A proton is moves on circular path with its constant angular spee… - Vidyadip

MCQ

A proton is moves on circular path with its constant angular speed then correct relation between its magnetic moment and angular momentum

A
$\vec M = - \frac{{e\vec L}}{{2{m_p}}}$
✓
$\vec M = \frac{{e\vec L}}{{2{m_p}}}$
C
$\vec M = \left( {\frac{{2e}}{{{m_p}}}} \right)\vec L$
D
$\vec M = - \left( {\frac{{2e}}{{{m_p}}}} \right)\vec L$

✓

Answer

Correct option: B.

$\vec M = \frac{{e\vec L}}{{2{m_p}}}$

b

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