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(B) The magnetic moment $\vec{M}$ of a current loop is given by $\vec{M} = I\vec{A}$,where $I$ is the current and $\vec{A}$ is the area vector.For a p

Vedclass · Accepted Answer

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

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(B) The magnetic moment $\vec{M}$ of a current loop is given by $\vec{M} = I\vec{A}$,where $I$ is the current and $\vec{A}$ is the area vector.For a proton of charge $e$ moving in a circular path of radius $r$ with speed $v$,the current $I$ is $I = \frac{e}{T} = \frac{e}{2\pi r / v} = \frac{ev}{2\pi r}$.The area of the loop is $A = \pi r^2$.Thus,the magnitude of the magnetic moment is $M = IA = \left(\frac{ev}{2\pi r}\right)(\pi r^2) = \frac{evr}{2}$.The angular momentum of the proton is $L = m_pvr$.Substituting $vr = L/m_p$ into the expression for $M$,we get $M = \frac{eL}{2m_p}$.Since the proton is positively charged,the direction of the magnetic moment is the same as the direction of the angular momentum vector. Therefore,$\vec{M} = \frac{e\vec{L}}{2m_p}$.

$A$ 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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