own energy.

The OP question was whether or not magnetic fields can do work on a wire loop. Whether the B field is from a permag or a wire loop is not the issue. "Can a B field do work on the rotor?", is the question. I feel that it is imperative we stay on track with the OP question.

As far as your refuting my other points, I say again, if something(s) I posted is questionable to you, feel free to ask me to clarify, or to offer your own explanation based on reliable info. I am not here to win anything, put anybody down, or cause trouble. Let me know what you want me to explain, clarify, & I'll do just that. Best regards.

Claude

 

[truesearch]

One reply to Per oni was As far as a permanent magnet motor goes this is a straw man you invented. Nobody said a permag "gives energy" to the motor. The input power source does that. Nobody in this thread would ever believe what you just presented. We're not that dumb. Please don't treat us like we are. BR.


very rude to say the least..."please don't treat US like we are"
How do you know how dumb I am ?

 

[Per Oni]

Maybe I should not have replied to your permag statement. You mentioned that if a permag did work it would lose its energy, to which I replied that it does not give up its own energy.Claude

So according to you at some point a permanent magnet receives extra energy? Does this extra energy result in it having a higher B-field?

The OP question was whether or not magnetic fields can do work on a wire loop. Whether the B field is from a permag or a wire loop is not the issue. "Can a B field do work on the rotor?", is the question. I feel that it is imperative we stay on track with the OP question.

It was you who started talking about 1001 other issues, not me. You are responsible for the statements you make. And according to me you made a lot of false statements. Show me 1 reliable reference which indicates that W=LI2/2 has anything to do with the output of motor power. Where on Earth does the emf come in? Without using emf it is impossible to calculate the power.

From 239: -

When the poles are aligned, I said that the torque is minimum. There is a force but the moment is zero. Look at a diagram, or draw one. The force is radial & does not spin the rotor.

When the poles are aligned B is maximum and since torque = radius x force, momentum or torque is at a maximum.

 

[cabraham]

One reply to Per oni was As far as a permanent magnet motor goes this is a straw man you invented. Nobody said a permag "gives energy" to the motor. The input power source does that. Nobody in this thread would ever believe what you just presented. We're not that dumb. Please don't treat us like we are. BR.


very rude to say the least..."please don't treat US like we are"
How do you know how dumb I am ?

Ok fine, I assumed you are smart, but you're asking me why I think that. Would you rather I assume otherwise? Anyway, I'll use 'me" in the future instead of "us". Is that fair? BR.

Claude

 

[cabraham]

So according to you at some point a permanent magnet receives extra energy? Does this extra energy result in it having a higher B-field?


It was you who started talking about 1001 other issues, not me. You are responsible for the statements you make. And according to me you made a lot of false statements. Show me 1 reliable reference which indicates that W=LI2/2 has anything to do with the output of motor power. Where on Earth does the emf come in? Without using emf it is impossible to calculate the power.

From 239: -
When the poles are aligned B is maximum and since torque = radius x force, momentum or torque is at a maximum.

If I discuss 1001 other issues it is in response to people who raised these 1001 issues. I would rather stay on track w/ the OP question.

Torque is radius x force x sine of angle. If the force is tangent to the radius, angle is 90 deg, sin 90 deg = 1. But when the poles are aligned, force is maximum, but angle is zero, as the force acts radially. Sin 0 deg = 0. There is maximum force but zero torque. Any motor text will clarify this for you. BR.

Claude

 

[Miyz]

1st bold: I'm glad you agree that B force produces torque. So we have consensus that B force, not E force, is, as you put it, "the origin of the torque". I agree with you completely. We will never argue over that. So what is the definition of "work". In linear motion (non-rotational) it is force times distance (dot product) times cosine of angle if both are constant. If one or both vary, work is the integral of force dot product w/ differential length over the path.

With rotational motion, work is the integral of torque dotted with angular displacement. A torque acting on the rotor multiplied by the radian angle measure the rotor has turned equals the work done on the rotor. If the torque was a constant 1.00 Newton-meter, & spins the rotor through an angle of 1.00 radian, then 1.00 joule of work has been done. So the magnetic force, B, exerts 1.00 N of force with a moment arm of 1.00 meter, for a torque of 1.00 N-m, spinning the rotor through 1.00 radian of angle, then B has done 1.00 joule of work on the rotor.

