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 follow these steps:

### Step-by-Step Solution

1. **Identify Given Values**:
- Oscillator frequency \( f = 12 \, \text{MHz} = 12 \times 10^6 \, \text{Hz} \)
- Dee radius \( R = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Charge of the proton \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of the proton \( m_p = 1.67 \times 10^{-27} \, \text{kg} \)

2. **Use the Cyclotron Frequency Formula**:
The relationship between the cyclotron frequency \( f_c \), charge \( q \), mass \( m \), and magnetic field \( B \) is given by:
\[ f_c = \frac{qB}{2\pi m} \]
Rearranging this formula to solve for the magnetic field \( B \):
\[ B = \frac{2\pi mf_c}{q} \]

3. **Substitute the Known Values**:
Plug in the values we have:
\[ B = \frac{2\pi (1.67 \times 10^{-27} \, \text{kg})(12 \times 10^6 \, \text{Hz})}{1.6 \times 10^{-19} \, \text{C}} \]

4. **Calculate the Magnetic Field**:
- First calculate the numerator:
\[ 2\pi (1.67 \times 10^{-27})(12 \times 10^6) \approx 3.78 \times 10^{-20} \]
- Now calculate the magnetic field \( B \):
\[ B = \frac{3.78 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 0.23625 \, \text{T} \]

5. **Final Calculation**:
- After performing the calculation:
\[ B \approx 0.786 \, \text{T} \]

### Conclusion
The magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron is approximately **0.786 T**.