A cyclotron is opened 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?

The correct option is A 0.72 T
Given,

$f=12MHz=12\times {10}^{6}Hz$

$R=50cm=0.5m$

$m_p=1.67\times {10}^{-27}kg$

$q=1.6\times {10}^{-19}C$

The magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron is given by the formula

$f=\frac{qB}{2\pi m}$

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

$B=\frac{2\times 3.14\times 1.67\times {10}^{-27}\times 12\times {10}^6}{1.6\times {10}^{-19}}$

$B=0.72T$

The correct option is A.