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A question about linear accelerator and relativistic momentum

  1. Aug 18, 2012 #1
    A 3.2km linear accelerator(linac) accelerates electrons constantly down the linac,each electron will have 50GeV of energy at the exit point. what is the speed of the electron after going 1m down the accelerator? After electrons exit the linac,magnets are used to curve the electron beams. the radius of curvature is 280m,what is the minimum magnetic field needed?

    I think i have a rough idea of how to do this question,but somehow my answers doesn't agree with the answer key!

    i think as 2as=v2,energy of electron after going down 1m should be (1/3200)*50GeV,which is 2.5x10-12joules,using eqn for relativistic kinetic energy
    KE=MC2/√1-V2/C2,V works out to be 0.99946c
    up to here my result agrees with the answer key
    then i calculated relativistic momentum of electron to be 8.31x10-21kgm/s using MV/√1-V2/C2.
    using eqn r=MV/Bq,sub in r=280m,i get B=1.86x10-4Tesla while the answer is 0.597T ???
     
  2. jcsd
  3. Aug 18, 2012 #2
    The question asks for the speed of the electron after going 1m which you have calculated but then asks for the magnetic field needed after the electron exits the LINAC which is 3.2km. You have used the velocity for the first part which won't be correct.
     
  4. Aug 18, 2012 #3
    yeah,thank you,i made a careless mistake here.but after i used the correct data i had v=0.9999c,relativistic momentum=8.89x10-26 and B field=1.98x10-9Tesla,which is even further off the answer of 0.597T,puzzling
     
  5. Aug 19, 2012 #4
    can anyone tell me if the eqn R=mev/Bq for circular motion of charged particles in a magnetic field is valid in this relativistic case,i suspect that's where the problem in my solution lies
     
  6. Aug 19, 2012 #5
    I did the question with that equation for R and got the given answer, can you show exactly what calculations you are doing at each step please - the value you have given for the momentum is incorrect (far too small)
     
  7. Aug 19, 2012 #6
    oh i can get the correct answer now,i think the problem of my previous calculation is i used v value 0f 0.9999c to calculate γ(p=γmv),which i think is not accurate enough. later i used γ value i get from part 1 to calculate and get the correct p and B

    Thank you very much for your help!
     
  8. Aug 19, 2012 #7
    oh i can get the correct answer now,i think the problem of my previous calculation is i used v value 0f 0.9999c to calculate γ(p=γmv),which i think is not accurate enough. later i used γ value i get from part 1 to calculate and get the correct p and B

    Thank you very much for your help!
     
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