Relation between torque and magnetic moment

[Happiness]

The magnetic moment $\vec{M}$ of a charged particle is defined as $\vec{M}=\frac{1}{2}q\,(\,\vec{r}\times\vec{v})$.

Starting with torque $\vec{N}=\vec{M}\times\vec{B}$, I arrive at a contradiction.

Consider a charge particle moving at a constant speed $v$ anticlockwise in a circle of radius $r$. A magnetic field $\vec{B}$ is applied parallel to the plane of this circle. Consider the instant when $\vec{B}$ is in the same direction as $\vec{r}$, the position vector of the particle, taking the center of the circle as the origin. Thus, $\vec{r}\times\vec{B}=0$.

Then, $\vec{N}=\frac{1}{2}q(\vec{r}\times\vec{v})\times\vec{B}$
$=\frac{1}{2}q[\vec{r}\times(\vec{v}\times\vec{B})-\vec{v}\times(\vec{r}\times\vec{B})]$
$=\frac{1}{2}\vec{r}\times\vec{F}-0$
$=\frac{1}{2}\vec{N}$, a contradiction.

What's wrong?

 

[Charles Link]

The magnetic moment is a current loop and the particle needs to travel in a circle with the charge distribution spread out over the complete ring. The different locations on the ring do not all experience the same torque. Meanwhile, please check your double cross product: I think $a \times (b \times c)=(a \cdot c)b-(a \cdot b)c$ . The definition of the magnetic moment that you have with the 1/2 factor is correct.