Difference between revisions of "Radius of Curvature Calculation"

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Below are my calculations done for determining the radius of curvature of an electron/positron in the magnetic field for the pair spectrometer.
 
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.
 
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>
+
<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
 
This equation can be rearranged to solve for r (and given that mv = momentum) to give the following equation
  
<math> r =  * v * B </math>
+
<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
 +
 
 +
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

Latest revision as of 08:57, 17 February 2009

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

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