A proton of eneergy 1 MeV described a circuler path in a plane at right angles to be a uniform magnetic field of induction `2pi xx 10^(-4)T`. The mass of the proton is `1.7 xx 10^(-27)kg`. The cyclotron frequency of the proton is very nearly equal to

Options

A. `10^(4) Hz`
B. `10^(5) Hz`
C. `10^(6) Hz`
D. `10^(7) Hz`

Explanation

To find the cyclotron frequency of a proton moving in a magnetic field, we can use the formula for cyclotron frequency, which is given by:

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

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

### Step 1: Identify the values
- Charge of the proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Magnetic field induction, \( B = 2\pi \times 10^{-4} \, \text{T} \)
- Mass of the proton, \( m = 1.7 \times 10^{-27} \, \text{kg} \)

### Step 2: Substitute the values into the formula
Now we can substitute these values into the cyclotron frequency formula:

\[ f = \frac{(1.6 \times 10^{-19} \, \text{C})(2\pi \times 10^{-4} \, \text{T})}{2\pi (1.7 \times 10^{-27} \, \text{kg})} \]

### Step 3: Simplify the equation
Notice that \( 2\pi \) in the numerator and denominator cancels out:

\[ f = \frac{(1.6 \times 10^{-19})(10^{-4})}{1.7 \times 10^{-27}} \]

### Step 4: Calculate the frequency
Now we can calculate the frequency:

\[ f = \frac{1.6 \times 10^{-23}}{1.7 \times 10^{-27}} = \frac{1.6}{1.7} \times 10^{4} \approx 0.941 \times 10^{4} \approx 9.41 \times 10^{3} \text{ Hz} \]

Thus, the cyclotron frequency of the proton is approximately \( 10^{4} \text{ Hz} \).

### Final Answer
The cyclotron frequency of the proton is very nearly equal to \( 10^{4} \text{ Hz} \).

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