a. A 65-cm-diameter cyclotron uses a $500\mathrm{V}$oscillating potential difference between the Dees. What is the maximum kinetic energy of a proton if the magnetic field strength is $0.75\mathrm{T}$?

b. How many revolutions does the proton make before leaving the cyclotron?

The cyclotron motion is defined as a particle travelling in an exceedingly uniform circular motion perpendicular to the magnetic field of force at a relentless speed. The radius of the circular motion produced by the magnetic flux is said thereto it by an equation within the form

$r_{\mathrm{cyc}}=\frac{mv}{qB}$

Where $q$denotes the particle's charge, $v$the particle's speed, $B$the magnetic flux and $m$the particle's mass.

To determine the mechanical energy, we must first determine the proton's velocity, thus we rearrange the equation for $v$to be

$v=\frac{qB{r}_{\mathrm{cyc}}}{m}$

Kinetic energy is half merchandise the product of mass and velocity squared.

$E=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\frac{qB{r}_{\mathrm{cyc}}}{m}\right)}^2$

$=\frac{q^2{B}^2{r}_{\text{cyc}}^2}{2m}$

Put the values into equation

$=\frac{{\left(1.6\times {10}^{-19}\mathrm{C}\right)}^2(0.75\mathrm{T}{)}^2{\left(32.5\times {10}^{-2}\mathrm{m}\right)}^2}{2\left(1.67\times {10}^{-27}\mathrm{kg}\right)}$

$=4.55\times {10}^{-13}\mathrm{J}$

$f_{\mathrm{cyc}}=\frac{qB}{2\pi m}$

Put the values in equation,

$=\frac{\left(1.6\times {10}^{-19}\mathrm{C}\right)(0.75\mathrm{T})}{2\pi \left(1.67\times {10}^{-27}\mathrm{kg}\right)}$

$=11.4\times {10}^6\mathrm{rev}$