Electron and proton revolutions per second in magnetic field

[FelixISF]

Homework Statement

An electron of speed v enters a region of magnetic field B directed normally to its velocity and is deflected into a circular path.
Find an expression for the number of revolutions per second the electron will make. If the electron is replaced by a proton, how does your answer change?

Homework Equations

revolutions per second = angular velocity = v/r
I also know, that the answer should be f=eB/2pi(m)
F=eBv sin(theta), where e=charge of an electron

The Attempt at a Solution

I figured, that one needs to replace v by 2(pi)r, which is angular velocity.
So, ma=eB2(pi)r




From then, I do not know how to go farther, please help with your suggestions or even solutions!




Thank you before hand!

 

[Cyosis]

In this case the magnetic field forces the particle into a circular orbit. There is a general name and expression for a force that keeps a particle in an orbit. Do you know what it is?

 

[FelixISF]

Well, this force is certainly known as centripetal force :)

I expect that I need to set up an equation of magnetic force and centripetal force and solve it for v/r which is the angular velocity or the revolutions per second.
The problem is, that I get the following answer:

v/r = eB/m...
now my book says, that the answer is f=eB/2(pi)m
where does the 2(pi) come from. any suggestions?

 

[Cyosis]

You're correct about the centripetal force and you're also correct that $\omega=v/r$. Now note that the answer in the book does not calculate omega, but it calculates f, frequency. Do you know how omega and f are related?

$$\frac{v}{r}=\frac{eB}{m}$$

Is correct.

 

[FelixISF]

omega = 2(pi)f

Thank you so much for your support, I finally got the answer.

I am new to the forum, and will recommend it among my friends, as I find it highly effective.

 

[Cyosis]

You're welcome.