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

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

$F=qvB$

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:

$\frac{v_p}{v_\alpha}=\frac{2eB}{eB}=2$