# Work done by the Magnetic Force

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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What's the "it"? Makes what independent of the choice of path? Grammatically, it sounds like you're asking whether the force itself is independent of the choice of path. Trivially the answer is no because different paths go through different fields. –  Mark Eichenlaub Oct 29 '11 at 14:14
@Mark Eichenlaub He means the work (it). –  Andyk Oct 29 '11 at 15:19
@Shrikant Giridhar en.wikipedia.org/wiki/Conservative_force –  Andyk Oct 29 '11 at 15:21
"For example, the magnetic force satisfies condition 2 (since the work done by a magnetic field on a charged particle is always zero), but does not satisfy condition 3, and condition 1 is not even defined (the force is not a vector field, so one cannot evaluate its curl). Accordingly, some authors classify the magnetic force as conservative, while others do not." –  Andyk Oct 29 '11 at 15:21
@Mark Eichenlaub I agree it does look like I am implying that the force is independent of path, sorry for that! And yes I meant work. Anyways, as the Wiki link says, the magnetic force is an unusual case and perhaps we can't pin it down under either category. –  Shrikant Giridhar Oct 29 '11 at 17:40

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.

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Well, the work is zero regardless of the path you take, so in this sense the magnetic force is (trivially) conservative.

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Work done by the total magnetic force,$F_m$ is zero.If we break it[the magnetic force] into two components[$F_m=F_1+F_2$] the work done in two separate directions may be non-zero individually, notwithstanding the fact that the total work is zero.The work done in some specified direction may be non zero. –  Anamitra Palit Dec 30 '11 at 21:58

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:

Let $V=V_1+V_2$

[We may choose $V_1$ and $V_2$so that $V_1,V_2$ and $B$ are not in the same plane]

$dW=0=q[(V_1+V_2)\times B](V_1+V_2)dt$

$=q[V_1 \times B]V_1+q[V_1 \times B]V_2+q[V_2 \times B]V_1+q[V_2 \times B]V_2$

$0=q[V_1 \times B]V_2+q[V_2 \times B]V_1$

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.

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Many forces (particularly those that depend on velocity) are not force fields. In these cases, the above three conditions are not mathematically equivalent. For example, the magnetic force satisfies condition 2 (since the work done by a magnetic field on a charged particle is always zero), but does not satisfy condition 3, and condition 1 is not even defined (the force is not a vector field, so one cannot evaluate its curl). Accordingly, some authors classify the magnetic force as conservative,[3] while others do not.[4] The magnetic force is an unusual case; most velocity-dependent forces, such as friction, do not satisfy any of the three conditions, and therefore are unambiguously nonconservative.

So it is just a matter of definitions.

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