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

[reasonhunt]

when a magnetic field exert force on a moving charge the work done by it is 0.But bar magnet do work.And i read that magnetic field is a non conservative field so work done by it in a closed path is not 0.But in case of a moving charge it is 0 and magnetic field also store energy so it have the ability to do work.but i am confused in one case magnetic field is unable to do work and in other it is able.
Also magnetic scalar potential is not defined.
So i have doubt whether magnetic field is conservative or non-conservative
and also is it necessary to have magnetic moment for magnetic field to do work
And i also want to know the physical significance of magnetic moment

 

[vanhees71]

As always the answer is in Maxwell's equations and in the Lorentz force law. First of all you have
\vec{\nabla} \times \vec{B}= \frac{1}{c} \partial_t \vec{E}+\frac{1}{c} \vec{j}.
This means if the right-hand side is not 0, the magnetic field cannot be a potential field, because it's curl doesn't vanish.

Further the force on a particle is
\vec{F}_{\text{mag}}=\frac{q}{c} \vec{v} \times \vec{B}.
The work done (no matter whether the force is conservative or not) is given by
\int_{t_1}^{t_2} \mathrm{d} t \vec{v} \cdot \vec{F}=0.
So the magnetic field does not do work.

Some time ago we had a monster discussion on this topic with a lot of confusion created by not sticking to the fundamental laws. I hope this time this won't happen again!

 

[cabraham]

The magnetic field itself does no work as it is normal to the displacement. But the magnetic component of Lorentz force does work, i.e. Fm = uXB, where u is the velocity vector.

I remember that long discussion we had. Those who claimed Fm does no work could not produce one picture or proof of any kind. They just assert themselves and expect whatever they say to stick. A bar magnet lifting a slab of iron is an example of work done by Fm.

Claude

 

[Doc Al]

The magnetic field itself does no work as it is normal to the displacement. But the magnetic component of Lorentz force does work, i.e. Fm = uXB, where u is the velocity vector.

Note that Fm is perpendicular to u. No work done.

I remember that long discussion we had. Those who claimed Fm does no work could not produce one picture or proof of any kind. They just assert themselves and expect whatever they say to stick. A bar magnet lifting a slab of iron is an example of work done by Fm.

Oh please.

In Introduction to Electrodynamics (4th Edition), Griffiths devotes a chapter to this very misconception. Check it out.

 

[vanhees71]

@cabraham:
NO! I gave the proof in my answer. I gave it already within this earlier thread. If you still don't believe the mathematics and Maxwell's Laws, I can't help you. We even discussed more complicated issues concerning how a electric motor works etc. The master equations for energy-momentum conservation in electrodynamics can really be found in any sufficiently complete textbook on electrodynamics.

 

[Delta2]

Simple question is who/what does the work in the case of a bar magnet (or even an electromagnet) attracting a piece of iron? Cause that piece gains kinetic energy...

 

[cabraham]

Note that Fm is perpendicular to u. No work done.


Oh please.

In Introduction to Electrodynamics (4th Edition), Griffiths devotes a chapter to this very misconception. Check it out.

But u is the electron velocity not the velocity of the moving mass. Fm does no work on the electron but does work on the loop in which the electron is conducting. A rotor in an induction motor is a prime example.

The loop is in the x-y plane, current is counter-clockwise. B field is in y direction tangent to electron. Electron velocity u is in x direction. Fm = uXB must lie in z direction. Fm is normal to electron velocity u, and cannot do work on electron. Likewise, B is normal to u. So far no work can be done. But Fm acts in the z direction on the loop, producing torque and spinning said loop. The angular displacement θ, times the torque is the work done by Fm.

If Fm does no work on the loop, then which force is doing the work spinning said loop?

https://www.physicsforums.com/showthread.php?p=4030846#post4030846

The above thread includes a hand drawn picture illustrating the forces in action. I will elaborate if needed. Best regards.

Claude

 

[reasonhunt]

@vanhees71
According to you magnetic field does no work then how it is able to store energy because to consume energy it is necessary to do work so how the energy stored in field can be used

i'm unable to clear my doubt regarding this matter.

 

[reasonhunt]

And the case given by you is in the case of a moving charge but when a closed loop carrying current is placed in the magnetic field it experiences torque and we are also able to calculate the work done by the magnetic field in rotating the closed loop and that's why we associate potential energy with the system.

 

[milesyoung]

If Fm does no work on the loop, then which force is doing the work spinning said loop?

From Matter and Interactions, 3rd Edition, Chabay and Sherwood, ISBN-13: 978-0470503478;
Excerpt from page 1448-1449 (PDF) in "Chapter 21: Magnetic Force":

Why Does a Current-Carrying Wire Move?

