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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

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  • $\begingroup$ Your fundamental misunderstanding is in what $I$ refers to; here it refers to current. See magnetic moment for more info. $\endgroup$
    – DanDan0101
    Aug 15 '20 at 16:23
  • $\begingroup$ Thank you, I updated my question, as I could still not even get close to that form. $\endgroup$
    – Genoma
    Aug 15 '20 at 16:29
  • $\begingroup$ This might help you. $\endgroup$
    – SarGe
    Aug 16 '20 at 9:07
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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.

enter image description here

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.

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  • $\begingroup$ Exactly what I was looking for, Thank you!The key here is to think about the problem in an xy plane. I apologize for the notation, I was not aware that tau is the globally recognized symbol for Torque. $\endgroup$
    – Genoma
    Aug 16 '20 at 15:59
  • $\begingroup$ I also edited the original question in order to apply vector notation. $\endgroup$
    – Genoma
    Aug 16 '20 at 16:47
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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 $\alpha$.

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