(i) Behaviour of revolving electron as a tiny magnetic dipole (ii) Proof of the relation $\vec{\mu} = -\frac{e}{2 m_e} \vec{L}$ (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 : $I = \frac{e}{T}$ $\text{ and } T = \frac{2 \pi r}{v}$ $\therefore I = \frac{e}{2 \pi r} v$ Magnetic moment of the loop, $\mu = I A$ $\mu = L A = \frac{e v}{2 \pi r} \pi r^2 = \frac{e v r}{2} = \frac{e \cdot m_e v r}{2 m_e}$ Orbital angular momentum of the electron $L = m_e v r$ $\vec{\mu} = \frac{-e}{2 m_e} \vec{L}$ -ve sign signifies that the angular momentum of the revolving electron is opposite in direction to the magnetic moment associated with it.