1540-initial data.sagews

 

ROUGH CALCULATION FOR A MAGNETIC TOF DEFLECTOR

As everything depends on everything I'll put some initial guesses and tweak a little until every relevant design parameter falls into an achievable range

initial data

Let's start with some initial calculations that will help us later. We want to deflect protons up to maxEnergy (MeVs) and expect currents about current (Amps). Using a classical approximation, $T = {1/2}mv^2$, $v = (2*T/m)^{1/2}$

maxEnergy_eVs = 10e6  # ev
protonMass = 1.672e-27 # kg
kineticEnergy = maxEnergy_eVs * 1.602e-19 
protonSpeed = sqrt(2 * kineticEnergy / protonMass)
"protons of 10MeVs are traveling at %s m/s " % protonSpeed

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{protons of 10MeVs are traveling at 4.37751978017659e7 m/s }

Which clearly falls outside the range of validity of the classical aproximation.  so we have to use the relativistic formula $E_c = {{mc^2}\over{\sqrt{1-{{v^2}\over{c^2}}}}} - mc^2$

mc2 = 3.0e8 * 3.0e8 * protonMass
i = 1 / ((kineticEnergy / mc2) + 1) 
g2 = 1 - (i ^ 2 )
g = sqrt(g2)
protonSpeed = g * 3.0e8
kineticEnergy, g2, g
"protons of 10MeVs are traveling at %s m/s (gamma = %s)" % (protonSpeed, g)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{protons of 10MeVs are traveling at 4.34292057133611e7 m/s (gamma = 0.144764019044537)}

current = 100e-9  # we expect nanoamps
electron_charge = 1.16e-19
particles_per_second = current / electron_charge

"we expect a current of %f nA. that means %s protons/s " % (current * 1e9, particles_per_second)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{we expect a current of 100.000000 nA. that means 8.62068965517241e11 protons/s }

geometric calculations

Let's call z the distance between colimators and beam entry point  and d the wanted maximum deflection at the target.

degrees =  (2 * 3.141592) / 360  # conversion from degrees to rads, to be able to input angles in degs
z = 0.2 # distances in meters
d = 0.05
theta = arctan(d/z)
"theta = %s degrees" % (theta / degrees)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{theta = 14.0362463880846 degrees}

from there, and giving a bending radius R we get the aprox. area in which the B field should be present and therefore the size of the coils. (see figure)

R = 0.2
a = 2 * R * (1-cos(theta))
b = R * sin(theta)
"required magnetic field area (width, length): %f cm x %f cm " % (a * 100, b * 100)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{required magnetic field area (width, length): 1.194300 cm x 4.850713 cm }

As we'll use circular coils (at least for this calculations), we get the radius as diagonal of the resulting box divided by two, and scale with a safety factor

c = (sqrt(a*a + b*b) / 2) * 1.5
"coil radius (aprox) %s m " % c

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{coil radius (aprox) 0.0374668084915988 m }

The required field to bend a particle with the given mass and speed is that for which the centrifugal force is the same as the magnetic bending force: ${{mv^2}\over{R}} = q * v * B$ , also $B = {{m * v}\over{q * R}}$

B = (protonMass * protonSpeed )/ (electron_charge * R)
"required field: %s Teslas ( %s Gauss) " % (B, B * 500000)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{required field: 3.12989792899740 Teslas ( 1.56494896449870e6 Gauss) }

beam and time constraints

let's suppose that the beam and the collimator have the same width w. As the beam position is given (aprox) by $x=R*sin(\omega*t)$, the speed at which the beam is deflected at the collimator distance is given by $\delta \omega (t) =  R * \omega * cos(\omega * t)$ and will be maximum at x = 0.$v(0) = R * \omega$. We want the beam to be switched on in switch_time. This gives a beam deflection speed of:

beam_width = 5e-3 # m
switch_time = 10e-9 # s
v_0 = beam_width / switch_time
"beam sweep speed (m/s) = %2f m/s" % v_0

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{beam sweep speed (m/s) = 500000.000000 m/s}

This beam speed is attainable with the given radius with a frequency ($\omega$) of:

omega = v_0 / R
hz = omega / 6.2830
"required frequency: %s rads/s. About %s kHz" % (omega, hz / 1000.0)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{required frequency: 2.50000000000000e6 rads/s. About 397.899092790068 kHz}

Which is to be taken into account for the design of the coils.

Other considerations

If (as supposed) collimator and beam are of the same width, during the on cycle, the number of protons arriving to the target is:

time_on = 2 * switch_time
n_cycle = current * time_on / electron_charge
"protons in each cycle : %f" % (n_cycle)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{protons in each cycle : 17241.379310}

Design of the coils

up to here, we have the following design parameters for the coils (see geometrical calculations):

"bending angle: +-%i deg, bending radius: %i mm" % (theta / degrees, R * 1000)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{bending angle: +-14 deg, bending radius: 200 mm}

"coil-radius: %f mm (section: %.1f x %.1f mm)" % (c * 1000, a * 1000, b * 1000)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{coil-radius: 81.840573 mm (section: 97.7 x 48.5 mm)}

"required field: %s Teslas ( %s Gauss) " % (B, B * 500000)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{required field: 3.15483322088589 Teslas ( 1.57741661044294e6 Gauss) }

To get this field we use the aproximation of field in the center of one coil-turn: $B={{\mu_0 * I} \over{ 2 * c}}$, being c the radius of the coil. This makes $I = { {2* c * B} \over { \mu_0 * nturns} }$. Now putting some reasonable values for the number of turns and the intensity...

turns = 500
mu_0 = 4 * 3.1415 * 1e-7
I = 2 * c * B / (mu_0 * turns * 2)
"required intensity for two coils with %s turns each: %s miliamps" % (turns, I * 1000.0)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{required intensity for two coils with 500 turns each: 410939.612369745 miliamps}

To calculate the electrical characteristics, we need to know the inductance of the coil: From a very crude approximation, $L = {{\mu_0 N^2 A }\over{l}}$ . This is for a toroidal coil and we should not expect from it more than the order of magnitude

coilSection = 3.1415 * c * c
l =  0.5 # this is a fake value. is the mayor length for a torus
L = mu_0 * turns * turns * coilSection / l
"self inductance (accurate only to an order of magnitude): %s H" % (L,)

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{self inductance (accurate only to an order of magnitude): 0.0132203041368704 H}

the important thing here is that we need a rough estimation of the self-inductance L to calculate the required capacity for the oscillator in the parallel-resonant circuit. The resonant frequency will be aprox LC. Also $\omega = {{1}\over{\sqrt{L * C}}}$;  $C = {{1}\over{\omega^2 * L}}$

capacitance = 1 / (omega * omega * L)
"req. capacitor for the resonant circuit: %s C (right only up to order of magnitude)" % capacitance

\newcommand{\Bold}[1]{\mathbf{#1}}\hbox{req. capacitor for the resonant circuit: 1.21025960025967e-11 C (right only up to order of magnitude)}

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