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I just have came across information that confuses me: I used to believe that Stokes' theorem of curl is valid everywhere, but in the course of learning magnetostatics I read about the magnetic force $(F=qv×B)$, and my professor told me that this force depends on the velocity of the moving charge so Stokes' theorem is not valid in this case: Instead of being $ ∮F\cdot \mathrm{d}r=0 $, we have $∇×F=0$.

I can't understand why Stokes' thorem is not valid for moving objects. What is the relation of velocity with Stokes' theorem?

ACuriousMind
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user101134
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2 Answers2

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Stokes' theorem is valid everywhere, in the sense that if you have a field $\vec{F}$ defined everywhere, then $\int_{\partial S} \vec{F}\cdot d\vec{s}=\int_S (\vec{\nabla}\times\vec{F})\cdot d\vec{a}$. The problem is, what is your field $\vec{F}$ that you want to define? Since the value of the force $\vec{F}$ doesn't depend on just your position in space but on the particle's velocity as well, it's not possible to unambiguously define a force field at every point in space, since once you get to that point, your particle might have any velocity (and thus any possible force).

I'll add that in the case of the magnetic field itself, you do have a field $\vec{B}$ defined everywhere in space, and so Stokes' theorem applies perfectly well to $\vec{B}$: $\int_{\partial S} \vec{B}\cdot d\vec{s}=\int_S (\vec{\nabla}\times\vec{B})\cdot d\vec{a}$. Combining this with the fact that $\vec{\nabla}\times\vec{B}=\vec{J}$, you get Ampere's Law. So it's not like Stokes' theorem fails for magnetism, it's really just that there is no force field $\vec{F}$ to apply Stokes' theorem to.

ACuriousMind
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Jahan Claes
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  • No actually I read that magnetic force does no work. So F.dr=0 so ∮F.dr=0 also so from stock's theorem ∮F.dr=∬(∇×F).ds =0. So ∇×F=0 but in text book it clearly says that ∇×F≠0. So do you think that stock's theorem is being valid here?? – user101134 Nov 02 '16 at 15:54
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    @user101134 The point is, you can't define the integral of $\vec{F}$ over a surface, because $\vec{F}$ depends on the path you take. You can certainly define $\vec{F}$ along a SPECIFIC path you take, but the step where you convert a line integral to a surface integral is invalid because $\vec{F}$ is not defined on that surface. I don't know where in your textbook it says $\nabla\times\vec{F}=0$, but without context I am going to say that equation makes no sense. – Jahan Claes Nov 02 '16 at 18:07
  • @user101134 And you have been shown that in general, the curl even of the magnetic component of the Lorentz force is not zero if the B-field changes in time. – ProfRob Nov 02 '16 at 21:25
  • $\mathbf{F} \cdot \operatorname{d} \mathbf{r} \neq 0$ for the general path, only for $\operatorname{d} \mathbf{r}$ that are confined to the plane defined by $\mathbf{v}$ and $\mathbf{B}$. The path a particle takes will always have $\operatorname{d} \mathbf{r} = \mathbf{v} \operatorname{d} t$, giving no work done regardless of if the path is open or closed. – Sean E. Lake Nov 03 '16 at 00:52
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Stokes theorem is purely mathematical and is valid for all smoothly differentiable vector fields. If you assume a constant velocity vector $\vec v$, Stokes theorem can also be applied to the magnetic force field $\vec F=q\vec v×\vec B$. It can also be applied in the case that both the vector $\vec v(\vec r)$ and the vector $\vec B(\vec r)$ is a given function of position $\vec r$.

If you consider a single point charge $q$ moving in a magnetic field $\vec B$ without any additional forces, the closed line integral along the point charges path $∮\vec F\cdot \mathrm{d}r$ will always be zero because magnetic fields don't do work on a moving charge, they only change the direction of the velocity.

freecharly
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