Question

If the operating magnetic field in a cyclotron for accelerating protons is 668 mT, then the angular frequency of the oscillator of the cyclotron is (Charge of proton \( = 1.6 \times 10^{-19} \,\mathrm{C} \), Mass of proton \( = 1.67 \times 10^{-27} \,\mathrm{kg} \))

• A. \( 9.6 \times 10^{7} \,\mathrm{rad/s} \)
• B. \( 3.2 \times 10^{7} \,\mathrm{rad/s} \)
• C. \( 12.8 \times 10^{7} \,\mathrm{rad/s} \)
• D. \( 6.4 \times 10^{7} \,\mathrm{rad/s} \)

Hint:

For cyclotron calculations, use the formula \( \omega = \frac{qB}{m} \) where \( \omega \) is the angular frequency, \( q \) is the charge, \( B \) is the magnetic field, and \( m \) is the mass of the particle.

Answer 1

The Correct Option is D

The angular frequency \( \omega \) of a cyclotron is given by: \[ \omega = \frac{qB}{m} \] where: - \( q = 1.6 \times 10^{-19} \,\mathrm{C} \) (charge of proton), - \( B = 668\,\mathrm{mT} = 668 \times 10^{-3}\,\mathrm{T} \), - \( m = 1.67 \times 10^{-27} \,\mathrm{kg} \) \[ \omega = \frac{1.6 \times 10^{-19} \times 668 \times 10^{-3}}{1.67 \times 10^{-27}} \approx 6.4 \times 10^{7} \,\mathrm{rad/s} \]