Torque due to magnetic field My textbook states That the torque $\vec{\tau}$ experienced by a current carrying loop due to a magnetic    field $\vec{B}$, is given by the equation $\vec{\tau} =\vec{M} \times \vec{B}$,where $\vec{M}$ denotes the magnetic moment of the current loop=$I\vec{A}$. However, it didnt specify about what point(s) is this equation valid. The most obvious choice seems to be the Center of mass of the loop.
For a uniform circular Loop, I was able to derive that this indeed holds true about the COM. However, we can extend this further for ANY  random point: 
Say the COM is $c$ and we wish to calculate the torque of magnetic field about a point say $p$. Take a current element $I\vec{dl}$. The force experienced by this Current element is $\vec{dF}$=$I\vec{dl} \times\vec{B}$. If the position vector Of point P wrt to the COM is $\vec{r_p}$, and that of the current element is $\vec{r_c}$, Then the torque, $\vec{dT}$ about point P is $(\vec{r_c}-\vec{r_p}) \times\vec{dF}$ = $\vec{r_c}\times\vec{dF}-\vec{r_p}\times\vec{dF}$. Integrating across the entire loop, The first integral becomes $\vec{M} \times \vec{B}$, while the second becomes zero.Which seems to suggest that the torque about ANY point is the same, and is $\vec{M} \times \vec{B}$.
1) How can we prove that for any , arbitrary loop,the Torue about COM is $\vec{M} \times \vec{B}$?
2)How do we show that the second integral cancels out? In the case I described above, the second integral had a term of $\int_{\theta=0} ^ {2\pi} \cos(\theta)\mathrm d\theta$ which became zero. Apart from "symmetry" how can we prove that the integral vanishes for a general case? 
3) 1) and 2) together imply whatever be the loop, the Torque due to magnetic field about every point is the same. Can we generalize this further to the torque in an electric field given by $\vec{P} \times \vec{E}?$
 A: $\overrightarrow{\tau} = \overrightarrow{M} X \overrightarrow{B}$ is valid only in uniform magnetic field. Any point can be taken as origin and torque will be same.
To prove it, first of all prove it for a rectangular loop. Handling rectangular loop is easier than circular loop. You will also find that net force is zero. This means that this torque is a couple. Torque of a couple does not depend on origin. This is a direct result from rotational mechanics.
Now we can argue that any loop is made up of large number of rectangular loops! We get the required result immediately!
This result can be extended to $\overrightarrow{\tau} = \overrightarrow{p} X \overrightarrow{E}$ in uniform electric field.
A: Do you remember that $\vec \tau=\vec r\times \vec F$?
I didn’t find direct derivation of torque due to magnetic field. (For electric field also) but we ignore electro- when we are thinking of torque due to magnetic field and vice-versa.
$$\vec F=Id\vec l \times \vec B$$
For now forget that RHS is vector (you can assume that angle between axis and magnetic field is perpendicular)
$$\vec F= IBdl$$
$$\vec \tau = \vec r\times \vec F$$
$$=rF=rIBdl$$
$$[rdl]=[L^2]=[\text{Area}]$$
$$\vec \tau=AIB$$
We know that $$\vec M=A\vec I$$
So, $$\vec \tau = MB$$
Since LHS is vector so RHS should be vector also. Assume that angle between magnetic moment and magnetic field is $90\deg$.
$$\vec \tau = ||M|| \ ||B|| \ \sin(\frac{\pi}{2})$$
Now again assume that angle between them is $\theta$ rather than $\frac{\pi}{2}$
$$\vec\tau = ||M|| \ ||B|| \ \sin\theta \tag{1}\label{1}$$
We know from vector identity,
$$\vec A\times \vec B= ||A|| \ ||B|| \ \sin\theta$$
So equation $\eqref{1}$ becomes:
$$\vec \tau=\vec M \times \vec B$$
