Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 9C (Classical Dynamics)

The canonical momentum is $p_i=\partial L/\partial {\dot{x}}_i=m{\dot{x}}_i+q{A}_i$. Its time derivative contains the total derivative along the trajectory: ${\dot{p}}_i=m{\ddot{x}}_i+q{\partial }_t{A}_i+q{\dot{x}}_j{\partial }_j{A}_i$. The Euler-Lagrange equation therefore givesThe magnetic field is $\mathbf{B}=\nabla \times \mathbf{A}$, so the last term is $(\dot{\mathbf{x}}\times \mathbf{B}{)}_i$. With the electric field $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$, this is precisely the Lorentz force law:

The specified vector potential gives $\mathbf{B}=(0,0,B)$. Put $\omega =qB/m$, retaining its sign. The Lorentz force law yields $\ddot{x}=\omega \dot{y}$, $\ddot{y}=-\omega \dot{x}$ and $\ddot{z}=0$. For initial velocities $(u,v,w)$ and $\omega \ne 0$, integration givesThe transverse motion is a circle centred at $(x_0+v/\omega ,y_0-u/\omega )$ with radius $\sqrt{u^2+v^2}/|\omega |$. Thus the full orbit is a helix around a line parallel to the magnetic field, with period $2\pi /|\omega |$ and signed axial advance $2\pi w/|\omega |$ per turn. The charge sign determines the sense of rotation. If $w=0$ it is a circle, if $u=v=0$ it is a straight line parallel to the field, and if $qB=0$ all motion is uniform rectilinear motion.