A cyclotron of Diameter 1 meter is used to accelerate protons. It is operated at Alternating Voltage of 10 MHz of Peak Value 5000 volts. Calculate the Magnetic Field needed and Energy of Emerging Beam.
\[ f_{\text{cyc}} = \frac{qB}{2\pi m} \]
where:
- \(f_{\text{cyc}}\) is the cyclotron frequency,
- \(q\) is the charge of the particle,
- \(B\) is the magnetic field strength,
- \(m\) is the mass of the particle.
In a cyclotron, the frequency of the alternating voltage (\(f_{\text{rf}}\)) is twice the cyclotron frequency:
\[ f_{\text{rf}} = 2 \cdot f_{\text{cyc}} \]
Let's proceed with the calculations.
Given:
- Diameter of the cyclotron (\(D\)) = 1 meter
- Operating frequency (\(f_{\text{rf}}\)) = \(10 \, \text{MHz}\)
- Peak voltage (\(V\)) = \(5000 \, \text{volts}\)
- Charge of a proton (\(q\)) ≈ \(1.602 \times 10^{-19} \, \text{C}\)
Step 1: Calculate Cyclotron Frequency (\(f_{\text{cyc}})\):
\[ f_{\text{cyc}} = \frac{f_{\text{rf}}}{2} \]
\[ f_{\text{cyc}} = \frac{10 \, \text{MHz}}{2} = 5 \, \text{MHz} \]
Step 2: Calculate Magnetic Field (\(B\)):
\[ B = \frac{2\pi m f_{\text{cyc}}}{q} \]
Now, the mass of a proton (\(m_p\)) ≈ \(1.673 \times 10^{-27} \, \text{kg}\).
\[ B = \frac{2 \pi \times 1.673 \times 10^{-27} \times 5 \times 10^6}{1.602 \times 10^{-19}} \]
\[ B \approx 1.06 \, \text{T} \]
Step 3: Calculate Energy of Emerging Beam (\(E\)):
\[ E = qV \]
\[ E \approx (1.602 \times 10^{-19}) \times (5000) \]
\[ E \approx 8.01 \times 10^{-16} \, \text{Joules} \]
So, the magnetic field needed is approximately \(1.06 \, \text{T}\), and the energy of the emerging beam is approximately \(8.01 \times 10^{-16} \, \text{Joules}\).
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