# Frequency of circular motion

## Homework Statement

The frequency of circular motion for a charged particle moving around in the presence of a uniform magnetic field does not depend on ...

b)The mass of the particle
c)The charge of the particle
d)The magnitude of the magnetic field
e)Actually, it depends on all of the above quantities

## The Attempt at a Solution

I believe the answer is d because the magnetic field alone cannot alter the KE of a particle because it is perpendicular to the particle velocity

OR

V = mv^2/R ----> QBr/m = 2piR/T (B is the magnitude of the magnetic field)
Where you solve for the period, cancel out the radius on both sides of the equation and take the inverse. Therefore the answer does not depend on the radius.

Which is correct?

rl.bhat
Homework Helper
Q*v*B= Q*ω*r*B = m*v^2/r = m*ω^2*r ( velocity v = rω)

Clarification

Q*v*B= m*v^2/r =
QB = mv^2/r
QB = m(f x wavelength) /r

Therefore

f = rQB/(m x wavelength)

Therefore, the frequency of circular motion for a charged particle moving around in the presence of a uniform magnetic field depends on e) all of the above quantities

rl.bhat
Homework Helper
v = f*λ relation is used in the propagation of waves in a medium, not in the circular motion.

I am confused now.How do I find the frequency then for circular motion

Q*ω*r*B = m*ω^2*r

I know from this equation the radius cancels but what does this have to do with the frequency

ideasrule
Homework Helper
Omega = 2*pi*f. Omega is the number of radians that go by each second, so omega/(2*pi) is the number of revolutions that can fit in each second.

Right. Thank you very much