Cyclotron frequency Watch

TheSK00T3R
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#1
Report Thread starter 5 years ago
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A beam of low speed protons are introduced into a cyclotron.
Show that:

f=(eB)/(2pim)

where f is the number of revolutions per second
e is the electronic charge
B is the uniform magnetic flux density within the cyclotron
m is the mass of a proton
( and pi is pi)


I tried the old 'try rearranging a whole bunch of potentially relevant equations', but to no avail.

What I've got so far:
T=2pi/w

To be honest I don't even know what's going on inside the cyclotron. Why is there an electric field as well as a magnetic one? Where exactly is the centripetal force coming from? Are Bev and Bil in the house?
Any leads will be much appreciated.
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Stonebridge
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#2
Report 5 years ago
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(Original post by TheSK00T3R)
A beam of low speed protons are introduced into a cyclotron.
Show that:

f=(eB)/(2pim)

where f is the number of revolutions per second
e is the electronic charge
B is the uniform magnetic flux density within the cyclotron
m is the mass of a proton
( and pi is pi)


I tried the old 'try rearranging a whole bunch of potentially relevant equations', but to no avail.

What I've got so far:
T=2pi/w

To be honest I don't even know what's going on inside the cyclotron. Why is there an electric field as well as a magnetic one? Where exactly is the centripetal force coming from? Are Bev and Bil in the house?
Any leads will be much appreciated.
Start with the fact that the frequency is 1/T where T is the period.

The period is 2Pi.r/v

You also know that the centripetal force mv2/r = Bev
Sub for v from this in the period
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TheSK00T3R
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#3
Report Thread starter 5 years ago
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(Original post by Stonebridge)
Start with the fact that the frequency is 1/T where T is the period.

The period is 2Pi.r/v

You also know that the centripetal force mv2/r = Bev
Sub for v from this in the period
Aha.
m.v^2/r=Bev
cancel the v's so m.v/r=Be
B.e.r/m=v
sub that instead of v:
T=2Pi.r/(B.e.r/m)=2Pi.r.m/B.e.r
r's cancel out, so T=2Pi.m/B.e
f=1/T=Be/2Pim
woohoo
Thanks!
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