# Integration of Torque for a Circular Current Loop in Magnetic Field [closed]

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

• Your fundamental misunderstanding is in what $I$ refers to; here it refers to current. See magnetic moment for more info. Aug 15 '20 at 16:23
• Thank you, I updated my question, as I could still not even get close to that form. Aug 15 '20 at 16:29

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.

• 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. Aug 16 '20 at 15:59
• I also edited the original question in order to apply vector notation. Aug 16 '20 at 16:47

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 $$\alpha$$.