Torque on a current loop in a uniform magnetic field

Question

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.

Answer 1

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.