## Exa11.1 : : Page-535(2011)
#calculate The optimum number of stages in the accelerator and the ripple voltage
import math
V_0 = 10**5; ## Accelerating voltage, volts
C = 0.02e-006; ## Capacitance, farad
I = 4*1e-003; ## Current, ampere
f = 200.; ## Frequency, cycles per sec
n = math.sqrt (V_0*f*C/I); ## Number of particles
delta_V = I*n*(n+1.)/(4.*f*C);
print'%s %.2f %s'%("\nThe optimum number of stages in the accelerator = ", n,"");
print'%s %.2f %s'%("\nThe ripple voltage = ", delta_V/1e+003,"kV");
## Result
## The optimum number of stages in the accelerator = 10
## The ripple voltage = 27.5 kV
## Exa11.2 : : Page-536 (2011)
#calculate The charging current and The rate of rise of electrode potential
import math
s = 15.; ## Speed, metre per sec
w = 0.3; ## Width of the electrode, metre
E = 3e+06; ## Breakdown strength, volts per metre
eps = 8.85e-12; ## Absolute permitivity of free space, farad per metre
C = 111e-12; ## Capacitance, farad
i = round (2*eps*E*s*w*10**6); ## Current, micro ampere
V = i/C*10**-12; ## Rate of rise of electrode potential, mega volts per sec
print'%s %.2f %s %.2f %s '%("\nThe charging current =",i," micro-ampere"and " \nThe rate of rise of electrode potential = ",V," MV/sec");
## Result
## The charging current = 239 micro-ampere
## The rate of rise of electrode potential = 2.15 MV/sec
## Exa11.3 : : Page-536 (2011)
#calculate The length of the final drift tube and The kinetic energy of the injected protons
import math
import scipy
from scipy import integrate
f = 200.*10**6; ## Frequency of the accelerator, cycle per sec
M = 1.6724e-27; ## Mass of the proton, Kg
E = 45.3*1.6e-13; ## Accelerating energy, joule
L_f = round (1./f*math.sqrt(2.*E/M)*100.); ## Length of the final drift tube, centi metre
L_1 = 5.35*10**-2; ## Length of the first drift tube, metre
K_E = (1./2.*M*L_1**2.*f**2.)/1.6e-13; ## Kinetic energy of the injected proton, MeV
E_inc = E/1.6e-13-K_E; ## Increase in energy, MeV
q = 1.6e-19; ## Charge of the proton, C
V = 1.49e+06; ## Accelerating voltage, volts
N = E_inc*1.6e-13/(q*V); ## Number of drift protons
def fun(n):
y=n**0.5
return y
l2=scipy.integrate.quad(fun,0,N)
L = 1./f*math.sqrt(2.*q*V/M)*l2[0]; ## Total length of the accelerator, metre
print'%s %.2f %s %.2f %s %.2f %s '%("\nThe length of the final drift tube = ",L_f," cm"and"\nThe kinetic energy of the injected protons = ",K_E," MeV"and"\nThe total length of the accelerator = ",L," metre");
## Result
## The length of the final drift tube = 47 cm
## The kinetic energy of the injected protons = 0.60 MeV
## The total length of the accelerator = 9.2 metre
## Exa11.5 : : Page-536 (2011)
#find energy of the emergine deutron and frequency of the dee voltage
import math
B = 1.4; ## Magnetic field, tesla
R = 88e-002; ## Radius of the orbit, metre
q = 1.6023e-019; ## Charge of the deutron, C
M_d = 2.014102*1.66e-27; ## Mass of the deutron, Kg
M_He = 4.002603*1.66e-27; ## Mass of the He ion, Kg
E = B**2*R**2*q**2/(2*M_d*1.6e-13); ## Energy og the emerging deutron, mega electron volts
f = B*q/(2.*math.pi*M_d)*10**-6; ## Frequency of the deutron voltage, mega cycles per sec
B_He = 2*math.pi*M_He*f*10**6/(2*q); ## Magnetic field required for He(++) ions, weber per square metre
B_change = B-B_He; ## Change in magnetic field, tesla
print'%s %.2f %s %.2f %s %.2f %s '%("\nThe energy of the emerging deutron = ",E," MeV" and "\nThe frequency of the dee voltage = ",f,"MHz" and"\nThe change in magnetic field = ",B_change," tesla");
