Question

The operating magnetic field for accelerating protons in a cyclotron oscillator having frequency of 12 MHz is ( q = $1.6\times {10}^{-19}$ C, $m_p=1.67\times {10}^{-27}$ kg and 1 MeV = $1.6\times {10}^{-13}$ J)

• A. 0.69 T
• B. 0.79 T
• C. 0.59 T
• D. 0.49 T

Answer 1

Answer: (B)

0.79 T

Given:

$v=12MHz$

$q=1.6\times 10-19C,m_p=1.67\times 10-27kg1MeV=1.6\times 10-13J$

$B=\frac{2\pi m_{p}v}{q}$

$B=\frac{2\times 3.14\times 1.67\times {10}^{-27}\times 12\times {10}^6}{1.6\times {10}^{-19}}=0.79T$