Can anybody please tell me if magnetism is a conservative force or if there is a field associated with it? How to reason? One thing I know is that the work done by a magnetic force is equal to 0.
Answer
The force q→v×→B acting on a charged particle is clearly not conservative because it depends on the velocity. Conservative forces are those that integrate to a fixed work – energy difference that is independent of the path – between two points so they may only depend on the location. We can't associate a potential with a force that is velocity-dependent.
On the other hand, the magnetic force acting e.g. on a small piece of ferromagnet deserves a special discussion. If the magnetic has magnetic moment →m, the energy of this magnet in an external magnetic field is simply U=−→m⋅→B
In general, whether or not →B is the gradient of something depends on curl→B and this is governed by one of Maxwell's equations, namely ∇×B=μ0(J+ε0∂E∂t)
However, this doesn't immediately translate to the conservative character of any magnetic force because the magnetic forces we have measured aren't proportional to →B as vectors. We would have →F=c→B which would be analogous to the electric force acting on charges but such a form of the force only applies to magnetic monopoles, something that may exist in Nature but only in the form of extremely heavy elementary particles that we haven't observed yet.
So because →F isn't c→B for any known force, the question whether →B is a gradient of something has no impact on the question whether a known force related to magnetism is conservative or not. For this reason, it isn't even too useful to consider potentials ΦB such that →B=−∇ΦB even though in some cases and regions, we could find such a potential (when the curl vanishes).
Instead, it is useful to write →B=curl→A where →A is called the vector potential. As long as there are no magnetic monopoles, →B may always be written in this way because the only obstruction that could prevent us from rewriting →B in this way would be a nonzero div→B but this divergence vanishes due to a simple Maxwell's equation.
However, the vector potential →A can't be "immediately" interpreted as the potential energy of a unit charge etc. Instead, it appears in the Lagrangian and the Hamiltonian (with various signs) in the combination →j⋅→A where →j is the electric current. In this sense, the vector potential is the potential energy per unit current (both of them are vectors).
This is a bit formal description because to get the actual energy, one would have to integrate over the whole paths of the currents etc. but it may be done. For example, if you have a small loop of electric current, it behaves like a magnet (electromagnet) and the contour integral ∮→dℓ⋅→A of →A over this loop is nothing else than →B⋅→dS – by Stokes' law – where →dS is the infinitesimal area of the loop with the normal direction added to turn it into a vector (a right hand rule is implicit here). So this tells you that the potential energy of the loop of current i.e. a small electromagnet is nothing else than the −→m⋅→B potential energy I mentioned at the beginning.
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