In reality, the torque changes as the rotor spins, so instead of simple multiplication, an integration must be done. Anyway, since we agree B produces torque, all we have to do is integrate said torque wrt angle, & we obtain the work done by B.

Poynting, Maxwell, Faraday, Ampere, etc. laws are all upheld here. Your equations & facts presented are certainly correct, but nowhere did you apply them to the question at hand. If you still disbelieve me, please draw a sketch indicating the quantities & show where the work is being done & by which force. I grow tired of asking but this is important. Otherwise, best regards.

Claude

Hey Claude,

Isn't it both the RPM and torque? I mean the rotational motion is generally caused by the magnetic force/fields right? So is it safe to say that its the cause of both RPM + Torque?

 

[cabraham]

Hey Claude,

Isn't it both the RPM and torque? I mean the rotational motion is generally caused by the magnetic force/fields right? So is it safe to say that its the cause of both RPM + Torque?

Torque = I * alpha, where I = moment of inertia, alpha = angular acceleration.

Of course if a torque is acting on the rotor, the rotor will rotate. Depending on the currents in the stator & rotor, the counter emf, & load, the speed will reach an equilibrium value in rpm. I would say that the magnetic force acting on the rotor produces torque except when poles are directly aligned, & torque results in rotor spinning.

Again, one who wishes to explore motors in detail should obtain good reference books on motors/generators. Anybody can read faster than I can type, & I only have limited time on the web. BR.

Claude

 

[cabraham]

Yes, definitely. I was trying to find a simplification that would make the calculations easier. My basic goal is to find the simplest calculation that represents a motor and look at the energy balances to determine what the work done on the rotor is equal to. Then we can look at how that changes and determine which terms in the EM energy equation represent the work on the rotor.

OK, so in general the following expression represents the conservation of energy for EM fields (equation 1033 here):

$$\nabla \cdot \left( E \times B \right) + \frac{\partial}{\partial t}\left( \frac{E^2}{2} + \frac{B^2}{2} \right) - E \cdot j = 0$$

The first term is the energy flux, the second term is the energy in the field, and the third term is usually interpreted to be the work done on matter. It is the meaning of this third term which is in dispute in this thread.

The total work done on the rotor in a motor can be given by
$$P=I^2 R+\tau \omega$$

The first term is the resistive losses which increases the thermal energy of the rotor and the second term is the mechanical work done by the rotor.

My claim is that E.j=P and therefore by energy conservation all of the work is done by the usual term.

If I understand it correctly, your claim is that E.j = I²R < P which implies that all of the mechanical work done by the rotor, τω, has come from the other terms in the energy conservation equation (whether the first or second is not important since both have B terms).

Is this, in your opinion, an accurate statement of the disagreement? If so, then we can in principle calculate P and I²R and compare them to E.j to determine the correct answer. Do you agree?

EDIT: I had to make some late corrections to how I had originally posted. The above equation for EM energy is the time derivative of the energy density, so we will have to do some integration to get energy or power. Also, I had accidentally written IR instead of I²/R.

E.J cannot be power, but rather power density in watt/meter³. To obtain the power from E.J a volume integral is needed.

E.J should definitely include I²R, as well as LI²/2. The integral of E.J should equal the input power (complex) VI. The total power from the input source must account for all the heating loss in the rotor, reactive power in the rotor inductance, as well as Tω mechanical power.

As far as the relation between input power & Tω goes, the input source energizes the rotor inductance, & we know that N∅ = LI. But ∅ = AB, where A is flux area, B is flux density, so that NAB = LI. Rearranging we get B = LI/NA.

Of course, B force produces torque which spins the rotor. So the input power source provides V & I to the stator, as well as to the rotor for certain types of motors. For an induction ac motor, or a permag dc motor, power is not directly applied to the rotor. In the stator we get E & B fields due to the input source providing power as well as counter emf from the rotor motion. The current I divided by the conductor area is J, the current density. So E.J should, as far as I can tell, account for the total power per unit volume.