There is a subtle point about the magnetic force that a magnetic field exerts on a current-carrying wire. The magnetic force only acts on moving particles, which in the case of a copper wire are the drifting electrons. The stationary positive atomic cores do not experience a magnetic force—so why does the entire wire move?

The answer to this puzzle is closely related to the Hall effect. In Figure 21.22, consider again a current-carrying metal wire in a magnetic field, with a growing number of electrons piling up on the bottom surface (and a deficiency of electrons accumulating on the top surface). Again, for clarity we don't show the usual surface charges along the wire; we just show the extra surface charges due to the magnetic deflection of the electrons.

The extra surface charges exert an electric force eE_\perp upward on the moving electrons, which in the steady state just balances the magnetic force evB. The extra surface charges also exert an electric force eE_\perp on the positive atomic cores, and this is an unbalanced force, because the atomic cores initially are not moving and hence are not subject to a magnetic force.

What is the direction of this electric force exerted by the surface charges on the atomic cores?

Evidently the positively charged atomic cores are pulled downward. The electric force due to the Hall effect surface charges forces the wire as a whole to move in the direction of \Delta\overrightarrow{F}_\mathrm{magnetic} = I\Delta\overrightarrow{l}\times\overrightarrow{B}, and the magnitude of the electric force is equal to this force. The overall effect is that in Figure 21.22 the atomic cores are dragged downward as the electrons are dragged downward. The motion of the wire is an electric side effect of the magnetic force on the moving electrons.

Another way of thinking about this is to say that the electrons are bound electrically to the metal wire—they can't simply fall out of the wire. Thus when the electrons are pushed down by the magnetic force, they drag the wire along with them. Note that ultimately it is the battery that supplies energy to the wire.

It should be absolutely perfectly clear from the Lorentz force law that the magnetic force does not do any work. An electric force ultimately does the work, as it must.

 

[Jano L.]

But Fm acts in the z direction on the loop, producing torque and spinning said loop.

Yes, the magnetic force $\mathbf F_m$ does have torque and does impart angular momentum to the loop in general.

The angular displacement θ, times the torque is the work done by Fm.

It is mechanical work on the loop alright, but it is not done by $\mathbf F_m$. Remember, you defined $\mathbf F_m$ above by saying that it acts on the electron in the loop. Proper way to calculate work of this force is to multiply it by displacement of the electron. Using displacement of the element of the rest of the loop (the lattice + nonconducting electrons) makes no sense for this force, because this element body has a different velocity than the electron inside it.

If a force acts on the electron, it cannot act on an element of the rest of the loop (ERL). One force cannot act on two distinguished bodies - if two distinct bodies experience force, even of the same magnitude and direction, there are two distinct forces. Total force acting on the ERL without the conducting electrons deserves its own symbol, say $\Delta \mathbf F_w$.

This force $\Delta \mathbf F_w$ and displacement of ERL $\mathbf u$ is what should be used to calculate net mechanical work done on the rest of the loop.

From experiments, we know the force $\Delta \mathbf F_w$ can be approximately expressed as

$$
\Delta \mathbf F_w \approx \Delta V\,\mathbf j \times \mathbf B_{ext},
$$
where $\mathbf B_{ext}$ is external magnetic field and $\mathbf j$ is current density of the conduction electrons in the wire of the loop.

This force does have magnetic field in its approximate expression, but it is not a magnetic force in the sense of the formula

$$
\int \mathbf j_r \times \mathbf B_{ext}\,dV,
$$
because the current density due to the rest of the loop $\mathbf j_r$ is very different from the density of conduction electrons $\mathbf j$ used in expression for $\Delta F_w$.

Therefore $\Delta \mathbf F_w$ have to be sums of electric forces or forces of non-electromagnetic nature. They are due to conduction electrons pushing or pulling on the rest of the loop due to their modified motion in the external magnetic field. These electrons exert electric and possibly non-electromagnetic forces on the rest of the loop and these can do work. Thus $\Delta \mathbf F_w$ are sums of internal forces. Internal forces do all the mechanical work in this case. Energy for this is transferred from the energy of magnetic field and the source of voltage.

Mechanical work done on the ERL is

$$
\Delta \mathbf F_w \cdot \mathbf u
$$
where $\mathbf u$ is velocity of the element, not of the electrons inside it, which is parallel to $\mathbf j$ and would give zero work. That's why the actual work can be non-zero; the force $\Delta \mathbf F_w$ acting on the element is not perpendicular to its velocity $\mathbf u$.

 

[milesyoung]

Look, I'm not going to spend 20+ pages chasing your red herrings.

After all this, if you're still not convinced, then I doubt me or anyone else is going to change your mind.

 

[pleco]

It should be absolutely perfectly clear from the Lorentz force law that the magnetic force does not do any work. An electric force ultimately does the work, as it must.

