# If the net force on a current loop in a magnetic field is zero, why is torque independent of choice of origin?

Im trying to show that the integral over a closed loop of a crossproduct stays the same if I choose a different origin with $\overrightarrow{r}=\overrightarrow{r}\prime+\overrightarrow{r_0}$ and $\oint{d\overrightarrow{F}}=0$

$$\overrightarrow{N} = \oint{\overrightarrow{r} \times d\overrightarrow{F}} = \oint{(\overrightarrow{r}\prime+\overrightarrow{r_0} )\times d\overrightarrow{F}} = \oint{\overrightarrow{r}\prime \times d\overrightarrow{F}} + \oint{\overrightarrow{r_0} \times d\overrightarrow{F}}$$

How do I show $\oint{\overrightarrow{r_0} \times d\overrightarrow{F}}= 0$, or why can I say $\oint{\overrightarrow{r_0} \times d\overrightarrow{F}} = \overrightarrow{r_0} \times\oint{ d\overrightarrow{F}}$?

To clarify, $\overrightarrow{dF}$ is the force on the current $I \overrightarrow{dl}$ in a loop in a uniform and constant magnetic field. So $\overrightarrow{dF} = I \overrightarrow{dl} \times \overrightarrow{B}$.

You are on the right way with your line of thought. The only part of the puzzle you are missing is, as you have correctly identified, the explanation of $$\oint{\overrightarrow{r_0} \times d\overrightarrow{F}} = \overrightarrow{r_0} \times\oint{ d\overrightarrow{F}}$$ It is actually very simple: $\overrightarrow{r_0}$ is a constant vector (it doesn't depend on coordinates) and as such you can factor it out of the integral. Rewriting the vectors in their component elements does nothing for you.

I am sure you know of the fact that you can factor constants out of integrals, but just for completeness, here is a link to the wikipedia entry for this rule called the Constant factor rule in integration.

You can extract the constant term out of integral directly, however keep the operation unchanged. See:

$$\oint \vec{r}_0\times \mathrm{d}\vec{F}=\oint\epsilon_{ijk}r_{0i}\mathrm{d}F_j\hat{e}_k=\epsilon_{ijk}r_{0j}\oint\mathrm{d} F_j \hat{e}_k=\vec{r}_0\times\oint\mathrm{d}F$$

Similarly:

$$\int \vec{B_0}\cdot \mathrm{d}\vec{S}= \vec{B_0}\cdot \int\mathrm{d}\vec{S}$$

• Ah so to see why I can take out r0, I should split it up in its scalars and then its easier to see. Thanks! I dont understand your notation, what is epsilon_ijk? And why is dFj in the direction of e-hat_k? – Leo Apr 10 '14 at 14:27
• @Leo This is Levi-Civita symbol, usually use to write vector product in a more compact way. – an offer can't refuse Apr 10 '14 at 14:41
• I accepted the other answer because yours went a bit too quickly for me. I didnt understand what you were saying at first. But thanks for also trying and making it more general by showing the same principle with a dot product. – Leo Apr 11 '14 at 6:54