Radius & Angular Speed of Proton in Cyclotron

[pat666]

Homework Statement

A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconducting electromagnet of the cyclotron produces a magnetic field of 3.5 T, perpendicular to the proton’s orbits.
a) When the protons have achieved a kinetic energy of 2.7 MeV, what is the radius of their circular orbit and what is their angular speed?
b) Repeat part a) when the protons have achieved their final kinetic energy of 5.4 MeV.

Homework Equations

R=mv/qB
f_c=qB/(2pi*m)

The Attempt at a Solution

google calc
2.7 megaelectron volts = 4.32587645 × 10^-13 joules
mass of a proton 1.67262158 × 10^-27 kilograms
elementary charge = 1.60217646 × 10-19 coulombs
K=1/2mv² not totally sure if I can use this -cant see why not either?
v=2.2743*10⁷
subbing all values in
R=0.068m
not sure what sort of range these are usually in so I can't tell if my answer seems right or not?

found f_c to be 5.33588*10⁷Hz so w=3.353*10⁸rad/s

I really just need to know if these values seem right and someone to check my procedure(and answers if your really nice). b is the same so i won't bother wasting time with that.
thanks for any help

 

[Mindscrape]

Yeah, that seems right, though I'll admit that I didn't do the numbers myself. I think you should be using w = v/R though.