Magnetic fields do no work? How come

[Doc Al]

Does the magnetic force affect the motion of the charge and thus the stretch of the spring? Sure. Does the magnetic field do work on the charge? No. All it does is change the direction of motion of the charge. The energy of the system--elastic potential energy + kinetic energy--remains the same.

 

[kmarinas86]

Does the magnetic force affect the motion of the charge and thus the stretch of the spring? Sure. Does the magnetic field do work on the charge? No. All it does is change the direction of motion of the charge. The energy of the system--elastic potential energy + kinetic energy--remains the same.

Ok, so basically the energy is not added by this static magnetic field to the system (obviously). Instead, it just facilitates changes of potential energy into kinetic energy and vice versa.

Imagine a large magnetic field external to our solarsystem that changes only over the course of thousands of years. This "effectively static" magnetic field could give us an impression of anomalous effect on the movement of bodies in our solarsystem, but in such a way that the total energy, GPE + KE, is conserved. This would imply that magnetic fields can be used to facilitate the unleashing of potential energy, but that such potential energy does not come from the magnetic field itself.

 

[Dadface]

Expressing it a bit differently the charge has kinetic energy before being deflected and it has exactly the same kinetic energy after being deflected.Some of this kinetic energy is changed to elastic potential energy but the kinetic energy is not gained from the magnetic field,it is gained from the electric field that caused the charge to move in the first place.

 

[JustinLevy]

A magnetic dipole in an external magnetic field with a gradient will feel a force.

So, two ideal magnetic dipoles at rest can extert a force on each other.

That means two neutrons can exert a magnetic force on each other while they are at rest. If you don't like composite particles, then the same can be said with electrons (but there is also an electric force as well then, but you can set it up such that the forces are in orthoganol directions to make it clear which force did what).

I would consider that work legitimately done by a magnetic field.
... unless you want to claim an electron is not a point particle.

 

[sweet springs]

Does the magnetic force affect the motion of the charge and thus the stretch of the spring? Sure. Does the magnetic field do work on the charge? No. All it does is change the direction of motion of the charge. The energy of the system--elastic potential energy + kinetic energy--remains the same.

Hm.. Let me pose one more question.

In the case of motion of charge in NO magnetic field

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◎ww●↑○○○○○
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｜○○●○○○○○○
｜○○●○○○○○○
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By magnetic flux x B, the Lorentz force →　and the elastic force ←　balance,

x B
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｜○○○○●○○○○
｜○○○○●○○○○
◎www ●↑○○○
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So what caused the stretch of the spring therefore increase of elastic energy?

Regards.

 

[Doc Al]

I'm not sure I understand your diagrams, but in any case.

By magnetic flux x B, the Lorentz force →　and the elastic force ←　balance,

No reason to think that those forces are balanced.

So what caused the stretch of the spring therefore increase of elastic energy?

The mass is moving.

 

[Relay]

Here's my answer to elliotr's origininal question. (Please be kind if it doesn't make sense.)

First and formost, nobody knows what magnetism is. We just know how it behaves.

Second is how one looks at a problem. I think that all textbooks state that magnetism can't do real work directly same as gravity. What I mean by this is that people percieve the Earth as the center for gravity and do not do the same for magnets. Each magnet is its own gravity well. On Earth we can use a ground state like a sidewalk and base calculations from that level. A cannon ball falling from a building converts its potention energy to kinetic energy. This is where it gets tricky. The cannot ball got it's energy from somewhere other than gravity. The person who took the cannon ball to the top of the building is the source of that energy. So when the cannon ball falls and breaks the sidewalk nobody questions that it did work and had energy. The same can be said about a paperclip falling up to a magnet. It will do work on the surface of the magnet as it hits, but the energy would not have come from the magnet. The energy came from many processes that created and placed the paperclip just prior to being pull (or pushed) up by the magnet.

We don't even know if magnetism is a pulling or pushing force. Anyhow, if you think about a magnetic well that things can fall into and be pulled from, it starts to make more sense. Well, that's my two cents worth...

 

[fantispug]

Dipoles have come up a couple of times in this discussion, but I still don't quite get it.

We can calculate the torque due to the magnetic force on a dipole, m, in a uniform magnetic field, B, and calculate a corresponding energy in rotating the dipole -m.B.

We are only considering the magnetic forces here, yet they have appeared to do work.

One way to construct this situation is using spinning spherical shells; a uniformly charged sphere rotating at constant angular velocity has a constant, uniform magnetic field inside and a dipolar magnetic field outside. So we can put one spinning shell inside another, using the outer shell to produce the uniform magnetic field and the inner shell to represent the dipole.

It turns out that the inner shell produces a torque on the outer shell, and so under a rotation of the dipole this produces a work +m.B.

Thus we find (at least in this case) the TOTAL work done by magnetic fields is +m.B - m.B = 0.

However the justification that the magnetic force does no work is done particle-by-particle. I would have thought this implied that there should be no magnetic work in rotating a dipole in a uniform magnetic field. Why is this wrong?

 

[Vanadium 50]

Fantispug, I think I covered this in Post #13. Is there something unclear that I should elaborate?

 

[sweet springs]

Hi.

The mass is moving.

Now I know my misunderstanding. Another resource to keep mass moving in constant velocity is necessary. Thank you for your teachings. I apologize the delay of my thanks. I was wondering of the following another situation.

On X-Y plane there exists around the origin a ring of radius a. This ring is made of charge, the line density of which is ρ, and is rotating with angular velocity ω. Let homogeneous magnetic field of B of direction x be applied. The parts of circle get force in Z direction of －ρaωB cosφ　dl ＝ －ρa^2 ωB cosφ dφ where φ is the direction angle in X-Y plane.

Does the ring start to rotate around the Y axis (stand up) therefore get rotation energy under the influence of magnetic force which keep perpendicular to the motion of charge i.e. rotating around Z and perhaps Y axes ?
If yes, does this Y rotation energy come from the work done by the magnetic field ,or just from the conversion of Z rotation energy ,therefore ω decrease, in conservation of energy ?

Regards.

 

[fantispug]

Hi Vanadium 50,
I don't understand in the particular case of a dipole in a magnetic field where the electric forces come in; I'm interested in this case at looking at a careful analysis of the forces just to understand how it works.

The calculation that I eluded to, using the torque on a magnetic dipole by a magnetic field, seems to imply that the magnetic field does the work. The electric field plays no role in this argument; there are no external electric fields and we have not included the effects of the electric fields generated by the dipole.

(In fact the electric fields generated by the dipole give an additional energy to the system of +m.B)

So how are the electric forces responsible for the work in this specific case?

 

[Vanadium 50]

What holds the dipole together?

 

[cabraham]

What holds the dipole together?

We're going in circles again. All pertinent forces have been thorughly examined. Refer to pages 1 through 4 and you will find good info from several contributors.

We pretty much reached a consensus that H fields do no direct work on a charged particle, but H exerts force while E provides tethering force yanking the lattice towards the free electrons.

E does indeed hold the lattice and electrons together. H moves the electrons in a direction normal to its velocity, and the lattice is yanked along due to the tethering nature of the E force.

At a mIcroscopic scale, H does not literally "do work" on an electron. At the mAcro scale, H determines the magnitude and direction of the wire deflection. H moves electrons which yank the lattice along due to E.

It is well understood now. Every e/m fields text makes it a point to emphasize the mutually inclusive nature of E & H. They are strongly inter-related, and under time-varying conditions, cannot exist independently. Under static conditions, either can exist alone.

We should all agree at this point.

Claude

 

[fantispug]

Ok, I think I get what you're saying now. It doesn't matter what's holding the dipole together; whether its electric forces, gluons or Achillies. If we're thinking of a dipole as a spinning spherical shell, the important thing to notice is that if it is conducting when we put it in a magnetic field the charge will redistribute so that it is no longer a dipole at all.

Consequently when we put a dipole in a magnetic field, turn on the magnetic field and assume that the charge does NOT redistribute, we need to supply special angle-dependent forces to prevent this redistribution. The difference in electromagnetic energy between redistributing and not redistributing the charge is supplied by the work done by this "special force". We claim that this is then exactly +m.B.

I've been trying to verify this in a specific example, but the maths is a little tricky so I'll see how we go; but the explanation seems physically plausible. Thanks.

 

[kmarinas86]

Under static conditions, either can exist alone.

Whether a condition is static is relative to the velocity of the obsever. I think a better term would be comoving.