Question

a)An $e^{+}e^{-}$ collider has a diameter of 8 km and produces beams of energy 45 GeV. Each beam consists of 10 bunches, each containing $2\times 10^{11}$ particles. The bunches have a cross sectional area of 0.02 $mm^{2}$ . What is the luminosity of the machine in units of $cm^{-2}s^{-1}$ ?

b)What would bee an average magnetic field of dipoles in the collider to keep electrons on orbit?(Assume dipole magnets populate the entire ring).

Answer 1

Energy = 45 Gev for luminosity = number of particle passing per unit time for velocity implies 45 Gev = m e c^2/squareroot 1 - v^2/c^2 implies 45 times G e v/0.511 M e v = 1/squareroot 1 - v^2/c^2 implies 1 - v^2/c^2 = 1/7.755 times 10^9 implies v = 0.9999999 c time period = 2 pi R/v Time period = 8.3733 times 10^-5 s passing in one sec = 1/T = 1.194 times 10^4 s^-1 each bunch has 2 times 10^11 particle than every sec total particle passing is = 2 times 10^11 times 1.194 times 10^4 s^-1 = 2.388 times 10^15 s^-1 passing particle per unit area per unit time implies 2.388 times 10^15/0.02 m m^2 luminosity = 1.194 times 10^23 particle/m^2 s luminosity = 1.194 times 10^19 particle/cm^2 s^1 implies m v^2/R = 2 v B implies B