In a velocity selector for ions, when the velocity of the ions is double the value for which the selector is set, how do the magnetic force $\mathrm{F}_{\mathrm{M}}$ and the electric force $\mathrm{F}_{\mathrm{E}}$ compare?
A) $\mathrm{F}_{\mathrm{M}}=4 \mathrm{~F}_{\mathrm{E}}$
B) $\mathrm{F}_{\mathrm{M}}=1.414 \mathrm{~F}_{\mathrm{E}}$
C) $\mathrm{F}_{\mathrm{M}}=\mathrm{F}_{\mathrm{E}} / 2$
D) $F_{M}=F_{E} / 4$
E) $F_{M}=2 F_{E}$

Question

Quizplus · Accepted Answer

The magnetic force $F_M$ on an ion in a velocity selector is given by $F_M = qvB$, where $q$ is the charge of the ion, $v$ is its velocity, and $B$ is the magnetic field strength. The electric force $F_E$ is given by $F_E = qE$, where $E$ is the electric field strength. In a velocity selector, these forces are set to be equal ($F_M = F_E$) for a specific velocity to allow ions with that velocity to pass through without deflection. If the velocity of the ions is doubled, the magnetic force becomes $F_M = q(2v)B = 2qvB$, which is twice the original magnetic force. Since the electric force $F_E$ remains constant (as it depends only on the charge and the electric field, which are unchanged), the new magnetic force is double the electric force ($F_M = 2F_E$).

In a Velocity Selector for Ions, When the Velocity of the Ions