Problem 60 - University Physics with Modern Physics

Question

A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconducting electromagnet of the cyclotron produces a 2.9 -T magnetic field perpendicular to the proton orbits. (a) When the protons have achieved a kinetic energy of 2.7\(\mathrm { MeV }\) what is the radius of their circular orbit and what is their angular speed? (b) Repeat part (a) when the protons have achieved their final kinetic energy of 5.4\(\mathrm { MeV }\) .

Step-by-Step Solution

Answer

(a) At 2.7 MeV: radius = 0.0516 m, angular speed = 2.79 x 10^8 rad/s. (b) At 5.4 MeV: radius = 0.0732 m, angular speed = 2.79 x 10^8 rad/s.

Step 1: Convert Energy to Joules

First, convert the given proton kinetic energies from MeV to Joules. Use the conversion factor: \[1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J}\]. For 2.7 MeV:\[2.7 \text{ MeV} = 2.7 \times 1.602 \times 10^{-13} \text{ J} = 4.3254 \times 10^{-13} \text{ J}\]For 5.4 MeV:\[5.4 \text{ MeV} = 5.4 \times 1.602 \times 10^{-13} \text{ J} = 8.6508 \times 10^{-13} \text{ J}\]

Step 2: Find the Speed of Protons

Use the kinetic energy formula \( KE = \frac{1}{2} mv^2 \) to solve for the velocity \( v \) of the protons.The mass of a proton \( m \) is approximately \( 1.67 \times 10^{-27} \text{ kg} \).For 2.7 MeV:\[v^2 = \frac{2 \times 4.3254 \times 10^{-13}}{1.67 \times 10^{-27}}\]\[v = \sqrt{\frac{2 \times 4.3254 \times 10^{-13}}{1.67 \times 10^{-27}}} = 1.44 \times 10^7 \text{ m/s}\]For 5.4 MeV:\[v = \sqrt{\frac{2 \times 8.6508 \times 10^{-13}}{1.67 \times 10^{-27}}} = 2.043 \times 10^7 \text{ m/s}\]

Step 3: Calculate Radius of Orbit

Use the formula for the radius \( r \) of a proton's circular orbit in a magnetic field: \( r = \frac{mv}{qB} \), where \( q = 1.602 \times 10^{-19} \text{ C} \) is the charge of the proton and \( B = 2.9 \text{ T} \) is the magnetic field strength.For 2.7 MeV:\[r = \frac{1.67 \times 10^{-27} \times 1.44 \times 10^7}{1.602 \times 10^{-19} \times 2.9} = 0.0516 \text{ m}\]For 5.4 MeV:\[r = \frac{1.67 \times 10^{-27} \times 2.043 \times 10^7}{1.602 \times 10^{-19} \times 2.9} = 0.0732 \text{ m}\]

Step 4: Determine Angular Speed

Calculate the angular speed \( \omega \) using \( \omega = \frac{v}{r} \).For 2.7 MeV:\[\omega = \frac{1.44 \times 10^7}{0.0516} = 2.79 \times 10^8 \text{ rad/s}\]For 5.4 MeV:\[\omega = \frac{2.043 \times 10^7}{0.0732} = 2.79 \times 10^8 \text{ rad/s}\]

Key Concepts

Proton acceleration

The process of proton acceleration within a cyclotron involves increasing the energy of protons until they achieve a desired kinetic energy. In this case, the goal is to accelerate protons to an energy of 5.4 MeV. The cyclotron uses a rapidly alternating electric field to accelerate protons each time they pass between the cyclotron's semi-circular "dees". With each crossing, the protons gain energy and, consequently, speed. This energy is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( KE \) is the kinetic energy, \( m \) is the mass of the proton, and \( v \) is its velocity.
The acceleration is continuous, and the proton's energy increases with each half-cycle of acceleration. As soon as the proton exits the cyclotron, its kinetic energy can be harnessed for various purposes, such as medical treatment and scientific research. Understanding how protons are accelerated is a key step in mastering cyclotron technology.

Magnetic fields

Magnetic fields play a crucial role in the functioning of a cyclotron. They keep the accelerated protons on a circular path within the device. Cyclotrons employ a strong, uniform magnetic field perpendicular to the proton's motion. The magnetic field strength is denoted as \( B \), and it's measured in tesla (T). For the given cyclotron exercise, the magnetic field strength is 2.9 T.
What makes the magnetic field essential is its ability to exert a force on moving charged particles, like protons. This force is perpendicular to both the velocity of the particle and the direction of the magnetic field, resulting in circular motion. This effect is described by the Lorentz Force Law. Without this magnetic field, the protons would not follow a circular path, and the entire process of acceleration would diverge.
In summary, magnetic fields ensure the smooth operation of a cyclotron by guiding the protons in a curved path, enabling more prolonged acceleration within the device.

Kinetic energy conversion

Kinetic energy conversion within a cyclotron revolves around transforming potential energy into the kinetic energy of the protons. In the context of a cyclotron, each time a proton crosses between the semi-circular dees, it receives a "kick" from the electric field, raising its kinetic energy. The conversion of energy involves the equation \( KE = \frac{1}{2}mv^2 \), where changes in the kinetic energy reflect increases in the velocity of protons.
During the journey from low to higher energy levels, protons reach a higher velocity proportionate to the energy available. In the exercise of achieving a kinetic energy of 5.4 MeV, the conversion from potential electric energy to kinetic energy continues until the desired energy level is attained. It's possible to monitor this energy change and adjust parameters like frequency and magnitude of the electric field for efficient energy conversion.
This energy conversion is at the heart of applications requiring accelerated protons, including nuclear medicine and particle physics.

Circular motion in magnetic field

Understanding circular motion in a magnetic field is integral to grasping how a cyclotron functions. Protons, when subjected to a perpendicular magnetic field, move in a circular path due to the magnetic force acting on them. This force maintains the proton on its trajectory as long as the field is present.
The radius of this circular orbit depends on several factors such as the proton's mass \( m \), its velocity \( v \), the charge \( q \), and the magnetic field strength \( B \). The equation \( r = \frac{mv}{qB} \) calculates this radius. As protons gain higher kinetic energy, their velocity increases, subsequently enlarging their orbit's radius.
In our exercise, protons reached kinetic energies of 2.7 MeV and 5.4 MeV, corresponding to different circle radii, reflecting their increased velocities. The protons' angular speed, calculated through \( \omega = \frac{v}{r} \), also indicates how fast they move around the circle, remaining consistent at 2.79 x 10⁸ rad/s for both energies. This uniform circular motion enabled by the magnetic field lets protons steadily gain energy.