Exploring the Work Done by Magnetic Fields: A Scientific Inquiry

[cabraham]

This is incorrect. The work done according to E.J is not arbitrarily small in this scenario. E.J is power. If you make E arbitrarily small then the time becomes arbitrarily large so that an arbitrarily small power is necessary to do the same finite work.

E.J is always correct, as shown by Poynting. Neither cabraham's 0 E.J nor your arbitrarily small E.J are counter examples. In both cases E.J is the correct value, I.e. the power is in fact 0 or arbitrarily small, and E.J correctly reflects that.

But E*J is a scalar and only describes total work supplied by power source. It does not explain work done on current loop which is a magnetic dipole, nor on work done lifting object.

Here is an illustrative example as to why E*J is a vague indicator of total work but cannot explain specific work on loop (or rotor of a rotating machine).

We have 2 identical induction motors. One is attached to a load, one has it rotor locked. The 1st motor is powered from the 120V/60Hz wall outlet and drives a load. The current is measured at 1.0 amp. The power is 120 watts (120V*1.0A=120W). The 2nd motor is rotor locked, and is powered from a variac connected to wall outlet. Being locked, at only 4.0 volts the current is 30 amps. But the power is still 120 watts just as in the case of the 1st motor.

In both cases E*J is 120 W per volume, both being identical in volume means that they are at equal input power. E*J is the total input power being converted but E*J does not itemize where the power is going. Motor 1 processes 120W and converts most of it to mechanical power in the form of rotor shaft torque times shaft speed. Some work is done energizing stator and rotor loops, some is lost as I²R and friction loss. Most of the power is transferred to the rotor.

Motor 2 has the same E*J value, but the rotor has 0 kinetic energy. All 120 watts is converted to heat.

To determine the work done spinning the rotor, we need to know more than just the E*J power. As I said many times, all work is ultimately done by the power source. Fields in the stator and rotor regions do work, giving up energy in the process. This energy is replenished by the source via E*J. Clearly E*J is not directly driving the rotor.

Claude

 

[Dale]

But E*J is a scalar and only describes total work supplied by power source. It does not explain work done on current loop which is a magnetic dipole, nor on work done lifting object.

I think that you have a misunderstanding about what E.J is mathematically. E.J is a scalar field which represents the power density of EM work at each point in space. It is not simply an overall number at the power source.

If energy is being transferred at a loop then E.J is non-zero at that loop.

E*J is the total input power being converted but E*J does not itemize where the power is going.

I am not sure if you are saying something correct here or not.

If by "where the power is going" you mean "is the power going here or there" then what you said is wrong. E.J is a scalar field and describes the power density at each location, specifically in locations where E.J is high then a lot of power is being transferred to matter at that location, and where E.J is low then very little is being transferred at that location.

If by "where the power is going" you mean "is the power going to Ohmic losses or mechanical work" then what you said is correct. E.J is the total power density and does not distinguish between Ohmic work or mechanical work. Both the Ohmic work and the mechanical work are parts of E.J, and the energy they transfer to matter is through their respective contributions to E.J.

If you would like a derivation of that, then I can certainly post it.

Clearly E*J is not directly driving the rotor.

Any power transferred at the rotor is given by E.J at the rotor. It doesn't matter if it is purely resistive or purely mechanical or some mix. If there is EM power transferred to the rotor in any form then there is E.J at the rotor in exactly the amount that covers all of the power transferred.

 

[johana]

In my opinion what is happening in the case of a bar magnet and iron is that the magnetic field of the bar magnet exerts force vxB on the orbiting electrons of the iron nuclei which does no work but alters their trajectory.

This example contradicts the idea work can not be done just because the force is perpendicular to velocity.

 

[Dale]

This example contradicts the idea work can not be done just because the force is perpendicular to velocity.

$\int f\cdot v \; dt =0$ if f is perpendicular to v. No example can contradict that without some other mistake which invalidates the example.

 

[Jano L.]

Electron pulls nucleus. What pulls electron?

External magnetic force and the forces due to nuclei.

 

[johana]

$\int f\cdot v \; dt =0$ if f is perpendicular to v. No example can contradict that without some other mistake which invalidates the example.

That's only for x direction. The force acts in y direction and displaces the ball in y direction. This displacement ds happens over a period of time t, so vertical velocity Vy = ds/t, and F.Vy > 0.

 

[johana]

External magnetic force and the forces due to nuclei.

Do electric and magnetic forces both act in the same direction? Which force acts in the direction two permanent magnets are moving towards each other?

 

[Dale]

That's only for x direction. The force acts in y direction and displaces the ball in y direction. This displacement ds happens over a period of time t, so vertical velocity Vy = ds/t, and F.Vy > 0.

Then it isn't the magnetic force because the magnetic force doesn't work that way.

 

[Jano L.]

