Calculation of the Cyclotron frequency (non-relativistic)

Calculation:

The Lorentz force $F_{Lorentz}$ is the centripetal force $F_{Zentripetal}$ and causes the particles path to bend in a circle:$$F_{Lorentz}=F_{Zentripetal}\Rightarrow q\cdot v\cdot B=\frac{m\cdot v^2}{r}\Rightarrow v=\frac{r\cdot q\cdot B}{m}$$ $$q=\text{charge of the particle}, v=\text{velocity of the particle}, B=\text{magnetic flux}, m=\text{mass of the particle}, r=\text{Radius of the circle}$$ With $v=\omega\cdot r=2\pi \cdot f\cdot r$ the Cyclotron frequency $f$ is$$2\pi \cdot f\cdot r=\frac{r\cdot q\cdot B}{m}\Rightarrow f=\frac{q\cdot B}{2\pi\cdot m}$$ $$\omega=\text{angular velocity}, f=\text{frequency}$$ and for the orbital period T$$T=\frac{1}{f}=\frac{2\pi\cdot m}{q\cdot B}$$ The cyclotron frequency f, the orbital period T and so the time of a particle in a "dee" are independent of the radius r. So is possible to keep the frequency of the alternating voltage constant.

The kinetic energy Ekin grows with every passage of the electric field in the gap between the dees by $q\cdot V$:$$E_{kin}(n)=E_{kin_0}+n\cdot q\cdot V=\frac{1}{2}\cdot m\cdot {v_0}^2+n\cdot q\cdot V$$ $$E_{kin_0}=\text{Kinetic energy when leaving the paricle source}, n=\text{Number of passages through the electric field}, V=\text{Voltage between dees}$$ The velocity v of the particle can be calculated with the initial speed v0 when it leaves the particle source and the number n of passages through the electric field between the "dees":$$\frac{1}{2}\cdot m\cdot v^2=\frac{1}{2}\cdot m\cdot {v_0}^2+n\cdot q\cdot V\Rightarrow v=\sqrt{\frac{2}{m}\left( \frac{1}{2}\cdot m\cdot {v_0}^2+q\cdot V\cdot n\right)}$$ If the initial speed is $v_0=0$ the follwing formula can be used:$$v=\sqrt{\frac{2}{m}\cdot q\cdot V\cdot n}$$