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

Per Oni

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

From 263: 284: Can you please stick a little longer to your own good resolution not to talk for other people?

cabraham

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

DaleSpam

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.

cabraham

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

DaleSpam

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.

DaleSpam

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: http://www.youtube.com/watch?v=3aPQqNt15-o

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

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!

DaleSpam

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

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

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 Im trying to keep up! How could I show any indications far beyond the simplicity I've understood about this matter?

I noticed.

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

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

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

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

[itex]ds = \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}[/itex]

This can be decomposed such that [itex] \ \ \ \ \ 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[/itex]

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

[itex]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[/itex]

This can be broken down to the following equations

[itex]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)[/itex]
[itex]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)[/itex]
[itex]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)[/itex]

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

[itex]{\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[/itex]

And likewise for [itex]F_y[/itex] and [itex]F_z[/itex]

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.

DaleSpam

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.

PhilDSP

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

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?