As the voltage source energizes the stator inductance, current builds up as does the B field. The B field spins the rotor. So E.J is definitely providing power, but the B force is producing torque. As I've stated so many times, it is the B force that is directly providing rotor torque, but E is interactive, providing power, along with J, to energize the B field. Ultimately no single quantity is responsible for the motor working.

If we have an input voltage with zero input current, it is impossible for the motor to output mechanical power since input electric power is zero. Likewise zero input voltage does not produce output power. Since E is directly related to V, & J related to I, as well as B related to J, a zero value for either 1 of these quantities means a zero output power.

I've stated numerous times, without E, well ----- forget it! E is all important, every bit as much so as B. Again, how do get 1 w/o the other under dynamic conditions? B is what directly spins the rotor, but E is right behind B providing B with the needed energy. What powers E, well that is the input power source which also powers J & B.

The input power source, i.e. wall outlet, is ultimately what makes everything happen. Did I help or just make matters worse?

Claude

 

[Miyz]

Torque = I * alpha, where I = moment of inertia, alpha = angular acceleration.

Of course if a torque is acting on the rotor, the rotor will rotate. Depending on the currents in the stator & rotor, the counter emf, & load, the speed will reach an equilibrium value in rpm. I would say that the magnetic force acting on the rotor produces torque except when poles are directly aligned, & torque results in rotor spinning.

Again, one who wishes to explore motors in detail should obtain good reference books on motors/generators. Anybody can read faster than I can type, & I only have limited time on the web. BR.

Claude

However, As we agreed that Torque is generated by the B fields... Is RPM also generated by it? Didn't really find the specific answer from many online sources... Logically RPM is generated due to the magnetic interaction isn't it? Just to be sure...

Thanks Claude.

 

[Darwin123]

Cabraham: I think you have defined 'gap' for us : "for all those electrons in the 'overlap region' there is zero energy gap". The Valence and conduction bands do not need to 'overlap completely' there just needs to be a 'contact' so that all energies in the conduction band are available to the valence band.
We seem to agree on the physics.
Does this mean that the statement from post 127:
"So far the naysayers have produced nothing. They talk a big game about Einstein, reference frames, etc., but cannot show me the fields working in a simple induction motor. Show me, please, how it is E force, & not B force that spins the rotor. So far all I get is people blowing smoke. Not 1 naysayer has addressed the motor operation question.
In a motor, we are not simply moving electrons from valence to conduction. We are exerting forces on wire loops resulting in torque & work being done. "
can be discounted as part of your explanation?...it seems logical to do so.
What did you think of my reference to the Hall effect in the conductors? Is this the physics explanation behind your 'tethering' analogy?
I have never met this 'tethering' analogy before.
Must dash to ebay now...

