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

[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,...).

 

[cabraham]

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,...).

Well, my belief that B does work was shown independent of the E.J equation. But one thing should be clarified. I've asked everybody involved on this thread to provide illustrations. It's hard to answer questions if I don't know the quantities you are discussing. At first I presumed E.J involved the rotor conduction loss as heat, since J = σE, which is Ohm's law in 3 dimensions. Without a diagram I wasn't sure which E field was being discussed.

But it later became apparent that the E in the E.J expression involves more then just I²R loss, but inductance L as well. So I then treated E.J as the total input power since the input voltage V across the stator winding results in current I which depends on R as well as X_L which equals Lω, i.e. I = V / (R + jωL). A diagram would have saved us a few pages since it would have clarified which E is under scrutiny, internal to the winding, or across the winding. Forgive me for saying it again, but unless participants post diagrams communication is more difficult.

I agree that we can use Maxwell's equations to describe a relation between E & B, E & J, J & B, P, L, R, etc. But in the end we must draw a picture, account for all forces, motion, etc. Then we can ascertain which force is doing what, where the energy is coming from & going to including mechanical, & gain a good insight into motor operation. I feel we have done that & arrived at a conclusion that is rational & supported by established science.

I gave credible reference texts earlier in the thread. Every motor text I've read stated unequivocally that B does work on the rotor. I will elaborate if there are still unanswered questions. BR.

Claude

 

[Dale]

I would prefer concise clear answers to questions over further elaboration.

 

[Dotini]

Possibly a bit off topic - if so please forgive me, but you guys are all too smart and studied-up to ignore any longer.

It is well known that thunderclouds possesses both magnetic and electric fields, which are measured to change and deplete when lightning is discharged. Presumably, this also includes those upward discharges resulting in jets, sprites, elves, etc. There are various theories as to the exact process of charge separation and electrification of the clouds, but there's no doubt about the electric fields and the lightning. Source for all this is The Lightning Discharge, by Martin Uman. Now to my questions.

Would it be fair to say that (a) the thunderclouds (electric fields) "do work" in the discharge of lightning, and that (b) thunderclouds (electric fields) constitute a "particle accelerator" in the sense that electrons and ions are launched high into the ionosphere?

Respectfully submitted,
Steve

 

[cabraham]

I would prefer concise clear answers to questions over further elaboration.

Then please give me a concise clear answer as to which force is doing work on the rotor & include a sketch showing the direction of said field/force. Does Maxwell support this field? I will add another sketch if requested. I've told you which force is acting, the nature of the attraction, diagrams, & mathematical relations between all relevant quantities.

You seem to be the only person left not happy w/ the thesis that B does the work. Remember that E is along the direction of J. For the rotor to spin there must be a force acting radially. The only force pointed in said direction is B. BR.

Claude

 

[Dale]

Then please give me a concise clear answer as to which force is doing work on the rotor & include a sketch showing the direction of said field/force. Does Maxwell support this field?

E.j does the work, with j being tangent to the rotor at all points. I believe that E is also tangent, but I would have to actually do a full derivation to be sure. Maxwell does support this field. E is just the regular E from Maxwell's equations involving all terms and not excluding any contribution.

Sorry about not including a sketch, but hopefully the clarity of the answer is sufficient without it.

Can you now please clearly answer my question of post 280? Feel free to include a "yes, but ..." or a "no, because ...", but at least let me know if you agree or disagree with my attempt to understand your position. From your previous response I cannot even tell that much.

 

[truesearch]

'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.'
You are close here to recognising the Hall effect...you have not referred to this in any of your explanations yet it is an established part of science.
The displaced electrons leave behind a + charge and an electric field exists across the width of a wire. This is called the Hall effect and it gives rise to an internal force that stops the electrons 'jumping out of the wire'
I wonder if this is what you have had in mind when you mentioned your 'tethering' forces along with the short nuclear force. I think these are unnecessary in the explanation of electric motors, they are certainly not standard textbook content.
Electric motors have been around since the 19th century and their operation is very satisfactorily described in standard textbooks.

 

[cabraham]

E.j does the work, with j being tangent to the rotor at all points. I believe that E is also tangent, but I would have to actually do a full derivation to be sure. Maxwell does support this field. E is just the regular E from Maxwell's equations involving all terms and not excluding any contribution.

Sorry about not including a sketch, but hopefully the clarity of the answer is sufficient without it.

Can you now please clearly answer my question of post 280? Feel free to include a "yes, but ..." or a "no, because ...", but at least let me know if you agree or disagree with my attempt to understand your position. From your previous response I cannot even tell that much.

