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 $\vec{P} \times \vec{E}?$

  • $\begingroup$ for a circulr loop yes it is always about the centre ,but it is not applicable to all kind of bodies ,point we choose in such kind of faishon that each $dl$ length experince same magnitude of force ! $\endgroup$ Apr 14, 2020 at 10:59
  • $\begingroup$ physics.stackexchange.com/questions/129068/… $\endgroup$ Apr 14, 2020 at 10:59
  • 2
    $\begingroup$ On any loop, the net force due to magnetic field is zero. So you can take torque about ANY point, every calculated torque would be the same. You can prove this for a system of particles by assuming their centre of mass and take torque about any point. Simplify your equations. You will get torque abot COM = torque about general point P $\endgroup$
    – ba-13
    Apr 14, 2020 at 11:11
  • 1
    $\begingroup$ @YuvrajSingh. I didnt quite understand what your comment meant. T=MxB is valid about the COM for a circular loop only?? $\endgroup$
    – satan 29
    Apr 14, 2020 at 11:36
  • 1
    $\begingroup$ If you see the derivation, we consider "couples" on the coil. It's easy to prove that torque due to a "couple" is same about any point, and hence your conclusion follows. $\endgroup$
    – user600016
    May 7, 2020 at 14:51

2 Answers 2


$\overrightarrow{\tau} = \overrightarrow{M} X \overrightarrow{B}$ 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.


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: $$\vec \tau=\vec M \times \vec B$$

  • $\begingroup$ (replying to the question) 1. Apart from "symmetry" how can we prove that the integral vanishes for a general case? What did you mean by general case? Did you mean $\int_0^\pi$ rather than $\int_0^{2\pi}$? 2. Can we generalize this further to the torque in an electric field given by $\vec P\times \vec E$ ? Our case is magnetostatics rather than electromagnetism in the question. $\endgroup$ Mar 25, 2022 at 17:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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