Problem P. 5362. (November 2021)
P. 5362. In a synchrocyclotron, the velocity dependence of the mass of elementary particles is compensated by decreasing the frequency of the electric field, which accelerates the particles. For example, if protons are accelerated, frequency of the applied voltage between the dees (D-shaped, metal, hollow half-discs) are varied from 25 MHz to 18.9 MHz in each cycle. Determine in this case
\(\displaystyle a)\) the magnitude of the magnetic induction;
\(\displaystyle b)\) the kinetic energy of the exiting protons.
(5 pont)
Deadline expired on December 15, 2021.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. \(\displaystyle a)\) A proton mozgásegyenlete kis sebességeknél:
| \(\displaystyle (1)\) | \(\displaystyle eBv= \frac{mv^2}{r}=mv\omega= mv\cdot 2\pi f_1,\) |
ahonnan a mágneses indukcióvektor nagysága:
\(\displaystyle B=2\pi f_1\frac{m}{e}=6{,}28\cdot\left(25\cdot 10^6~{\rm s}^{-1}\right)\frac{ 1{,}67\cdot 10^{-27}~{\rm kg} } {1{,}6\cdot 10^{-19}~{\rm As}}=1{,}64~\rm T.\)
\(\displaystyle b)\) A felgyorsított, \(\displaystyle v\) sebességgel mozgó protonok mozgásegyenlete:
| \(\displaystyle (2)\) | \(\displaystyle eBv= \frac{mv}{\sqrt{1-v^2/c^2}}\cdot 2\pi f_2.\) |
Az (1) és (2) egyenletekből következik, hogy
\(\displaystyle \frac1{\sqrt{1- {v^2}/{c^2}}} =\frac{f_1}{f_2}=\frac{25~\rm MHz}{18{,}9~\rm MHz}=1{,}32,\)
vagyis a kilépő protonok mozgási (kinetikus) energiája:
\(\displaystyle E_{\rm m}=\frac{mc^2}{\sqrt{1- {v^2}/{c^2}}}-mc^2=0{,}32\,mc^2=0{,}32\cdot 938~\rm MeV=300~\rm MeV.\)
Statistics:
15 students sent a solution. 5 points: Bencz Benedek, Kertész Balázs, Nemeskéri Dániel, Somlán Gellért, Téglás Panna, Toronyi András, Vágó Botond. 4 points: Albert Máté, Barna Benedek, Kürti Gergely, Magyar Gábor Balázs. 3 points: 2 students. 1 point: 1 student.
Problems in Physics of KöMaL, November 2021