If the cyclotron oscillator frequency is 16 MHz, then what should be the operating magnetic field for accelerating the proton of mass `1.67 xx 10^(-27) kg`?

Options

A. `0.334 pi T`
B. `3.34pi T`
C. `33.4 pi T`
D. `334 pi T`

Answer

To solve the problem, we need to find the operating magnetic field (B) for a cyclotron oscillator given its frequency (f) and the mass of the proton (m). The relevant formula that relates these quantities is:

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

Where:
- \( f \) is the frequency of the cyclotron oscillator,
- \( q \) is the charge of the proton,
- \( B \) is the magnetic field,
- \( m \) is the mass of the proton.

### Step 1: Write down the known values
- Frequency \( f = 16 \text{ MHz} = 16 \times 10^6 \text{ Hz} \)
- Mass of the proton \( m = 1.67 \times 10^{-27} \text{ kg} \)
- Charge of the proton \( q = 1.6 \times 10^{-19} \text{ C} \)

### Step 2: Rearrange the formula to solve for B
We can rearrange the formula to solve for the magnetic field \( B \):

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

### Step 3: Substitute the known values into the equation
Now we substitute the known values into the rearranged formula:

\[ B = \frac{2\pi (1.67 \times 10^{-27} \text{ kg})(16 \times 10^6 \text{ Hz})}{1.6 \times 10^{-19} \text{ C}} \]

### Step 4: Calculate the numerator
First, calculate the numerator:

\[ 2\pi (1.67 \times 10^{-27})(16 \times 10^6) \]

Calculating \( 1.67 \times 16 = 26.72 \):

\[ 2\pi \times 26.72 \times 10^{-21} \]

Now, calculate \( 2\pi \):

\[ 2\pi \approx 6.2832 \]

Thus, the numerator becomes:

\[ 6.2832 \times 26.72 \times 10^{-21} \approx 1.68 \times 10^{-20} \]

### Step 5: Calculate the denominator
The denominator is simply:

\[ 1.6 \times 10^{-19} \]

### Step 6: Divide the numerator by the denominator
Now we can calculate \( B \):

\[ B = \frac{1.68 \times 10^{-20}}{1.6 \times 10^{-19}} \]

This simplifies to:

\[ B = 0.105 \text{ T} \]

### Step 7: Final result
Thus, the operating magnetic field \( B \) required for the cyclotron oscillator to accelerate the proton is approximately:

\[ B \approx 0.105 \text{ T} \]