$A$ proton is accelerating in a cyclotron where the applied magnetic field is $2 \,T$. If the potential gap is effectively $100 \,kV$, then how many revolutions does the proton have to make between the "dees" to acquire a kinetic energy of $20 \,MeV$?

Assertion : Cyclotron is a device which is used to accelerate the positive ion.
Reason : Cyclotron frequency depends upon the velocity.

$A$ cyclotron's oscillator frequency is $10 \; MHz$. What should be the operating magnetic field for accelerating protons? If the radius of its 'dees' is $60 \; cm$,what is the kinetic energy (in $MeV$) of the proton beam produced by the accelerator? $(e = 1.60 \times 10^{-19} \; C, m_p = 1.67 \times 10^{-27} \; kg, 1 \; MeV = 1.6 \times 10^{-13} \; J)$

Discuss the motion of a charged particle in a uniform magnetic field with initial velocity perpendicular to the magnetic field. $OR$ Explain the principle of a cyclotron.

An alternating electric field of frequency $f$ is applied across the radius $R$ of a cyclotron to accelerate protons (mass $m$). The operating magnetic field $B$ used and kinetic energy $(K.E.)$ of the proton beam produced by it are respectively ($e=$ charge on proton).

$A$ cyclotron is used to accelerate protons. If the operating magnetic field is $1.0\,T$ and the radius of the cyclotron 'dees' is $60\,cm$,the kinetic energy of the accelerated protons in $MeV$ will be.
[use $m_{p} = 1.6 \times 10^{-27}\,kg, e = 1.6 \times 10^{-19}\,C$]