= Solution
Take $\mathbf B=B\hat{\mathbf z}$. The magnetic <Lorentz force> gives
$$
\dot v_x=\Omega v_y,\qquad \dot v_y=-\Omega v_x,\qquad \dot v_z=0,\qquad \Omega=\frac{qB}{m}.
$$
Thus $v_x+iv_y=(v_x(0)+iv_y(0))e^{-i\Omega t}$, while $v_z$ is constant. Integrating the transverse velocity gives a circle of radius $r_L=v_\perp/|\Omega|$ about a fixed guiding centre; adding $z(t)=z_0+v_\parallel t$ gives a <helical motion in a uniform magnetic field>. The <magnetic force> does no work, so the speed and kinetic energy remain constant. The handedness reverses with the sign of $q$. If $v_\parallel=0$ the helix degenerates to a circle; if $v_\perp=0$ it is a straight line parallel to the field.
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