A) d. 10T
When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by
[tex]F=qvB[/tex]
where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.
This force acts as a centripetal force, keeping the particle in a circular motion - so we can write
[tex]qvB = \frac{mv^2}{r}[/tex]
which can be rewritten as
[tex]v=\frac{qB}{mr}[/tex]
The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):
[tex]\frac{2\pi r}{T}=\frac{qB}{mr}[/tex]
So, we get:
[tex]T=\frac{2\pi m r^2}{qB}[/tex]
We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.
B) [tex]a. f/10[/tex]
The frequency of revolution of a particle in uniform circular motion is
[tex]f=\frac{1}{T}[/tex]
where
f is the frequency
T is the period
We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:
T' = 10 T
Then its frequency of revolution will be:
[tex]f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}[/tex]
For the charged particle that enters in a magnetic field (B) and moves with a velocity (v) in a circular motion, we have:
A. The period of the second particle that is more massive is 10T (option d).
B. The frequency of revolution of the more massive particle is f/10 (option a).
We can find the period and frequency of the particle with the equations deduced from the Lorentz force. Â Â Â Â Â Â
Part A
The period of the charged particle that enters in a magnetic field (B) and moves with a velocity (v) describing a circular path is given by:
[tex] T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} [/tex]
Where: Â Â
r: is the radius = mv/qB
v: is the velocity
m: is the mass
q: is the charge of the particle
B: is the magnetic field
The period of the first particle is:
[tex] T_{1} = \frac{2\pi m_{1}}{qB} [/tex]
If the second particle is 10 times more massive than the first particle, the period is:
[tex] T_{2} = \frac{2\pi m_{2}}{qB} = \frac{2\pi (10m_{1})}{qB} = 10(\frac{2\pi m_{1}}{qB}) = 10T_{1} [/tex]
Hence, the period for the second particle that is ten times as massive is 10T. The correct option is d. Â
Part B
The frequency can be calculated as follows:
[tex] f = \frac{1}{T} [/tex]
The frequency of revolution of the first particle is f:
[tex] f_{1} = \frac{1}{T_{1}} [/tex]
and for the more massive particle we have:
[tex] f = \frac{1}{T_{2}} = \frac{1}{10T_{1}} = \frac{1}{10}*\frac{1}{T_{1}} = \frac{1}{10}*f_{1} [/tex] Â
Therefore, the frequency of revolution of the more massive particle is f/10. The correct option is a.
You can find more about the velocity of a charged particle in a circular motion here https://brainly.com/question/16088547?referrer=searchResults Â
I hope it helps you!
