Magnet

Specs:

TESLA ENGINEERING: 7 Degree Bend Angle Dipole

[math]B_{max}=0.35 T[/math]

Current: [math]138 A[/math]

Resistance: [math]0.116 Ohm[/math]

Voltage: [math]V=IR=16.008 V[/math]

Water Flow: [math]1.2 L/minute[/math]

CALCULATIONS

Radius of Curvature

Below are my calculations done for determining the radius of curvature of an electron/positron in the magnetic field for the pair spectrometer.

The Lorentz force is the centripetal force acting upon the electron/positron in the magnetic field which gives the following equation.

[math]\frac{(m*v^2)}{r} = q * v * B [/math]

This equation can be rearranged to solve for r (and given that mv = momentum) to give the following equation

[math]r = \frac{\rho}{q*B}[/math]

For this sample equation 1 MeV will be used to determine the radius of curvature per MeV of the incident beam.

Energy = 1 MeV = [math] 1 * 10^6 \frac{J}{C} * 1.6022 * 10^{-19}C = 1*10^6\frac{kg * m^2}{s^2 * C} * 1.6022 * 10^{-19}C [/math]

Magnetic Field (B) = 0.35 Tesla = [math]0.35\frac{kg}{C*s}[/math]

Charge of an electron (q) = [math]1.6022 * 10^{19} Coulombs[/math]

To get the momentum of the positron/electron the energy of the particle is divided by the speed of the particle (ie [math] 2.99 * 10^8 \frac {m}{s})[/math]

Substituting the numbers above into the radius equation gives the following

[math] \frac{(1*10^6 \frac{kg*m^2}{s^2*C}) * (1.6022 * 10^{-19}C)}{(2.99*10^8\frac{m}{s})*(0.35\frac{kg}{C*s})*(1.6022 * 10^{-19}C)} = 0.009556 meters = 0.9556 cm[/math]

So basically the radius of curvature for the electrons/positrons is ~0.9556cm per MeV

So for a 7 MeV electron/positron pair the radius of curvature for either particle in a 0.35 T field would be 6.69cm

Calculating the Magnetic Field Needed:

Lorentz Force equation: F=q(v×B)

Electron moves through the magnetic field B accelerated by force F proportional to the component of velocity perpendicular to the field B and velocity v. Moves with constant kinetic energy and speed due to the fact that the magnetic field never does work on the particle since the always moves perpendicular to the force.

Magnetic force: [math]F=e*v*B[/math]

The radius of the arc can be through: [math](m*v^2)/R=e*v*B[/math]

giving: [math]R=m*v/e*B[/math]

The length of the circular arc is S and the deflection angle is found as: sin(θ)=S/R

For small θ, and large R, the arc length S will be approx L, giving: sin(θ)=L/R=L*e*B/m*v

Giving θ=sin^(-1)(c*B*L/p)

The displacement is found as: d=R-R*cos(θ)=m*v/e*B*(1-cos(θ))

Table: Data for B=0.0078 T.

Momentum	Radius of Curvature	Bend Angle	Bend Angle	Displacement @ end of magnet
P (MeV)	R (m)	θ (radians)	θ (degrees)	d (cm)
1	0.43	0.70	40.02	10.01
2	0.86	0.33	18.76	4.54
3	1.28	0.22	12.38	2.98
4	1.71	0.16	9.25	2.22
5	2.14	0.13	7.39	1.78
6	2.57	0.11	6.15	1.48
7	2.99	0.09	5.27	1.27
8	3.42	0.08	4.61	1.11
9	3.85	0.07	4.10	0.98
10	4.28	0.06	3.69	0.89
11	4.70	0.06	3.35	0.80
12	5.13	0.05	3.07	0.74
13	5.56	0.05	2.84	0.68
14	5.99	0.05	2.63	0.63
15	6.41	0.04	2.46	0.59
16	6.84	0.04	2.30	0.55

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