Calculations for accelerating particles in a Cyclotron

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SUMMARY

The forum discussion centers on the calculations for determining the kinetic energy (KE) of particles in a cyclotron, specifically addressing discrepancies in expected results. The formula used is KE=mv^2/2=q^2B^2r^2/2m, leading to a calculated value of 16.45 MeV, while the expected answer is 165 keV. The confusion arises from a potential error in the radius value used, with one participant suggesting that the official answer may have incorrectly used r = 53 mm instead of r = 53 cm. Additionally, the discussion touches on the effects of electric field strength on particle exit time from the cyclotron.

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  • Understanding of kinetic energy calculations in particle physics
  • Familiarity with cyclotron operation principles
  • Knowledge of electromagnetic theory, specifically magnetic fields and forces
  • Proficiency in unit conversions, particularly between millimeters and centimeters
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  • Review the derivation of kinetic energy formulas in cyclotron physics
  • Study the impact of electric field strength on particle motion in cyclotrons
  • Learn about the significance of significant figures in scientific calculations
  • Explore common errors in radius measurements and their effects on particle energy calculations
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Students and professionals in physics, particularly those focusing on particle acceleration, cyclotron design, and electromagnetic theory. This discussion is beneficial for anyone looking to deepen their understanding of kinetic energy calculations in high-energy physics contexts.

sss1
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Homework Statement
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Relevant Equations
mv^2/2, f=qB/2pim
For this question part d, KE=mv^2/2=q^2B^2r^2/2m (I rearranged B=mv/qr for v and subbed into mv^2/2). q^2b^2r^2/2m=2F_cyc^2r^2m(pi)^2
But when I subbed the values in I got 16.45MeV but the answer says 165keV instead. I'm not sure what went wrong?

What's a good explanation for part e also?
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sss1 said:
But when I subbed the values in I got 16.45MeV but the answer says 165keV instead. I'm not sure what went wrong?
I got the same answer as you. (Except I rounded mine to an appropriate number of significant figures!)

It looks like the official answer is wrong - probably because whoever did it used r =53mm instead of r=53cm.

sss1 said:
What's a good explanation for part e also?
As you probably know, ther rules here require you to show your own thoughts/attempt before guidance is offered.
 
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Steve4Physics said:
As you probably know, ther rules here require you to show your own thoughts/attempt before guidance is offered
Well if the e field is strong, it’ll only take a few cycles until the particle leaves? Whereas if if the e field is weak, it’ll take a long time to leave the cyclotron, so the KE is the same regardless?
 
sss1 said:
Well if the e field is strong, it’ll only take a few cycles until the particle leaves? Whereas if if the e field is weak, it’ll take a long time to leave the cyclotron, so the KE is the same regardless?
You need to do better than that. What is an expression for the kinetic energy of the particle just before it exits? Is there an ##E## in it?
 
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