A cyclotron in which protons are accelerated has a flux density of `1.57t`. The variation of frequency of electric field is `(` in `Hz)`

Options

A. `4.8xx10^(8)` cycles/sec
B. `2.5xx10^(7)` cycles/sec
C. `4.8xx10^(6)` cycles/sec
D. `8.4xx10^(8)` cycles/sec

Answer

To find the variation of frequency of the electric field in a cyclotron where protons are accelerated, we can use the formula for the frequency of the electric field in a cyclotron:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnetic flux density, \( B = 1.57 \, \text{T} \)
- Charge of a proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of a proton, \( m = 1.67 \times 10^{-27} \, \text{kg} \)

2. **Use the Formula for Frequency:**
The frequency \( f \) of the electric field in a cyclotron is given by the formula:
\[ f = \frac{qB}{2\pi m} \]

3. **Substitute the Values into the Formula:**
\[ f = \frac{(1.6 \times 10^{-19} \, \text{C})(1.57 \, \text{T})}{2\pi (1.67 \times 10^{-27} \, \text{kg})} \]

4. **Calculate the Denominator:**
First, calculate \( 2\pi m \):
\[ 2\pi (1.67 \times 10^{-27}) \approx 3.14 \times 1.67 \times 10^{-27} \approx 3.14 \times 1.67 \times 10^{-27} \approx 5.24 \times 10^{-27} \]

5. **Calculate the Numerator:**
Now calculate the numerator \( qB \):
\[ (1.6 \times 10^{-19})(1.57) \approx 2.512 \times 10^{-19} \]

6. **Combine the Results:**
Now, substitute the calculated values into the frequency formula:
\[ f = \frac{2.512 \times 10^{-19}}{5.24 \times 10^{-27}} \approx 4.79 \times 10^{7} \, \text{Hz} \]

7. **Final Result:**
The variation of frequency of the electric field is approximately:
\[ f \approx 4.79 \times 10^{7} \, \text{Hz} \]