Question

Deduce an expression for magnetic dipole moment of an electron revolving around a nucleus in a circular orbit. Indicate the direction of magnetic dipole moment? Use the expression to derive the relation between the magnetic moment of an electron moving in a circle and its related angular momentum?

Answer 1

Magnetic moment of an electron moving in a circle:
Consider an electron revolving around a nucleus $$(N)$$ in circular path of radius $$r$$ with speed $$v$$. The revolving electron is equivalent to electric current
$$I=\dfrac{e}{T}$$
where $$T$$ is period of revolution $$=\dfrac{2\pi r}{v}$$
$$I=\dfrac{e}{2\pi r/v}=\dfrac{ev}{2\pi r}$$ ...(i)
Area of current loop ( electron orbit ), $$A=\pi r^2$$
Magnetic moment due to orbital motion,
$$M_l=IA =\dfrac{ev}{2\pi r}\times (\pi r^2)=\dfrac{evr}{2}$$ ....(ii)
This equation gives the magnetic dipole moment of a revolving electron. The direction of magnetic moment is along the axis.
Relation between magnetic moment and angular momentum
Orbit angular momentum of electron
$$L=m_evr$$ ...(iii)
where $$m_e$$ is mass of electron,
Dividing (ii) by (iii), we get
$$\dfrac{M_l}{L}=\dfrac{ev r/2}{m_evr}=\dfrac{e}{2m_e}$$
Magnetic moment $$\vec M_l=-\dfrac{e}{2m_e}\vec L$$ ...(iv)
This expression of magnetic moment of revolving electron in terms of angular momentum of electron.
In vector form $$\vec M_l=-\dfrac{e}{2m_e}\vec L$$ ...(v)