Exploring the Work Done by Magnetic Fields: A Scientific Inquiry

[Dale]

How did you get zero from "qv x B"?

It is zero from (qv x B).v. This follows from the standard vector algebra identities for the scalar triple product:

$(q \mathbf{v} \times \mathbf{B}) \cdot \mathbf{v}$
$q \mathbf{v} \cdot (\mathbf{v} \times \mathbf{B})$ by commutativity of dot product
$q \mathbf{B} \cdot (\mathbf{v} \times \mathbf{v})$ by circular shift property of scalar triple product
$q \mathbf{B} \cdot (\mathbf{0})$ by anti-commutivity of the cross product
$0$

Basically, the cross product gives a vector perpendicular to v, and the dot product of v and a vector perpendicular to v is 0.

 

[pleco]

No one is arguing that a magnetic field can't change the trajectory of a point charge in motion. It just can't change its SPEED, i.e. it can't change its kinetic energy because the magnetic force on it can't do any work.

Work is force times displacement, speed is not a part of the equation. Anyway, do you think a free electron would not change its speed when passing next to a permanent magnet where this external B field maximum gradient density is not located perpendicularly to is trajectory, but this time more in front or behind it?

 

[WannabeNewton]

You seem to have a very dramatic misunderstanding of the basics of forces. Constraint forces do not do work. This is because constraint forces always act orthogonal to the instantaneous displacement. The magnetic part of the Lorentz force is a constraint. It is given by $\vec{F} = q \frac{d\vec{x}}{dt}\times \vec{B}$ so clearly $\vec{F} \cdot d\vec{x} = 0$. If this is not obvious to you then you should take a step back and review basic vector algebra.

There is no need to confuse yourself by considering magnetism. Does the tension in a string due to an attached mass moving uniformly in a circle do work? Does the normal force on a wheel rolling without slipping down an incline do work? Clearly not as these forces act orthogonal to the direction of motion. Their goal is to keep the particle(s) constrained to move along a certain trajectory hence the name constraint forces; in the case of the wheel the normal force keeps the wheel on the incline and in the case of the string the tension keeps the mass moving in a circle.

 

[pleco]

Bending a trajectory does not do work.

That's the first argument that makes some sense. What equation are you referring to and why would it have precedence over W= F*s?

 

[milesyoung]

Work is force times displacement, speed is not a part of the equation. Anyway, do you think a free electron would not change its speed when passing next to a permanent magnet where this external B field maximum gradient density is not located perpendicularly to is trajectory, but this time more in front or behind it?

Regardless of source and gradient, the magnetic force won't change the speed of the particle.

Here I use only the definition of work and the expression for the magnetic force:
$$W = \int_{t_1}^{t_2} \mathbf{F_\mathbf{mag}} \cdot \mathbf{v} \, \mathrm{d}t = \int_{t_1}^{t_2} \left( q \mathbf{v} \times \mathbf{B} \right) \cdot \mathbf{v} \, \mathrm{d}t = \int_{t_1}^{t_2} 0 \, \mathrm{d}t = 0$$
It's a very simple equation. You're saying this is somehow wrong?

 

[pleco]

It is zero from (qv x B).v. This follows from the standard vector algebra identities for the scalar triple product:

$(q \mathbf{v} \times \mathbf{B}) \cdot \mathbf{v}$
$q \mathbf{v} \cdot (\mathbf{v} \times \mathbf{B})$ by commutativity of dot product
$q \mathbf{B} \cdot (\mathbf{v} \times \mathbf{v})$ by circular shift property of scalar triple product
$q \mathbf{B} \cdot (\mathbf{0})$ by anti-commutivity of the cross product
$0$

Where did you get that .v at the end of the first line, what is it? That's extra. There is charge velocity vector, there is B field vector, and cross product between them, thus force vector is perpendicular to both.

Basically, the cross product gives a vector perpendicular to v, and the dot product of v and a vector perpendicular to v is 0.

It's just a geometrical rotation defined via unit vector, it doesn't change the magnitude. If the force had zero magnitude Ampere's force law wouldn't work.

http://en.wikipedia.org/wiki/Ampère's_force_law

 

[pleco]

You seem to have a very dramatic misunderstanding of the basics of forces. Constraint forces do not do work. This is because constraint forces always act orthogonal to the instantaneous displacement. The magnetic part of the Lorentz force is a constraint. It is given by $\vec{F} = q \frac{d\vec{x}}{dt}\times \vec{B}$ so clearly $\vec{F} \cdot d\vec{x} = 0$. If this is not obvious to you then you should take a step back and review basic vector algebra.

What is significance of F.dx = 0? Dx is not the displacement vector we are interested in. Only displacement in the direction of the force matters.

There is no need to confuse yourself by considering magnetism. Does the tension in a string due to an attached mass moving uniformly in a circle do work? Does the normal force on a wheel rolling without slipping down an incline do work? Clearly not as these forces act orthogonal to the direction of motion. Their goal is to keep the particle(s) constrained to move along a certain trajectory hence the name constraint forces; in the case of the wheel the normal force keeps the wheel on the incline and in the case of the string the tension keeps the mass moving in a circle.

I don't know. What equation are you referring to and why would it have precedence over W= F*s?

 

[Dale]

Where did you get that .v at the end of the first line, what is it?

P=F.v=dW/dt

 

[Dale]

That's the first argument that makes some sense. What equation are you referring to and why would it have precedence over W= F*s?

It doesn't have precedence, it is the time derivative of that equation: P=dW/dt=d(F.s)/dt=F.ds/dt=F.v

Since W=F.s then P=F.v

This is not even EM or fields. This is just very basic Newtonian mechanics. This stuff should be second nature before you venture into EM. There simply is no way that you are prepared to understand EM given these questions.

 

[Dale]

What is significance of F.dx = 0? Dx is not the displacement vector we are interested in. Only displacement in the direction of the force matters.

That is exactly what F.dx means, the force in the direction of the displacement! Wow! You really need to go back and learn basic vectors and Newtonian mechanics before you try to tackle fields and EM.

I am going to go ahead and close the thread. The OP has not checked in recently and we have now wandered off topic into basic mechanics quantities.

reasonhut, if you still have unanswered questions on EM then I encourage you to open another thread that hopefully will not get side tracked as badly.

pleco, I encourage you to open a separate thread on the basic mechanics and vector questions that you have.

EDIT: I decided to reopen this thread to continue with the EM topics for reasonhut. pleco, please open a separate thread for your mechanics questions

 

[USeptim]

Perhaps the magnetic force will not do work on charges because it creates a force orthogonal to the v, however, if v changes the trajectory of charges will change and therefore they can move to a another region with a different potential besides, changing particles trajectory will change the field they generate.

So the magnetic field will cause a work indirectly by changing the electric fields as well as particle’s position.

 

[haruspex]

For the Lorentz force it's all the same whether those are a pair of parallel traveling electrons, or two parallel electron beams, or two current carrying wires, or a single electron moving next to a permanent magnet, or electron beam passing by permanent magnet, or current carrying wire placed next to a permanent magnet. F= qv x B always gives the right answer.

No, the difference is that in a wire the electrons are constrained to keep moving in the same direction unless the wire itself moves. And if the wire does move then the electrons move in a different way than they would in an unconstrained beam.

B field vectors are not the same thing as force vectors, they are perpendicular.

I didn't say they were. I distinguished the field direction, the direction of motion of the electrons, and the direction of the acceleration (the force direction). The three are mutually perpendicular.

Electrons initial drifting direction along the wire is not the displacement we are interested in. We are interested only in lateral component, towards or away from the other wire, and that displacement is naturally in the same direction as Lorentz force acting on those electrons.

You are confusing direction of displacement with direction of acceleration (= direction of force). Think about my stone on a string analogy. As the stone whirrs around, the force is radial, so the acceleration is radial. But there's no actual displacement in the radial direction - the radius does not change. As long as velocity and acceleration are perpendicular there's no speed change and no work done.

 

[Dale]

Perhaps the magnetic force will not do work on charges because it creates a force orthogonal to the v, however, if v changes the trajectory of charges will change and therefore they can move to a another region with a different potential

If they move to a region with a different potential then, by definition, there will be an E field in that direction. So, again, the work will be E.J as stated in Poynting's theorem.

