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 and  K MeV is the kinetic energy of the proton beam produced by the accelerator.$\left(e=1.60\times {10}^{-19}C,m_p=1.67\times {10}^{-27}kg\right)$  .  Then what is the value of 2K.

We are given that frequency of the oscillator, $v=10MHz={10}^{7}Hz$ , radius of the dees, $R=60cm=0.60m$, mass of the proton, $m_p=1.67\times {10}^{-27}kg$ , charge on the proton, $q=e=1.60\times {10}^{-19}C$ . If B is the operating magnetic field for accelerating electrons,

$v=\frac{qB}{2\pi m_p} or B=\frac{2\pi m_{p}v}{q}$

or $B=\frac{2\times 3.14(1.67\times {10}^{-27}){10}^7}{1.60\times {10}^{-19}}T$

Kinetic energy of the proton,  $K_{\max }=\frac{q^2{R}^2{B}^2}{2m_p}$
or $K_{\max }=\frac{{(1.6\times {10}^{-19})}^2{\left(0.60\right)}^2{\left(0.66\right)}^2}{2(1.67\times {10}^{-27})}J$
 

$=1.2\times {10}^{-12}J$

or $K_{\max }=\frac{1.2\times {10}^{-12}}{1.602\times {10}^{-13}}MeV=7.5MeV$