Problem 12 - University Physics with Modern Physics

Question

The magnetic field in a cyclotron that accelerates protons is 1.30 T. (a) How many times per second should the potential across the dees reverse? (This is twice the frequency of the circulating protons.) (b) The maximum radius of the cyclotron is 0.250 \(\mathrm{m}\) . What is the maximum speed of the proton? (c) Through what potential difference would the proton have to be accelerated from rest to give it the same speed as calculated in part (b)?

Step-by-Step Solution

Answer

(a) 6.30x10^7 reversals/s, (b) 3.11x10^7 m/s, (c) 9426 V

Step 1: Determine Circulating Frequency

The frequency of proton circulation in a cyclotron is given by the formula \( f = \frac{qB}{2\pi m} \), where \( q \) is the charge of the proton \( (1.6 \times 10^{-19} \text{ C}) \), \( B \) is the magnetic field \( (1.30 \text{ T}) \), and \( m \) is the mass of the proton \( (1.67 \times 10^{-27} \text{ kg}) \). Substitute the known values:\[ f = \frac{(1.6 \times 10^{-19} \text{ C})(1.30 \text{ T})}{2\pi (1.67 \times 10^{-27} \text{ kg})} = \frac{2.08 \times 10^{-19}}{1.05 \times 10^{-26}} \approx 3.15 \times 10^{7} \text{ Hz}\]

Step 2: Calculate Potential Reversal Frequency

The potential across the dees should reverse at twice the frequency of proton circulation. Therefore, multiply the frequency by 2:\[ 2f = 2 \times 3.15 \times 10^{7} = 6.30 \times 10^{7} \text{ reversals per second} \]

Step 3: Determine Maximum Speed of Proton

The maximum speed \( v_{max} \) of a proton in a cyclotron is determined by the maximum radius \( r \) and the magnetic field \( B \). The formula for speed is \( v_{max} = qBr/m \). Use the maximum radius \( 0.250 \text{ m} \):\[ v_{max} = \frac{(1.6 \times 10^{-19} \text{ C})(1.30 \text{ T})(0.250 \text{ m})}{1.67 \times 10^{-27} \text{ kg}} = \frac{5.2 \times 10^{-20}}{1.67 \times 10^{-27}} \approx 3.11 \times 10^{7} \text{ m/s} \]

Step 4: Compute Potential Difference

The potential difference \( V \) required to accelerate a proton from rest to a given speed is calculated using the kinetic energy formula: \( K = \frac{1}{2}mv^2 \) and relates to the potential difference by \( K = qV \). Solve for \( V \):\[ V = \frac{mv^2}{2q} = \frac{(1.67 \times 10^{-27})(3.11 \times 10^{7})^2}{2 \times 1.6 \times 10^{-19}} \approx 9426 \text{ V} \]

Key Concepts

Proton Circulation

Proton circulation in a cyclotron involves the movement of protons in a circular path under the influence of a magnetic field. The speed and frequency at which these protons circulate depend on various factors such as the magnetic field strength and the charge and mass of the protons.
To determine the frequency of proton circulation, we use the formula: \[ f = \frac{qB}{2\pi m} \] Here, \( f \) is the frequency, \( q \) is the charge of a proton, \( B \) is the magnetic field, and \( m \) is the mass of the proton.
In our cyclotron example, the magnetic field is 1.30 T, leading to a circulation frequency of roughly \( 3.15 \times 10^{7} \) Hz. This means that protons make approximately 31.5 million revolutions per second within the cyclotron. Understanding this phenomenon is essential for manipulating the behavior of charged particles in electromagnetic fields.

Magnetic Field

The magnetic field is central to a cyclotron's operation, serving as a guiding force for charged particles like protons. It is measured in teslas (T), and in the context of our cyclotron, the field strength is 1.30 T.
The magnetic field acts perpendicularly to the motion of charged particles, exerting a force that causes them to move in circular paths. This deflection ensures that particles remain within the cyclotron as they gain energy and move outward in spirals.
A cyclotron utilizes a constant magnetic field to keep the particles in stable, circular motion, which allows for consistent acceleration. The strength of this field directly influences the rate of changes in direction control of circulating particles.

Potential Difference

In a cyclotron, the potential difference is vital for accelerating protons to high speeds. It is the voltage needed to give particles sufficient kinetic energy.
Part (c) of our exercise illustrates calculating this potential difference, where we found that the proton requires an approximate 9426 V to reach the desired maximum speed.
The potential difference between the dees (the hollow electrodes in a cyclotron) is reversed periodically, aligning with the particle’s path, ensuring continuous acceleration as the particles break paths touching electric fields. This potential difference reversal happens at a frequency twice that of proton circulation, crucial to maintaining consistent acceleration through the cyclotron.

Kinetic Energy Formula

The kinetic energy formula is essential in determining how much energy a proton gains in the cyclotron. Kinetic energy \( K \) is given by \[ K = \frac{1}{2}mv^2 \] where \( m \) is the mass of the proton and \( v \) is its velocity.
By setting this kinetic energy equal to the electrical energy from a potential difference \( qV \), we derive the potential needed to accelerate the proton: \[ V = \frac{mv^2}{2q} \] This relationship allows us to compute the energy transitions as particles accelerate through the machine. Understanding kinetic energy is crucial for adjusting parameters to achieve desired velocities, as seen in our calculation of around 9426 V for reaching maximum calculated speeds.