Is the Magnetic Field Conservative or Non-Conservative in Doing Work?

[milesyoung]

Let's just start with this:

F= qv X B clearly says magnetic force lines are perpendicular to magnetic field lines. F= ma clearly says acceleration, and consequently displacement, can only be in the direction of the force. Where do you see a problem?

Acceleration happens in the direction of the net force on the point charge. Any motion the particle has is in the direction of its velocity. Just consider where the particle is at some time t and at some time t + \delta t later. If you keep reducing \delta t then it should be clear that the direction of its displacement vector will tend to the direction of its velocity.

Don't take my word for it. Do some simple math:

The work done on the point charge by the magnetic force between some time t_1 and t_2 is given by:
<br /> 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<br />

 

[pleco]

When you are dealing with fields there are some additional complexities. Basically, fields are distributed over both space and time. So when dealing with fields the appropriate quantity of interest for work and energy is power density.

I don't see how is this any different from calculating Coulomb's attraction and work done by electric force, or work done by gravity on a falling object. There is a force vector, there is a displacement vector, and multiplied together they represent work done.

When you look at power density the corresponding quantity to force . displacement is force density . velocity. When you take the dot product of the continuous Lorentz force law (force density) with velocity you find that the B field drops out leaving only J.E.

Take a single electron moving through B field, nearby permanent magnet for example. There isn't any external electric field at all, E becomes zero and all that is left to interact is external B field and electron's magnetic B field defined by J= p*v, or more specifically by Biot-Savart law and q*v:

In this scenario there is absolutely no any electric force interaction what so ever, there is only one electric field of the electron with nothing to interact with. So the only force that can possibly be to displace the electron and bend its trajectory is the magnetic force. That's how cathode ray tube works.

Yes, power density is the correct measure for the work done by a field.

I think the question is about the standard definition of work as force times displacement.

 

[milesyoung]

In this scenario there is absolutely no any electric force interaction what so ever, there is only one electric field of the electron with nothing to interact with. So the only force that can possibly be to displace the electron and bend its trajectory is the magnetic force. That's how cathode ray tube works.

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.

 

[pleco]

That gives the right answer for force, but not for power (work over time).

Power = (force * displacement) / time

That would be F.v = (qv x B).v = 0.

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

 

[Dale]

I don't see how is this any different from calculating Coulomb's attraction and work done by electric force, or work done by gravity on a falling object. There is a force vector, there is a displacement vector, and multiplied together they represent work done.

It is not different, it is just the generalization for fields. If you take power density and integrate over space then you get power, and if you integrate power over time then you get work. It is not different, just generalized. This is very standard stuff for fields.

Take a single electron moving through B field, nearby permanent magnet for example. There isn't any external electric field at all, E becomes zero and all that is left to interact is external B field and electron's magnetic B field defined by J= p*v, or more specifically by Biot-Savart law and q*v:
...
In this scenario there is absolutely no any electric force interaction what so ever, there is only one electric field of the electron with nothing to interact with. So the only force that can possibly be to displace the electron and bend its trajectory is the magnetic force. That's how cathode ray tube works.

Bending a trajectory does not do work.

 

[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:
<br /> 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<br />
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?