(a) If N is the number of revolutions then
(2πRN)(4f) = c
The factor 4 arises due to the fact that the duty cycle is over a quarter of a **.
N = c/8π Rf = 3 × 108/8π × 0.9 × 50 = 2.65 × 105
(b) Radius, R = 0.9 m
Tmax = BRec = 1.2 × 0.9 × 1.6 × 10−19 × 3 × 108
= 5.184 × 10−11J
= 5.184 × 10−11/1.6 × 10−13 = 324MeV
(c) The average energy gained per revolution
ΔT = Tmax/N = 324 MeV/2.65 × 105
= 122.2 × 10−5 MeV
= 1.222 keV
