Question

Deuterons in a cyclotron travel in a circle with radius 32.0 cm just before emerging from the dees. The frequency of the applied alternating voltage is 9.00 MHz. Find (a) the magnetic field and (b) the kinetic energy and speed of the deuterons upon emergence.

Short answer

(a) Magnetic field, \( B = 0.115 \) T; (b) Kinetic energy, \( KE = 5.47 \times 10^{-13} \) J.

Step-by-step solution

Understand Cyclotron Motion

A cyclotron accelerates charged particles in a magnetic field, causing them to move in circular paths. The force to keep the deuterons in a circular path is provided by the magnetic Lorentz force, which is balanced by the centripetal force needed to keep the particle moving in a circle. The formula that represents this balance is: \( qvB = \frac{mv^2}{r} \), where \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field, \( m \) is the mass, and \( r \) is the radius of the path.

Calculate the Magnetic Field

To find the magnetic field \( B \), rearrange the formula: \( B = \frac{mv}{qr} \). We need the velocity \( v \), which we calculate using \( v = 2\pi rf \), where \( f \) is the frequency. The charge of a deuteron \( q = 1.6 \times 10^{-19} \) C, the radius \( r = 0.32 \) m, and the mass \( m = 2 \times 1.67 \times 10^{-27} \) kg. Plug these into the rearranged formula after calculating \( v \).

Determine Velocity

Calculate the velocity \( v \) using the frequency relation: \( v = 2\pi rf = 2\pi \times 0.32 \times 9 \times 10^6 \). Compute this to find \( v = 18.1 \times 10^6 \) m/s.

Substitute Values to Get Magnetic Field

Substitute \( m = 3.34 \times 10^{-27} \) kg, \( v = 18.1 \times 10^6 \) m/s, \( q = 1.6 \times 10^{-19} \) C, and \( r = 0.32 \) m into \( B = \frac{mv}{qr} \) to find \( B \). The resulting magnetic field \( B = 0.115 \) T (Tesla).

Calculate Kinetic Energy

Once the magnetic field is determined, find the final kinetic energy \( KE \) of the deuterons using the formula \( KE = \frac{1}{2}mv^2 \). With \( m = 3.34 \times 10^{-27} \) kg and \( v = 18.1 \times 10^6 \) m/s, compute \( KE = \frac{1}{2} \times 3.34 \times 10^{-27} \times (18.1 \times 10^6)^2 \) to get \( KE = 5.47 \times 10^{-13} \) J (joules).

Short Answer Verification

Verify that the computed values make sense within the context of the exercise. The magnetic field calculated helps produce the given frequency to keep the particles in circular motion, and the kinetic energy being a small value is consistent with small particles like deuterons.

Key concepts

Magnetic Field

In a cyclotron, the magnetic field plays a crucial role in keeping charged particles such as deuterons on a circular path. This magnetic field is perpendicular to the motion of the deuterons, providing the centripetal force needed to maintain their circular trajectory.
It functions by exerting a force on the moving deuterons, which is known as the Lorentz force. This force is represented by the equation:

• \( qvB = \frac{mv^2}{r} \)

In this equation, \( q \) is the charge of the deuteron, \( v \) is its velocity, \( m \) is the mass, \( B \) is the magnetic field, and \( r \) is the radius of the circular path. By manipulating these variables, the magnetic field can be calculated and defined, ensuring the deuterons move in an intended circular motion.

Kinetic Energy

Kinetic energy describes the energy possessed by an object in motion, like the deuterons in a cyclotron. As they accelerate through the alternating voltage, their speed increases, leading to a rise in their kinetic energy.
The kinetic energy (

• \( KE \)

) of a deuteron is calculated using the formula:

• \( KE = \frac{1}{2}mv^2 \)

Here, \( m \) is the mass of the deuteron and \( v \) is its velocity. Once the velocity is determined through calculations based on cyclotron frequency and path radius, this formula allows us to calculate the deuteron's kinetic energy upon exiting the cyclotron.
This energy is significantly small due to the tiny mass and size of deuterons, yet it is critical for understanding their behavior in the magnetic field.

Lorentz Force

The Lorentz force is a fundamental force responsible for the circular motion of charged particles like deuterons in a magnetic field within a cyclotron. It acts perpendicular to both the velocity of the deuterons and the direction of the magnetic field.
This force can be represented mathematically by the formula:

• \( F = qvB \)

Where \( F \) is the Lorentz force, \( q \) is the charge of the deuteron, \( v \) is its velocity, and \( B \) is the magnetic field. This force ensures that the deuterons continue moving in a circular path, balancing the centripetal force required in such motion.
Understanding Lorentz force helps in predicting how the deuterons will behave as they traverse through the cyclotron, allowing scientists and engineers to harness their motion productively.

Deuterons

Deuterons are nuclei of deuterium, an isotope of hydrogen. They consist of one proton and one neutron, endowing them with a unique set of properties beneficial in nuclear science and technology.
In the cyclotron, deuterons are accelerated using a magnetic field and an alternating voltage, causing them to process along a circular path. Their small size and charge \( q = 1.6 \times 10^{-19} \) C make them ideal for experimentation and application within a cyclotron setting.
The study of deuterons in physics has helped elucidate concepts in nuclear reactions and particle acceleration, proving essential in the advancement of both theoretical and experimental physics.

Circular Motion

Circular motion is the movement of an object along the circumference of a circle or rotation along a circular path. In a cyclotron, deuterons experience this type of motion while moving within a magnetic field.
This motion is sustained by the balance between the centripetal force and the Lorentz force. The centripetal force, often calculated as \( \frac{mv^2}{r} \), is needed to keep the deuteron moving in a circle, whereas the Lorentz force provides the necessary pull to maintain this path.
Understanding circular motion in this context allows for the practical application of physics principles, assisting in the advancement of devices that rely on precision in particle acceleration, such as cyclotrons.