A cyclotron designed to accelerate protons has an outer radius of 0.318 m. The p
ID: 1262906 • Letter: A
Question
A cyclotron designed to accelerate protons has an outer radius of 0.318 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 638 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.880 T.
(a) Find the cyclotron frequency for theprotons in this cyclotron
(b) Find the speed at which protons exit the cyclotron.
(c) Find their maximum kinetic energy.
(d) How many revolutions does a proton make in the cyclotron?
(e) For what time interval does one proton accelerate?
Explanation / Answer
a)
centripatal force =magnetic force
mv2/r =qvB
=>v=qBr/m =(1.6*10-19)*0.88*0.318/(1.67*10-27)
v=2.68*107 m/s
Time period
T=2pi*r/v =2pi*0.318/(2.68*107)
T=7.45*10-8 s
frequency
f=1/T =1/7.45*10-8
f=1.34*107 Hz or 13.42 MHz
b)
speed of proton
v=2.68*107 m/s or 26.8 Mm/s
c)
Maximum Kineitc energy
KEmax=(1/2)mv2=(1/2)*(1.67*10-27)*(2.68*107)2
KEmax=6*10-13 J
d)
The kinetic energy K is built up from 2N passes through V .Number of revolutions is
N=KE/2qV =6*10-13/2*(1.6*10-19)*638
N=2938 rev
e)
time interval does one proton accelerate is
dt=N/f =2938/1.34*107
dt=2.2*10-4 s or 220 us
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.