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Since the magnetic force is a no work force, $dW=\vec F\cdot d\vec r=0$ for $\vec F(\vec r)=q(\vec v(\vec r) \times \vec B(\vec r))$, therefore $\oint \vec F \cdot d\vec r=0$ by Stoke's theorem. Therefore, $\vec\nabla \times \vec F=0$. If this conclusion is true then expanding $\vec\nabla\times[\vec v(\vec r) \times \vec B(\vec r)]$ explicitly using the relevant vector identity one must also obtain zero. But it seems to be non-zero in general. So my question is whether $\vec\nabla\times\vec F=0$ or not?

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I) It is true that for Lorentz magnetic force

$$\tag{1} {\bf F}~=~q{\bf v}\times {\bf B},$$

that hence the corresponding (infinitesimal) work along the particle's trajectory

$$\tag{2} dW~=~{\bf F}\cdot d{\bf r}~=~{\bf F}\cdot {\bf v}dt~=~q({\bf v}\times {\bf B})\cdot {\bf v}dt~=~0$$

is zero, simply because the triple product vanishes.

II) The velocity ${\bf v}$ is typically not a function of position ${\bf r}$, cf. e.g this Phys.SE post, so therefore the traditional notion of a conservative force does not apply. For a generalization of the notion of conservative force to velocity-dependent forces, see this Phys.SE answer.

III) If we artificially imagine a system where the velocity ${\bf v}({\bf r})$ is a function of position ${\bf r}$ only, then it is easy to find examples where the force (1) is not rotation-free, or equivalently, not a gradient field. This is not in contradiction with eq. (2), since eq. (2) does only apply for the actual path of the particle; not necessarily for an arbitrary virtual path.

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@ Qmechanic- If we take $\vec v$ to be independent of $\vec r$, then can we not consider $\vec v \times \vec B(\vec r)$ to be a legitimate vector field and find out its curl? As a mathematical problem can we not calculate curl of $\vec v \times \vec B(\vec r)$ directly and show it to be zero? What about the conclusion from Stoke's theorem about the curl of $\vec F$, now? – SRS May 3 '14 at 12:00

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