# Rotating square coil in constant magnetic field

1. Apr 17, 2009

### snufkin

1. The problem statement, all variables and given/known data

There is one square coil of area A that rotates with speed $$\omega$$ around axis y. The magnetic field is uniform and points to the z direction. Show that the torque is $$\frac{A^2B^2N^2\omega sin\omega t}{R}$$.

2. Relevant equations
$$T = |\underline{m} \times \underline{B}|$$
$$\underline{m} = N I A$$
Thus $$T = NIABcos(\omega t)$$

$$EMF = NIR$$
$$EMF = - N\frac{d \Phi}{d t}$$

3. The attempt at a solution

Now I have read many similar questions on this forums, and I think I am close to fully understand the problem itself, but there is a piece missing, and I don't see from my initial equations where I am going wrong.

I calculate I via the EMF, substitute it into T. But with this I end up with one N, whereas the supposed solution that I need to prove in the exercise is $$T = \frac{A^2B^2\omega sin^2\omega t}{R}$$.

I am assuming one of the initial equations wrong, I suspect the EMF section, because I not understand it. Any help would be appreciated.

Last edited: Apr 17, 2009
2. Apr 17, 2009

### snufkin

Thinking of V that is probably wrong, the number of loops has nothing to do with the potential, so $$V = IR$$.

So EMF is $$-N \cdot \frac{d \Phi}{d t}$$ because we take $$\Phi$$ at each loop, and just sum those up. I am wondering now why can we actually do that, since in a different alignment the flux in each loop must be different.

3. Apr 18, 2009

### Redbelly98

Staff Emeritus
The flux is the same for each loop, so yes EMF = -N dΦ/dt