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

Just like depicted force vector acting on wire 1 points towards wire 2, so does force vector acting on wire 2 points to wire 1, and thus two wires attract. As always the movement is in the direction of the force, in this case it's Lorentz force. It is important to distinguish field lines and force lines. While electric field and force lines coincide, magnetic field lines and force lines are perpendicular.

The magnetic force only acts on moving particles, which in the case of a copper wire are the drifting electrons.

Magnetic force only acts between magnetic fields. It acts on protons, electrons and even neutrons, because all of them have their spin magnetic moments, whether they move or not.

It's mostly free electrons in wires that we say interact with external magnetic fields because protons, neutrons and other electrons usually prefer to stick to their molecular arrangement since those forces are greater for them.

All these electric and magnetic fields superimpose and on average neutralize, in an average material. But when free electrons are drifting additional magnetic fields are produced and then there is an excess of magnetic field with the same orientation relative to only one particular direction of the drift.

The electric force due to the Hall effect surface charges forces the wire as a whole to move in the direction of \Delta\overrightarrow{F}_\mathrm{magnetic} = I\Delta\overrightarrow{l}\times\overrightarrow{B}, and the magnitude of the electric force is equal to this force.

They calculated magnetic force, renamed it to electric force and asserted they are equal, out of the blue. Magnetic and electric force can hardly be equal, but if they were equal then we could equally say for either is the one which does the work.

 

[Pythagorean]

An analogy I came up with during my undergrad is a ramp. A ramp does no work, but it translates horizontal work to vertical motion, and vertical (as done by gravity) to horizontal motion.

Obviously the coupling between magnetic and electrical fields is more complicated than that of gravity and ramps, but it illustrates how you can get forces on an object from the ramp without the ramp doing work.

 

[Dale]

I hope this time this won't happen again!

That is also my hope.

 

[pleco]

This looks to me like a weird semantic argument like when people say the same word but mean different things, only this time they say different words but still think they are talking about the same thing.

Can magnetic fields do work? Of course they can not, no field can. Forces do work, not fields. They are different things, with different units. They don't even compare. Work is not field, but force times displacement.

 

[Dale]

According to you magnetic field does no work then how it is able to store energy because to consume energy it is necessary to do work so how the energy stored in field can be used

The key theorem regarding energy conservation in EM is called Poynting's theorem. It says:

$$-\frac{du}{dt}=\nabla \cdot S + J \cdot E$$

Where u is the energy density of the field, $u=\frac{1}{2}(E\cdot D + B \cdot H)$ and S is the flux of the EM energy $S=E\times H$.

So Poynting's theorem says that the EM energy in the field at a point can only decrease (-du/dt) if the EM energy flows out to other parts of the field (∇.S) or does work on matter (J.E). The work done on matter is always J.E, but the magnetic field can decrease as part of u decreasing and the magnetic field is always involved in moving EM energy from one place to another as part of S. So the work done is always through the E field, but the energy required can come from the magnetic field.

 

[pleco]

https://www.physicsforums.com/showthread.php?t=229056
https://www.physicsforums.com/showthread.php?t=347539

I found these two threads and I see what the problem is, in the case of two parallel wires.

There are two forces, magnetic force acting on drifting electrons and electric force between electrons and protons.

Magnetic force displaces drifting electrons in the direction of the Lorentz force, therefore magnetic force does that work on electrons.

Sure electrons are responsible for dragging protons along, but whether electric force does any work as well depends on whether there is any displacement between electrons and protons.

There is probably some relative displacement between electrons and protons as magnetic force is trying to yank or blow away electrons from protons. That is _away from protons. So if there is any displacement at all it will be in the opposite direction of the electric force and in the direction of magnetic force, thus magnetic force would be the one doing the work again.

Protons basically just add to resistance, that is mass, but we don't need protons at all. We can bend electron beams with magnetic fields, so then without a doubt we can see magnetic force can indeed do work.

 

[Dale]

See Poyntings theorem. Work done by EM is always J.E. Bending an electron beam is not a counter example since no work is done.

 

[pleco]

See Poyntings theorem. Work done by EM is always J.E. Bending an electron beam is not a counter example since no work is done.

There is a displacement of electrons in the direction of the magnetic force whether it is an electron beam in a magnetic field or a current carrying wire, so why do you say no work is done?

 

[haruspex]

There is a displacement of electrons in the direction of the magnetic force

No, it's orthogonal to the magnetic force.

 

[pleco]

No, it's orthogonal to the magnetic force.

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


Are you saying something is wrong with this diagram or the equation from Wikipedia? Which one do you think is going in the wrong direction, force or displacement? Can you explain?

 

[Dale]

There is a displacement of electrons in the direction of the magnetic force whether it is an electron beam in a magnetic field or a current carrying wire, so why do you say no work is done?

Because I accept Poyntings theorem.

