# B Is the magnetic field conservative or nonconservative?

1. Nov 27, 2016

### Isaac0427

I've gotten so many different answers off the internet on whether the magnetic field is conservative or nonconservative using the three requirements stated here. The part at the bottom of that article doesn't really clear things up for me.

Thanks!

2. Nov 27, 2016

### Cutter Ketch

In a force field the value of the field at any location in space determines the magnitude and direction of the force a test particle will experience. That is not true for a point charge in a magnetic field. The force experienced by a point charge depends not just on its charge and location, but also how fast it is moving and in what direction. So the magnetic field isn't a force field and the definition of conservative in the Wikipedia article does not apply.

So in this case we need to define conservative for ourselves. What do we want conservative to mean? For me the most meaningful definition is the idea that movement over a closed path does no work. The Wikipedia article said that the magnetic field met that criteria for a point charge.

Now instead of a point charge make the test particle a magnetic dipole. Here the force on the dipole IS given by the magnitude of the field and there is no hysteresis. So I would say a magnetic field is conservative with respect to magnetic dipoles.

What can you do? It doesn't fit in the defined box.

3. Nov 27, 2016

### Cutter Ketch

The above quickly edited for a brain fart I hope no one saw and I won't repeat. If you did, sorry.

4. Nov 27, 2016

$\nabla \times B=4 \pi J_{total}/c+(1/c)dE/dt$. (Ampere's law/Maxwell's equation c.g.s. units) Even in the steady state (where $dE/dt=0$ ), when there are any kind of currents in conductors or magnetic currents (e.g. magnetic materials where $\nabla \times M=J_m/c$), $\nabla \times B$ is non-zero so that the magnetic field $B$ is non-conservative.

5. Nov 27, 2016

### Isaac0427

Thank you for the helpful answers, but once again I have two completely opposite answers. I'd ordinarily agree with @Charles Link but I question that because of what people like @Cutter Ketch say. The B field shouldn't fit any of the mathematical definitions unless there is no current, and you can't have a B field without a current, right (that could be wrong, so please correct me. I believe that the A potential has to do with current, and without an A potential you have no B field)? What's going on here. To be clear, I am not talking about the force, I am talking about B.

6. Nov 27, 2016

You can have regions in space where there are no magnetic materials or currents, and for these regions, in the steady state, the magnetic field $B$ will be conservative/can be considered conservative, and can be written as $B=-\nabla \Omega_m$ where $\Omega_m$ is a magnetic potential function. It really is better physics to be able to speak in general terms about the magnetic field $B$, instead of finding cases where it can be considered conservative. editing...You might ask, what applications might this have? For a single electric charge, the force is proportional to the velocity, but for a magnetic dipole, or for a permanent magnet which is a conglomeration of magnetic dipoles, you can write $U=-m \cdot B$ and you could compute the forces on the magnet in the magnetic field. Under certain conditions (e.g. fixed magnetic moments, i.e. where the magnetization $M$ of the permanent magnet is not affected by the external field), conservative force laws may be applicable.

Last edited: Nov 27, 2016
7. Nov 27, 2016

### Isaac0427

Ok, so what I am getting at is that B is nonconservative and the magnetic force is neither conservative nor nonconservative (it does not follow condition three but it does follow the others). Correct?

8. Nov 27, 2016

Please read the "edit" that I added to post #6. The force on a permanent magnet in a magnetic field can be written as an integral of such gradients(condition 3 in your OP) of $U=-M \cdot B$ . In the way I just wrote it $U$ is the magnetic energy per unit volume.

Last edited: Nov 27, 2016
9. Nov 28, 2016

The magnetic force is a atypical case; most velocity-dependent forces, such as friction, do not satisfy any of the three conditions, and therefore are unambiguously nonconservative. The magnetic field is NOT conservative in the presence of currents or time-varying electric fields.

10. Nov 28, 2016

### dmankit

Hello

Magnetic field itself is neither non-conservative nor conservative.Magnetic fields can passed from closed paths, but these are not conservative. We can say, a field is conservative when force on a test particle moving around a closed path does no net work.
Thanks

11. Nov 28, 2016

### Isaac0427

In the case that the magnetic field is conservative, would the magnetic force also be conservative?

In the case that a current exists, would it be a reasonable convention to call the magnetic force nonconservative as it does not observe all three conditions?

EDIT-- Sorry, there were 2 typos in there.

Last edited: Nov 28, 2016
12. Nov 28, 2016

### Isaac0427

Also, if a field is conservative then the field's respective force is conservative, and if a field is nonconservative its respective force is nonconservative, correct?

Last edited: Nov 28, 2016
13. Nov 28, 2016

In general, with static magnetic fields, you will always have currents. If the region of interest lies outside these currents, and the object of interest consists of magnetic dipoles, then it would appear that you can have a conservative field and a conservative force. If the magnetic field induces additional (magnetic) dipoles in the material, I think you will find the field and forces are non-conservative and hysteresis losses will occur. $\\$ If you are looking at the effect of the magnetic field on charged particles, it is clearly non-conservative. The purpose of calling it conservative is to be able to write a potential function or similar expression like you do for a charged particle in an electric field where you can assign a voltage to a location and that voltage is independent of the path the particle took to get there. In travelling from one location to another the energy it acquires or loses can be computed from the potential function. A similar calculation applies for the gravitational potential.

Last edited: Nov 28, 2016
14. Nov 29, 2016

### vanhees71

A magnetic field is in general not conservative. To keep the answer simple, let's look only at the magnetostatic case, i.e., you have currents and charges as well as the electromagnetic field time-indepenent. Then the Maxwell equations decouple for electric and magnetic field components. For the magnetic components you get
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}.$$
As you see, wherever there is a current density, the static magnetic field cannot be a gradient field. Due to the first equation, it's rather a solenoidal field, i.e., there is a vector potential with
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Of course, in regions outside of current distributions, there is a (local!) gradient field, because there $\vec{\nabla} \times \vec{B}=0$. Sometimes it can be of advantage to use a magnetic scalar potential in such regions. For a very good treatment, see

A. Sommerfeld, Lectures on Theoretical Physics, Vol. III (electromagnetism)