# Is magnetic Field conservative in nature?

Is magnetic field conservative in nature?

Magnetic field lines do go in closed paths but that's not the definition of conservative.

Rather, a field is conservative when the force on a test particle moving around any closed path does no net work. But magnetic fields only act on moving charges, and at right angles to the motion, so the work is always zero and the concept doesn't properly apply.

Also, if there were magnetic monopoles, they would try to follow the magnetic field the way electric charges try to follow the electric field lines.

Now we consider the magnetic field due to magnetic monopoles, which is going to be conservative.

So Is there a precise answer? or the nature of the magnetic field depends on the way you produce it? If the latter is the case then, is the magnetic field created by monopoles conservative?

## 1 Answer

A field $\mathbf{F}$ is conservative if and only if $\nabla \times \mathbf{F}=0$. From Maxwell's equations we know that

$$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0\epsilon_0 \frac{\partial\mathbf{E}}{\partial t}.$$

Hence, the magnetic field is only conservative in the absence of free currents and time varying electric fields.

• I like this answer, but the Maxwell-Ampère equation that you quote applies to each point. Do we insist that for a field to be conservative there are no currents or time-varying electric fields at $any$ point? I'm considering the field around a long wire carrying a steady current. Curl B is zero everywhere except in the wire, because J is zero everywhere except in the wire. Yet we'd say, wouldn't we, that the magnetic field is non-conservative? – Philip Wood Jul 14 '18 at 16:44
• Still, the magnetic field created by monopoles is conservative, Right? – NK Khiwaal Jul 15 '18 at 4:51
• "Do we insist that for a field to be conservative there are no currents or time-varying electric fields at any point?" Yes. – Chris Aug 16 '18 at 15:12
• @NKKhiwaal As far as we know today there exist no magnetic monopoles since the magnetic field is sourcefree. – EuklidAlexandria Aug 20 '18 at 10:55
• There are non conservative fields such that $\vec\nabla\times \vec F=0$. – Diracology Sep 11 '18 at 13:21