Radius of Curvature Calculation

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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