"What did you think of my reference to the Hall effect in the conductors? Is this the physics explanation behind your 'tethering' analogy?"
I am not sure that I understand your use of the word "tethering". I think that you are referring to the idea that the nonmagnetic forces that hold the charge carriers in the wire do the work. If this is what you mean, then the answer to the second question is "yes".
The voltage in the Hall effect is caused by the forces that keep the electric charge in the conducting plate. There would be no voltage if there wasn't an "edge" to the plate
The Hall effect is caused by an accumulation of electric charge at the edge of the plate. The electric charge carriers are pushed in the direction of current by the electric field in the circuit. This electric field does work, because the force is in the direction of motion. However, the electric field perpendicular to the circuit does no work at the stationary state because the forces are balanced.
In the bulk of the plate, the force by the electric field in the direction of the voltage drop precisely balances the force by the magnetic field opposite to the direction of the voltage drop. So in the bulk of the plate, there is no component of motion for the electric charge carriers perpendicular to the current. In the bulk of the plate, there are also no electric charge density. The electric field is continuous. Therefore, charge density of both moving charges and stationary charges is zero.
At the edge of the plate, there is a discontinuity of the electric field. Therefore, there is an electric charge distributed at the edge of the plate. The electric charge at one edge is equal in magnitude but opposite in sign from the electric charge on the other edge. If there wasn't this electric charge at the edge, there couldn't be a Hall voltage.
The electric charges do not cancel out at the edge. There is a discontinuity in electric field. Outside the plate, the electric field is zero. In the bulk, the electric field is that necessary to balance the magnetic force. Thus, the electric field has to give way to the surface force at the edge.
The only way such a charge can accumulate is if there is a "surface force" at the edge that prevents the electric charge from leaving the plate entirely. This surface force is equal and opposite the force of the magnetic field.
One can "tap" into the Hall voltage by putting electrodes on opposite sides of the plate. Electric current will flow from one electrically charged edge to the other. However, what does the work in this case is the electric field caused by the charge carriers that were caused by the surface force.
The magnetic field does cause the electric field that does the work. However, the magnetic field does not directly do the work. It changes the electric charge distribution, which changes the electric field which does the work.
Perhaps it would help if I qualify the statements better. A magnetic field can not "directly" do work on an electric charge. However, there are several ways that a magnetic field can "indirectly" do work on an electric charge.
A magnetic field can change charge distributions that create an electric field, for example. The electric field can directly do work. The magnetic field indirectly did the work.
A magnetic field can put a stress on a rotating body which is electrically charged. The change in electric charge distribution can directly do work on the electric charges in this rotating body. The magnetic field did not directly do work on the electric charges. So again, the magnetic field indirectly did work on the electric charges.
What has to be recognized here is that no every change in the distribution of electric charges involves directly doing work. The magnetic field can redistribute electric charges without doing work. The redistributed charges create an electric field, and that directly does the work.
What the Poynting theorem shows is that the work is directly determined by the dot product of the electric current and the electric field. However, this does not mean that the magnetic field does not play a role. The magnetic field can determine the direction of the electric current. The electric current can determine the direction of the electric field.
I suggest the following. In a motor or other dynamic process involving electric current, the magnetic field does not have any effect on work in the first few moments of operation. As the dynamic process approaches a stationary state, the magnetic field establishes an angle between the electric field and the electric current. Thus, the electric field ends up doing work on the electric charges.
The electric field that directly does the work. However, the magnetic field supervises the work.
That is how I think of electric motors and electric generators. The electric field does all the work, and the magnetic field supervises the work.
The magnetic field is a force supervisor! The magnetic field is all torque and no action!

 

[Dale]

E.J cannot be power, but rather power density in watt/meter³. To obtain the power from E.J a volume integral is needed.

Yes, I already mentioned that in my edit.

E.J should definitely include I²R, as well as LI²/2. The integral of E.J should equal the input power (complex) VI. The total power from the input source must account for all the heating loss in the rotor, reactive power in the rotor inductance, as well as Tω mechanical power.

I am not sure if I understand what you are saying here. It sounds to me like you are now agreeing with me that E.j does account for all of the work done on the rotor, both the resistive and the mechanical.

If so, then that leaves no work left over to be accounted for by B. You cannot have E.j account for all of the work, B also do work, and energy be conserved.

Also, I think with the interaction from the stator field that the rotor field is too complicated to simplify as a just an inductance, L, as you correctly pointed out to me earlier. And since the stator is not part of the circuit you cannot account for it with a mutual inductance either. The L energy you are talking about is included in the second term of the standard energy conservation equation I posted above.

 

[cabraham]

