1
$\begingroup$

I'm trying to prove the result the net torque on a current loop in a uniform magnetic field is $\vec{\tau}_{net}=\vec{\mu}\times \vec{B}$ where $\vec{\mu}=I\frac{1}{2}\oint(\vec{r}\times d\vec{r})$ starting from Lorentz force $$\vec{\tau}_{net}=\sum_i\vec{\tau}_i=\sum_i \vec{r}_i\times (I\vec{r}_i\times\vec{B})\\=I\int\vec{r}\times(d\vec{r}\times\vec{B})$$ now I need to get $\vec{B}$ ouside the integral. I tried Jacobi identity $$\vec{r}\times(d\vec{r}\times\vec{B})=(\vec{r}\times d\vec{r})\times\vec{B}-d\vec{r}\times(\vec{B}\times\vec{r})$$ unless I made a mistake, it remains to prove $d\vec{r}\times(\vec{B}\times\vec{r})=\frac{1}{2}(\vec{r}\times d\vec{r})\times\vec{B}$. I tried triple vector product identity with no luck.

$\endgroup$
2
  • 1
    $\begingroup$ The proof you want can only be made under the integral, but it is difficult. You're better off starting $a\times(b\times c)=b(a\cdot c)-c(a\cdot b)$,, but that proof is tricky. It is done in textbooks, chapter 7 in mine, but is not easy. $\endgroup$ Aug 16, 2023 at 18:28
  • $\begingroup$ @JerroldFranklin That turned out to be a nice exercise. It is a shame that freshmen texts do it for a square loop and make a shaky argument that it can be generalized. I also couldn't find it in advanced textbooks (obviously a remiss from my part). Thanks for pointing me to your book, it is an amazing text! $\endgroup$
    – Eris
    Aug 16, 2023 at 23:59

1 Answer 1

0
$\begingroup$

After Franklin's note that it must be done under the integral, obviously Stokes' theorem would help to turn it into a surface integral. Starting by introducing an arbitrary vector $\vec{k}$ \begin{eqnarray*} \oint d\vec{r}\cdot(\vec{k}\times(\vec{B}\times\vec{r}))&=&\int_S d\vec{A}\cdot(\vec{\nabla}\times(\vec{k}\times(\vec{B}\times\vec{r})))\\ \vec{k}\cdot \oint (\vec{B}\times\vec{r})\times d\vec{r}&=&\int_Sd\vec{A}\cdot\Big(\vec{\nabla}\times((\vec{k}\cdot\vec{r})\vec{B}-(\vec{k}\cdot\vec{B})\vec{r})\Big)\\ &=&\int_Sd\vec{A}\cdot\Big((\vec{k}\cdot\vec{r})\underbrace{\vec{\nabla}\times \vec{B}}_{=0}+\vec{\nabla}(\vec{k}\cdot\vec{r})\times\vec{B}\Big)-\vec{k}\cdot\vec{B}\int_Sd\vec{A}\cdot\vec{r}\\ &=&\int_S d\vec{A}\cdot(\vec{k}\times\vec{B})-\vec{k}\cdot\vec{B}\int_Sd\vec{A}\cdot \vec{r}=\vec{k}\cdot(\vec{B}\times\int_S d\vec{A}-\int_Sd\vec{A}\cdot \vec{r}) \end{eqnarray*} as $\vec{k}$ is arbitrary, $\oint (\vec{B}\times\vec{r})\times d\vec{r}=-\vec{S}\times\vec{B}-\int_Sd\vec{A}\cdot \vec{r}$. Now completing the computation, \begin{equation*} \vec{\tau}_{net}=\oint (\vec{r}\times\vec{dr})\times{B}+\oint(\vec{B}\times\vec{r})\times d\vec{r}=2\vec{S}\times\vec{B}-\vec{S}\times\vec{B}-\vec{C}\\=\vec{S}\times\vec{B}-\vec{C} \end{equation*} where $\vec{C}\equiv \vec{B}\int_Sd\vec{A}\cdot \vec{r}$. I just wanted to conclude it with the same logic I started but of course the approach of Franklin's Classical Electromagnetism is simpler; by the triple vector product identity, noticing $\oint \vec{r}\cdot d\vec{r}=0$ and the vector identity $\oint d\vec{r}\phi=\int_Sd\vec{A}\times\vec{\nabla}\phi$.

Edit: there was a mistake in my derivation. Now it seems an additional term $\vec{C}$ shows up.

$\endgroup$
4
  • 1
    $\begingroup$ The expression $\int_S d\vec{A}\cdot \vec{r}$ can't be transformed that way (already units don't match, the expression is a volume, while the replacement is a length) and it is easily seen from geometrical picture that it is not, in general, zero. $\endgroup$ Aug 17, 2023 at 12:42
  • 1
    $\begingroup$ So the r.h.s. has contribution $\text{const} . \vec{B}$ where the constant depends on the choice of origin of the coordinate system (where $\vec{r} = 0$). $\endgroup$ Aug 17, 2023 at 12:59
  • $\begingroup$ The step in question is Stokes' Theorem done backwards and wrong. The correct step does depend on the origin of $\bf r$. For a plane surface loop, the origin would be at the center and the integral would be zero. For a non-planar loop, the origin could be picked at a point that makes the integral zero. This means that, except for a planar loop, the derivation needs some external (heavenly) help. $\endgroup$ Aug 22, 2023 at 19:29
  • $\begingroup$ Now I'm more confused. It appears this additional term is zero independently of B only around the centroid (where the first moment of area is zero)? But why did it not show up with the derivation starting by the triple vector product identity? was that assumption somehow tacitly implied at any step? $\endgroup$
    – Eris
    Aug 23, 2023 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.