Re post #280, here is my answer. Yes, B does work. Is E.J = P? Well, again, I assume you infer that the "E" is the total E across the rotor R & L. If you mean the E across the whole rotor winding, then E.J is the power density in W/m³. If "E" is the electric field inside the wire, then E.J is basically the rotor heat loss per volume.

Re your position that E & J are tangential to the rotor, please see my sketch. To spin the rotor a torque is needed. This requires a force radial to the loop, not tangential. If a loop is in the x-y plane, & B in along the z axis, with J & E tangential, there is no radial force. Another loop (2) is carrying current & coupled to loop 1. Its B field is in the z direction. The Lorentz force acting on the e- in loop 1 has 2 parts. The E force moves the e- along the tangent. The uXB force acts radially. B is in z direction, u is tangent. Do the cross product & the force due to B is radial, which results in force. If the 2 loops are coplanar, their B fields both align with z axis, & radial force exists but no moment & no torque. Rotor remains still. But say that there is 45 degrees spatial angle between 2 loops, rotor & stator.

The radial force is skewed at an angle of 45 degrees to the x-y plane. I showed the direction in my diagram. This results in a torque which spins the rotor. I'll add a diagram tonight when I'm home. I will illustrate all forces explicitly. BR.

Claude

 

[cabraham]

'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.'
You are close here to recognising the Hall effect...you have not referred to this in any of your explanations yet it is an established part of science.
The displaced electrons leave behind a + charge and an electric field exists across the width of a wire. This is called the Hall effect and it gives rise to an internal force that stops the electrons 'jumping out of the wire'
I wonder if this is what you have had in mind when you mentioned your 'tethering' forces along with the short nuclear force. I think these are unnecessary in the explanation of electric motors, they are certainly not standard textbook content.
Electric motors have been around since the 19th century and their operation is very satisfactorily described in standard textbooks.

I've been saying that motors have been around since 19th century & that their operation is well described. But the operation is described as magnetic fields spinning the rotor, not electric. As far as tethering goes, we can just regard the rotor winding as having internal bonding forces which maintain the integrity of the loop structure. The B force moves the electrons & the internal bond in the atomic structure assures that the other particles move with the electrons. BR.

Claude

 

[Dale]

I am tired of asking for clear confirmation and getting long-winded obfuscations instead, so I will simply assume that I am understanding your position and encourage you to politely correct me if I accidentally misunderstand.

Is E.J = P? Well, again, I assume you infer that the "E" is the total E across the rotor R & L. If you mean the E across the whole rotor winding, then E.J is the power density in W/m³. If "E" is the electric field inside the wire, then E.J is basically the rotor heat loss per volume.

E is the E field from Maxwell's equations. I don't know why you think that there is any ambiguity about what E is. It might be difficult to calculate, but what it is should be clear.

Since there is no j outside the wire then E.j is 0 outside of the wire, so the only E corresponding to a non-zero E.j is the E inside the wire. So from this and your comments above it seems like you think that E.j≠P.

If E.j≠P then energy is not conserved. The E and B fields have a certain amount of energy density given by (E²+B²)/2. If that energy changes then it must either have gone to EM fields elsewhere as described by \nabla \cdot (E \times B) or it must have gone to matter as described by E.j. There is nowhere else for energy to go. http://farside.ph.utexas.edu/teaching/em/lectures/node89.html

If matter obtains more power P from EM than the integral of E.j then there is energy in the matter which appears out of nowhere without a corresponding decrease in the EM energy. So, your position is incompatible with the conservation of energy.

Re your position that E & J are tangential to the rotor, please see my sketch. To spin the rotor a torque is needed. This requires a force radial to the loop, not tangential. If a loop is in the x-y plane, & B in along the z axis, with J & E tangential, there is no radial force. Another loop (2) is carrying current & coupled to loop 1. Its B field is in the z direction. The Lorentz force acting on the e- in loop 1 has 2 parts. The E force moves the e- along the tangent. The uXB force acts radially. B is in z direction, u is tangent. Do the cross product & the force due to B is radial, which results in force. If the 2 loops are coplanar, their B fields both align with z axis, & radial force exists but no moment & no torque. Rotor remains still. But say that there is 45 degrees spatial angle between 2 loops, rotor & stator.

The radial force is skewed at an angle of 45 degrees to the x-y plane. I showed the direction in my diagram. This results in a torque which spins the rotor. I'll add a diagram tonight when I'm home. I will illustrate all forces explicitly.

There is no doubt that the B field provides the torque. You are welcome to draw a diagram if you wish, but it is not under any dispute.

 

[Per Oni]

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

You are right on this single point. I mixed up the conductor of the op pic with the pole.

