A charged particle is moving in a uniform magnetic field in a circular path. The energy of the particle is doubled. If the initial radius of the circular path was R, the radius of the new circular path after the energy is doubled will be :

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.With a suitable diagram briefly explain the working of a cyclotron and obtain an expression for cyclotron frequency.

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?

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