An electric field perpendicular to a magnetic field is used as a velocity selector for $\mathrm{Li}^{+}$ions, selecting the velocity $\mathrm{v}_{1}$. Next, $\mathrm{Li}^{++}$ions are sent through the same velocity selector, selecting velocity $\mathrm{v}_{2}$. If neither the electric nor magnetic fields have changed between the two processes, how do $\mathrm{v}_{1}$ and $\mathrm{v}_{2}$ compare?

Question

Quizplus · Accepted Answer

The velocity $v$ selected by a velocity selector is given by $v = E/B$, where $E$ is the electric field strength and $B$ is the magnetic field strength. Since neither $E$ nor $B$ changes between the two processes, the selected velocity $v$ remains the same regardless of the charge state of the ions passing through. Therefore, $\mathrm{v}_{2}=\mathrm{v}_{1}$.

An Electric Field Perpendicular to a Magnetic Field Is Used Li+\mathrm{Li}^{+}Li+

An Electric Field Perpendicular to a Magnetic Field Is Used $\mathrm{Li}^{+}$