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The ill effectsassociatedwith thevariatiion of the periodof revolution of the particleina cyclotron due tothe increase of itsenergyare eleminated by slow monitoring(modulating ) the frequency of accelertingfield. Accordingto whatlaw omega (t) shouldthis frequencecy by mointoredif the masgneticinductionis equalto B and theparticleacquiresan energyDelta Wper revolution ? The charge of theparticle is q and its mass is m. |
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Answer» Solution :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 theinstantanceous 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 crossingof either `Dee`. But,n `W_(n) = m_(0) c^(2) + (n Delta W)/(2)`, therebeing two crossings of theDees 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) (piDelta 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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