Question

A proton (charge e), traveling perpendicular to a magneticfoeld, experiences the same force as an alpha particle (charge 2e)which is also traveling perpendicular to the same field. The ratioof their speeds vproton/valpha is

a.0.5
b. 1
c. 2
d. 4
e. 8

Answer 1

c) 2

Step-by-step 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

`F=qvB`

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

`q=e` (charge)

`v=v_p` (speed of the proton)

B is the magnetic field

So the force experienced by the proton is

`F_p = ev_p B` (1)

- An alpha particle, which has

`q=2e` (charge)

`v=v_\alpha` (speed of the alpha particle)

B is the magnetic field

So the force experienced by the alpha particle is

`F_\alpha = (2e) v_\alpha B`

Here we are told that the force experienced by the two particles is the same, so:

`F_p = F_\alpha`

And so we get:

`ev_p B = 2ev_\alpha B`

Solving for the ratio between their speed, we find:

`(v_p)/(v_\alpha)=(2eB)/(eB)=2`