# Torque experienced by a coplanar loop of current in a uniform magnetic field

There are a lot of posts on this already, but apparent all of them just consider some special case. I am now struggling with this more general case.

Let there be a magnetic field with strength $$B$$. Let $$C$$ be a loop of wire with current $$I$$ in it. $$C$$ lies on a plane, the normal of which makes an angle $$\theta$$ with the magnetic field lines. $$C$$ encloses an area $$A$$.

My goal is to compute the torque. The formula for this is $$\boldsymbol\tau=I\int_C \mathbf r\times (d\mathbf r \times \mathbf B),$$ and ideally, I should get the answer $$AB\sin \theta$$. However, I am facing two difficulties: I don't know how to choose the origin; the torque clearly depends on the choice of origin, so it is impossible to obtain the answer $$AB\sin \theta$$ if I don't know where the origin is. Correction: torque as a vector, of course, does not depend on the choice of origin, so if we can evaluate it as an vector, then it's OK. Secondly, when evaluating the integral, I find it difficult to get the factor $$A$$ out of it. I know that $$A=\frac12 \left|\int_C \mathbf r \times d\mathbf r\right|,$$ but the problem is that cross products are not associative - so I cannot get an $$A$$ out of the formula for $$\boldsymbol\tau$$.

How could I overcome those problems?

• Torque computed as a vector doesn’t depend on the choice of origin. You’re perhaps thinking about computing the torque about some chosen point, which is a different thing. – Bob Jacobsen Oct 26 '19 at 3:59
• @BobJacobsen Oh, I am confusing myself. Actually, I just want to compute it somehow. Could you tell me how to compute it as a vector? Thank you. – Ma Joad Oct 26 '19 at 7:38
• hint: $\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )=(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {a} \cdot \mathbf {b} )\mathbf {c}$ (see: en.wikipedia.org/wiki/Triple_product#Vector_triple_product ) – hyportnex Oct 26 '19 at 17:45
• Ok. Now I get the answer. – Ma Joad Oct 27 '19 at 0:59

$$\def\r{\mathbf r}\def\B{\mathbf B}\boldsymbol\tau=I\int_C \mathbf r\times (d\mathbf r \times \mathbf B)\\ =I\int_C (\r.\B)d\r-(\r.d\r)\B\\ =I\int_C (\r.\B)d\r-I\B\int_C\frac12 d(\r.\r)\\ =I\int_C (\r.\B)d\r-0$$ since $$C$$ is a closed curve. In the following expressions, summation conventions apply. Let $$\def\e{\mathbf e}\r=r_i\e_i$$. $$\nabla\times \e_i=0$$. $$\int_C (\r.\B)d\r=\int_C (\r.\B)\e_idr_i\\ =\sum_i\int_{\Sigma} \nabla\times ((\r.\B)\e_i) .\mathbf n dA\\ =\sum_i\int_{\Sigma} (\nabla(\r.\B)\times \e_i+(\r.\B)(\nabla\times \e_i) ).\mathbf n dA\\ =\sum_i\int_{\Sigma} (\B\times \e_i) .\mathbf n dA =\sum_i(\B\times \e_i) .\left(\int_{\Sigma}\mathbf n dA\right)\\ =\sum_iA\mathbf n.(\B\times \e_i)=A\sum_i\e_i.(\mathbf n\times\B)=A(\mathbf n \times \B)\\ =\mathbf A\times \mathbf B.$$