A proton moves on a circular path with a cons | vedclass

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(B) The magnetic moment $M$ of a charge $e$ moving in a circular path of radius $r$ with speed $v$ is given by $M = I A$,where $I = \frac{e}{T}$ and $

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

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(B) The magnetic moment $M$ of a charge $e$ moving in a circular path of radius $r$ with speed $v$ is given by $M = I A$,where $I = \frac{e}{T}$ and $T = \frac{2\pi r}{v}$.Thus,$M = \frac{e}{2\pi r / v} 	imes \pi r^2 = \frac{evr}{2}$.The angular momentum $L$ of the proton is $L = m_p v r$,which implies $vr = \frac{L}{m_p}$.Substituting this into the expression for $M$,we get $M = \frac{e}{2} \left( \frac{L}{m_p} ight) = \frac{eL}{2m_p}$.Since the proton is a positively charged particle,the direction of the magnetic moment $\vec M$ is the same as the direction of the angular momentum $\vec L$ (given by the right-hand rule).Therefore,the vector relation is $\vec M = \frac{e}{2m_p} \vec L$.

$A$ proton moves on a circular path with a constant angular speed. What is the correct relation between its magnetic moment $\vec M$ and angular momentum $\vec L$?

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