The operating magnetic field for accelerating protons in a cyclotron oscillator having frequency of 12 MHz is `(q= 1.6 xx 10^(-19) C, m_(p) = 1.67 xx 10^(-27) kg and 1 MeV = 1.6 xx 10^(-13J))`

Options

A. 0.69T
B. 0.79T
C. 0.59T
D. 0.49T

Explanation

To find the operating magnetic field for accelerating protons in a cyclotron oscillator with a given frequency, we can use the formula that relates the frequency of the cyclotron (\( \mu_c \)), the charge of the proton (\( q \)), the mass of the proton (\( m_p \)), and the magnetic field (\( B \)):

\[ \mu_c = \frac{qB}{2\pi m_p} \]

From this formula, we can rearrange it to solve for the magnetic field \( B \):

\[ B = \frac{2\pi m_p \mu_c}{q} \]

### Step-by-step Solution:

1. **Identify the given values:**
- Frequency (\( \mu_c \)) = 12 MHz = \( 12 \times 10^6 \) Hz
- Charge of proton (\( q \)) = \( 1.6 \times 10^{-19} \) C
- Mass of proton (\( m_p \)) = \( 1.67 \times 10^{-27} \) kg

2. **Substitute the values into the formula:**
\[ B = \frac{2\pi (1.67 \times 10^{-27} \, \text{kg}) (12 \times 10^6 \, \text{Hz})}{1.6 \times 10^{-19} \, \text{C}} \]

3. **Calculate the numerator:**
- First, calculate \( 2\pi \):
\[ 2\pi \approx 6.28 \]
- Now calculate the product:
\[ 2\pi m_p \mu_c = 6.28 \times (1.67 \times 10^{-27}) \times (12 \times 10^6) \]
- Calculate \( 1.67 \times 12 = 20.04 \):
\[ 6.28 \times 20.04 \times 10^{-21} \approx 1.26 \times 10^{-20} \]

4. **Calculate the denominator:**
\[ q = 1.6 \times 10^{-19} \]

5. **Now divide the numerator by the denominator:**
\[ B = \frac{1.26 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 0.7875 \, \text{T} \]

6. **Round the answer:**
\[ B \approx 0.79 \, \text{T} \]

### Final Answer:
The operating magnetic field \( B \) is approximately **0.79 Tesla**.