What would the magnetic-field magnitude have to be for the maximum energy to which a proton can be accelerated to be twice that calculated in part A? The magnetic poles of a small cyclotron produce a magnetic field with magnitude 0.89 T. The poles have a radius of 0.38 m which is the maximum radius of the orbits of the accelerated particles Express your answer using two significant figures Submit Request Answer PartE For B = 0.89 T , what is the maximum energy to which alpha particles (q = 3.20 x 10 19 C. m-6.65 x 10 27 kg) can be accelerated by this cyclotron? Express your answer using two significant figures. ?? Submit Request Answer
Solution
in cyclotron radius of the orbit is
r = (m*v) / (q*B)
v = (q*B*r/m)
v = (1.6 x 10^-19 x 0.89 x 0.38) / (1.67 x 10^-27)
v = 3.24 x 10^7 m/s
A)
Maximum Energy is
Emax = 0.5*m*v^2
Emax = 0.5 x 1.67 x 10^-27 x (3.24 x 10^7)^2
Emax = 8.77 x 10^-13 J
B)
Emax = 8.77 x 10^-13 / (1.6 x 10^-19)
Emax = 5.48 x 10^6 eV
C)
T = (2*pi*r)/v
T = (2 x 3.142 x 0.38) / (3.24 x 10^7)
T = 7.34 x 10^-8 sec
.
What would the magnetic-field magnitude have to be for the maximum ene.docx
1. What would the magnetic-field magnitude have to be for the maximum energy to which a proton
can be accelerated to be twice that calculated in part A? The magnetic poles of a small cyclotron
produce a magnetic field with magnitude 0.89 T. The poles have a radius of 0.38 m which is the
maximum radius of the orbits of the accelerated particles Express your answer using two
significant figures Submit Request Answer PartE For B = 0.89 T , what is the maximum energy
to which alpha particles (q = 3.20 x 10 19 C. m-6.65 x 10 27 kg) can be accelerated by this
cyclotron? Express your answer using two significant figures. ?? Submit Request Answer
Solution
in cyclotron radius of the orbit is
r = (m*v) / (q*B)
v = (q*B*r/m)
v = (1.6 x 10^-19 x 0.89 x 0.38) / (1.67 x 10^-27)
v = 3.24 x 10^7 m/s
A)
Maximum Energy is
Emax = 0.5*m*v^2
Emax = 0.5 x 1.67 x 10^-27 x (3.24 x 10^7)^2
Emax = 8.77 x 10^-13 J
B)
Emax = 8.77 x 10^-13 / (1.6 x 10^-19)
Emax = 5.48 x 10^6 eV
2. C)
T = (2*pi*r)/v
T = (2 x 3.142 x 0.38) / (3.24 x 10^7)
T = 7.34 x 10^-8 sec
