Physics, 25.10.2019 19:43 maya4321

Acharged particle enters a uniform magnetic field b with a velocity v at right angles to the field. it moves in a circle with period t.
part a
if a second particle, with the same electric charge but ten times as massive, enters the field with the same velocity v, what is its period?
a. t/10
b. t
c. 5t
d. 10t

part bif the frequency of revolution (the number of revolutions per unit time) of the lighter particle is f, what is the frequency of revolution of the more massive particle? a. f/10b. fc. 5fd. 10f

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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.

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:

$T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$

Where:

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:

$T_{1} = \frac{2\pi m_{1}}{qB}$

If the second particle is 10 times more massive than the first particle, the period is:

$T_{2} = \frac{2\pi m_{2}}{qB} = \frac{2\pi (10m_{1})}{qB} = 10(\frac{2\pi m_{1}}{qB}) = 10T_{1}$

Hence, the period for the second particle that is ten times as massive is 10T. The correct option is d.

Part B

The frequency can be calculated as follows:

$f = \frac{1}{T}$

The frequency of revolution of the first particle is f:

$f_{1} = \frac{1}{T_{1}}$

and for the more massive particle we have:

$f = \frac{1}{T_{2}} = \frac{1}{10T_{1}} = \frac{1}{10}*\frac{1}{T_{1}} = \frac{1}{10}*f_{1}$

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 link

I hope it helps you!

Answer from: spooderfaxy7813

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

F=qvB

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

$qvB = \frac{mv^2}{r}$

which can be rewritten as

$v=\frac{qB}{mr}$

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

$\frac{2\pi r}{T}=\frac{qB}{mr}$

So, we get:

$T=\frac{2\pi m r^2}{qB}$

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) a. f/10

The frequency of revolution of a particle in uniform circular motion is

$f=\frac{1}{T}$

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:

$f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}$

Acharged particle enters a uniform magnetic field b with a velocity v at right angles to the field.

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