(B) The relation $\vec{\mu}_{l} = -(e/2m)\vec{L}$ is in accordance with the classical result.
Derivation:
Consider an electron of mass $m$ and charge $-e$ moving in a circular orbit of radius $r$ with speed $v$ in time period $T$.
The orbital magnetic moment $\mu_{l}$ is given by $\mu_{l} = iA$,where $i = -e/T$ is the current and $A = \pi r^2$ is the area of the orbit.
Thus,$\mu_{l} = (-e/T)(\pi r^2)$.
The orbital angular momentum $L$ is given by $L = mvr$. Since $v = 2\pi r/T$,we have $L = m(2\pi r/T)r = 2\pi mr^2/T$.
Taking the ratio,$\mu_{l}/L = [(-e/T)(\pi r^2)] / [2\pi mr^2/T] = -e/2m$.
Therefore,$\vec{\mu}_{l} = -(e/2m)\vec{L}$.
This shows that $\vec{\mu}_{l}$ and $\vec{L}$ are antiparallel. The relation for spin $\mu_{s} = -(e/m)S$ is purely quantum mechanical and has no classical analogue.