So the magnetic field will cause a work indirectly by changing the electric fields as well as particle’s position.

I am fine with that wording. I think it is appropriate to think of B as indirectly doing work since it does have energy and that energy can be transferred to matter eventually. E.J is the quantity that directly measures the work done, but B can influence both E and J and thereby indirectly do work.

 

[Delta2]

Ehm hm, fine till now it should be clear to all (at least it is to me) that the electric field does the work in cases where laplace force is involved.

However in the case where we have a bar magnet that is attracting a piece of iron, can someone elaborate how the electric field does the work?

 

[OmCheeto]

Ehm hm, fine till now it should be clear to all (at least it is to me) that the electric field does the work in cases where laplace force is involved.

However in the case where we have a bar magnet that is attracting a piece of iron, can someone elaborate how the electric field does the work?

I'm just a layman, but it looks like the argument is that electric charge motion is the cause of magnetism, and therefore, everything results from that.

Magnetism, at its root, arises from two sources:

Electric current (see electron magnetic dipole moment).
Nuclear magnetic moments of atomic nuclei. These moments are typically thousands of times smaller than the electrons' magnetic moments, so they are negligible...

Though, I've been watching these threads for years, and have never figured out who was right or wrong.

I liked Vanadium's comment:

I don't like teaching the meme "magnetic fields do no work." It is true, but it is not useful.
...

I also liked WannabeNewton's analogy:

There is no need to confuse yourself by considering magnetism. Does the tension in a string due to an attached mass moving uniformly in a circle do work? Does the normal force on a wheel rolling without slipping down an incline do work? Clearly not as these forces act orthogonal to the direction of motion. Their goal is to keep the particle(s) constrained to move along a certain trajectory hence the name constraint forces; in the case of the wheel the normal force keeps the wheel on the incline and in the case of the string the tension keeps the mass moving in a circle.

Though, it appears to me that the problem with these threads is that the question is stated as, or evolves into, an abstraction that does not mimic reality.

It's true, WBN's incline does no work. But the incline would not exist unless there were something to incline against. So work was done somewhere, and it involved a system of things, beyond just the incline and wheel.

My guess is that because magnetic monopoles haven't been discovered yet, everything comes about because of the simple motion of charges. And the masses that complete the system, of course. And the interacting virtual photons. And the ...

 

[Dale]

Ehm hm, fine till now it should be clear to all (at least it is to me) that the electric field does the work in cases where laplace force is involved.

However in the case where we have a bar magnet that is attracting a piece of iron, can someone elaborate how the electric field does the work?

A permanent magnet or a magnetized piece of iron each have a J from the magnetism (Ampere's law). As soon as the iron begins to move there is a ∂B/∂t and therefore there is an E (Faraday's law). The work done is J.E.

 

[Dadface]

Just one point for now (which was referred to in post 8 ). When dealing with electric circuits reference is often made to the energy stored in the magnetic fields of inductive parts of circuits. If, for example, the field collapses there is a conversion of energy. Are people here suggesting that the work is not done at the expense of the collapsing magnetic field? If so what is the source of the work and what terminology should be used when teaching this topic?

 

[cabraham]

A permanent magnet or a magnetized piece of iron each have a J from the magnetism (Ampere's law). As soon as the iron begins to move there is a ∂B/∂t and therefore there is an E (Faraday's law). The work done is J.E.

Faraday's Law? The force lifts the iron while speed is zero, meaning that Faraday induction is zero. Cause must precede effect. Fm does the work lifting the magnet. Fm acts upward, object displaces upward. F*u is non-zero.

Claude

 

[USeptim]

Just one point for now (which was referred to in post 8 ). When dealing with electric circuits reference is often made to the energy stored in the magnetic fields of inductive parts of circuits. If, for example, the field collapses there is a conversion of energy. Are people here suggesting that the work is not done at the expense of the collapsing magnetic field? If so what is the source of the work and what terminology should be used when teaching this topic?

The EM force is always computed on the charges, therefore the power is J•E.

However, the EM field evolves with time and so it does the density of energy of both electric and magnetic fields but the transfer of energy-momentum is always evaluated only on charges, as the force done by the EM field on charges.