## Result
## The energy of the emerging deutron = 36.4 MeV
## The frequency of the dee voltage = 10.68 MHz
## The change in magnetic field = 0.01 tesla
## Exa11.6: : Page-537 (2011)
#calculate effective reduction in magnetic field and charge in orbit radius
import math
K_E = 7.5*1.6023e-13; ## Kinetic energy, joule
r = 0.51; ## Radius of the proton's orbit, metre
E = 5*10**6; ## Electric field, volts per metre
m = 1.67e-27; ## Mass of the proton, Kg
q = 1.6023e-19; ## Charge of the proton, C
v = math.sqrt(2.*K_E/m); ## Velocity of the proton, metre per sec
B_red = E/v; ## The effective reduction in magnetic field, tesla
B = m*v/(q*r); ## Total magnetic field produced, tesla
r_change = r*B_red/B; ## The change in orbit radius, metre
print'%s %.2f %s %.2f %s '%("\nThe effective reduction in magnetic field = ",B_red," tesla "and "\nThe change in orbit radius = ",r_change," metre ");
## Result
## The effective reduction in magnetic field = 0.132 tesla
## The change in orbit radius = 0.087 metre
## Exa11.7 : : Page-537 (2011)
#calculate enegy of the elctron and average enegy gained per revolution
import math
B = 0.4; ## Magnetic field, tesla
e = 1.6203e-19; ## Charge of an electron, C
R = 30*2.54e-02; ## Radius, metre
c = 3e+08; ## Capacitance, farad
E = B*e*R*c/1.6e-13; ## The energy of the electron, mega electron volts
f = 50.; ## Frequency, cycles per sec
N = c/(4*2*math.pi*f*R); ## Total number of revolutions
Avg_E_per_rev = E*1e+006/N; ## Average energy gained per revolution, electron volt
print'%s %.2f %s %.2f %s '%("\nThe energy of the electron = ",E," MeV"and "\nThe average energy gained per revolution = ",Avg_E_per_rev," eV");
## Result
## The energy of the electron = 92.6 MeV
## The average energy gained per revolution = 295.57 eV
## Note: Wrong answer is given in the textbook
## Average energy gained per revolution : 295.57 electron volts
## Exa11.8 : : Page-537 (2011)
#calculate peak current and average current and duty cycle
import math
R = 0.35; ## Orbit radius, metre
N = 100e+06/480.; ## Total number of revolutions
L = 2*math.pi*R*N; ## Distance traversed by the electron, metre
t = 2e-06; ## Pulse duration, sec
e = 1.6203e-19; ## Charge of an electron, C
n = 3e+09; ## Number of electrons
f = 180.; ## frequency, hertz
I_p = n*e/t; ## Peak current, ampere
I_avg = n*e*f; ## Average current, ampere
tau = t*f; ## Duty cycle
print'%s %.2e %s %.2e %s %.2e %s'%("\nThe peak current = ",I_p," ampere" and "\nThe average current = ",I_avg," ampere "and "\nThe duty cycle =",tau,"");
## Result
## The peak current = 2.4e-004 ampere
## The average current = 8.75e-008 ampere
## The duty cycle = 3.6e-004
## Exa11.9 : : Page-538 (2011)
#calculate maximum frequncy of the dee voltage and kinetic energy of the deutron
import math
q = 1.6023e-19; ## Charge of an electron, C
B_0 = 1.5; ## Magnetic field at the centre, tesla
m_d = 2.014102*1.66e-27; ## Mass of the deutron, Kg
f_max = B_0*q/(2*math.pi*m_d*10**6); ## Maximum frequency of the dee voltage, mega cycles per sec
B_prime = 1.4310; ## Magnetic field at the periphery of the dee, tesla
f_prime = 10**7; ## Frequency, cycles per sec
c = 3e+08; ## Velocity of the light, metre per sec
M = B_prime*q/(2*math.pi*f_prime*1.66e-27); ## Relativistic mass, u
K_E = (M-m_d/1.66e-27)*931.5; ## Kinetic energy of the particle, mega electron volts
print'%s %.2f %s %.2f %s '%("\nThe maximum frequency of the dee voltage = ",f_max," MHz"and "\nThe kinetic energy of the deuteron = ",K_E," MeV");
