Q. 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 $T$
Explanation
(a) The oscillator frequency should be same as proton’s cyclotron frequency
Cyclotron frequency, $v_{c} = \frac{qB}{2\pi m}$
or $B = \frac{2\pi\ mv_{c}}{q}$
$= \frac{2 \times 3.14 \times 1.67 \times 10^{- 27} \times 12 \times 10^{6}}{1.6 \times 10^{- 19}}$
$= 78.6 \times 10^{- 2}T = 0.8\ T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.