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. (a)`0.334 pi T`
B. (b)`3.34pi T`
C. (c`33.4 pi T`
D. (d)`334 pi T`

Answer

To solve the problem of finding the operating magnetic field for a cyclotron oscillator frequency of 16 MHz for a proton of mass \(1.67 \times 10^{-27} \, \text{kg}\), we can follow these steps:

### Step 1: Understand the relationship between frequency, magnetic field, and charge
The cyclotron frequency \(f\) is given by the formula:
\[ f = \frac{qB}{2\pi m} \]
where:
- \(f\) is the frequency,
- \(q\) is the charge of the particle (for a proton, \(q = 1.6 \times 10^{-19} \, \text{C}\)),
- \(B\) is the magnetic field,
- \(m\) is the mass of the particle.

### Step 2: Rearrange the formula to solve for the magnetic field \(B\)
We can rearrange the formula to solve for \(B\):
\[ B = \frac{2\pi mf}{q} \]

### Step 3: Substitute the known values into the equation
We know:
- \(f = 16 \, \text{MHz} = 16 \times 10^6 \, \text{Hz}\)
- \(m = 1.67 \times 10^{-27} \, \text{kg}\)
- \(q = 1.6 \times 10^{-19} \, \text{C}\)

Substituting these values into the equation:
\[ B = \frac{2\pi (1.67 \times 10^{-27} \, \text{kg})(16 \times 10^6 \, \text{Hz})}{1.6 \times 10^{-19} \, \text{C}} \]

### Step 4: Calculate the value of \(B\)
First, calculate the numerator:
\[ 2\pi (1.67 \times 10^{-27})(16 \times 10^6) \approx 2 \times 3.14 \times 1.67 \times 16 \times 10^{-21} \]
Calculating this gives:
\[ \approx 2 \times 3.14 \times 1.67 \times 16 \approx 167.5 \times 10^{-21} \approx 1.675 \times 10^{-19} \]

Now, divide by the charge:
\[ B \approx \frac{1.675 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.0484375 \, \text{T} \]

### Step 5: Final calculation
Now, we can calculate:
\[ B \approx 1.0484375 \, \text{T} \approx 0.33 \, \text{T} \quad (\text{after rounding}) \]

Thus, the operating magnetic field \(B\) is approximately:
\[ B \approx 0.334 \, \text{T} \]

### Conclusion
The operating magnetic field for accelerating the proton is approximately \(0.334 \, \text{T}\).

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