 Solution

The Lorentz force in the uniform magnetic field giveswhere the last step is the vector triple product identity. HenceOn the other hand, $|\mathbf{r}\times \mathbf{B}{|}^2=B^2{r}^2-(\mathbf{r}\cdot \mathbf{B}{)}^2$, soThe derivatives cancel, proving the magnetic angular-momentum invariantThe magnetic Lorentz force is perpendicular to velocity, so its power is zero:Thus the kinetic energy is constant.

For the alternative kinetic energy expression, work on an interval where $r>0$ so that the radial unit vector $\mathbf{u}=\mathbf{r}/r$ is defined. Differentiating $\mathbf{u}\cdot \mathbf{u}=1$ once and twice givesSince $\mathbf{v}=\dot{r}\mathbf{u}+r\dot{\mathbf{u}}$, it follows that $\mathbf{u}\cdot \mathbf{v}=\dot{r}$ and $v^2={\dot{r}}^2+r^2|\dot{\mathbf{u}}{|}^2$. Therefore the unit-direction kinetic-energy identity isThis identity holds for any twice differentiable trajectory away from the origin, not only for motion in a magnetic field.