Introduction to accelerator and their applications: Exercize 1 Gabriele Chiodini - INFN Lecce - May 2015
PhD lessons in Physics for Università del Salento 2015 (20 hours, 4 CFD)
Unswer 1: b Unswer 2: a Unswer 3: a Unswer 4: a Unswer 5: b Unswer 6: a Unswer 7: b Unswer 8: b Unswer 9: b Unswer 10: c Unswer 11: b Unswer 12: c Unswer 13: a Unswer 14: b
!
Solution 1E2=(m0c2)2+(pc)2
E2/(m0c2)2=1+(pc)2/(m0c2)2 p=mv=γm0βc
E2/(m0c2)2=1+(γβ)2
γ2 =1/(1−β2 ), γ2 − γ2β2 =1, 1+γ2 −=γ2β2 Than E/(m0c2)= γ
β=p/(γm0c) =pc/(γm0c2)
Than replacing in the previous relation we obtain: β=pc/E.
!
Solution 2E=T+m0c2=1+0.9383=1.9383GeV
(pc)2=E2-(m0c2)=(1.938)2-(0.938)2=2.8766GeV/c.
!
Solution 3The guide field B=1 T reaches is peak value at T=1/f=5ms.
The peak magnetic flux concatenated with the donut is equal to the area multiplied by the magnetic field peak averaged on the area, this is tw times the peak of the guide field (rule 2 to 1):
flux=2B x πρ2. where ρ is the donut radius.
The potential voltage along the donut is given by the flux variation in the unit time:
V/turn=flux/T=2 x 1 x π x 0.12/5E-3=12.6Volts/turn.
The number of turn n during the time T is n=cT/(2πρ)=3E8*5E-3/
(2π0.1)=2.4E6 turn.
The final energy is 12.6 x 2.4E6 = 30.24E6 eV = 30.24 MeV.
! !
! !
! 1
Introduction to accelerator and their applications: Exercize 1 Gabriele Chiodini - INFN Lecce - May 2015
PhD lessons in Physics for Università del Salento 2015 (20 hours, 4 CFD)
Solution 4
frev=qB/(2πm)=1.6E-19 x 1.2 / (2π x 1.7E-27)=18MHz
If the magnetic field double the frequency double too, but there is not dependency of the frequency with the radius.
!
Solution 5In the magnetic rigidity formula there is not dependency on mass, than the maximum momentum is equal for protons and deuterons.
Let’s calculate the proton momentum from the proton energy.
T=1GeV.
(pc)2=E2-(m0c2)2=(T+m0c2)2-(m0c2)2=(1+0.938)2-(0.938)2= 2.875GeV2; than pc=1.696GeV.
In the same way for deuterons (A=2 but p the same)
(pdeuc)2=E2-(2m0c2)2=(Tdeu+2m0c2)2-(2m0c2)2=2.875GeV2;
than (Tdeu+2m0c2)2=(2m0c2)2+2.875GeV2=3.52GeV2+2.875GeV2
=6.39GeV2 than Tdeu+2m0c2 =2.53GeV, finally Tdeu=0.653GeV.
From exercize 1 we obtain E for protons with T=1GeV and β=pc/E than v=βc.
β=1.696/1.938=0.875.
Let’s calculate the rotation frequency.
f= v/(2πρ)=βc/(2πρ)=0.875 x 2.9979E8 / (2π x 25)=1.67 MHz.
Repeting the calculation for deuterons:
β=1.696/(0.653+2x0.938)=0.67.
f=βc/(2πρ)=0.67 x 2.9979E8 / (2π x 25)=1.28 MHz.
!
Answer 6Being 10kGauss = 1Tesla the bending field about B=0.5 x 10E-4 Tesla = 5 x 10E-5 Tesla.
An electron with kinetic energy T owns a relativistic energy E = mc^2 + T and a relativistic moment of
p^2=E^2-mc^2=(T+mc^2)^2-(mc^2)^2= T^2+2Tmc^2.
p^2=10*10 keV^2 + 2*10*500 keV^2=100.5^2 keV^2
p = 1 0 0 . 5 k e V, f r o m t h e m a g n e t i c r i g i d i t y f o r m u l a p(GeV)=0.3R(m)B(T),
R(m)=3.3p(GeV)/B(T)=3.3 * 10E-4GeV / 5 x 10E-5 Tesla = 6 m.
The direction of the magnetic field is the one that goes from the south pole to the north pole of a compass needle free to rotate, and the field generated by a dipole goes from the north pole to the south pole (the needle north pole) and the geographic north pole is the south pole of the Earth's magnetic field.
The Earth's magnetic field out of the geographic south pole and enters the geographic north pole. The strength of the Lorentz force must bend the electrons towards the center then the electrons must move in a counterclockwise direction.
! 2
