2 Guiding Particles and Keeping Them Focused

Discussion Topics…

  • Electrostatic Devices
  • Magnetic Devices
  • Rigidity
  • Solenoids, and when (not) to use them
  • Global vs. Local Coordinates
  • Quadrupoles as Lenses
  • Doublets, Triplets; FODO System

Homework Problem 1:

theta = 45*pi/180
rho = 0.25 # m
Wmax = 20 # MeV

Suppose we wish to bend a proton’s trajectory by 45 degrees, with a bending radius of 0.25 m, by (a) electrostatic deflection or (b) by magnetic deflection.

For (a) plot the value of the electric field required along a circular (equipotential) trajectory within a cylindrical bend as a function of the proton’s kinetic energy, up to 20 MeV. What path length is required? Neglect edge effects. Note: A practical engineering limit for an electric field in this problem is about \(E_{max} \approx\) 20 MV/m.

#F = eE = gamma m v^2/R = gamma E0 beta^2 / R 
#  = gamma beta^2 E0/R = (beta*gamma)^2/gamma E0/R
#  = (gamma^2 - 1)/gamma E0/R  =  (gamma-1)(gamma+1)/gamma E0/R
#  = (W/E0)(W/E0+2)/(1+W/E0) E0/R;  x == W/E0
E0 = 938       # MeV
E  = function(x) {x*(2+x)/(1+x)*E0/rho}  # MV/m
curve(E(x/E0),0,Wmax,
   xlab="kinetic energy (MeV)", ylab="E (MV/m)")
abline(h=20, lty=2)

Le = rho*theta
Le # m
## [1] 0.1963495

For (b) plot the magnetic field strength required in an electromagnet as a function of the proton’s kinetic energy. Again, neglect edge effects. Note: A practical engineering limit for a magnetic field in this problem is about \(B_{max} \approx\) 2 T.

# Brho = p/e = pc/ec = 10/2.9979*pc[GeV]
# pc = E0 sqrt(gamma^2-1) = E0 sqrt(W/E0 * (2+W/E0))
# B = Brho/rho = 10/2.9979*(E0/1000)/rho * sqrt(W/E0 * (2+W/E0))
B = function(x){ 10/2.9979*(E0/1000)/rho*sqrt(x*(2+x)) } # Tesla
curve(B(x/E0),0,Wmax,
   xlab="kinetic energy (MeV)", ylab="B (T)")
abline(h=2, lty=2)

Homework Problem 2:

E0 = 0.511 # MeV
W = 200    # MeV
pc = sqrt(((W+E0)/E0)^2 - 1)*E0/1000 # GeV
Brho = 10/2.9979*pc   #  T-m
F = 4       # m
Leff = 0.2  # m
d    = 0.6  # m
  1. Consider a beam of electrons, each with kinetic energy 200 MeV. A quadrupole lens is used to focus this beam, with a focal length of 4 m. What magnetic gradient, \(B'\), is required if the effective length of the magnet is 0.2 m? Is a thin lens approximation justified for use in this situation?
# 1/F = G Leff / Brho
Grad = Brho/F/Leff
Grad   # T/m
## [1] 0.836045
Leff/F
## [1] 0.05
ifelse(Leff/F < 0.1, "yes", "no")
## [1] "yes"

  1. We wish to focus the beam both horizontally and vertically using a “doublet”. Consider a second quadrupole magnet with the same parameters as the one used above, but with its electrical polarity reversed creating a “defocusing” lens. If these two thin lenses are now separated by 0.6 m, center-to-center, with the beam passing through the horizontally focusing magnet first and then the horizontally defocusing magnet,

    1. what is the resulting focal length in the horizontal plane?
    2. What is it in the vertical plane?
    3. How do these compare with the theoretical thin lens doublet focal length?
    4. Can you create a system of these two magnets that has the same focal length in both planes? (Justify your answer.)

In the above, measure focal length from the mid-point between the two quadrupole magnets.

MF = matrix(c(  1,  0,
              -1/F, 1), nrow=2, ncol=2, byrow=TRUE)
Md = matrix(c(  1,  d,
                0,  1), nrow=2, ncol=2, byrow=TRUE)
MD = matrix(c(  1,  0,
               1/F, 1), nrow=2, ncol=2, byrow=TRUE)
x0 = array(c(1,0))
x0
## [1] 1 0
Outx = MD %*% Md %*% MF %*% x0
Outy = MF %*% Md %*% MD %*% x0
Fx = d/2 - Outx[1]/tan(Outx[2])
Fy = d/2 - Outy[1]/tan(Outy[2])

Fx
## [1] 22.95604
Fy
## [1] 30.95229
F^2/d
## [1] 26.66667
r0 = 1.1
x0 = c(1,0,1,0)
Out = function(r){
  MF = matrix(c(  1,     0,    0,     0, 
                -1/F,    1,    0,     0,
                  0,     0,    1,     0,
                  0,     0,   1/F,    1 ), nrow=4, ncol=4, byrow=TRUE)
  Md = matrix(c(  1,     d,    0,     0, 
                  0,     1,    0,     0,
                  0,     0,    1,     d,
                  0,     0,    0,     1 ), nrow=4, ncol=4, byrow=TRUE)
  MD = matrix(c(  1,     0,    0,     0, 
                 1/F*r,  1,    0,     0,
                  0,     0,    1,     0,
                  0,     0, -1/F*r,   1 ), nrow=4, ncol=4, byrow=TRUE)
  MD %*% Md %*% MF %*% x0
}
Fcn = function(r){
  V = Out(r)
  Fx = d/2 - V[1]/tan(V[2])
  Fy = d/2 - V[3]/tan(V[4])
abs(Fx-Fy)
}
F2d = optimize(Fcn,c(0.5,1.5))
rFin = F2d$minimum
Vf = Out(rFin)
Fx = d/2 - Vf[1]/tan(Vf[2])
Fy = d/2 - Vf[3]/tan(Vf[4])
paste("Increase strength of D quad by factor ", round(rFin,4), ".")
## [1] "Increase strength of D quad by factor  1.023 ."
paste("Common focal length will be Fx =", round(Fx,2), "m, Fy = ", round(Fy,2), "m.")
## [1] "Common focal length will be Fx = 26.35 m, Fy =  26.35 m."