= Solution
Write $\boldsymbol v=\dot{\boldsymbol r}$. The original PDF contains the cross product $\dot{\boldsymbol r}\times\boldsymbol r$; the TeX aid drops its <velocity> dot. Since $\boldsymbol v\times\boldsymbol v=0$, differentiation and the equation of motion give
$$
\frac d{dt}\left[m(\boldsymbol v\times\boldsymbol r)\cdot\boldsymbol B\right]
=m(\ddot{\boldsymbol r}\times\boldsymbol r)\cdot\boldsymbol B
=-e[(\boldsymbol v\times\boldsymbol B)\times\boldsymbol r]\cdot\boldsymbol B.
$$
The radial restoring <force> contributes no <torque>. The vector triple-product identity yields
$$
[(\boldsymbol v\times\boldsymbol B)\times\boldsymbol r]\cdot\boldsymbol B
=B^2\boldsymbol v\cdot\boldsymbol r
-(\boldsymbol v\cdot\boldsymbol B)(\boldsymbol r\cdot\boldsymbol B).
$$
Meanwhile
$$
\frac d{dt}\left[\frac e2|\boldsymbol r\times\boldsymbol B|^2\right]
=e(\boldsymbol r\times\boldsymbol B)\cdot(\boldsymbol v\times\boldsymbol B)
=e\left[B^2\boldsymbol r\cdot\boldsymbol v
-(\boldsymbol r\cdot\boldsymbol B)(\boldsymbol v\cdot\boldsymbol B)\right].
$$
The <derivatives> cancel, proving
$$
\boxed{m(\dot{\boldsymbol r}\times\boldsymbol r)\cdot\boldsymbol B
+\frac e2|\boldsymbol r\times\boldsymbol B|^2=\mathrm{constant}.}
$$
This is the <magnetic axial angular-momentum invariant>. Its first term is the negative of the ordinary <angular momentum> projected on $\boldsymbol B$; retaining that sign is essential.
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