(i) Behaviour of revolving electron as a tiny magnetic dipole
(ii) Proof of the relation $\overrightarrow{\mathit{\mu }}=-\frac{\mathit{e}}{2{\mathit{m}}_{\mathit{e}}}\overrightarrow{\mathit{L}}$
(iii) Significance of negative sign
Electron, in circular motion around the nucleus, constitutes a current loop which behaves like a magnetic dipole.
Current associated with the revolving electron :

$\mathit{I}=\frac{\mathit{e}}{\mathit{T}}$

$\therefore \mathit{I}=\frac{\mathit{e}}{2\mathit{\pi }\mathit{r}}\mathit{v}$

Magnetic moment of the loop, $\mathit{\mu }=\mathit{I}\mathit{A}$
$\mathit{\mu }=\mathit{L}\mathit{A}=\frac{\mathit{e}\mathit{v}}{2\mathit{\pi }\mathit{r}}\mathit{\pi }{\mathit{r}}^2=\frac{\mathit{e}\mathit{v}\mathit{r}}{2}=\frac{\mathit{e}\cdot {\mathit{m}}_{\mathit{e}}\mathit{v}\mathit{r}}{2{\mathit{m}}_{\mathit{e}}}$
Orbital angular momentum of the electron
$\mathit{L}={\mathit{m}}_{\mathit{e}}\mathit{v}\mathit{r}$
$\overrightarrow{\mathit{\mu }}=\frac{-\mathit{e}}{2{\mathit{m}}_{\mathit{e}}}\overrightarrow{\mathit{L}}$
-ve sign signifies that the angular momentum of the revolving electron is opposite in direction to the magnetic moment associated with it.