A cyclotron is used to obtain 2 Me V protons. If the frequency is 5MHz and potential is 20kV. The magnetic field necessary for resonance is

Options

A. 2.32T
B. 0.327T
C. 0.528T
D. 0.389T

Answer

To find the magnetic field necessary for resonance in a cyclotron that produces 2 MeV protons with a frequency of 5 MHz and a potential of 20 kV, we can follow these steps:

### Step 1: Understand the relationship between frequency, magnetic field, and charge
In a cyclotron, the frequency \( f \) of the charged particle in a magnetic field \( B \) is given by the formula:
\[ f = \frac{qB}{2\pi m} \]
where:
- \( f \) is the frequency (in Hz),
- \( q \) is the charge of the particle (in Coulombs),
- \( B \) is the magnetic field (in Tesla),
- \( m \) is the mass of the particle (in kg).

### Step 2: Rearrange the formula to solve for \( B \)
We can rearrange the formula to express \( B \) in terms of \( f \), \( q \), and \( m \):
\[ B = \frac{2\pi mf}{q} \]

### Step 3: Identify the values needed for calculation
- The frequency \( f = 5 \text{ MHz} = 5 \times 10^6 \text{ Hz} \)
- The charge of a proton \( q = 1.6 \times 10^{-19} \text{ C} \)
- The mass of a proton \( m = 1.67 \times 10^{-27} \text{ kg} \)

### Step 4: Substitute the values into the equation
Now we can substitute the known values into the equation for \( B \):
\[ B = \frac{2 \pi (1.67 \times 10^{-27} \text{ kg})(5 \times 10^6 \text{ Hz})}{1.6 \times 10^{-19} \text{ C}} \]

### Step 5: Calculate the magnetic field
Calculating the numerator:
\[ 2 \pi \times 1.67 \times 10^{-27} \times 5 \times 10^6 \approx 5.24 \times 10^{-20} \]
Now, calculating the magnetic field:
\[ B = \frac{5.24 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 0.327 \text{ Tesla} \]

### Final Answer
The magnetic field necessary for resonance is approximately \( 0.327 \text{ Tesla} \).

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