Larmor precession

Let z-axis be the direction of the magnetic field

and theta is the angle between the spin axis and the magnetic field $\mu = IA$ $I = \frac{q}{2\pi /\Omega }=\frac{\Omega q}{2\pi} , A = \pi r^{2}$ $\therefore \mu = q \Omega r^{2} /2$ $\vec{\tau} = \vec{\mu}\times \vec{B}$ so, torque is in the plane perpendicular to the field $\tau = q \Omega B r^{2}sin\theta /2$ and $\vec{\tau} = \frac{\mathrm{d}\vec{L} }{\mathrm{d} t}$ angular momentum in z-direction is constant, but the direction of angular momentum perpendicular to the field changes. so, $\frac{\mathrm{d} L_{r}}{\mathrm{d} t} = L_{r}sin\theta \omega$ and, $L_{r}=mr^{2}\Omega$ thus, $q \Omega B r^{2}sin\theta /2\ = mr^{2} \Omega sin\theta \omega$ \ $\omega = \frac{qB}{2m}$

This phenomena is well known as Larmor precession you can find more about it here: http://en.wikipedia.org/wiki/Larmor_precession

Note that torque is always perpendicular to the $p_m$ which is parallel to the $L$ and it's telling us that $L=const$ . So on the picture bellow we can easy see relations:

$L\sin{\theta}=\frac{Mdt}{d\phi}$

$M=p_mB \sin\theta$

$\frac{L}{p_m}=\frac{2m}{q}$

$d\phi = \omega dt$

From these simple equations you will find that:

$\omega = \frac{qB}{2m}$