A cyclotron in which flux density is 1.4 T is employed to accelerate protons. How rapidly should the field between the dees be reversed if mass of protoon be taken as `1.6 xx 10^(-27)` kg ?

Options

A. `(49)/(22) xx 10^(5)` Hz
B. `(49)/(22) xx 10^(7)` Hz
C. `(49)/(22) xx 10^(4)` Hz
D. `(49)/(22) xx 10^(6)` Hz

Answer

To solve the problem of how rapidly the magnetic field in a cyclotron should be reversed to accelerate protons, we can follow these steps:

### Step 1: Understand the relationship between frequency, charge, magnetic field, and mass
The frequency \( f \) of a charged particle moving in a magnetic field is given by the formula:
\[ f = \frac{qB}{2\pi m} \]
where:
- \( q \) is the charge of the proton,
- \( B \) is the magnetic flux density,
- \( m \) is the mass of the proton.

### Step 2: Identify the values provided in the question
From the question, we have:
- Charge of the proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Magnetic flux density, \( B = 1.4 \, \text{T} \)
- Mass of the proton, \( m = 1.6 \times 10^{-27} \, \text{kg} \)

### Step 3: Substitute the values into the frequency formula
Now, substituting the known values into the frequency formula:
\[ f = \frac{(1.6 \times 10^{-19} \, \text{C}) \times (1.4 \, \text{T})}{2\pi (1.6 \times 10^{-27} \, \text{kg})} \]

### Step 4: Calculate the numerator
Calculating the numerator:
\[ 1.6 \times 10^{-19} \times 1.4 = 2.24 \times 10^{-19} \, \text{C T} \]

### Step 5: Calculate the denominator
Calculating the denominator:
\[ 2\pi \times 1.6 \times 10^{-27} \approx 10.053 \times 10^{-27} \, \text{kg} \]
(using \( \pi \approx 3.14 \))

### Step 6: Calculate the frequency
Now, substituting back into the frequency equation:
\[ f = \frac{2.24 \times 10^{-19}}{10.053 \times 10^{-27}} \approx 2.23 \times 10^{7} \, \text{Hz} \]

### Step 7: Conclusion
The frequency at which the magnetic field should be reversed is approximately:
\[ f \approx 2.23 \times 10^{7} \, \text{Hz} \]