Torque due to magnetic force on a loop

Question

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?

Answer 1

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.

Answer 2

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.