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

Answers: 1

Show answers

Answers

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.

Answer from: Quest

Ibelieve your answer is b.hope this : )

Answer from: Quest

elly's long hair:

a. if the salon only takes cash, then this isn't the problem. it doesn't matter how much she has in her checking account. she can't get a haircut until she has the cash.

b. if she has overdraft protection, then there's no problem. she can go ahead and write a check that her account can't cover, then sometime later, pay the bank some outrageous usury for loaning her a few bucks for a few days, but with beautiful hair.

c. if elly has some principles, and doesn't want to write a bad check, then now she has a problem. her desire for the new hairstyle is in conflict with her honesty and integrity.

d. if the bank will not process a check she writes, then even if she had a million $$ in her account, she couldn't use any of it. the amount she has in her account is not the problem. and if i may suggest, she might as well close her checking account, and put the money where it'll work for her. it's not doing her a bit of good in a checking account, since she can't write any checks.

justin can't pay the full balance:

a. he can't even handle the credit limit he has now. under no circumstances should he ask for a higher limit.

b. he can ask for the grace period to be extended. but that's no solution, the bank will almost certainly refuse anyway, and there are much better options. he knew the rules of his credit card when he started using it.

c. he definitely should pay the interest on the balance, plus as much of the balance as he can cover at the same time.

d. he definitely should pay the minimum otherwise they could shut down his card. he should also pay as much more as he possibly can. he should also remember the awkward situation he's in, stop buying stuff on credit that he can't afford, and avoid adding any more debt.

i'd say that justin's best best option isn't on this list of choices. what he has to do: pay as much as he can at least the minimum payment plus as much more as he can and then avoid adding more debt.

Other questions on the subject: Physics

Physics, 22.06.2019 14:40, akitchen10

A5 foot by 5 foot box culvert is buried 6 feet beneath the surface of the ground. a crosssection of this very long culvert (into and out of the page) is illustrated below. assume that the soil in which the culvert is buried has the following properties: jt = 120 pcf, w = 12%, ko = 0.50, ka = 0.33, and kp = 3.00. calculate the total horizontal force per unit length of culvert (fh) felt on one of the vertical faces of the culvert.

Answers: 1

continue

Physics, 22.06.2019 23:50, jazzzzhands21

An alpha particle with kinetic energy 12.5 mev makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude l = p0 b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.40×10−12 m. (assume that the lead nucleus remains stationary and that it may be treated as a point charge. the atomic number of lead is 82. the alpha particle is a helium nucleus, with atomic number 2.) what is the distance of closest approach?

Answers: 2

continue

Physics, 23.06.2019 06:10, danielbamf1414

Question 1(multiple choice worth 2 points) in order to be considered work, the components that must be present are • a force and a movement in the same direction as the force an action which causes movement and an opposite reaction an action which causes no movement or reaction a force and a movement in the opposite direction of the force

Answers: 1

continue

Physics, 23.06.2019 14:00, Dinojacob

If the wavelength is tripled, while the speed of the wave is held constant, what happens to the frequency?

Answers: 1

continue

You know the right answer?

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