In electrostatics it’s demonstrated that for a continuous charge distribution the integral of the charges’ potential energy is equal to the energy stored in the electric field (there is no magnetic field of course). So it suggests that the charges’ potential is stored on the field.

If the same is applied to the general case (electrodynamics), the changes of the energy stored in the field would be consequence of the changes in the charges’ four-potential and so we would not have to worry about them.

However I haven’t seen any that charges four-potential energy is stored on the fields in electrodynamics.

 

[johana]

Faraday's Law? The force lifts the iron while speed is zero, meaning that Faraday induction is zero. Cause must precede effect. Fm does the work lifting the magnet. Fm acts upward, object displaces upward. F*u is non-zero.

Claude

Which is stronger force between magnet 1 and magnet 2, electric or magnetic? Are they both acting in the same direction?

 

[Dale]

Faraday's Law? The force lifts the iron while speed is zero, meaning that Faraday induction is zero

Which is the exact correct value it should have since the work done while the speed is 0 is also 0. Work is E.J is 0.

Fm does the work lifting the magnet. Fm acts upward, object displaces upward. F*u is non-zero

E.J is non zero in that scenario also.

E.J includes all of the work done by EM on matter including mechanical work done on currents, mechanical work done on charges and ohmic work on conductors.

 

[Dale]

Are people here suggesting that the work is not done at the expense of the collapsing magnetic field? If so what is the source of the work and what terminology should be used when teaching this topic?

I am not suggesting that. I would either state that the energy in the B field can be used to do work indirectly through the B fields influence on E and J or I would say that it is part of the work included in E.J.

 

[johana]

A toy car is moving in x direction with uniform velocity, it carries electrically charged ping-pong ball on a spring. Under the table we can turn on uniform magnetic or electric field separately. What is the difference between electric and magnetic force pulling down the ball in y direction? How can electric force do work when only magnetic field is turned on under the table?

 

[Dale]

I have not worked this problem or an equivalent problem in detail, so I cannot say for sure. If I had to guess then I would suspect that the energy would actually come from the KE of the ball, not the EM field at all.

However, if there is any work done by the fields then there must be E.J, per Poyntings theorem. The ball itself provides both E and J, so that is where I would look if the KE is insufficient.

 

[johana]

I have not worked this problem or an equivalent problem in detail, so I cannot say for sure. If I had to guess then I would suspect that the energy would actually come from the KE of the ball, not the EM field at all.

However, if there is any work done by the fields then there must be E.J, per Poyntings theorem. The ball itself provides both E and J, so that is where I would look if the KE is insufficient.

- "It is often claimed that the magnetic force can do work to a non-elementary magnetic dipole, or to charged particles whose motion is constrained by other forces, but this is incorrect."
http://en.wikipedia.org/wiki/Magnetic_field


I guess that's one of the hardest examples. Wikipedia links to this paper for explanation:
http://academic.csuohio.edu/deissler/PhysRevE_77_036609.pdf

Can you summarize in simple terms what their explanation really is?

 

[Jano L.]

Faraday's Law? The force lifts the iron while speed is zero, meaning that Faraday induction is zero.
Claude

Which is the exact correct value it should have since the work done while the speed is 0 is also 0. Work is E.J is 0.

In this matter, cabraham is right. If the iron body undergoes finite displacement in magnetic field in the direction of attraction arbitrarily slowly, finite work is done on the iron. This cannot be done by arbitrarily small force due to the arbitrarily small induced electric field, but has to be done by a force with magnitude comparable to that of the magnetic force. The power is almost zero if the motion is very slow, but the net work done is not and so the working force is not zero either.

Since magnetic force does not work, the work done has to be due to electrical (other than induced) or non-electromagnetic forces of the moving charged particles acting on the rest of the iron of the mentioned order of magnitude.

In the presence of current-conducting body in magnetic field, the charge carriers are being pushed away from their natural path by magnetic force. The charge carriers translate this push via electric or other internal forces on the rest of the body (nuclei + non-conducting electrons). Since the rest of the body also moves, net work can be done on it by these internal forces.

