Question

The approximate operating magnetic field for accelerating protons in a cyclotron oscillator having frequency of $12\text{MHz}$ is:

• A. $0.790\text{T}$
• B. $0.690\text{T}$
• C. $0.395\text{T}$
• D. $0.345\text{T}$

Answer 1

The correct option is A $0.790\text{T}$

Given,
frequency of cyclotron
$f=12\text{MHz}=12\times {10}^6\text{Hz}$

We know that,
charge of proton, $q_p=1.6\times {10}^{-19}\text{C}$
mass of proton, $m=1.67\times {10}^{-27}\text{kg}$

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

$\Rightarrow B=\frac{2\pi mf}{q}=\frac{2\times 3.14\times (1.67\times {10}^{-27})\times 12\times {10}^6}{1.6\times {10}^{-19}}$

$\therefore B=78.69\times {10}^{-2}=0.786\approx 0.79\text{T}$

Hence, option (a) is the correct answer.