Can someone please help me with this problem too. (look at details)?

Question

A charged particle with a charge-to-mass ratio of q/m = 5.9 108 C/kg travels on a circular path that is perpendicular to a magnetic field whose magnitude is 0.70 T. How much time does it take for the particle to complete one revolution?

Answer 1

The frequency of a charged particle (mass m, and charge q) spinning around a Magnetic field line can be solved by setting the centripetal force of the rotating particle equal to the magnetic force acting on it.

m * v^2 / r = B * q * v
Where r is the radius of rotation and v is the velocity at which the particle of moving in the magnetic field (perpendicular to the field).

Re-arranging this equation, we get,
v / r = B * q / m, which is the angular frequency of the particle, ω.

ω = B * q / m
Since w = 2 * π * f, we can solve for the ordinary frequency of the particle,

f = (B * q) / (2 * π * m)

And the period of rotation (the time it takes to complete one loop) is equal to 1 over the frequency.
So the period would be,
T = 1/ f = (2 * π * m) / (B * q)

You are told the magnitude of the magnetic field is .7 Teslas.
You are also told that the ration of the charge to mass of the particle is 5.9108 C/kg, taking the inverse will give us the ratio of mass to charge.

T = 2 * pi * (ratio of mass to charge) / B
T = 2 * (3.1415) * (.1692 kg / C) / (.7 Teslas)
Period = 1.52 seconds


http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/cyclot.html