# Are magnetic fields 'conservative'

## Main Question or Discussion Point

surely $$\vec{\nabla} \times \vec{B} \neq 0$$ in general
but the work done by magnetic field on any charge is 0 hence is independent of the path taken
So can we call such a field conservative

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pam
Be careful in using "surely" and "general". Curl B is not zero in the presence of current.
The usual textbook definition of a "conservative field" is that
$$\oint{\bf B}\cdot{\bf dr}=0$$, which is not true if the path circles any current.

want to follow up a question. i read in wiki and it says that non-conservative force is due to negligence of certain degrees of freedom. Is this a well accepted concept? if yes, then what sort of degrees are neglected in case of magnetic force so that it becomes non-conserving? thankyou http://en.wikipedia.org/wiki/Conservative_force

want to follow up a question. i read in wiki and it says that non-conservative force is due to negligence of certain degrees of freedom. Is this a well accepted concept? if yes, then what sort of degrees are neglected in case of magnetic force so that it becomes non-conserving? thankyou http://en.wikipedia.org/wiki/Conservative_force
I'm thinking wiki is just wrong.

surely $$\vec{\nabla} \times \vec{B} \neq 0$$ in general
but the work done by magnetic field on any charge is 0 hence is independent of the path taken
So can we call such a field conservative
Indeed, if you have a static magnetic field, you can have a scalar potential (search for magnetic scalar potential on Google). This potential is multivalued, and undefined at points with current. Thus actually a magnetic field is kinda conservative --- if the field is static and the region you're interested in has no currents. However, it's not a very useful point of view, and doesn't actually simplify the algebra --- multivalued functions aren't very friendly.

pam
However, it's not a very useful point of view, and doesn't actually simplify the algebra --- multivalued functions aren't very friendly.
That is Griffith's, somewhat naive, point of view, but the scalar potential does actually simplify the algebra, and can be quite useful, especially for permanent magnets.
The multi-valued part is no real problem. Are you also going to exclude logarithms and roots?

That is Griffith's, somewhat naive, point of view, but the scalar potential does actually simplify the algebra, and can be quite useful, especially for permanent magnets.
The multi-valued part is no real problem. Are you also going to exclude logarithms and roots?
I believe the idea isn't that the scalar potential isn't simpler vs direct manipulation of fields, but rather that multivalued scalar potential isn't simpler than vector potential, plus the latter is generally true with no restrictions.