 Solution

Take $\mathbf{B}=B\hat{\mathbf{z}}$. The magnetic Lorentz force givesThus $v_x+i{v}_y=(v_x(0)+i{v}_y(0))e^{-i\Omega t}$, 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.