Question

A cyclotron?s oscillator frequency is 10 MHz and the operating magnetic field is 0.66 T. If the radius of its dees is 60 cm, then the kinetic energy of the proton beam produced by the accelerator is

• A. 9 MeV
• B. 10 MeV
• C. 7 MeV
• D. 11 MeV

Answer 1

Answer: (C)

7 MeV

Answer 2

Radius of circular path: $R= \frac{\sqrt{2mk}}{qB}$
$k = \frac{q^2 B^2 R^2}{2m}\,\,\,\,\,\,\dots(i)$
Cyclotron frequency $v = \frac{qB}{2\pi m}$
or $q^2 B^2 = 4 \pi^2 m^2 v^2\,\,\,\,\,\,\dots(2)$
Using (2) in (1)
$k = \frac{1}{2m} (4 \pi^2 m^2 v^2 ) R^2$
$k = 2 \pi^2 mv^2 R^2$ $\dots$ joule
$k = \frac{2 \pi^2 mv^2 R^2}{e} \dots eV$
$= \frac{2\times10 \times \left(1.67 \times 10^{-27}\right)\times \left(10\times 10^{6}\right)^{2} \times \left(0.6\right)^{2} }{1.6\times 10^{-19}}eV$
$\simeq 7.2 \times10^{6} eV$
$= 7.2 \,MeV$