## Result
## The maximum frequency of the dee voltage = 11.44 MHz
## The kinetic energy of the deuteron = 171.6 MeV
#calculate frequency of the applied electric field and magnetic field intensity
## Exa11.10 : : Page-538 (2011)
import math
e = 1.6023e-19; ## Charge of an electron, C
E = 70*1.6e-13; ## Energy, electron volts
R = 0.28; ## Radius of the orbit, metre
c = 3e+08; ## Velocity of light, metre per sec
B = E/(e*R*c); ## Magnetic field intensity, tesla
f = e*B*c**2/(2*math.pi*E); ## Frequency, cycle per sec
del_E = 88.5*(0.07)**4*10**3/(R); ## Energy radiated by an electron, electron volts
print'%s %.2f %s %.2f %s %.2f %s '%("\nThe frequency of the applied electric field = ",f," cycles per sec" and"\nThe magnetic field intensity = ",B," tesla"and "\nThe energy radiated by the electron =",del_E," eV");
## Result
## The frequency of the applied electric field = 1.705e+008 cycles per sec
## The magnetic field intensity = 0.832 tesla
## The energy radiated by the electron = 7.6 eV
#calculate kinetic energy of the accelerated nitrogen
## Exa11.11 : : Page-538 (2011)
import math
E = 3.; ## Energy of proton synchrotron, giga electron volts
m_0_c_sq = 0.938; ## Relativistic energy, mega electron volts
P_p = math.sqrt(E**2-m_0_c_sq**2); ## Momentum of the proton, giga electron volts per c
P_n = 6*P_p; ## Momentum of the N(14) ions, giga electron volts
T_n = math.sqrt(P_n**2+(0.938*14.)**2)-0.938*14; ## Kinetic energy of the accelerated nitrogen ion
print'%s %.2f %s'%("\nThe kinetic energy of the accelerated nitrogen ion = ",T_n," MeV");
## Result
## The kinetic energy of the accelerated nitrogen ion = 8.43 MeV
## Exa11.12 : : Page-539 (2011)
#calculate maximum magnetic flux density and maximum frequency of the accerlating voltage
import math
e = 1.6e-19; ## Charge of an electron, C
R = 9.144; ## Radius, metre
m_p = 1.67e-027; ## Mass of the proton, Kg
E = 3.6*1.6e-13; ## Energy, joule
L = 3.048; ## Length of the one synchrotron section, metre
T = 3; ## Kinetic energy, giga electron volts
c = 3e+08; ## Velocity of the light, metre per sec
m_0_c_sq = 0.938; ## Relativistic energy, mega electron volts
B = round (math.sqrt(2*m_p*E)/(R*e)*10**4); ## Maximum magnetic field density, web per square metre
v = B*10**-4*e*R/m_p; ## Velocity of the proton, metre per sec
f_c = v/(2*math.pi*R*10**6); ## Frequency of the circular orbit, mega cycles per sec
f_0 = 2*math.pi*R*f_c*10**3/(2*math.pi*R+4*L); ## Reduced frequency, kilo cycles per sec
B_m = 3.33*math.sqrt(T*(T+2*m_0_c_sq))/R; ## Relativistic field, web per square metre
f_0 = c**2*e*R*B*1e-004/((2*math.pi*R+4*L)*(T+m_0_c_sq)*e*1e+015); ## Maximum frequency of the accelerating voltage, mega cycles per sec
print'%s %.2f %s %.2f %s '%("\nThe maximum magnetic flux density = ",B_m," weber/Sq.m"and "\nThe maximum frequency of the accelerating voltage = ",f_0," MHz");
## Result
## The maximum magnetic flux density = 1.393 weber/Sq.m
## The maximum frequency of the accelerating voltage = 0.09 MHz
## Answer is given wrongly in the textbook
#calculate energy of the proton
## Exa11.13 : : Page-539 (2011)
import math
E_c = 30e+009; ## Energy of the proton accelerator, GeV
m_0_c_sq = 0.938*10**6; ## Relativistic energy, GeV
E_p = (4*E_c**2-2*m_0_c_sq**2)/(2*m_0_c_sq) ; ## Energy of the proton, GeV
print'%s %.2f %s'%("\nThe energy of the proton = ",E_p/1e+009," GeV");
print("wrong answer in the textbook")
## Result
## The energy of the proton = 1.92e+006 GeV
## Wrong answer given in the textbook