From 263:

Anyway, I'll use 'me" in the future instead of "us".

284:

You seem to be the only person left not happy w/ the thesis that B does the work.

Can you please stick a little longer to your own good resolution not to talk for other people?

 

[cabraham]

I am tired of asking for clear confirmation and getting long-winded obfuscations instead, so I will simply assume that I am understanding your position and encourage you to politely correct me if I accidentally misunderstand.
E is the E field from Maxwell's equations. I don't know why you think that there is any ambiguity about what E is. It might be difficult to calculate, but what it is should be clear.

Since there is no j outside the wire then E.j is 0 outside of the wire, so the only E corresponding to a non-zero E.j is the E inside the wire. So from this and your comments above it seems like you think that E.j≠P.

If E.j≠P then energy is not conserved. The E and B fields have a certain amount of energy density given by (E²+B²)/2. If that energy changes then it must either have gone to EM fields elsewhere as described by \nabla \cdot (E \times B) or it must have gone to matter as described by E.j. There is nowhere else for energy to go. http://farside.ph.utexas.edu/teaching/em/lectures/node89.html

If matter obtains more power P from EM than the integral of E.j then there is energy in the matter which appears out of nowhere without a corresponding decrease in the EM energy. So, your position is incompatible with the conservation of energy.

There is no doubt that the B field provides the torque. You are welcome to draw a diagram if you wish, but it is not under any dispute.

Of course B field provides the torque. I'm glad you have no doubt about that point. When a torque T turns the rotor through an angle of β. is the work done not equal to Tβ?

Now we turn to E.J I don't believe that E.J is zero when E is taken as the composite value considering R & L both. A dot product of 2 vectors does not vanish because the 2 vectors do not coincide. Again, if I take E as the composite value, accounting for IR as well as L*dI/dt, E.J accounts for the power density. Integrating over volume gives power. Outside the wire, I don't believe that the E.J value vanishes.

E.J represents input power V*I, & output power plus heat loss is V*I*PF, where PF = power factor. Energy is conserved using my computation. Power source energizes stator. A portion of the power is converted to heat via I²R. A portion is energizing LI²/2. Then from LI²/2, energy is transferred to the rotor in the form of Tβ. Is all the energy accounted for? Usually the rotor winding still has some LI²/2 energy remaining. This reactive energy is returned to the input source. The next cycle delivers more energy & reactive power etc.

Most ac motors, synchronous or induction type, have a power factor less than unity. A typical value is 0.8 lagging. I see your point. You are taking the "E" in E.J as being only that "E" internal to the wire. Then E.J is what you take to be the total power inputted to the system. You then claim that E.J is the rotor heat loss plus the rotor mechanical power.

I believe that E.J is the total power, but only if "E" is the total E across L as well as R. Visualize a winding. The E field is across the entire coil & includes the voltage drop across R (IR) & L (L*dI/dt). That is the E I refer to. I now understand what you're saying.

So here is my reply. Outside the wire, when we consider total E including inductance, the dot product E.J is not necessarily zero. Please review the dot product math. E & J do not have to occupy the same physical space to have a dot product. I will confirm tonight when I get home with my ref texts. BR.

Claude

 

[Dale]

Of course B field provides the torque. I'm glad you have no doubt about that point. When a torque T turns the rotor through an angle of β. is the work done not equal to Tβ?

See my previous responses on this topic in post 272 and 277.

Outside the wire, I don't believe that the E.J value vanishes.

Really? What is the current density outside the wire?

I believe that E.J is the total power, but only if "E" is the total E across L as well as R. Visualize a winding. The E field is across the entire coil & includes the voltage drop across R (IR) & L (L*dI/dt). That is the E I refer to. I now understand what you're saying.

You seem to be mixing up the E field of Maxwell's equations and the voltage of circuits.

Outside the wire, when we consider total E including inductance, the dot product E.J is not necessarily zero. Please review the dot product math. E & J do not have to occupy the same physical space to have a dot product. I will confirm tonight when I get home with my ref texts.

Please do so, this is flat wrong.

 

[cabraham]

See my previous responses on this topic in post 272 and 277.

Really? What is the current density outside the wire?

You seem to be mixing up the E field of Maxwell's equations and the voltage of circuits.


Please do so, this is flat wrong.

Current density outside wire is zero. That does not make E.J zero. You seem to be confucing dot product with line integral. The path of intagration for E.J is the volume which includes the wires & the magnetic core. Two vectors parallel have a non-zero dot product. They do not have to coincide. E is non-zero outside the wire. J is non-zero inside the wire. Their dot product is non-zero.

