Work done by the Magnetic Force

Question

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?

Answer 1

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.

This is explained here: Does a static electric field and the conservation of momentum give rise to a relationship between $E$, $t$, and some path $s$?

Answer 2

Work done by net magnetic force is zero. But one of its components may not be zero. As we had seen when block slips from an angle on incline plane, its normal was zero, but its horizontal component helped it to move.

Answer 3

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

Answer 4

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$</p>

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.

Answer 5

From here: http://en.wikipedia.org/wiki/Conservative_force

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.