A cyclotron is operated at an oscillator frequency of 12 MHz and has a dee radius R 50 cm. What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron?

Options

A. 0.78 T
B. 0.65 T
C. 0.39 T
D. 0.12 T

Answer

To find the magnitude of the magnetic field needed for a proton to be accelerated in a cyclotron, we can use the formula:

\[ B = \frac{2 \pi m f}{q} \]

where:
- \( B \) is the magnetic field,
- \( m \) is the mass of the proton,
- \( f \) is the frequency of oscillation,
- \( q \) is the charge of the proton.

### Step-by-Step Solution:

1. **Identify the given values**:
- Frequency \( f = 12 \, \text{MHz} = 12 \times 10^6 \, \text{Hz} \)
- Radius \( R = 50 \, \text{cm} = 0.5 \, \text{m} \) (not needed for this calculation)
- Mass of the proton \( m_p = 1.672 \times 10^{-27} \, \text{kg} \)
- Charge of the proton \( q = 1.602 \times 10^{-19} \, \text{C} \)

2. **Substitute the values into the formula**:
\[ B = \frac{2 \pi (1.672 \times 10^{-27} \, \text{kg}) (12 \times 10^6 \, \text{Hz})}{1.602 \times 10^{-19} \, \text{C}} \]

3. **Calculate the numerator**:
- First, calculate \( 2 \pi \):
\[ 2 \pi \approx 6.2832 \]
- Now calculate the product:
\[ 6.2832 \times 1.672 \times 10^{-27} \times 12 \times 10^6 \]
- This gives:
\[ \approx 1.267 \times 10^{-19} \, \text{kg} \cdot \text{Hz} \]

4. **Calculate the magnetic field \( B \)**:
\[ B = \frac{1.267 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 0.79 \, \text{T} \]

5. **Final result**:
- The magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron is approximately:
\[ B \approx 0.79 \, \text{Tesla} \]