 Solution

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: