Physics - Magnetism!!! Help please!?

Question

A 70.0 -cm-diameter cyclotron uses a 460 oscillating potential difference between the dees.

What is the maximum kinetic energy of a proton if the magnetic field strength is 0.740 ?

How many revolutions does the proton make before leaving the cyclotron?

The worked solutions would be nice, but even a formula would be soo helpful. I dont know which ones to use?? Please help!

Answer 1

First you have to solve for the velocity, using the equation:
v= (qBr/m)
where B=0.740T, r=(70cm/2)=0.35m q= 1.6E-19 C
mass(proton) 1.67262158 E -27 kg

so your v = 2.59 x 10^7 ms^-1
then you plug this into Ek = 1/2 (mv^2)
Ek= 5.36 x 10-13 Joules is the answer to the first part

As for the second part: You use the formula
Ek= 2Ne(V) and you are solving for N.

N = Ek/(2qV)
N= 5.36 x 10-13 Joules /(2(1.6E-19 C)*(460V))
N= 3490 Revolutions

Answer 2

Momentum of Proton = mv = RqB
R = radius
q = charge on particle
B = magnetic field
m = mass
v = velocity

We need to calculate 'v' first-:

v = R q B / m

v = 0.35 x 1.6x10^-19 x 0.740 / 1.6 x 10^-27
v = 2.59 x 10^7 ms^-1 (about 8.6% of the speed of light)

Kinetic Energy = 1/2 m v^2
( m =mass, v = velocity)

K.E. = 0.5 x 1.6x10^-27 x (2.59x10^7)^2
Max K.E. = 5.36 x 10-13 Joules or 3.354 MeV (Mega Electron Volts)

Not sure about how many revolutions it would make......