Question

A medical cyclotron used in the production of medical isotopes
accelerates protons to 6.5 MeV. The magnetic field in the
cyclotron is 1.2 T.
a. What is the diameter of the largest orbit, just before the protons
exit the cyclotron?
b. A proton exits the cyclotron 1.0 ms after starting its spiral
trajectory in the center of the cyclotron. How many orbits
does the proton complete during this 1.0 ms?

Answer 1

(a) The kinetic energy of the proton is given by, \(K E=\frac{1}{2} m v^{2}\)

\(v=\sqrt{\frac{2(K E)}{m}}\)

\(v=\sqrt{\frac{2(6.5 \mathrm{MeV})\left(1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)}{1.672 \times 10^{-27} \mathrm{~kg}}}=3.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\)

The radius of the orbit is given by, \(r=\frac{m v}{B q}=\frac{1.672 \times 10^{-27} \mathrm{~kg}\left(3.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\right)}{(1.2 \mathrm{~T})\left(1.6 \times 10^{-19} \mathrm{C}\right)}=0.30 \mathrm{~m}\)

Hence, the diameter of the largest orbit just before the proton leaves the orbit is \(0.60 \mathrm{~m}\).

(b) The time required to complete one revolution is given by, \(T=\frac{2 \pi m}{B q}=\frac{2 \pi\left(1.672 \times 10^{-27} \mathrm{~kg}\right)}{(1.2 \mathrm{~T})\left(1.6 \times 10^{-19} \mathrm{C}\right)}=5.48 \times 10^{-8} \mathrm{~s}\)

The number of revolutions made by the proton is given by, \(T n=t\)

\(n=\frac{t}{T}=\frac{1 \mathrm{~ms}}{5.48 \times 10^{-8} \mathrm{~s}}=1.8 \times 10^{4}\)