Question

A cyclotron's oscillator frequency is 10MHz10MHz. What should be the operating magnetic field fro accelerating protons? If the radius of its dees is 60cm60cm, what is the kinetic energy (in MeVMeV) of the proton beam produced by the acceleration?

• A. $5$
• B. $6.5$
• C. $7.5$
• D. $12.6$

Answer 1

Answer: (C)

$7.5$

Magnetic field Cyclotron's oscillator frequency should be same as the proton's revolution frequency (in circular path)

$\therefore F=\frac{Bq}{2\pi rm}$

or $B=\frac{2\pi mf}{q}$

substituting the values in $SI$ units, we have

$B=\frac{\left(2\right)\left(22/7\right)(1.67\times {10}^{-27})(10\times {10}^6)}{1.6\times {10}^{-19}}$

$=0.67T$

Kinetic energy Let final velocity of proton just after having the cyclotron is $v$. Then radius of dee should be equal to

$R=\frac{mv}{Bq}$ or $v=\frac{BqR}{m}$

Kinetic energy proton,

$K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\frac{BqR}{m}\right)}^2=\frac{B^2{q}^2{R}^2}{2m}$

Substituting the values in SI units, we have

$K=\frac{\left(0.67{)}^2\right(1.6\times {10}^{-10}{)}^2(0.60{)}^2}{2\times 1.67\times {10}^{-37}}$

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

$=\frac{1.2\times {10}^{-12}}{(1.6\times {10}^{-10})\left({10}^6\right)}MeV$

$=7.5MeV$