The magnetic part of the Lorentz force acts perpendicular to the charge's velocity, and consequently does zero work on it. Can we extrapolate this statement to say that such a nature of the force essentially makes its corresponding work independent of the choice of path, and hence that the magnetic force is conservative?
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Not really, because the magnetic force is velocity dependent, not solely position dependent, so you can't extrapolate from knowing the integral along a path is zero to the conclusion that the force is the gradient of a potential.
What you can do is make an analog of the potential argument for the momentum components, so that the magnetic field is the curl of a vector potential. This argument can be made physically for conservation of momentum around a space-time loop, much like the conservation of energy follows from the integral of the force along a space-loop.
Well, the work is zero regardless of the path you take, so in this sense the magnetic force is (trivially) conservative.
Work done by the magnetic force is indeed zero.
$F_m=q[V \times B]$
$dW=q[V \times B].dr =q[V\times B]Vdt=0$
Consequently for work done along some arbitrary path from A to B$$=\int_A^B dW=0$$
Work done by the magnetic force is independent of path[and equal to zero for all such paths]
However we may have the following interesting consideration:
[We may choose $V_1$ and $V_2 $so that $V_1,V_2$ and $B$ are not in the same plane]
$=q[V_1 \times B]V_1+q[V_1 \times B]V_2+q[V_2 \times B]V_1+q[V_2 \times B]V_2$
$0=q[V_1 \times B]V_2+q[V_2 \times B]V_1$
The quantities $q[V_1 \times B]V_2 $and $q[V_2 \times B]V_1$ may not be zero individually[considering tha fact that$ V_1,V_2$ and $B$ do not lie in the same plane according to our choice] though their sum is zero.We might think of using either of them for some technological purpose.
So it is just a matter of definitions.