"What did you think of my reference to the Hall effect in the conductors? Is this the physics explanation behind your 'tethering' analogy?"
I am not sure that I understand your use of the word "tethering". I think that you are referring to the idea that the nonmagnetic forces that hold the charge carriers in the wire do the work. If this is what you mean, then the answer to the second question is "yes".
The voltage in the Hall effect is caused by the forces that keep the electric charge in the conducting plate. There would be no voltage if there wasn't an "edge" to the plate
The Hall effect is caused by an accumulation of electric charge at the edge of the plate. The electric charge carriers are pushed in the direction of current by the electric field in the circuit. This electric field does work, because the force is in the direction of motion. However, the electric field perpendicular to the circuit does no work at the stationary state because the forces are balanced.
In the bulk of the plate, the force by the electric field in the direction of the voltage drop precisely balances the force by the magnetic field opposite to the direction of the voltage drop. So in the bulk of the plate, there is no component of motion for the electric charge carriers perpendicular to the current. In the bulk of the plate, there are also no electric charge density. The electric field is continuous. Therefore, charge density of both moving charges and stationary charges is zero.
At the edge of the plate, there is a discontinuity of the electric field. Therefore, there is an electric charge distributed at the edge of the plate. The electric charge at one edge is equal in magnitude but opposite in sign from the electric charge on the other edge. If there wasn't this electric charge at the edge, there couldn't be a Hall voltage.
The electric charges do not cancel out at the edge. There is a discontinuity in electric field. Outside the plate, the electric field is zero. In the bulk, the electric field is that necessary to balance the magnetic force. Thus, the electric field has to give way to the surface force at the edge.
The only way such a charge can accumulate is if there is a "surface force" at the edge that prevents the electric charge from leaving the plate entirely. This surface force is equal and opposite the force of the magnetic field.
One can "tap" into the Hall voltage by putting electrodes on opposite sides of the plate. Electric current will flow from one electrically charged edge to the other. However, what does the work in this case is the electric field caused by the charge carriers that were caused by the surface force.
The magnetic field does cause the electric field that does the work. However, the magnetic field does not directly do the work. It changes the electric charge distribution, which changes the electric field which does the work.
Perhaps it would help if I qualify the statements better. A magnetic field can not "directly" do work on an electric charge. However, there are several ways that a magnetic field can "indirectly" do work on an electric charge.
A magnetic field can change charge distributions that create an electric field, for example. The electric field can directly do work. The magnetic field indirectly did the work.
A magnetic field can put a stress on a rotating body which is electrically charged. The change in electric charge distribution can directly do work on the electric charges in this rotating body. The magnetic field did not directly do work on the electric charges. So again, the magnetic field indirectly did work on the electric charges.
What has to be recognized here is that no every change in the distribution of electric charges involves directly doing work. The magnetic field can redistribute electric charges without doing work. The redistributed charges create an electric field, and that directly does the work.
What the Poynting theorem shows is that the work is directly determined by the dot product of the electric current and the electric field. However, this does not mean that the magnetic field does not play a role. The magnetic field can determine the direction of the electric current. The electric current can determine the direction of the electric field.
I suggest the following. In a motor or other dynamic process involving electric current, the magnetic field does not have any effect on work in the first few moments of operation. As the dynamic process approaches a stationary state, the magnetic field establishes an angle between the electric field and the electric current. Thus, the electric field ends up doing work on the electric charges.
The electric field that directly does the work. However, the magnetic field supervises the work.
That is how I think of electric motors and electric generators. The electric field does all the work, and the magnetic field supervises the work.
The magnetic field is a force supervisor! The magnetic field is all torque and no action!

How can B produce torque yet no work? Work = torque times angle. If B makes torque & the rotor moves through an angle, then B makes work as well as torque. Regarding the "edge" buildup of charge, I don't see how the torque can be due to E instead of B. Of course when B acts on the e- in the rotor loop the e- migrate towards the periphery of said loop. But the force of attraction to the stator magnetic dipole is magnetic not electric.

The rotor electrons have a velocity that is tangential to the loop. The B field acts normal to the velocity & the resulting Lorentz force is normal to both velocity & B. The fact that the e- are now out at the edge does not change the fact that their velocity is still tangential thus the Lorentz force is still radial. Of course the e- at the edge produce their own local E field, since e- each as individuals possesses their own E field. But since they are moving wrt the stator, the force of atttraction between stator & rotor is B force, not E. However there is a component of e- velocity directed radially, very small in comparison to the loop current. This would result in a Lorentz force tangentail to the loop. But these 2 forces are orders of magnitude, at least 2, apart.

