= Solution
In Cartesian components, $\partial L/\partial v_i=mv_i+eA_i$ and $\partial L/\partial x_i=ev_j\partial_iA_j-e\partial_i\phi$. The <Euler-Lagrange equation> therefore gives
$$
m\dot v_i=e[-\partial_i\phi-\partial_tA_i+v_j(\partial_iA_j-\partial_jA_i)]
=e[E_i+(\mathbf v\times\mathbf B)_i].
$$
Thus \b[the minimally coupled particle <Lagrangian> yields the <Lorentz force>].
For an arbitrary smooth gauge function $\chi$, take $\mathbf A\to\mathbf A+\nabla\chi$ and $\phi\to\phi-\partial_t\chi$. Both fields are unchanged, while
$$
L\to L+e\left(\partial_t\chi+\mathbf v\cdot\nabla\chi\right)
=L+e\frac{d\chi}{dt}.
$$
The <action> changes only by an endpoint term, so fixed-endpoint variation gives the same motion. <Canonical momentum> shifts by $e\nabla\chi$, whereas <kinetic momentum> $m\mathbf v$ and the <Lorentz force> are gauge invariant. This is the relevant <gauge invariance> of the classical particle dynamics.
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