# Homework Help: Magnetic force: conservative or not?

1. Jun 5, 2007

### loom91

Hi,

I was thinking, velocity dependent forces such as friction are considered non-conservative. The Lorentz force on a point charge moving in a constant magnetic field depends on its velocity, so it seems the magnetic force is non-conservative. Other criteria such as vanishing curl are also not met.

Yet, since the Lorentz force is always perpendicular to the velocity, it can never change the kinetic energy of the charged particle. It does no net work and the particle's energy does not dissipate when in the field. Is this not a characteristic of a conservative force?

Thanks.

Molu

2. Jun 5, 2007

### chaoseverlasting

It would be non conservative. As for the particle's energy, its being accelerated (centripital acceleration) and it should radiate energy, so its energy dissipates.

3. Jun 5, 2007

### jackiefrost

If an external force does work on a charge in a magnetic field, can the magnetic field perform that same amount of work on the charge that would tend to restore the charge to its previous energy state? Can we (theoretically) move a charge, in a magnetic field, between any two arbitrary points, and always do the same amount of work regardless of our choice of path?

4. Jun 5, 2007

### chaoseverlasting

No, because if you move the particle along the direction of the magnetic field, then there is no magnetic force acting on the particle and there is no work done against the field.... conversely, if you move the particle perpendicular to the field, then there will me a magnetic force acting on the particle and you will have to do work against it...

5. Jun 6, 2007

### loom91

Not really. It is easily proven using a little vector math that a force that is always perpendicular to velocity can never do any work (and therefore no work can be done against it). An external agent does not have to perform any work to move a charge in a magnetic field. Force does not equal work.

Molu

6. Jun 6, 2007

### loom91

But that has nothing to do with the magnetic field itself. It's a different effect.

Molu

7. Jun 6, 2007

### loom91

I don't really get your first question. The answer to the second question seems to be yes. The presence of the magnetic field makes no contribution to the work because the force vector is always perpendicular to the velocity vector and hence the elementary displacement vector. This means at each point of the path the elementary work, which is the dot product of the force vector and the elementary displacement vector, is zero.

Consider a particle being made to go around in a loop by an external force in free space. Also let this external force be conservative in nature, so that it is possible to define a potential function for the particle. Now if a magnetic field is suddenly turned on, there will be no change to the potential function because of the magnetic fields inability to perform work. The applied force will definitely have to change in order to keep the trajectory unchanged, but since the displacement change will be zero, the work can not change.

Molu

8. Jun 6, 2007

### jackiefrost

Sorry - I was not clear. I guess, in my questions, I was just trying to probe your ideas regarding a couple of the usual requisites for what we mean by "conservative force" in relation to magetism. The first question was supposed to be about whether the magnetic force has "restorative" capabilities. That is, whether work done against it is recoverable as stored energy that it can then use to perform work in the reverse manner. Yeah, I know - that's not much better than the first time I asked. Another apparent requiste, path independence, you commented on.

Actually, I'm a confused by what constitutes a conservative force field in physics vs a conservative vector field in mathematics. We have all these inter-related criteria like curl free, path independent, can be derived from the negative gradient of some scalar potential, closed loop work equals zero, etc... I mean like, which one(s) actually captures the essence of "conservative" in the physical context??? I'm tempted to just pick curl free and say the hell with it. Especially after pouring through some of the previous discussions on this topic - see https://www.physicsforums.com/archive/index.php/t-8259.html<br /t-132584.html"

Good night - I'm outta here :zzz:

Last edited by a moderator: Apr 22, 2017
9. Jun 7, 2007

### loom91

I can't see how curl-free can be a universal criteria as not all forces can be written as a function of purely space coordinates. To me the most fundamental definition seems to be a force which permits the concept of a scalar quantity, the potential, to model the particle's motion instead of the vector force.

Molu

Last edited by a moderator: Apr 22, 2017
10. Jun 7, 2007

### jackiefrost

Well, I pretty much agree... I just picked curl-free almost (but not quite) at random as a demonstration of my frustration with trying to isolate these inter-related concepts and settle on one as a "universal criteria". I sometimes get the feeling like I'm playing a self-defeating game of semantics in this endeavor. However, I agree that the existence of a scalar potential function for some force field whose gradient represents the force, does point to a crucial aspect of the actual physical nature of the force. Potential is crucial since what we're really alluding to in the word "conservative" is conservation of energy. That is, whether work done against the force on a unit "body" (whatever the force is defined as acting on) in moving it between two points can be represented by a difference in scalar potential coupled with whether the force can then perform the negative of that potential difference on the "body". OK - yes - I think this should always(?) be true for a force F when F= grad f (where f is a scalar potential function).

So, I think I agree with you more than when I began this post . Whether or not F = grad f seems to automatically take care of a lot of the essentials of what we mean by calling a force or field "conservative".