Question

If the cyclotron oscillator frequency is $16$ MHz, then what should be the operating magnetic field for accelerating the proton of mass $1.67 \times 10^{-27}$ kg?}

• A. $0.334\pi$ T
• B. $3.34\pi$ T
• C. $33.4\pi$ T
• D. $334\pi$ T
• E. $3340\pi$ T

Hint:

Cyclotron frequency depends only on magnetic field and particle properties, not on radius.

Answer 1

The Correct Option is B

Concept:
Cyclotron frequency: \[ f = \frac{qB}{2\pi m} \]

Step 1: Rearrange formula.
\[ B = \frac{2\pi m f}{q} \]

Step 2: Substitute values.
\[ m = 1.67 \times 10^{-27}, \quad f = 16 \times 10^6, \quad q = 1.6 \times 10^{-19} \]

Step 3: Calculate.
\[ B = \frac{2\pi \cdot 1.67 \cdot 16}{1.6} \times 10^{-27+6+19} \] \[ = \frac{53.44\pi}{1.6} \times 10^{-2} \approx 3.34\pi \]