# Torque on a current loop in a uniform magnetic field

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.

• 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. Aug 16, 2023 at 18:28
• @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!
– Eris
Aug 16, 2023 at 23:59

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.
• 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. Aug 17, 2023 at 12:42
• 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 $\vec{r} = 0$). Aug 17, 2023 at 12:59
• 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. Aug 22, 2023 at 19:29