A particle of mass $m$ and charge $q$ has position vector $\mathbf{r}(t)$ and moves in a constant, uniform magnetic field $\mathbf{B}$ so that its equation of motion is

$$m \ddot{\mathbf{r}}=q \dot{\mathbf{r}} \times \mathbf{B}$$

Let $\mathbf{L}=m \mathbf{r} \times \dot{\mathbf{r}}$ be the particle's angular momentum. Show that

$$\mathbf{L} \cdot \mathbf{B}+\frac{1}{2} q|\mathbf{r} \times \mathbf{B}|^{2}$$

is a constant of the motion. Explain why the kinetic energy $T$ is also constant, and show that it may be written in the form

$$T=\frac{1}{2} m \mathbf{u} \cdot\left((\mathbf{u} \cdot \mathbf{v}) \mathbf{v}-r^{2} \ddot{\mathbf{u}}\right)$$

where $\mathbf{v}=\dot{\mathbf{r}}, r=|\mathbf{r}|$ and $\mathbf{u}=\mathbf{r} / r$.

[Hint: Consider u $\cdot \dot{\mathbf{u}} .]$