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(C) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since the magnetic field $\vec{B}$ is perpendicular to the

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$\vec{	au} \cdot \vec{L} = 0$

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(C) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since the magnetic field $\vec{B}$ is perpendicular to the plane of the orbit,the force $\vec{F}$ acts towards the center of the circular path (centripetal force).The torque $\vec{	au}$ about the nucleus is given by $\vec{	au} = \vec{r} 	imes \vec{F}$.Since the force $\vec{F}$ is directed towards the center,the position vector $\vec{r}$ and the force vector $\vec{F}$ are collinear (anti-parallel).Therefore,$\vec{	au} = \vec{r} 	imes \vec{F} = 0$.Since the torque is zero,the dot product of the torque vector with any vector,including the angular momentum vector $\vec{L}$,must be zero: $\vec{	au} \cdot \vec{L} = 0$.

An electron revolves around the nucleus in a circular path with angular momentum $\vec{L}$. $A$ uniform magnetic field $\vec{B}$ is applied perpendicular to the plane of its orbit. If the electron experiences a torque $\vec{\tau}$,then

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