# Torque due to magnetic force on a loop

I read that the torque due to the force exerced by a uniform magnetic field $\mathbf{B}$ on a closed conducting rectangular loop is $$\boldsymbol{\tau}=I\mathbf{S}\times \mathbf{B}$$where $\mathbf{S}$ is a vector whose orientation is determined by the right hand rule applied to the flow of the electric current, whose intensity is $I$, in the wire and whose norm $S$ is the area of the loop.

Then the book says that, by approximating any plane loop with rectangles, we get the same result $\boldsymbol{\tau}=I\mathbf{S}\times \mathbf{B}$ for any plane loop. What does approximating mean in this context, translating it into rigourous mathematical language? I heartily thank any answerer.

I have got an idea, which follows. If $\boldsymbol{\ell}:[a,b]\to\mathbb{R}^3$ is a parametrisation of the loop, I think that the torque with respect to the origin (but it is the same with respect to any point because the resultant force is null) is $$\boldsymbol{\tau}=\int_\gamma \boldsymbol{\ell}\times (I\,d\boldsymbol{\ell}\times\mathbf{B}):=\int_a^b \boldsymbol{\ell}(s)\times \left(I\,\frac{d\boldsymbol{\ell}(s)}{ds}\times\mathbf{B}\right)ds$$ I suspect that the appromixation talked about means that, if $\boldsymbol{\ell}_n:[a,b]\to\mathbb{R}^3$ is a parametrisation of the contour (red in the figure) of the union of rectangular loops (whose interior sides are purple) such that the series $\{\boldsymbol{\ell}_n\}$ uniformly converges to $\boldsymbol{\ell}$, then $$\boldsymbol{\tau}_n:=\int_a^b \boldsymbol{\ell}_n(s)\times \left(I\,\frac{d\boldsymbol{\ell}_n(s)}{ds}\times\mathbf{B}\right)ds\xrightarrow{n\to\infty}\int_a^b\boldsymbol{\ell}(s)\times \left(I\,\frac{d\boldsymbol{\ell}(s)}{ds}\times\mathbf{B}\right)ds$$ Am I right? If I am, how can such convergence be proved?

Using small sections from any loop we get:

$\tau = I \Delta r \times(\Delta l \times B)$

Now use rule for triple cross product:

$A \times B \times C = B(A.C) - C(A.B)$ //back cap

Now it's clear that you are looking at projections and because it is a planar loop, so its projections can be modeled as a rectangle.

As you know, take more of them and you won't worry about their width as the force will pass through the line of rotation giving zero torque.

• Thank you very much for your answer! I suspect I did not make myself understood: I do realise how that rectangles can be chosen to approach the loop's enclosed area, but how do we rigourously see that the torque exerced on the loop is the limit of the torque exerced on the loops made of rectangles? I've added an edit about what I think that "approximating" means in that context and my doubts about how and why the approximation works... – Self-teaching worker Dec 15 '15 at 14:51
• Can you clarify more about what do you mean by $l_{n}$ and $l$?? – I.Omar Dec 15 '15 at 15:40
• $\boldsymbol{\ell}$ (what you call $\Delta r$, say) is a parametrisation (i.e. the position of the point of the loop w.r.t. the origin) of the loop (black in the figure) and $\boldsymbol{\ell}_n$ is a parametrisation of the red polygonal chain. – Self-teaching worker Dec 15 '15 at 23:30

When you decompose the loop into small rectangles, you can complete a circuit around each rectangle as the current will be equal and opposite in all the interior segments. Each small rectangle then experiences the torque you say. The other important point is that when adding up the torques to get the overall torque you don't care where each small torque is applied. Many people intuitively think a torque applied out at the edge counts for more than one applied at the center. That is true for forces generating torques, because of the $\vec F \times \vec r$ definition of a torque, but if you already have torques you just add them up.

• Thank you so much! I understand that the resultant torque due to the force acting on the interior segments is 0, but I do not understand how do we rigourously see that the torque is the limit of the torques exerced on the contour (parametrised by $\boldsymbol{\ell}_n$, with the notation of the OP) of the union of rectangles... – Self-teaching worker Dec 15 '15 at 15:07
• @Self-teachingDavide: When we do the integration, the torque on the interior loops is not zero, it is $I\Delta S \times B$. They all contribute. This is how we add up all the small $\Delta S$'s to get the one big $S$. The currents cancel in the interior, so our total current distribution matches. – Ross Millikan Dec 15 '15 at 15:33
• Thank you again! Yes, I had understood that: it is the resultant toque that is 0, because for any force acting on an interior segment (part of the purple net in the figure I've added) belonging to a rectangular loop there is an opposite force acting in the same point on the side (flown through by an oppositely oriented current) of an adjacent rectangular loop – Self-teaching worker Dec 15 '15 at 19:51
• The resultant force is zero, but the resultant torque is not. The current flows inside the loop cancel because one little rectangle will have the flow one direction while the neighboring one will have it the other. The torques, however, are all in the same direction because of the cross product. They are nonzero and add. That is why the total torque is proportional to the area of the loop. – Ross Millikan Dec 15 '15 at 20:20
• Yes, I've well understood that. What I don't understand is why the toque acting on any loop (like the black one in the figure of the original post), even if it hasn't got a right angled profile similar to the red approximation (see figure), can be considered as a limit of the torques acting on the union of such rectangular loops. It doesn't appear trivial at all to my mathematically uneducated eyes... – Self-teaching worker Dec 15 '15 at 23:39