Integration of Torque for a Circular Current Loop in Magnetic Field I am trying to derive the formula for Torque on a circular current loop inside a magnetic field.
I know the formula is:
$\tau = IAB\sin{\theta}$
Where I is the current, B is the magnetic field and A is the Area.
My attempt so far:
$d\vec{F} = I\,d\vec{s}\times \vec{B} = IB\,ds\cdot\sin{\alpha}$
Now, if the formula for Torque is: $\tau=bF\sin{\theta}$, and $b = r\sin{\alpha}$, then
$d\tau = r\cdot sin{\alpha}\cdot IB\sin{\theta}ds\cdot \sin{\alpha} = rIBsin{\theta}\cdot\sin^2{\alpha}\,ds$
Ultimately, if I take the integral of this last equation, I cannot exactly understand how to integrate $\sin{\alpha}^2\,ds$.
I guess that my underlying misunderstanding lies here: I can tell what the integral of $d\vec{s}\times \vec{B}$ will be, since I know the diameter of the circle. However, I think there is no way to express $\sin{\alpha}$ with respect to $ds$.
Am I getting this wrong? Thank you
 A: You didn't use vector notations so it seems to be quite terrible. Also, you've used $M$ for torque (it should be $\tau$) rather than for magnetic moment (which are generally accepted symbols).
Proof:
A circular loop lies in $x-y$ plane with raduis $r$ and center at origin $O$. It is carrying a constant current in anti-clockwise direction. There is uniform magnetic field $\vec B$ directed along positive $x$-axis.

Consider an element $d\vec s$ on the ring at an angle $\theta$ subtending an angle $d\theta$ at the origin. Torque on this element is given by
$$\begin{align}d\tau&=\vec r\times d\vec F=\vec r\times(Id\vec s\times\vec B)\\ 
&=I(r\cos\theta\ \hat i+r\sin\theta\ \hat j)\times\bigg((-rd\theta\sin\theta\ \hat i+rd\theta\cos\theta\ \hat j)\times(B_0\ \hat i)\bigg)\\
\tau&=I\bigg(\int_0^{2\pi}B_0r^2\cos^2\theta\ d\theta\ (\hat j)-\int_0^{2\pi}B_0r^2\sin\theta\cos\theta\ d\theta\ (\hat i)\bigg)\\
&=I(\pi r^2)B_0\ \hat j=(I\pi r^2\ \hat k)(B_0\ \hat i)\\
&=\vec M\times\vec B
\end{align}$$

Note: I've skipped the calculation part. Also, you can also take $\vec B=B_x\ \hat i+B_y\ \hat j +B_z\ \hat k$, I've taken only $x$-component for simplicity. The result will reamin the same. Same with the shape of conductor, doesn't matter whether square or circle.
A: I solved this by realizing that ds is actually $2r\cdot sin(d\alpha/2)\cdot sin(\alpha)$ by the length chord formula.
In short, by actually writing $d\vec{s}\times \vec{B}$ in terms of $\alpha$.
