Solving Cyclotron Movement Problem with 500V and 0.790B

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SUMMARY

The discussion focuses on calculating the number of revolutions a proton makes in a cyclotron with a 500V oscillating potential difference and a magnetic field strength of 0.790 T. Key equations include the relationship between velocity, radius, magnetic field, charge, and mass of the proton, specifically using v/r = Bq/m and kinetic energy equations. Participants seek clarification on deriving the number of revolutions from the maximum speed and kinetic energy calculated.

PREREQUISITES
  • Cyclotron physics principles
  • Understanding of electromagnetic fields
  • Basic mechanics involving kinetic energy
  • Familiarity with particle motion in magnetic fields
NEXT STEPS
  • Study the derivation of motion equations in cyclotrons
  • Learn about the Lorentz force and its application in particle accelerators
  • Explore the concept of escape velocity in magnetic confinement systems
  • Investigate the effects of varying magnetic field strengths on particle trajectories
USEFUL FOR

Students in physics, particularly those studying electromagnetism and particle dynamics, as well as educators looking for practical examples of cyclotron applications.

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Homework Statement



A 61.0 cm diameter cyclotron uses a 500 V oscillating potential difference between the dees. The magnetic field strength is 0.790 ?

How many revolutions would a proton make before leaving the cyclotron?

Homework Equations



v/r=Bq/m Ke=1/2mv^2

The Attempt at a Solution



I am completely stumped on this problem, I have calculated the max speed, and therefor max kinetic energy. But cannot figure out how to derive number of revolutions from there.

Any hints would be greatly appreciated.

Thanks
 
Physics news on Phys.org
Canadian,

Why would the electron leave the cyclotron,
Describe first how this device works.
How does the electron reach the escape speed?

Michel
 

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