As far as B doing work on a charge, if an e- is moving in free space, then B is always normal to its velocity, so that the e- cannot have its KE changed, no work is done by B. If an e- is conducting in a wire loop, B can only act radially on the loop if B is normal to the plane of the loop, hence no work is done, the e- KE does not change.

But if the rotor loop carries a current as does the stator loop, the Lorentz force acts radially moving the e- radially. But the e- is attached to the stationary lattice proton through E force internal to the atom. So the whole array of protons gets yanked radially. Likewisr, neutrons are attached to protons via SN force, & they move radially. So the B force is moving the whole loop. The moment of inertia of the loop times its angular velocity squared is the loop's rotational KE.

B did the work, but caould not have done so alone. If protons were not bonded to e- by E, & neutrons were not bonded by SN, the electrons would fly off the wire leaving protons & neutrons behind. The e- would not have their KE changed since B does not do that.

Anyway, that appears to be what happens. I'm willing to hear alternate explanations, but frankly to me, this is the only scenario that remotely makes sense.

As far as "all torque no action goes", that phrase is a logical contradiction. Whatever peoduces torque is doing work as long as the rotor spins. The only way any field or force can produce torque without doing work is if the rotor was locked. The torque would be there, but the rotor would not move so that work done is zero, not counting torsional energy storage.

For a locked rotor, your "all torque no action" statement would be correct. Once the rotor moves the source of torque has done work. That source would be B. BR.

Claude

 

[Dale]

How can B produce torque yet no work? Work = torque times angle. If B makes torque & the rotor moves through an angle, then B makes work as well as torque.

Even if the rotor moves, the B field does not. That is another reason why the more general thermodynamic definition of work is used when dealing with fields.

 

[cabraham]

Even if the rotor moves, the B field does not. That is another reason why the more general thermodynamic definition of work is used when dealing with fields.

Why would B have to move to do work? E does work on e- w/o E moving. Please examine your comment. BR.

Claude

 

[cabraham]

Yes, I already mentioned that in my edit.

I am not sure if I understand what you are saying here. It sounds to me like you are now agreeing with me that E.j does account for all of the work done on the rotor, both the resistive and the mechanical.

If so, then that leaves no work left over to be accounted for by B. You cannot have E.j account for all of the work, B also do work, and energy be conserved.

Also, I think with the interaction from the stator field that the rotor field is too complicated to simplify as a just an inductance, L, as you correctly pointed out to me earlier. And since the stator is not part of the circuit you cannot account for it with a mutual inductance either. The L energy you are talking about is included in the second term of the standard energy conservation equation I posted above.

No inconsistency at all. E does the work to energize L. A B field is inevitable due to current in L. In order to energize L, I is necessary since W = LI²/2. But to change I, we need V, per V = L*dI/dt. But V = integral E dot dl, so that transferring energy from input source to L is done via E. B increases as I increase. Then B exerts a force on the e- in the rotor loop. Normally, the e- KE cannot change if e- were unbound. But they are bound. Although the B force moves the e-, their KE remains unchanged. Their velocity changes in direction, not magnitude.

But when the B force grabs the e- yanking them radially, the internal bonding forces yank the protons as well as neutrons, which are originally stationary. Their KE went from 0 to Iω²/2. So while B is unable to change the KE of the e-, it did change the KE of the lattice protons & neutrons since their velocity changed in not only direction, but magnitude as well. But the energy in B was transferred from the power source through E, the energy is conserved.

Claude

 

[vanhees71]

Very good! Claude, I agree fully with your posting #274.

Nevertheless, I'll try to get a full calculation, including mechanical and electromagnetic energy/work, on the somewhat simpler example of the theory of oldfashioned ammeters, as you can find in any lab description for undergraduates (at least in Germany, where it's called "Grundpraktikum"). However nobody does an analysis on energy/power balance.

 

[cabraham]

Thanks vanhees71, for a great discussion, & for your making great contributions to this discussion, particularly you provided great insight re the math involved. One thing we all hopefully learned is that although the motor was invented in the 19th century, it is an incredibly fascinating device! Who would think that so much is involved when we turn on our fan & watch the blades spin? You really know your stuff.

