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 ???