The magnetic poles of a small cyclotron produce a magnetic field with magnitude
ID: 1696398 • Letter: T
Question
The magnetic poles of a small cyclotron produce a magnetic field with magnitude 0.85 T. The poles have a radius of 0.40 m. which is the maximum radius of the orbits of the accelerated panicles. What is the maximum energy to which protons (q = 1.60 times 10*19 C. m = 1.67 times 10-27kg) cab be accelerated by this cyclotron? Give your answer in electron volts and in joules. What is the time for one revolution of a proton orbiting at this maximum radius? What would the magnetic field magnitude have to be for the maximum energy to which a proton can be accelerated to be twice that calculated in part (a)? For B = 0.85 T. what is the maximum energy to which alpha particles (q = 3.20x 1-19* C m = 6.65x10-27 kg) can be accelerated by this cyclotron? How does this compare to the maximum energy for protons?Explanation / Answer
a ) Maximum kinetic energy E_m = R^2 q^2 B^2 / 2m
= ( 0.40 m)^2 * ( 1.6 *10^-19C)^2 * ( 0.85T)^2 / 2* 1.67*10^-27 kg
= 8.86*10^-13J
= 8.86* 10^-13 J / 1.6*10^-19C
= 5.53 MeV
b ) time period t = 2 m / q B
= 2 * 1.67 *10^-27 kg / 1.6 *10^-19 C * 0.85 T
= 7.71*10^-8s
c ) if energy is twice
E_m = R^2 q^2 B^2 / 2m
2 (8.86*10^-13J) = ( 0.40 m)^2 * ( 1.6 *10^-19C)^2 *B / 2* 1.67*10^-27 kg
B = 1.2 T
d )
E_m = R^2 q^2 B^2 / 2m
= ( 0.40 m)^2 * ( 3.20 *10^-19C)^2 * ( 0.85T)^2 / 2* 6.65*10^-27 kg
= 8.90*10^-13J
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.