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

[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.

 

[Miyz]

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.

Well... Thats because my knowledge about this matter is basic you and Claude are speaking an entirely different language I do not understand what you are all saying although I am trying to keep up! How could I show any indications far beyond the simplicity I've understood about this matter?

I am torn on the subject, as should be obvious.

I noticed.

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.

Well I didn't avoid that I just don't have any clue as I stated above. What makes sense to me the most is looking at the forces that actually are doing work. However, I speak nothing I do not understand of. I honestly don't know anything about this and its new for me. I'm only catching up and reading you're posts.

But again and again. I don't look at the energy conservation alone but... I look at the forces as well.
Because that matter is confusing as hell. I'd rather study it step by step to finally understand and discuss about it. Till then I'm a spectator.

It's wise for me to admit I know nothing of this matter and just let the experts do their work instead. If I did you'd seen me post a lot. All my post on E.J are nothing useful why? As I said before its new for me.

 

[cabraham]

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.

But E & B are from interacting loops. Maybe E¹, E², B¹, & B² are more appropriate.

E.J is no problem, & is consistent w/ what I've presented. You acknowledge B as producing torque, yet you deny its work contribution. For the 4thtime please draw a sketch showing the fields doing work. All work is ultimately done by input power source. We have 2 loops. Each loop has E & B. It is the interaction that makes motors run. An e/m have not ionizing an H atom is too simplistic dealing w/ motors. Please draw us a pic. Thanks.

Claude

 

[cabraham]

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

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.

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

Look up dot product of 2 parallel vectors, non-zero if both are non-zero.

Your equation could be expressed in terms of J & B, so what? It doesn't prove that E is irrelevant.

But the E that transfers energy must include inductive, external to wire. If the stator winding is powered from 120 V rms ac, let's say the current is 1.0 amp, the winding ersistance is 1.0 ohm. Te voltage drop inside the wire is 1.0 volt, & the other 119 volts appears across the inductance as well as core loss, leakage reactance, etc.

You claim that E.J inside the wire accounts for all work. But if the wire is superconduction E is zero inside. You should reexamine that whole theory. I say with firm conviction that E.J must be based on total E, not just inside wire. We seem to have come to a stand still since you insist E is inside, I say E is both outside & inside. Until this is resolved it is pointless to argue. BR.

Claude

 

[PhilDSP]

I think we can show that the work done is clearly zero. The general expression for work is

$W = \int_{\ i}^f F \cdot ds \ \ \ \ \$ where F is the applied force and ds is the displacement from initial $i$ to final $f$ position

$ds = \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}$

This can be decomposed such that $\ \ \ \ \ W = {\int_{\ x_i}}^{x_f} F_x \ dx + {\int_{\ y_i}}^{y_f} F_y \ dy + {\int_{\ z_i}}^{z_f} F_z \ dz$

According to the Lorentz force equation the force produced by the magnetic field is

$F \ \ = \ \ q \ (v \times B) \ \ = \ \ q \ \langle v_z B_y - v_y B_z, \ \ v_x B_z - v_z B_x, \ \ v_y B_x - v_x B_y \rangle$

This can be broken down to the following equations

$F_x \ = \ q(v_z B_y - v_y B_z) \ = \ q\ (\frac{\partial z}{\partial t} B_y - \frac{\partial y}{\partial t} B_z)$
$F_y \ = \ q(v_x B_z - v_z B_x) \ = \ q\ (\frac{\partial x}{\partial t} B_z - \frac{\partial z}{\partial t} B_x)$
$F_z \ = \ q(v_y B_x - v_x B_y) \ = \ q\ (\frac{\partial y}{\partial t} B_x - \frac{\partial x}{\partial t} B_y)$

Taking the integral with respect to x in the first equation gives

${\int_{\ x_i}}^{x_f} F_x \ dx \ \ = \ \ {\int_{\ x_i}}^{x_f} q(\frac{\partial z}{\partial t} B_y - \frac{\partial y}{\partial t} B_z) \ dx \ \ = \ \ 0$

And likewise for $F_y$ and $F_z$

So the work done by the magnetic field is zero (because only the direction of the velocity changes not its magnitude). However, we're assuming in using the Lorentz force equation that it applies to an instantaneously steady magnetic field. There are no parameters in the force equation for dealing with a changing magnetic field.

 

[Dale]

Look up dot product of 2 parallel vectors, non-zero if both are non-zero.

True, but j is 0 outside of the wire, therefore the dot product is 0 outside of the wire.

I say with firm conviction that E.J must be based on total E, not just inside wire. We seem to have come to a stand still since you insist E is inside, I say E is both outside & inside.