This is incorrect. The work done according to E.J is not arbitrarily small in this scenario. E.J is power. If you make E arbitrarily small then the time becomes arbitrarily large so that an arbitrarily small power is necessary to do the same finite work.

If J is total current density, I agree that the quantity $\int\int E.JdVdt$ is not small. My example was about calculating work done on the non-conducting rest of the body only. I'm sorry I did not clarify that.

The problem with talking about total power being E.J (with E being total value of induced electric field and J being total current density) is that this is only warranted from mechanics when the same field E gives accurately the net working force both on the current-forming charges ($J_{cur}$) and the rest of the body (nuclei...) ($J_{rest}$).

However, the net working force on the rest of the body cannot be given by electric field E, because it would imply the work done on this body would be

$$
\int\int E.J_{rest}\,dVdt.
$$
In this integral, both of the factors in the integrand are proportional to the (arbitrarily small) velocity of the body and thus net work done would be arbitrarily small (integration cancels only one of them). The resulting value of work would come out wrong; in fact, the non-conducting rest of the body gets finite amount of work independently of the speed of displacement in magnetic field (this work can be extracted and measured by mechanical forces as in electromotor).

Similarly, assuming E gives net working force on the current forming charges, the work done on them would be

$$
\int\int E.J_{cur}\,dVdt
$$
which has only one factor (E) proportional to velocity of the body and thus comes out finite as you say, but actually, this is again wrong; we know that the net work done on the current-forming charges in the body is always negligible since these almost do not change their (negligible) kinetic energy - they just keep circling around. It is all the other way around; it is the net work done on the rest of the body that should come out as finite even for quasi-static motions.

So, induced electric field is not the one to use in the formula $E.J_{rest}$ for work on the non-conducting part of the body and thus there is no mechanical justification for using it to calculate net work done on the body by

$$
E(J_{cur} + J_{rest}) = E.J
$$
either. This formula can still give power correctly if some other quantity is put for E. It has to be non-vanishing even for quasi-static motions and it has to be such that it gives close to zero work on the current forming charges.

One familiar kind of field that seems to have these attributes is static electric field. It does not depend on velocity and it gives zero net work for circular motion of charges. Or it can be some non-electromagnetic force holding the system together.

Of course, the question arises, where is this new force from. But the answer is easy - there are always forces between the current-forming charges and the rest of the body since they keep together - the "constraint" forces. When the body stands still these internal forces do no work, but when it moves, they may work on the body (internal forces may do net work even in mechanics).

E.J is always correct, as shown by Poynting.

The expression E.J and the Poynting theorem (the equation) may be mathematically correct, but whether the expression E.J correctly gives the work done on the matter per unit time and unit volume is a more difficult question. That this is so is assumed in the derivation of the energy conservation theorem in EM theory; it does not follow from the Poynting theorem, which is a statement on behaviour of the fields, with no implication on the work or energy of matter on its own.

 

[Dale]

My example was about calculating work done on the non-conducting rest of the body only. I'm sorry I did not clarify that.

EM fields don't ever do any work on an uncharged non-conducting body.

The problem with talking about total power being E.J (with E being total value of induced electric field and J being total current density) is that this is only warranted from mechanics when the same field E gives accurately the net working force both on the current-forming charges ($J_{cur}$) and the rest of the body (nuclei...) ($J_{rest}$).

You seem to be confusing force and power. They are two different things. E does not have to give the "net working force" for E.J to be the total power density. Most of the rest of your post is irrelevant because it follows this same mistake.

Look, E.J is the total power density of work done on matter. It comes from Poynting's theorem and it is derived from Maxwell's equations and the Lorentz force equation. It applies any time that those equations apply, i.e. for any classical EM scenario. You don't have do investigate "net working forces" or any of the other red herrings you are looking into. You know from first principles that if the system obeys the laws of classical EM then the work done on matter is E.J.

However, I do accept your criticism that induced E per Faraday's law may be insufficient to account for the total E.J. Maxwell's equations are larger than Faraday's law. As you indirectly mention, you can easily get E fields from the Hall effect and so forth, so you need to consider all of EM, not just part of it, in order to correctly calculate E. That is fine. The point is that even in such a scenario there are E fields and there is current density.

The expression E.J and the Poynting theorem (the equation) may be mathematically correct, but whether the expression E.J correctly gives the work done on the matter per unit time and unit volume is a more difficult question. That this is so is assumed in the derivation of the energy conservation theorem in EM theory; it does not follow from the Poynting theorem, which is a statement on behaviour of the fields, with no implication on the work or energy of matter on its own.

Do you have any current professional reference for this? Frankly, this sounds completely speculative. The math of Poynting's theorem proves that there is a conserved quantity with units of energy density. You can get that assuming only Maxwell's equations and the Lorentz force law. As far as I know no additional assumptions are made in the derivation, it is just manipulation of those equations.

