Question

A cyclotron is operated at an oscillator frequency of $ 12\, MHz $ and has a dee radius $ R = 50\,cm $ . What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron ?

• A. 0.72 T
• B. 0.65 T
• C. 0.39 T
• D. 0.12 T

Answer 1

Answer: (A)

Given, $f= 12 \,MHz = 12\times 10^{-6}\, Hz$
$R = 50\,cm= 520 \times 10^{-2}\, m$
$B = \frac{2\pi mf}{q}$
$ = \frac{(12\times 10^{6}) \times 2 \times \pi \times (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})}$
$ = 0.788 \approx 0.72\,T$