A cyclotron's oscillator frequency is `10 MHz` . The operating magnetic field for accelerating protons is

Options

A. `0.66T`
B. `0.12T`
C. `1.67T`
D. `0.36T`

Explanation

To find the operating magnetic field for accelerating protons in a cyclotron with an oscillator frequency of \(10 \, \text{MHz}\), we can use the formula that relates frequency, charge, mass, and magnetic field. The formula for the frequency \(f\) in a cyclotron is given by:

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

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

### Step 1: Convert the frequency to standard units
The given frequency is \(10 \, \text{MHz}\). We convert this to Hertz:

\[ f = 10 \, \text{MHz} = 10 \times 10^6 \, \text{Hz} = 10^7 \, \text{Hz} \]

### Step 2: Identify the charge and mass of the proton
The charge of a proton \(q\) is:

\[ q = 1.6 \times 10^{-19} \, \text{C} \]

The mass of a proton \(m\) is:

\[ m = 1.67 \times 10^{-27} \, \text{kg} \]

### Step 3: Rearrange the formula to solve for \(B\)
We need to rearrange the formula to solve for the magnetic field \(B\):

\[ B = \frac{2\pi mf}{q} \]

### Step 4: Substitute the known values into the formula
Now, substituting the known values into the equation:

\[ B = \frac{2 \pi (1.67 \times 10^{-27} \, \text{kg})(10^7 \, \text{Hz})}{1.6 \times 10^{-19} \, \text{C}} \]

### Step 5: Calculate the magnetic field \(B\)
Calculating the numerator:

\[ 2 \pi (1.67 \times 10^{-27})(10^7) \approx 3.34 \times 10^{-20} \, \text{kg} \cdot \text{Hz} \]

Now, divide by the charge of the proton:

\[ B = \frac{3.34 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 0.20875 \, \text{T} \]

### Step 6: Final result
Thus, the operating magnetic field for accelerating protons is approximately:

\[ B \approx 0.209 \, \text{T} \]