A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e)which is also traveling perpendicular to the same field.The ratio of their speeds, v_proton/v_alpha is: A)0.5 B)1 C)2 D)4 E)8

Question

Quizplus · Accepted Answer

The force experienced by a charged particle moving in a magnetic field is given by $F = qvB$, where $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field strength. For the proton, $F_p = e \cdot v_p \cdot B$, and for the alpha particle, $F_{\alpha} = 2e \cdot v_{\alpha} \cdot B$. Given that $F_p = F_{\alpha}$, we can set the equations equal to each other: $e \cdot v_p \cdot B = 2e \cdot v_{\alpha} \cdot B$. Simplifying, we find $v_p / v_{\alpha} = 2$, meaning the proton's speed is twice that of the alpha particle.

A Proton (Charge E), Traveling Perpendicular to a Magnetic Field