1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Particle Physics help (urgent, deadline soon)

  1. Nov 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Estimate the magnetic field required to maintain protons at 7TeV in the LHC at CERN. The LHC has a circumference of 27 km. State any assumptions or approximations you make.

    2. Relevant equations
    lorentz force, assuming magnetic field is always perpendicular to the direction of motion: F=qvB (q=charge of the proton) such that we can use the
    Formula for centripetal force: F=1/r mv^2
    ---equating the two yields the cyclotron formula: B=mv/qr
    relate momentum to mass velocity: p=mv then
    B=p/qr is the equation we need ( I think)

    3. The attempt at a solution
    I'm not completely sure as how to use the 7TeV energy given, this probably includes mass so I use the equation: E=p^2+M^2 assuming c=1 and all units are as such. I immediately notice that the mass of the proton is 938 MeV, this is a negligible amount compared to the 7TeV, so I suggest we just ignore the mass of the proton for this problem and treat all energy as kinetic energy such that: B=p/qr=(7* (5.36 x 10^-16 )kg m /s)/[(1.6022*10^-19 C)(27 * 10^3 m)]=0.867Teslas.

    Am I making any mistakes here or do you guys get similar answers?
  2. jcsd
  3. Nov 16, 2015 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You're off by an order of magnitude according to this article:


    At full power, the LHC is supposed to produce 8.3 T

    In your calculation of B = p/qr, if r is supposed to be the radius of the collider, then 27 km is stated to be its circumference, not its radius.
  4. Nov 16, 2015 #3
    ah yeah... reading.. I always suck at that. B=p/qr=(7* (5.36 x 10^-16 )kg m /s)/[(1.6022*10^-19 C)((27/(2 pi) *10^3) m)]=5.449 T does that look better? This is after assuming perfect conditions such as the LHC is perfectly round, the magnetic field is always perpendicular. It gets pretty close to the 8.3 Teslas, but still not very satisfactory tbh.
    Last edited: Nov 16, 2015
  5. Nov 16, 2015 #4


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    Just to enlighten me: if everything is in SI units, what is "7" ? and what happened to c=1 ?

    [edit] Ah, I should have read 7 TeV/c , shouldn't I ? But did they give you a momentum of 7 TeV/c or did they give you an energy of 7 TeV/c2 ?
  6. Nov 16, 2015 #5
    I just did that for the Pythagorean equation E^2=p^2+m^2, basically I say everything is energy, just as long as I remember all the units and conversion factors correctly. Well the 7 is because there is 7TeV(/c), not 1TeV(/c), the number behind it is the conversion factor to convert Mev(/c) to regular momentum units of kgm/s. ......SI in particle physics xD.... with their microbarns and such
  7. Nov 16, 2015 #6


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    then perhaps the
    confused me ?

    [edit] found it. Sorry for barging in. Thank you for reminding me...:rolleyes:
  8. Nov 16, 2015 #7
    should be E^2
  9. Nov 22, 2015 #8
    I think this one is solved, others also got 5.4ish teslas
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted