Cyclotron accelerating protons to 20 MeV with potential gap 100 kV

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SUMMARY

The discussion focuses on calculating the number of turns required for a cyclotron to accelerate protons to a kinetic energy of 20 MeV using a potential gap of 100 kV and an applied magnetic field of 2 T. The kinetic energy gained per turn is determined by the formula qV, where q is the charge of the proton and V is the potential gap. The conclusion indicates that relativistic effects become significant at 20 MeV, suggesting that the calculation must account for these effects to determine the exact number of turns needed.

PREREQUISITES
  • Understanding of cyclotron physics and operation
  • Familiarity with the concepts of kinetic energy and potential difference
  • Knowledge of relativistic effects in particle acceleration
  • Basic principles of electromagnetism, specifically magnetic fields
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This discussion is beneficial for physics students, particle accelerator engineers, and anyone interested in the principles of particle acceleration and cyclotron technology.

bhargavdjoshi
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Homework Statement


A Cyclotron is accelerating proton, where the applied magnetic field is 2 T, the potential gap is 100 KV, then how much turn are required for a kinetic energy of 20 MeV...

(a) 300
(b) 200
(c) 150
(d) 100
 
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20MeV is around when relativistic effects take place. The voltage alternates periodically. After one turn (or half a turn--semicircle...I'm not sure) the proton gains a kinetic energy of qV, V being the potential gap. Using that value, find out how many turns it takes to total 20MeV.
 

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