 Solution

In classical charged particle in a uniform magnetic field motion, the Lorentz force and Newton's second law giveAssume $\mathbf{B}\ne 0$ and choose an orthonormal coordinate system with $\mathbf{B}=B{\mathbf{e}}_z$, $B=|\mathbf{B}|$. Write $\Omega =qB/m$, retaining the sign of the electric charge. The velocity equations areThus the parallel velocity is constant, and the perpendicular velocity rotates at constant angular frequency. Choosing the time origin and perpendicular axes conveniently givesIntegrating the velocity, one obtainsThe first two coordinates describe a circle of radius $m{v}_{\perp }/(|q|B)$, while the third advances uniformly. This is helical motion in a uniform magnetic field: the helix has axis parallel to $\mathbf{B}$ through the guiding centre. Its cyclotron frequency isand its signed angular velocity vector about that axis is $-q\mathbf{B}/m$. The minus sign follows from the ordering $\mathbf{v}\times \mathbf{B}$ in the Lorentz force. The helix degenerates to a circle when $v_{\parallel }=0$ and to a straight line when $v_{\perp }=0$. If $q=0$ or $B=0$, there is only straight-line motion. In a relativistic extension with fixed Lorentz factor $\gamma$, the frequency becomes $|q|B/(\gamma m)$; the classical result above uses $m\mathbf{v}$ as momentum.