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I have read about magnetostatics and read about Lorentz force which is sum of electric force and magnetic force so it is a electromagnetic force and the mathematical expression of Lorentz force is $F=q(E+v\times B)$. So, if I want to calculate work done due to this force, I get \begin{align} \mathbf F \cdot \mathrm d\mathbf r & = q\mathbf E \cdot \mathrm d\mathbf r+ q (\mathbf v×\mathbf B)\cdot \mathrm dr \\ & =q\mathbf E \cdot \mathrm d\mathbf r+q (\mathbf v×\mathbf B)\cdot \mathbf v \:\mathrm dt \\ & =q\mathbf E \cdot \mathrm d\mathbf r+0 \\ & =q\mathbf E \cdot \mathrm d\mathbf r, \end{align} So it is only due to electrical force. However, the work done due to magnetic force, which is $\mathbf F=q(\mathbf v×\mathbf B)$, is zero.

So we can write, $$∮\mathbf F\cdot \mathrm d \mathbf r=0 .$$ Hence, from Stokes' theorem, we have $$∮\mathbf F \cdot \mathrm d \mathbf r=∬(∇×\mathbf F ) \cdot \mathrm d\mathbf s=0$$

So we can say $∇×\mathbf F=0$, but I read in books that $∇×\mathbf F≠0$.

But it gives no explanation please explain why is this happened how can curl of $\mathbf F$ be non zero?

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The force due to the electromagnetic field is only conservative for static fields. Faraday's law states that $\nabla \times \vec{E}= -\partial B/\partial t$. Thus, only when the B-field (and E-field) are time independent is the curl of the Lorentz force zero.

In fact, for the general case one finds that $$ \nabla \times \vec{F} = - q \frac{d \vec{B}}{d t},$$ which I showed in answer to this closely related question.

The reason that your treatment above does not betray this problem is that you haven't taken into account that whilst the particle moves around in a closed loop, the electric field can change and hence $$ \oint q\vec{E}(t) \cdot d\vec{r} \neq 0 $$

Also note that if the fields are not static then you cannot treat the magnetic and electric parts of the Lorentz force separately since a changing magnetic field is associated with a changing electric field. I would guess then that where you have seen that $F$ is always conservative it refers to static fields and where it says it is non-conservative it refers to the entire Lorentz force in electrodynamics.

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  • $\begingroup$ But here you said for Lorentz force which is F=qE+qv×B but I told only about F=qv×B (only magnetic force) then there is no E so there is no ∇×E so ∇×F=0 $\endgroup$ – user101134 Nov 1 '16 at 16:49
  • $\begingroup$ @user101134 If $\vec{B}$ varies then there is an E-field that varies with time. If the fields are time variable then you cannot separate them as you have attempted and the Lorentz force (which is not only the magnetic component) is not conservative. $\endgroup$ – Rob Jeffries Nov 1 '16 at 16:56
  • $\begingroup$ Ok then if B is not a time depended field then ∇×F=0 right?? $\endgroup$ – user101134 Nov 1 '16 at 18:14
  • $\begingroup$ By the way in ∇×F there are total 5 terms $\endgroup$ – user101134 Nov 1 '16 at 18:41
  • $\begingroup$ By the way in ∇×F there are total 5 terms. ∇×F=q∇×E+q(∇×(v×B)) . =q∇×E+q[(B.∇)v+(v(∇.B))-(∇.V)B -(v.∇)B] but where are the second and fifth term in your calculation? They are not zero $\endgroup$ – user101134 Nov 1 '16 at 18:47

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