To solve this problem, we need to apply some fundamental concepts from physics, particularly those related to cyclotrons, magnetic fields, and kinetic energy. Let's break it down step by step.
Understanding the Cyclotron
A cyclotron is a type of particle accelerator that uses a magnetic field to bend the path of charged particles, such as protons, allowing them to gain energy as they spiral outward. The kinetic energy (KE) of the proton after completing 100 rounds is given as 1 MeV (mega-electron volts).
Key Values and Constants
- Charge of a proton (q) = 1.6 x 10-19 C
- Mass of a proton (m) = 1.67 x 10-27 kg
- 1 MeV = 1.6 x 10-13 J
Calculating the Velocity of the Proton
The kinetic energy of the proton can be expressed as:
KE = (1/2)mv2
Rearranging this gives us:
v = sqrt((2 * KE) / m)
Substituting the values:
KE = 1.6 x 10-13 J
v = sqrt((2 * 1.6 x 10-13) / (1.67 x 10-27))
Calculating this, we find:
v ≈ 1.38 x 107 m/s
Finding the Magnetic Field Intensity
The radius of the circular path in a cyclotron is given by the formula:
r = (mv) / (qB)
Where:
- r = radius of the path
- B = magnetic field strength
We can rearrange this to find B:
B = (mv) / (qr)
Given that the radius of the outermost track is 3 m, we substitute the values:
B = (1.67 x 10-27 kg * 1.38 x 107 m/s) / (1.6 x 10-19 C * 3 m)
Calculating this gives:
B ≈ 3.24 x 10-2 T or 0.0324 T
Magnetic Induction Across the D's
The magnetic induction (B) across the D's is essentially the same as the magnetic field intensity we just calculated, since the magnetic field is uniform in the region between the D's. Therefore, the magnetic induction across the D's is also:
B ≈ 0.0324 T
Summary of Results
To summarize, the intensity of the magnetic field between the two D's of the cyclotron is approximately 0.0324 T, and the magnetic induction across the D's is the same value, 0.0324 T. This demonstrates how the cyclotron effectively accelerates protons using magnetic fields, allowing them to achieve high velocities and energies.