Voltage of circuits is merely the line integral of E dot dl.

Your previous responses were that B cannot do work because it does not move which makes no sense. Then you claim E does work but don't mind that E does not move. This is too bizarre.

Claude

 

[Dale]

Current density outside wire is zero. That does not make E.J zero.

Yes, it does. x.0=0 for any vector, x.

You seem to be confucing dot product with line integral. The path of intagration for E.J is the volume which includes the wires & the magnetic core. Two vectors parallel have a non-zero dot product. They do not have to coincide. E is non-zero outside the wire. J is non-zero inside the wire. Their dot product is non-zero.

E.j is a power density at every point in space. It is, in fact, zero everywhere outside the wire. Therefore, there is zero work done on matter outside the wire (since there is no matter there). This should be obvious.

Your previous responses were that B cannot do work because it does not move which makes no sense. Then you claim E does work but don't mind that E does not move. This is too bizarre.

Work is a transfer of energy. According to the equation I posted E transfers energy and B does not.

 

[Dale]

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,...).

It is too bad that you aren't making this argument. It seems like a great argument.

 

[Dadface]

The only approximate constants I see relevant to the E.j discussions are the supply voltage(V)the circuit resistance,the appropriate linear dimensions of the coil circuit and the B field of the magnets(not the resultant B field,this changes because there is a field component due to the current carrying coil as well as the field component due to the magnets)

When the motor starts turning and picking up speed the following changes occur:

1.The back emf starts to increase and the resultant E starts to decrease.

2.The current starts to decrease with a corresponding decrease of j.

3.The resultant B field changes due to:

a.The field set up due to the reducing current flowing through the coil circuit.

b.The coil turning,this resulting in a non stationary and changing B field(changes due to this effect occur also for a constant current and angular velocity)

As has been said before it is the work done against back emf that appears as mechanical work

We could write that at a rotational speed where the resultant E has a value of E' and the current density a value of j' the power density has a value given by:

P=E'.j'

{I think that the whole thing is expressed most easily by the equation I first posted:

VI=EbI+I squared R (VI= input power,EbI= output power and Eb=back emf).Remember that E and I change as the speed changes.}

E' can be written roughly as (V-Eb)/dl and j' as I/dA(l=length A =area).From this E'.J' is given by

p=(VI-EbI)/volume

In other words what E.j stands for depends on what E and j are taken to be

V.j represents input power
Eb.j represents mechanical output power
E'.j' represents power losses

 

[Per Oni]

There’s a lot of talk of time varying B fields to explain work being done.

I want to show you a motor which is in fact an even simpler one then the one of the op. Whereas the op shows a 2-pole motor it is also possible to make an 1-pole motor, the so called homo-polar motor. http://en.wikipedia.org/wiki/Homopolar_motor
There are some lovely u-tube demos:

Now, a 2 pole motor has all the same principles of power output and energy considerations then a 1 pole. Of course reversing of the current each revolution in a 2-pole brings its own interference problems etc as previously has been pointed out. But for a 1 pole there’s no: dI/dt, hence no: U=L dI/dt, and as you can see from u-tube it works perfectly.

 

[Miyz]

It is too bad that you aren't making this argument. It seems like a great argument.

Dale, you sill don't agree that B fields/forces do work on a loop?
If not I'll use simple logic and simple illustrations for you to imagine. I think you're seriously missing something and you're only! At on point! You need to open up the horizon a bit!


You were the 1st who agreed with full confidence now you've change you're opinion and fighting strong against you're 1st one. Umm I need to shack you back to you're original opinion!

 

[Dale]

You need to open up the horizon a bit!

I am sorry, but this is very funny advice coming from you. You are very closed-minded, and have shown no indication of even considering alternative viewpoints.

I am torn on the subject, as should be obvious. I personally would like to be able to say that magnetic fields do work. But the math says that they don't on point charges. So I thought that dealing with non point charges was a loophole, but vanhees71's paper closed that loophole. And looking deeper into the math shows that the power density is E.j for a continuous distribution also. So my loophole is slammed shut.

Nobody else has shown another loophole that holds up, and both you and cabraham seem to avoid the discussion of the energy conservation equation entirely.

 

[andrien]

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

these kinds of simple manipulation are not right.I would like to know about that B.Because so far I know E.J is rate at which electromagnetic field does work on the charges per unit volume and if B is magnetic field acting on charges in that volume,then for the case of an electromagnetic wave shinning on a hydrogen atom of certain frequency it will never be able to ionize the atom because in case of light E.B=0.Electric and magnetic fields are perpendicular to each other in light however strong it's frequency is.