1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Bob Jacobsen Oct 26 '19 at 3:59
  • $\begingroup$ @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. $\endgroup$ – Ma Joad Oct 26 '19 at 7:38
  • $\begingroup$ hint: ${\displaystyle \mathbf {a} \times (\mathbf {b} \times \mathbf {c} )=(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {a} \cdot \mathbf {b} )\mathbf {c} }$ (see: en.wikipedia.org/wiki/Triple_product#Vector_triple_product ) $\endgroup$ – hyportnex Oct 26 '19 at 17:45
  • $\begingroup$ Ok. Now I get the answer. $\endgroup$ – Ma Joad Oct 27 '19 at 0:59
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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