Question

A cyclotron’s oscillator frequency is 20 MHz. The operating magnetic field for accelerating protons is
(Charge of proton = \( 1.6 \times 10^{-19} \) C, mass of proton = \( 1.67 \times 10^{-27} \) kg)

• A. 0.66 T
• B. 1.1 T
• C. 0.33 T
• D. 1.31 T

Hint:

The cyclotron frequency depends on the charge-to-mass ratio of the particle and the magnetic field, not on the particle’s speed, making \( f = \frac{q B}{2 \pi m} \) a key formula.

Answer 1

The Correct Option is D

Step 1: Use the cyclotron frequency formula.
The cyclotron frequency \( f \) is given by: \[ f = \frac{q B}{2 \pi m}, \] where \( q = 1.6 \times 10^{-19} \) C (charge of proton), \( m = 1.67 \times 10^{-27} \) kg (mass of proton), and \( f = 20 \) MHz = \( 20 \times 10^6 \) Hz. Solve for the magnetic field \( B \): \[ B = \frac{2 \pi m f}{q}. \]
Step 2: Substitute the values and calculate.
\[ B = \frac{2 \pi (1.67 \times 10^{-27}) (20 \times 10^6)}{1.6 \times 10^{-19}}. \] First, compute the numerator: \[ 2 \pi (1.67 \times 10^{-27}) (20 \times 10^6) \approx 6.2832 \times 1.67 \times 10^{-27} \times 2 \times 10^7 \approx 2.098 \times 10^{-19}. \] Now divide by \( q \): \[ B = \frac{2.098 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.311 \, \text{T}. \] This rounds to 1.31 T, matching option (4). Final Answer: The magnetic field is \( \boxed{1.31 \, \text{T}} \).