What is the relationship between the current and the magnetic moment of a current carrying circular loop? Use the expression to derive the relation between the magnetic moment of an electron moving in a circle and its related angular momentum?

Let us assume that an electron of mass 'm_e' and charge 'e' revolves in a circular orbit of radius 'r' around the positive nucleus in anticlockwise direction. 

 
The angular momentum of the electron due to its orbital motion is given by 
                         L = m_evr    ...(i) 

Let the period of orbital motion of the electron be T.
Then, the electron crosses any point on its orbit after every T seconds.

Therefore, orbital motion of electron is equivalent to a current. 
                      $\mathrm{I} = \mathrm{e}.\left(\frac{1}{\mathrm{T}}\right)$
The period of revolution of the electron is given by
                      $\mathrm{T} = \frac{2\mathrm{\pi r}}{\mathrm{v}}$ 

$\therefore$                 $\mathrm{I} = \mathrm{e}\left(\frac{1}{2\mathrm{\pi r}/\mathrm{v}}\right) = \frac{\mathrm{ev}}{2\mathrm{\pi r}}$ 

The area of the electron orbit, $\mathrm{A} = {\mathrm{\pi r}}^2$ 

The magnetic dipole moment of the atom is

                  $\boxed{\mathrm{M} = \mathrm{IA}}$
          
                  $\mathrm{M} = \frac{\mathrm{ev}}{2\mathrm{\pi r}}\times {\mathrm{\pi r}}^2 = \frac{\mathrm{evr}}{2}$            ...(ii)
Now, using the equation (i) we have 

                          $\mathrm{M} = \left(\frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}\right)\mathrm{L}$                     [From I]

In vector rotation
                          $\overrightarrow{\mathrm{M}} = -\left(\frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}\right)\overrightarrow{\mathrm{L}}$ 

which, tells us that the magnetic dipole moment vector is directed in a direction opposite to that of angular momentum vector.