Question

The magnitude of the magnetic field required to accelerate protons (mass $=1.67 \times 10^{-27} kg$ ) in a cyclotron that is operated at an oscillator frequency $12\, MHz$ is approximately

• A. 0.8 T
• B. 1.6 T
• C. 2.0 T
• D. 3.2 T

Answer 1

Answer: (A)

The cyclotron frequency, $v=\frac{q B}{2 \pi m}$
Magnetic field $B =\frac{2 \pi m v}{q}$
$B =\frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 12 \times 10^{6}}{1.6 \times 10^{-19}}$
$=78.6 \times 10^{-2} T$
$B =0.8\, T$