Answer:
c) 2
Explanation:
When a charged particle is moving in a region with a magnetic field, it experiences a force perpendicular to the direction of motion, therefore it starts moving with a circular motion.
The force experienced by the particle which is moving perpendicular to the field is given by
[tex]F=qvB[/tex]
where
v is the speed of the particle
q is the charge of the particle
B is the strength of the magnetic field
Here we have:
- A proton, which has
[tex]q=e[/tex] (charge)
[tex]v=v_p[/tex] (speed of the proton)
B is the magnetic field
So the force experienced by the proton is
[tex]F_p = ev_p B[/tex] (1)
- An alpha particle, which has
[tex]q=2e[/tex] (charge)
[tex]v=v_\alpha[/tex] (speed of the alpha particle)
B is the magnetic field
So the force experienced by the alpha particle is
[tex]F_\alpha = (2e) v_\alpha B[/tex]
Here we are told that the force experienced by the two particles is the same, so:
[tex]F_p = F_\alpha[/tex]
And so we get:
[tex]ev_p B = 2ev_\alpha B[/tex]
Solving for the ratio between their speed, we find:
[tex]\frac{v_p}{v_\alpha}=\frac{2eB}{eB}=2[/tex]