Once you have Poynting's equation then it is merely a matter of interpreting the meaning of the terms. E.J is energy which leaves the field and does not go elsewhere in the field and is the only term involving matter, so what else could it possibly represent other than energy that goes into matter? Be sure to provide suitable references for any claim you make.

 

[johana]

Then it isn't the magnetic force because the magnetic force doesn't work that way.

Fm = qv x B. If the charge is moving in x direction to the right, and magnetic field is in z direction out of the page, then the force will act downwards in y direction. Velocity in x direction is a part of the cause, velocity in y direction is the effect.

 

[Dale]

Fm = qv x B.

Exactly. Which is why this:

F.Vy > 0.

Cannot be a magnetic force.

Don't get confused by your picture. Actually work out the math. Any energy that goes into the spring must come either from the KE or from external mechanical work.

 

[Jano L.]

That doesn't make any sense at all. EM fields don't ever do any work on an uncharged non-conducting body.

By the "non-conducting rest of the body" I mean all particles of the body minus the charged particles of conduction current. Since these are negatively charged, the "non-conducting rest of the body" is charged positively.

Look, E.J is the total power density of work done on matter.

Why do you think that? (I assume you're not taking that as a postulate).

To find power transferred to matter, we should look into definition of power. Total power given to matter in unit volume $V$ as defined in mechanics is

$$
\sum_{k\in V} \mathbf F_{-k} \cdot \mathbf v_k,
$$

where $\mathbf F_{-k}$ is total force acting on the particle $k$ in volume $V$ and $\mathbf v_k$ is velocity of this particle.

True, if all the forces $\mathbf F_{-k}$ can be expressed as
$$
\mathbf F_{-k}=q_k\mathbf E + q_k \frac{\mathbf v_k}{c} \times \mathbf B...(*)
$$
where $\mathbf E,\mathbf B$ are macroscopic electric and magnetic field throughout the volume, the power can be re-expressed as
$$
\mathbf E\cdot \mathbf j,
$$
where $\mathbf j = \sum_k q_k \mathbf v_k$ defines current density. For example, rarified gas of charged particles in external electric field in vacuum satisfies this.

However, there are many situations in macroscopic theory when the total force acting on the particle is not given by the formula (*) with macroscopic fields, but considerably deviates from it. For example, in condensed matter there are internal forces holding the particles of the body together; or chemical electromotive forces enabling flow of electric currents. These are complicated internal forces that are not determined by the macroscopic electromagnetic field $\mathbf E,\mathbf B$.

In such cases, even if total force on the unit volume was given by
$$
\mathbf F = \rho \mathbf E + \frac{\mathbf j}{c} \times \mathbf B,
$$
and its product with average velocity of the particles in the volume gave
$$
\mathbf E\cdot \mathbf j,
$$
this does not mean this product would give net power, because net power is not defined as net force times average velocity

$$
\mathbf F\cdot \langle \mathbf v_k\rangle,
$$
but as sum of individual powers of the particles
$$
\sum_{k\in V} \mathbf F_{-k} \cdot \mathbf v_k.
$$
In extreme case, this means one can have net power transferred to matter even if $\mathbf E\cdot \mathbf j$ is zero. This seems to be the case, for example, when two magnets approach each other quasi-statically by mutual attraction or when when matter is heated by thermal radiation, for example. Work on matter does not seem to require non-zero $\mathbf j\cdot \mathbf E$, not only when the work is due to contact forces, but also when it is due to electromagnetic forces.

[E.J gives net power] It comes from Poynting's theorem and it is derived from Maxwell's equations and the Lorentz force equation. It applies any time that those equations apply

The Lorentz force equation gives net electromagnetic force on matter in unit volume as
$$
\rho\mathbf E + \mathbf j/c\times \mathbf B.
$$
I do not think we can derive your statement about E.J from this force law, since $\mathbf E$ does not appear in the combination $\mathbf j\cdot \mathbf E$ but in the combination $\rho \mathbf E$. Even if we multiply by velocity of the matter in the volume $\mathbf v$, the product $\rho \mathbf v$ is not generally equal to current density $\mathbf j$ since the former can be zero while the latter is not, such as in current-conducting metal.

You don't have do investigate "net working forces" or any of the other red herrings you are looking into.

I believe this is the point of the discussion - to understand which force does work on which body. Some people do not like to discuss this question, probably because the standard expressions like E.J are not sufficient to answer it. Luckily we do not need to confine ourselves to this expression but can use knowledge of mechanics - including the concept of working forces - to address it.

However, I do accept your criticism that induced E per Faraday's law may be insufficient to account for the total E.J. Maxwell's equations are larger than Faraday's law.

Yes, that was one of my points.

 

[Dale]

In extreme case, this means one can have net power transferred to matter even if $\mathbf E\cdot \mathbf j$ is zero.

This is expressly contrary to Poynting's theorem.

I asked for references which you have not provided. This thread is closed in order to avoid unnecessary bans.