Calculate the magnetic field in which the cyclotron dees should be placed to accelerate protons. The applied frequency is $8$ megacycles/second. Take proton mass $1.67 \mathrm{x} {10}^{-27} \mathrm{kg}.$

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$\left.=3\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1}\right]$

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

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?

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