In short, the elusive working forces are internal forces. Probably that's why the question of work in magnetic field is so confusing and keeps threads going on.

 

[Dale]

In this matter, cabraham is right. If the iron body undergoes finite displacement in magnetic field in the direction of attraction arbitrarily slowly, finite work is done on the iron. This cannot be done by arbitrarily small force due to the arbitrarily small induced electric field, but has to be done by a force with magnitude comparable to that of the magnetic force. The power is almost zero if the motion is very slow, but the net work done is not and so the working force is not zero either.

E.J is not force, it is power. In the scenario you proposed the power is arbitrarily low, as is E.J. It correctly accounts for all the work, Ohmic and mechanical.

As I have said several times in this thread and others, I don't have a problem with saying that magnetic fields do work indirectly through their contribution to E.J. Magnetic fields have energy and that energy can be transferred to matter. Per Poyntings theorem all of the work done on matter is E.J, but due to their relationships B definitely contributes to E.J.

 

[Dale]

reference is often made to the energy stored in the magnetic fields of inductive parts of circuits. If, for example, the field collapses there is a conversion of energy.

If you look at Poyntings theorem there are exactly 3 possibilities if the magnetic field collapses. First, the energy lost from the B field could go into the E field, keeping the energy density constant. Second, the energy could move to another location in the field through ∇.ExH. Third, the energy can go to matter through E.J.

 

[Delta2]

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.

Since the orbiting electrons are quantum mechanically binded to the nuclei (and the external magnetic field doesn't provide energy to them to make em jump in higher energy) but their trajectory is altered, the electrons "drag" the nuclei along their trajectory but this dragging is happening with the help of the electrostatic Coulomb force between the orbiting electrons and the nuclei. So it is the electrostatic coulomb force that does the work.

 

[johana]

the electrons "drag" the nuclei along their trajectory but this dragging is happening with the help of the electrostatic Coulomb force

Doing it "together" still means the magnetic force does at least some if not most of the work. I don't think the two forces even act in the same direction. In what direction electrons want to move due to magnetic force and in what direction is electric force trying to drag them?

 

[Jano L.]

E.J is not force, it is power. In the scenario you proposed the power is arbitrarily low, as is E.J. It correctly accounts for all the work, Ohmic and mechanical.

As I have said several times in this thread and others, I don't have a problem with saying that magnetic fields do work indirectly through their contribution to E.J. Magnetic fields have energy and that energy can be transferred to matter. Per Poyntings theorem all of the work done on matter is E.J, but due to their relationships B definitely contributes to E.J.

My point was that E in E.J is not induced electric field, because induced electric field can be made arbitrarily small and the work done according this expression would be arbitrarily small too. If we use this expression at all, E is more like static electric field of the electrons acting on the nuclei (or some other internal working non-electromagnetic force).

 

[Jano L.]

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.

Since the orbiting electrons are quantum mechanically binded to the nuclei (and the external magnetic field doesn't provide energy to them to make em jump in higher energy) but their trajectory is altered, the electrons "drag" the nuclei along their trajectory but this dragging is happening with the help of the electrostatic Coulomb force between the orbiting electrons and the nuclei. So it is the electrostatic coulomb force that does the work.

I agree, your explanation is similar to mine. Work is done by internal forces of the electrons acting on the nuclei (or in macroscopic parlance, by forces of the current carriers acting on the rest of the body).

 

[johana]

Electron pulls nucleus. What pulls electron?

 

[Delta2]

Electron pulls nucleus. What pulls electron?

The magnetic field force vxB pulls the electron but that pull doesn't do work as Vanhees71 explains in post No2 at the start of thread. Allow me to say Vanhees explanation is "rock solid" as to why magnetic field cannot do work on matter.

However i have to admit that my explanation on how the work is done by the electric coulomb force is abit vague and incomplete. We need a treatment on how an external constant magnetic field changes the wave function of an orbiting electron around a nucleus. Such a treatment requires deep knowledge of Quantum Mechanics and possibly QED, i am not very good on neither of those unfortunately :(.

 

[Dale]

My point was that E in E.J is not induced electric field, because induced electric field can be made arbitrarily small and the work done according this expression would be arbitrarily small too.

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.