Also deserving mention is that E.J does involve B. Since J = I/A_w, where A_w is wire area, & NA_cB = LI = LJA_w where A_c is the area of the magnetic core, we have B = LJA_w/NA_c, or we can write J = BNA_c/LA_w.

Hence E.J = (E.B)(NA_c/LA_w).

If we wish to examine the work done by the power source energizing the overall system, then "E.J" is equal to "E.B" multiplied by a constant. So it is apparent that both E & B are involved. Of course anybody familiar with e/m field theory already knows that. Bit it is good to examine these questions now & then. I thank all who participated. Even my harshest critics helped me gain a better understanding & I thank them. I apologize if I came across as rude, & assure all that I sometimes get carried away. I don't have anything personal against anybody here, not even my critics. Feel free to ask for clarification. Thanks to all.

Claude

 

[Dale]

Why would B have to move to do work? E does work on e- w/o E moving.

There are two definitions of work. The less general mechanical definition (f.d) and the more general thermodynamic definition (energy transfer). Since E and B do not move it is not clear that the mechanical definition of work even applies, but the thermodynamic definition clearly applies. The thermodynamic definition has E doing work and B not because E transfers energy and B does not.

 

[Dale]

Normally, the e- KE cannot change if e- were unbound. But they are bound. Although the B force moves the e-, their KE remains unchanged. Their velocity changes in direction, not magnitude.

But when the B force grabs the e- yanking them radially, the internal bonding forces yank the protons as well as neutrons, which are originally stationary. Their KE went from 0 to Iω²/2. So while B is unable to change the KE of the e-, it did change the KE of the lattice protons & neutrons since their velocity changed in not only direction, but magnitude as well. But the energy in B was transferred from the power source through E, the energy is conserved.

Internal forces cannot do work on a system. If the B field cannot do work on the electrons, then it cannot do work on the rotor.

I'm afraid that with all the talk about internal forces I still don't understand what you think is equal to what in the energy conservation equation that I posted earlier. Do you think that E.j = P at all times or not? If not, then what do you think it is equal to? If so, then how can you balance energy in the energy conservation equation?

 

[cabraham]

Internal forces cannot do work on a system. If the B field cannot do work on the electrons, then it cannot do work on the rotor.

I'm afraid that with all the talk about internal forces I still don't understand what you think is equal to what in the energy conservation equation that I posted earlier. Do you think that E.j = P at all times or not? If not, then what do you think it is equal to? If so, then how can you balance energy in the energy conservation equation?

Your questions have already been asked & answered. Maybe a different example is needed. A steel ball is connected to a rubber ball by a short rope. An overhead electromagnet is turned on & the steel ball is lifted up to the magnet & the rubber ball is yanked along with it. Ordinarily the magnet cannot do work lifting the rubber ball. But it can lift the steel ball & the rope makes it possible to lift both. The B force does all the work. The rope being internal does no work.

B force on the rotor is similar, not exactly equivalent. Although B does no work on an e-, it can change its direction. It displaces an e-. But in doing so the internal force binding proton to electron resulted in protons being displaced. I agree that internal E & SN force cannot do work. The external force, B, did the work.

As far as E.J goes, see my post above. E.J is merely E.B(NAc/LAw). It is apparent that E & B are both involved. You claim that since all work is accounted for in E.J, which excludes B, then only E, not B, is doing work. But remember that J is related to B, as is L & I. The E field component that energizes inductance L energizes B as well. Hence B is as involved as is E. BR.

Claude

 

[Dale]

As far as E.J goes, see my post above. E.J is merely E.B(NAc/LAw). It is apparent that E & B are both involved. You claim that since all work is accounted for in E.J, which excludes B, then only E, not B, is doing work. But remember that J is related to B, as is L & I. The E field component that energizes inductance L energizes B as well. Hence B is as involved as is E.

So it sounds like you agree that E.j=P but believe that it still makes sense to claim that B does work since B is related to j via Maxwell's equations. Is this a correct statement of your position?

I.e. you could solve Maxwell's equations for j in terms of B and substitute into the energy conservation equation to get P = E.j = f(E,B,...).