A charged particle going around in a circle can be considered to be...

$i = \dfrac{q}{T} = \dfrac{qv}{2\pir}$

$M = i \times \pi{r}^{2}$

$= \dfrac{2v}{2\pir} \times \pi{r}^{2}$

$= \dfrac{1}{2}qv \times \dfrac{mv}{qB}$

Direction of $\overset{\to}{M}$ is opposite of $\overset{\to}{B}$ therefore

$\overset{\to}{M} = \dfrac{- {mv}^{2}\overset{\to}{B}}{2{B}^{2}}$