Maybe an example helps. Let $B$ be a constant magnetic field. Then we can take $A=\frac12B×x$. Now
$$\frac{(p+qA)^2}{2m}=\frac{p^2}{2m}+\frac{q}{2m}(p⋅A+A⋅p)+\frac{q^2A^2}{2m},$$ and $$p⋅A+A⋅p=l⋅B$$
where $l=x×p$. Thus
$$\frac{(p+qA)^2}{2m}=\frac{p^2}{2m}+\frac{q}{2m}l⋅B+\frac{q^2A^2}{2m}.$$
Here we recognise the $l⋅B$-term as the Zeeman term.
If we now calculate the derivative of $p$ according to
$$\frac{dp}{dt}=i[p,H],$$
then we obtain the Lorentz force.
This is the quantum version. In the classical case the commutator is replaced by the Poisson bracket.