Weren't you going to look at a textbook last night?

A field at a location far from the wire cannot deliver power to a wire. That energy must first be transported to the location of the wire, and then it can be delivered to the wire.

It is obviously nonsense to claim that the E field a light year away from a wire can do any work on the wire now. For the same reason that is nonsense, it is also nonsense that the field a foot away or a millimeter away can do any work on the wire now. Only the fields at the location of the matter at a given time can deliver power to the matter at that time.

 

[PhilDSP]

Another thing to consider is that work and torque share the same dimensions and measurement units: $\frac{ML^2}{T^2}$ with measurements in $N \cdot m$.

In a pure numeric sense they're equivalent, but in terms of semantics work refers to force exercised over a linear displacement while torque refers to force exercised over a rotational displacement.

Can one simply say that a magnetic field produces torque? In order to use it to do work you need to find a mechanical means of converting the rotary motion into linear motion?

 

[gabbagabbahey]

I say with firm conviction that E.J must be based on total E, not just inside wire.

The inner product between two vector fields is a scalar field; its value, just like the two vector fields, depends on position. If either of the two vector fields is zero at a specific location, their inner product will also be zero there.

Maxwell's equations, and the Lorentz Force Law both deal with vector fields. If you don't understand how to take the inner product of two vector fields, you don't fully understand Maxwell's equations.

Now, as for whether Magnetic field/forces do work on any classical system, the answer is fundamentally no. This assumes only that you take Maxwell's equations and the Lorentz force law as the basis for classical electrodynamics (which 99.9% of the current physics community likely does).

In many cases, the work done on an object depends on the value of the magnetic field applied to it, but this does not mean that the magnetic field is directly doing the work (and it isn't!).

 

[Dotini]

Now, as for whether Magnetic field/forces do work on any classical system, the answer is fundamentally no. This assumes only that you take Maxwell's equations and the Lorentz force law as the basis for classical electrodynamics (which 99.9% of the current physics community likely does).

Do the followers of Gauss-Ampere-Weber make up the remaining 0.1%?

Respectfully submitted,
Steve

 

[gabbagabbahey]

Do the followers of Gauss-Ampere-Weber make up the remaining 0.1%?

Respectfully submitted,
Steve

I wouldn't count them as part of the physics community. I was thinking more along the lines of actual physicists who have just spent too much time staring at equations and now believe that all of physics can be described by tiny Leprechauns made of Ether throwing God particles at each-other in a giant game of dodge-ball.

/sarcasm

 

[Dadface]

Now, as for whether Magnetic field/forces do work on any classical system, the answer is fundamentally no. This assumes only that you take Maxwell's equations and the Lorentz force law as the basis for classical electrodynamics (which 99.9% of the current physics community likely does).

In many cases, the work done on an object depends on the value of the magnetic field applied to it, but this does not mean that the magnetic field is directly doing the work (and it isn't!).

I don't think that a field of any type does work but a force does.When F=Bev is summed for all of the current carrying electrons in a wire at 90 degrees to a B field the total force is given by F=BIL.If free to move in a vacuum these electrons follow circular paths and no work is done.Within the confines of the wire,however,any resultant motion of the whole conductor is in the direction of the force.I think it's safe to say that movement of the wire results in work being done.But is it being suggested that F=BIL,which comes from summing F=Bev is not a magnetic force?

 

[cabraham]

The inner product between two vector fields is a scalar field; its value, just like the two vector fields, depends on position. If either of the two vector fields is zero at a specific location, their inner product will also be zero there.

Maxwell's equations, and the Lorentz Force Law both deal with vector fields. If you don't understand how to take the inner product of two vector fields, you don't fully understand Maxwell's equations.

Now, as for whether Magnetic field/forces do work on any classical system, the answer is fundamentally no. This assumes only that you take Maxwell's equations and the Lorentz force law as the basis for classical electrodynamics (which 99.9% of the current physics community likely does).

In many cases, the work done on an object depends on the value of the magnetic field applied to it, but this does not mean that the magnetic field is directly doing the work (and it isn't!).

Good grief, you think I don't how to take the inner product of 2 vectors? Just who are you? Physicist, EE? How much education - Ph.D., MS, BS? You talk like I'm an 18 year old just out of HS. If I've erred, show me where & offer correction. You state your position that mag fields do no work, & offer nothing in terms of proof.

You are just 1 more talker who says "Claude you're wrong!" but then fails to follow through & show why I'm wrong. Prove your case, until you do all you offer is emoty word.

Clauide

 

[gabbagabbahey]

I don't think that a field of any type does work but a force does.

Force is a field.

When F=Bev is summed for all of the current carrying electrons in a wire at 90 degrees to a B field the total force is given by F=BIL.If free to move in a vacuum these electrons follow circular paths and no work is done.Within the confines of the wire,however,any resultant motion of the whole conductor is in the direction of the force.

Yes, this is due to the fact that there are other internal forces at play inside the wire. In this case, the only forces that directly contribute to the work on the wire are the internal electric forces that maintain the current (these of course are created by a battery or other power source providing a potential difference along the wire. cf. Example 5.3 from Griffith's Introduction to Electrodynamics 3rd edition)

I think it's safe to say that movement of the wire results in work being done.But is it being suggested that F=BIL,which comes from summing F=Bev is not a magnetic force?

The general idea is that since all of Classical Electrodynamics (including all composite force laws such as F=IBL) can be derived from The Lorentz Force Law, Maxwell's equations and some assumptions about the composition of certain macroscopic bodies (like a bar magnet, etc), one can treat $\mathbf{F}_{ \text{mag} } = q \mathbf{v} \times \mathbf{B}$ (the magnetic part of the Lorentz force law) as being the fundamental magnetic force and $\mathbf{F}_{ \text{e} } = q \mathbf{E}$ as being the fundamental electric force. All other EM forces are treated as composites. So, when one says that magnetic forces do no work, one is unambiguosly referring to $\mathbf{F}_{ \text{mag, net } } = \int dq \mathbf{v} \times \mathbf{B}$ as the magnetic force on an object (where the integral is over the object's volume/charge distribution)

 

[gabbagabbahey]

Good grief, you think I don't how to take the inner product of 2 vectors? Just who are you? Physicist, EE? How much education - Ph.D., MS, BS? You talk like I'm an 18 year old just out of HS. If I've erred, show me where & offer correction. You state your position that mag fields do no work, & offer nothing in terms of proof.

You are just 1 more talker who says "Claude you're wrong!" but then fails to follow through & show why I'm wrong. Prove your case, until you do all you offer is emoty word.

Clauide

I am only going by your previous posts.

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

As Dale Spam has said, outside the wire the current density $\mathbf{J}(\mathbf{r})$ is zero, so the dot product of it with any vector field will also be zero there. Do you really need me to prove this to you?

 

[cabraham]

True, but j is 0 outside of the wire, therefore the dot product is 0 outside of the wire.

Weren't you going to look at a textbook last night?

A field at a location far from the wire cannot deliver power to a wire. That energy must first be transported to the location of the wire, and then it can be delivered to the wire.

It is obviously nonsense to claim that the E field a light year away from a wire can do any work on the wire now. For the same reason that is nonsense, it is also nonsense that the field a foot away or a millimeter away can do any work on the wire now. Only the fields at the location of the matter at a given time can deliver power to the matter at that time.

I'll find the textbook. But here is a flaw in your reasoning. If only the interior of the wire matters, E is zero inside a superconductor, herein SC. So E.J = 0, & you insist E.J = 0 outside the wire. So we have 0 power at the inpout hence 0 work done. But SC motors have been built & affirmed. They draw current from an input voltage source, store energy in fields, tranfer enrgy to output as rotor spins. Yet E.J = 0 for a SC.

How do you explain that one? Remember that the energy in a motor that transfers from stator to rotor is chiefly in the air gap. If the energy was all confined to the wire, how does it transfer? The stator L has a flux which is in the stator iron core, links to the rotor iron core via the air gap. This flux is energy, LI²/2. This energy does not appear in the E.J product inside the wire.

An example which is simpler to visualize is a 120 volt xfmr secondary driving a heater. The secondary winding resistance is 0.10 Ω, & the heater R is 11.9 Ω. The total is 12 Ω. Current is 10 amps. The power inside the secondary winding is 1.0V * 10A = 10W. If we only consider the inside of the sec winding, we get 10W, instead of 1200W total. Why is that?

The voltage across the sec terminals depends on path taken. Outside the Cu it is 119V, inside it is 1.0V. To get the total power you need to consider both. When both are accounted for we get 10 + 1190 = 1200W, the right answer. Once again we appear to agree on almost everything but for the inside/outside question. I will search for the text & post. BR.

Claude

 

[gabbagabbahey]

I'll find the textbook. But here is a flaw in your reasoning. If only the interior of the wire matters, E is zero inside a superconductor, herein SC.

The applied magnetic field will penetrate the superconductor up to the London penetration depth. The E-field will also be, in general, non-zero up to that depth.