Simple cyclotron

Problem

A proton cyclotron has $B = 1{,}5$ T and maximum radius $R_\text{max} = 0{,}5$ m. Calculate (a) the exit velocity of the proton, (b) the kinetic energy in MeV, (c) the rotation frequency.

Solution

(a) From the maximum radius $R = \dfrac{m_p v}{qB}$ the exit velocity is obtained: $v = \frac{qBR}{m_p} = \frac{1{,}6\cdot10^{-19}\cdot 1{,}5\cdot 0{,}5}{1{,}67\cdot10^{-27}} \approx 7{,}2\cdot10^7\;\text{m/s}$ $\ev{v \approx 7{,}2\cdot 10^7\;\text{m/s}}$

(b) The corresponding kinetic energy: $K = \tfrac12 m_p v^2 \approx 4{,}3\cdot10^{-12}\;\text{J} = \frac{4{,}3\cdot10^{-12}}{1{,}6\cdot10^{-13}} \approx 27\;\text{MeV}$ $\ev{K \approx 27\;\text{MeV}}$

(c) The cyclotron frequency is independent of the radius: $f_c = \frac{qB}{2\pi m_p} = \frac{1{,}6\cdot10^{-19}\cdot 1{,}5}{2\pi\cdot 1{,}67\cdot10^{-27}} \approx 2{,}29\cdot10^7\;\text{Hz}$ $\ev{f_c \approx 22{,}9\;\text{MHz}}$

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Topics: Magnetismo Concepts: Moto circolare uniforme · Energia cinetica · Forza di Lorentz