MCQ
A proton is moves on circular path with its constant angular speed then correct relation between its magnetic moment and angular momentum
- A$\vec M = - \frac{{e\vec L}}{{2{m_p}}}$
- B$\vec M = \frac{{e\vec L}}{{2{m_p}}}$
- C$\vec M = \left( {\frac{{2e}}{{{m_p}}}} \right)\vec L$
- D$\vec M = - \left( {\frac{{2e}}{{{m_p}}}} \right)\vec L$