= Solution
Take the scalar product of <Newton's second law> with the <velocity> $\dot{\mathbf r}$. The <cross product> is perpendicular to its first factor, so
$$
\frac d{dt}\left(\frac12m|\dot{\mathbf r}|^2\right)
=\dot{\mathbf r}\cdot\mathbf F
=-\dot{\mathbf r}\cdot\nabla\phi
+\dot{\mathbf r}\cdot(\dot{\mathbf r}\times\mathbf B)
=-\frac d{dt}\phi(\mathbf r).
$$
Consequently
$$
\boxed{E=\frac12m|\dot{\mathbf r}|^2+\phi(\mathbf r)\quad\text{is constant}.}
$$
This is <conservation of energy>: the sum of <kinetic energy> and <potential energy> stays fixed. The magnetic-type <Lorentz force> $\dot{\mathbf r}\times\mathbf B$ does no <work>, because it is perpendicular to the instantaneous <velocity>. It may turn the trajectory but cannot directly change its speed, regardless of the spatial variation of $\mathbf B$.
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