As you mentioned earlier fields obviously don't do work because the units are wrong. Forces also obviously don't do work because the units are wrong. For EM, the thing which does work and has the right units is E.J.

 

[milesyoung]

Are you saying something is wrong with this diagram or the equation from Wikipedia? Which one do you think is going in the wrong direction, force or displacement? Can you explain?

This wasn't directed at me, but I hope you won't mind if I comment:

I would just ask you to first convince yourself that the magnetic force on a charged particle can't do any work - that part should be clear from the Lorentz force law.

So why is it that when we sum up all the contributions of the magnetic force on the charged particles in a current-carrying conductor, the resulting force can seemingly change the kinetic energy of the conductor?

The answer is that it can't and it doesn't. When you start looking into what actually does the work on the conductor, you'll find an electrical force acting on it with the same magnitude and direction as that of the resulting magnetic force on the charged particles.

I would ask you to go back to the Lorentz force law now and then ask yourself how it could ever be any different?

I understand that you must now think me mad because, in the Wikipedia article and in other material, they clearly talk about a magnetic force on the current-carrying conductor in the same direction as its displacement. There's nothing wrong with that picture, because if you calculate the work that's seemingly done by the magnetic force on the conductor, you'll eventually find that this energy comes from whatever is supplying the current.

 

[haruspex]

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


Are you saying something is wrong with this diagram or the equation from Wikipedia? Which one do you think is going in the wrong direction, force or displacement? Can you explain?

This isn't exactly the right diagram for the thread. It becomes right if you replace the top wire by a stream of unconstrained electrons, i.e. they are free to change direction.
The magnetic field is shown in the diagram by the B arrow. The electrons are moving in the I direction, at right angles to B, and the resulting acceleration is at right angles to both. No work done.
It's just like a pendulum rotating in a horizontal circle. The acceleration is always perpendicular to the velocity. There is never any actual displacement in the direction of the acceleration.

 

[pleco]

Because I understand Poyntings theorem.

Is there something wrong with the standard approach of calculating force vectors and displacement vectors, then simply using Work = Force * displacement formula?

As you mentioned earlier fields obviously don't do work because the units are wrong. Forces also obviously don't do work because the units are wrong.

Force in Newtons times displacement in meters yields joules. The SI unit of work is joule.

For EM, the thing which does work and has the right units is E.J.

J is a measure of the magnetic field: p*v, it's volume integral of Biot-Savart law from Lorentz force equation. J.E is power density, with units of watts per unit volume.

- "J.E is the density of Electric power dissipated by the Lorentz force acting on charge carriers."
http://en.wikipedia.org/wiki/Poynting's_theorem

 

[pleco]

I would just ask you to first convince yourself that the magnetic force on a charged particle can't do any work - that part should be clear from the Lorentz force law.

Do you disagree the force vector acting on the drifting electrons in the top wire 1 is not pointing in the direction of the bottom wire 2, as the diagram shows? Is that not the same direction in which those electrons will be accelerated and displaced towards off from their drifting course?

So why is it that when we sum up all the contributions of the magnetic force on the charged particles in a current-carrying conductor, the resulting force can seemingly change the kinetic energy of the conductor?

What equation are you talking about?

I would ask you to go back to the Lorentz force law now and then ask yourself how it could ever be any different?

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?

I understand that you must now think me mad because, in the Wikipedia article and in other material, they clearly talk about a magnetic force on the current-carrying conductor in the same direction as its displacement. There's nothing wrong with that picture, because if you calculate the work that's seemingly done by the magnetic force on the conductor, you'll eventually find that this energy comes from whatever is supplying the current.

The question is not where the energy came from, but whether magnetic force can do work or not.

 

[Dale]

Is there something wrong with the standard approach of calculating force vectors and displacement vectors, then simply using Work = Force * displacement formula?

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.

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.

J is a measure of the magnetic field: p*v, it's volume integral of Biot-Savart law from Lorentz force equation. J.E is power density, with units of watts per unit volume.

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

- "J.E is the density of Electric power dissipated by the Lorentz force acting on charge carriers."
http://en.wikipedia.org/wiki/Poynting's_theorem

Yes.

 

[pleco]

This isn't exactly the right diagram for the thread. It becomes right if you replace the top wire by a stream of unconstrained electrons, i.e. they are free to change direction.

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.

The magnetic field is shown in the diagram by the B arrow. The electrons are moving in the I direction, at right angles to B, and the resulting acceleration is at right angles to both. No work done.

B field vectors are not the same thing as force vectors, they are perpendicular. To calculate work done you need to multiply the force with the displacement. Force vector, not field vector.

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.

 

[Dale]

F= qv x B always gives the right answer.

That gives the right answer for force, but not for power (work over time). That would be F.v = (qv x B).v = 0.

 

[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 ...