# Question #01eef

Apr 18, 2017

I'm assuming the physical set up looks like this:

Assuming the current, $I$, is moving CCW as indicated, then the effect of the B field, again acting in the direction shown, will be to create a torque $\vec{\tau}$ that spins the loop about it's axis as shown. In this set-up, from the right hand rule, the force vectors on the RHS of the blue vertical axis will act out of the page, and vice versa:

So we do that calculus thing and linearise a small element of the circular loop:

The force $\vec{F}$ on a current-carrying wire of length and orientation $\vec{L}$, due to a magnetic field $\vec{B}$ (BTW this is all derived from the Lorentz Force Law), is:

$\vec{F} = I \left(\vec{L} \times \vec{B}\right)$

From the drawing: $\left\mid \vec{L} \right\mid = R d \theta$

So, again from the right hand rule, the force $d \vec{F}$ on this small element is:

$d \vec{F} = I \left\mid \vec{L} \right\mid \left\mid \vec{B} \right\mid \sin \left(\frac{\pi}{2} - \theta\right) \setminus \hat{z}$, i.e. acting out of the screen/page.

$= I \setminus R \setminus d \theta \setminus B \setminus \cos \theta \setminus \hat{z}$

The associated torque $d \vec{\tau}$ is:

$d \vec{\tau} = \vec{r} \times d \vec{F}$, where $\vec{r}$ is the small element's position vector relative to its axis of rotation .

In this case: $\vec{r} = R \cos \theta \setminus \hat{x}$

And so:

$d \vec{\tau} = R \cos \theta \setminus \hat{x} \times I \setminus R \setminus d \theta \setminus B \setminus \cos \theta \setminus \hat{z}$

$= - I B {R}^{2} {\cos}^{2} \theta \setminus d \theta \setminus \hat{y}$

$\implies \vec{\tau} = - I B {R}^{2} {\int}_{0}^{2 \pi} {\cos}^{2} \theta \setminus d \theta \setminus \hat{y}$

Simple maths sub and...

$= - \pi I B {R}^{2} \setminus \hat{y}$

That means that:

$\implies \left\mid \vec{\tau} \right\mid = \pi I B {R}^{2}$

$= \pi \left(28 \times {10}^{- 3}\right) \left(0.9\right) {\left(\frac{1.3}{2}\right)}^{2} = 0.033 N m$