A proton is accelerated in cyclotron with dees of radius 40 cm has a operating frequency of 12 MHz. Calculate the magnitude of magnetic field. Given mass of proton, `m_(P) = 1.67 xx 10^(-27) ` kg and charge on proton ` q_(P) = 1.6 xx 10^(-19) C`.

Answer

To calculate the magnitude of the magnetic field in a cyclotron, we can use the formula that relates the frequency of the charged particle in a magnetic field. The formula is given by:

\[ f = \frac{qB}{2\pi m} \]

Where:
- \( f \) is the frequency (in Hz),
- \( q \) is the charge of the particle (in Coulombs),
- \( B \) is the magnetic field (in Tesla),
- \( m \) is the mass of the particle (in kg).

### Step 1: Convert the frequency from MHz to Hz
Given the operating frequency of the cyclotron is \( 12 \, \text{MHz} \):
\[ f = 12 \times 10^6 \, \text{Hz} \]

### Step 2: Substitute the known values into the formula
We know:
- Charge of proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of proton, \( m = 1.67 \times 10^{-27} \, \text{kg} \)

Substituting these values into the formula:
\[ 12 \times 10^6 = \frac{(1.6 \times 10^{-19})B}{2\pi(1.67 \times 10^{-27})} \]

### Step 3: Rearranging the formula to solve for \( B \)
Rearranging the equation to isolate \( B \):
\[ B = \frac{(12 \times 10^6) \cdot 2\pi(1.67 \times 10^{-27})}{1.6 \times 10^{-19}} \]

### Step 4: Calculate \( B \)
Now we can calculate \( B \):
1. Calculate \( 2\pi(1.67 \times 10^{-27}) \):
\[ 2\pi(1.67 \times 10^{-27}) \approx 1.05 \times 10^{-26} \]

2. Now multiply by \( 12 \times 10^6 \):
\[ (12 \times 10^6)(1.05 \times 10^{-26}) \approx 1.26 \times 10^{-19} \]

3. Finally, divide by \( 1.6 \times 10^{-19} \):
\[ B \approx \frac{1.26 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 0.7875 \, \text{T} \]

### Step 5: Final Result
The magnitude of the magnetic field \( B \) is approximately:
\[ B \approx 0.7875 \, \text{T} \]