Teaching Modern Physics - QuarkNet - Fermilab

However, to do it the “normal way,” one uses:
F = m v 2 /r = q v B → p = q r B
p = (1.6 × 10 -19 C) (0.25 m) (2 T) = 0.8 × 10 -19 kg m/s
Converting to “particle physics momentum”:
pparticle = pnormal × [3 × 10 8 m/s] / [1.6 × 10 -19 J/eV] =
1. 5 × 10 8 eV/c = 0.15 GeV/c
An alternate approach is to use:
p = q r B = pparticle physics × [q/c] → pparticle physics = c r B
(which will give a momentum in units of eV/c.)
To make the more useful GeV/c, multiply by the conversion for GeV/eV (10 -9 ).
This gives a nice and final result:
pparticle physics = 0.3 r B
where r is in meters, B is in Tesla
and pparticle physics is in GeV/c.
4. The Fermilab accelerator is designed to accelerate protons in a circular orbit with radius 1
km and to a momentum p. The new LHC accelerator accelerates protons to a momentum
seven times that of the Fermilab accelerator. If the radius of circular motion followed by
a proton in the LHC is 4.3 km, calculate the ratio of the magnetic field of the Tevatron to
the LHC.
F = m v 2 /r = q v B → p = q r B → B = p/(q r)
Thus,
BLHC/BTevatron = [p/(q r)]LHC / [p/(q r)]Tevatron = [pLHC / pTevatron] [rTevatron / rLHC]
BLHC = [7/1] [1/4.3] = 1.63 BTevatron
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