Radius of Curvature Calculation: Difference between revisions
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<math>r = \frac{\rho}{q*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. | For this sample equation 1 MeV will be used to determine the radius of curvature per MeV of the incident beam. | ||
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Charge of an electron = <math>1.6022 * 10^{19} Coulombs</math> | Charge of an electron = <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> | |||
Revision as of 08:42, 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]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ 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
Charge of an electron = [math]\displaystyle{ 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]\displaystyle{ 2.99 * 10^8 \frac{m}{s} }[/math]