= Solution
The <Lorentz force> is
$$
\boxed{\mathbf F=q(\mathbf E+\mathbf v\times\mathbf B)}.
$$
For $\mathbf E=0$, the magnetic part is perpendicular to the <velocity>, so its instantaneous <power> $\mathbf F\cdot\mathbf v$ vanishes. The <speed> and <kinetic energy> therefore remain constant. Decompose the initial <velocity> into $v_\parallel$ along the uniform <magnetic field> and $v_\perp$ perpendicular to it. The parallel component is unchanged, while the perpendicular component rotates uniformly.
For a nonrelativistic particle with $qB\ne0$, the resulting helical motion has
$$
\boxed{r_L=\frac{mv_\perp}{|q|B},\qquad\omega_c=\frac{|q|B}{m},\qquad T=\frac{2\pi m}{|q|B},\qquad h=v_\parallel T}.
$$
Here $r_L$ is the <Larmor radius>, $\omega_c$ the <cyclotron frequency> in angular units, and $h$ the axial advance per turn. Opposite signs of <electric charge> give opposite senses of rotation. Pure perpendicular motion gives a circle; pure parallel motion gives a straight line. If $q=0$ or $B=0$, there is no magnetic deflection. At relativistic <speeds>, replace $m$ in the radius and period by $\gamma m$; $\gamma$ stays constant because the magnetic <force> does no work.
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