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Distance travelled in a Cyclotron

  • Thread starter Uku
  • Start date
  • #1
Uku
82
0

Homework Statement


Known data:
[tex]v_{final}=3.99*10^{7}m/s[/tex]
[tex]f_{osc}=12MHz[/tex]
[tex]B=1.6T[/tex]
[tex]R=0.53m[/tex]
[tex]V_{acc}=80kV[/tex]
[tex]q_{deut}=1.602*10^{-19}C[/tex]
[tex]m_{deut}=3.344*10^{-27}kg[/tex]
answer according to Halliday and Resnick x=240m

I got it all, and I have to find how much distance was travelled by the deuteron in the cyclotron, from starting of acceleration to the end of it.

The Attempt at a Solution


From the potential across the dees I get

[tex]v_{gained_per_turn}=\sqrt{\frac{2E}{m_{deut}}}=2768586.602m/s[/tex]

Since I know the final speed I can find the amount of turns the deuteron was in the cyclotron

[tex]n=\frac{v_{final}}{v_{gained_per_turn}}=14.41turns[/tex]

Since f is constant, so is T and I can find the amount of time spent in the cyclotron:

[tex]t_{spent}=n*T=n*\frac{1}{f_{osc}}=1.2*10^{-6}s[/tex]

The acceleration must be constant, the accelerating voltage is not changing

[tex]a=\frac{v_{gained_per_turn}}{T}=3.32*10^{13}\frac{m}{s^{2}}[/tex]

Now

[tex]x=\frac{1}{2}a*t^{2}=23.92m[/tex]

The answer in the book is 240m, which differs from mine ten times, smells like a power error?
Or am I doing something wrong?
 

Answers and Replies

  • #2
142
64
you don't need to find the time and stuff..
Change in Kinetic energy, ΔK=2qΔV (ΔV is the difference in potential and factor 2 because the particle is accelerated twice in one cycle)
Final Kinetic energy is Kf=(qBR)2/2m (use B=2πfoscm)
then no. of cycles, n=Kf/ΔK
Total distance covered = n×2πRavg (where Ravg = R/√2)
You'll get around 244m
 

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