What is the working principle of a cyclotron?

In a cyclotron, magnetic field of $1.4 \mathrm{Wb} {\mathrm{m}}^{-2}$ is used. To accelerate protons, how rapidly should the electric field between the Dees be reversed?

$\left(\pi =3.142, M_p=1.67\times {10}^{-27} \mathrm{kg}, e=1.6\times {10}^{-19} \mathrm{C}\right)$

Derive an expression for cyclotron frequency with the help of a neat diagram.

If the maximum value of accelerating potential provided by a radio frequency oscillator is $12\mathrm{kV}$. The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is: 

$\left[{\mathrm{m}}_{\mathrm{p}}=1.67\times {10}^{-27} \mathrm{kg}, \mathrm{e}=1.6\times {10}^{-19} \mathrm{C}\right.$, Speed of light$\left.=3\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1}\right]$

Cyclotron is a device used to accelerate charged particles.

$\left(\mathrm{b}\right)$ A cyclotron oscillator frequency is $10 \mathrm{MHz}$. What should be the operating magnetic field for accelerating protons?

Cyclotron is a device used to accelerate charged particles.With a suitable diagram briefly explain the working of a cyclotron and obtain an expression for cyclotron frequency.

A cyclotron is used to accelerate protons $\left({}_1{}^1\mathrm{H}\right)$, Deuterons $\left({}_1{}^2\mathrm{H}\right)$ and $\alpha$- particles $\left({}_2{}^4\mathrm{He}\right)$. While exiting under similar conditions, the minimum K.E. is gained by