InterviewSolution
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InterviewSolution
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The ill effects associated with the variatiion of the period of revolution of the particle in a cyclotron due to the increase of its energy are eleminated by slow monitoring (modulating ) the frequency of accelerting field. According to what law `omega (t)` should this frequencecy by mointored if the masgnetic induction is equal to `B` and the particle acquires an energy `Delta W` per revolution ? The charge of the particle is `q` and its mass is `m`. |
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Answer» The basic condition is the relatistiv equaction, `(mv^(2))/(r) = Bqv`, or, `mv = (m_(0) v)/(sqrt(1 - v^(2)//c^(2))) = Bqr`. Or calling, `omega = (Bq)/(m)`, we get, `omega = (omega_(0))/(sqrt(1 + (omega_(0)^(2) r^(2))/(c^(2)))), omega_(0) = (Bq)/(m_(0)) r` is the radius of the instantanceous orbit, The time of acceleration is, `t = sum_(n = 1)^(N) (1)/(2 v_(n)) = sum_(n = 1)^(N) (pi)/(omega_(n)) = sum_(n)^(N) (pi W_(n))/(q Bc^(2))` `N` is the numbr of crossing of either `Dee`. But,n `W_(n) = m_(0) c^(2) + (n Delta W)/(2)`, there being two crossings of the Dees per revolution. So, `t = sum (pi m_(0) c^(2))/(qB c^(2)) + sum (pi Delta W_(n))/(2q B c^(2))` `= N (Pi)/(omega_(0)) + (N (N + 1))/(4) (pi Delta W)/(qB c^(2)) = N^(2) (pi Delta W)/(4 q Bc^(2)) (N gt gt 1)` Also, `r = r_(N) (v_(N))/(omega_(N)) = (c)/(pi) (del t)/(del N) = (Delta W)/(2q Bc) N` Hence finally, `omega = (omega_(0))/(sqrt(1 + (q^(2) B^(2))/(m_(0)^(2) c^(2)) xx (Del W^(2))/(4 q^(2) B^(2) c^(2))) N^(2))` `= (omega_(0))/(sqrt(1 + ((Delta W)^(2))/(4 m_(0)^(2) c^(4)) xx (4 q Bc^(2))/(pi Delta W) t)) = (omega_(0))/(sqrt(1 + at))`, `a = (q B Delta W)/(pi m_(0)^(2) c^(2))` |
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