Teaching Modern Physics - QuarkNet - Fermilab

Teacher’s Key
Circular Motion Problems
1. A proton in the Fermilab Tevatron (a circular particle accelerator) has a momentum of
980 GeV/c. If the radius of the Tevatron is 1 km (it is), calculate the magnetic field
necessary to accomplish this.
Need to convert “particle physics” momentum to “normal” momentum. Then it is a
simple case of using the formula for circular motion.
p = 980 GeV/c = (980 × 10 9 eV) (1.602 × 10 -19 J/eV)/(3 × 10 8 m/s) =
5.2 × 10 -16 kg m/s
F = m v 2 /r = q v B → p = q r B → B = p/(q r) = 3.3 T
2. The Fermilab accelerator is designed to use magnetic fields to make protons orbit in a
circular path with a radius of 1 km. The magnetic field strength can be changed. If
protons are injected into the accelerator with an initial energy of 120 GeV and are
accelerated to a final energy of 980 GeV, calculate the ratio of the final to initial
magnetic field.
F = m v 2 /r = q v B → p = q r B → B = p/(q r)
Thus,
Bfinal/Binitial = [p/(q r)]final / [p/(q r)]initial = 980/120
B980 GeV = 8.2 B120 GeV
3. A particle physics detector consists of a circular region of radius 50 cm filled with air,
surrounded by an extended region filled with metal. The air-filled region contains a
magnetic field with strength 2 T, directed out of the page in the attached diagram. A
proton is emitted from the center of the circle with momentum p in the plane of the paper
and directed radially outwards from the center. Calculate the maximum p for which the
proton will not hit the metal region.
This problem is mildly tricky, as one can do it a hard way and an easy way. The first
thing to realize is that if the radius of the magnetic field region is 50 cm, then a particle
originating at the center of the cylinder will have to travel with a radius of curvature of
25 cm to fit fully within the magnetic region. The next thing one has to do is to convert
the “particle physics momentum,” with its units of GeV/c to “normal momentum.”
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