Problem 33 - Fundamentals of Physics

Question

A positron with kinetic energy \(2.00 \mathrm{keV}\) is projected into a uniform magnetic field \(\vec{B}\) of magnitude \(0.100 \mathrm{~T},\) with its velocity vector making an angle of \(89.0^{\circ}\) with \(\vec{B}\). Find (a) the period, (b) the pitch \(p\), and (c) the radius \(r\) of its helical path.

Step-by-Step Solution

Answer

T = 3.57 x 10^{-10} s; p = 1.65 x 10^{-4} m; r = 1.51 x 10^{-3} m.

Step 1: Determine Relevant Formulas

To solve this problem, we need to apply the physics of charged particles in magnetic fields. The relevant formulas include:1. **Period (T):** This is given by \( T = \frac{2\pi m}{qB} \), where \( m \) is the mass of the positron, \( q \) is the charge of the positron, and \( B \) is the magnetic field.2. **Pitch (p):** The pitch is given by \( p = v_{\parallel}T \), where \( v_{\parallel} \) is the component of velocity parallel to \( B \) and \( T \) is the period.3. **Radius (r):** The cyclotron radius is calculated by \( r = \frac{mv_{\perp}}{qB} \), where \( v_{\perp} \) is the component of velocity perpendicular to \( B \).

Step 2: Calculate the Total Velocity

The kinetic energy of the positron is given by \( KE = \frac{1}{2}mv^2 \).Convert kinetic energy to joules: \( 2.00 \text{ keV} = 2.00 \times 10^3 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-16} \text{ J} \).The mass of a positron \( m = 9.11 \times 10^{-31} \text{ kg} \), solve for velocity \( v \) using \( v = \sqrt{\frac{2KE}{m}} \).Plugging in values:\( v = \sqrt{\frac{2 \times 3.2 \times 10^{-16}}{9.11 \times 10^{-31}}} \approx 2.65 \times 10^7 \text{ m/s} \).

Step 3: Determine Velocity Components

The velocity \( v \) of the positron makes an angle \( 89.0^{\circ} \) with the magnetic field \( \vec{B} \).Calculate \( v_{\parallel} \):\( v_{\parallel} = v \cos(\theta) \approx 2.65 \times 10^7 \times \cos(89.0^{\circ}) = 4.63 \times 10^5 \text{ m/s} \).Calculate \( v_{\perp} \): \( v_{\perp} = v \sin(\theta) \approx 2.65 \times 10^7 \times \sin(89.0^{\circ}) \approx 2.65 \times 10^7 \text{ m/s} \).

Step 4: Compute the Period (T)

The period is given by \( T = \frac{2\pi m}{qB} \).The charge of a positron is \( q = 1.6 \times 10^{-19} \text{ C} \).Substitute \( m = 9.11 \times 10^{-31} \text{ kg}, \) \( q = 1.6 \times 10^{-19} \text{ C}, \) and \( B = 0.100 \text{ T} \):\(T = \frac{2\pi \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 0.100} \approx 3.57 \times 10^{-10} \text{ s} \).

Step 5: Calculate the Pitch (p)

The pitch is the axial distance traveled in one period: \( p = v_{\parallel}T \).Plugging in \( v_{\parallel} = 4.63 \times 10^5 \text{ m/s} \) and \( T = 3.57 \times 10^{-10} \text{ s} \):\(p = 4.63 \times 10^5 \times 3.57 \times 10^{-10} \approx 1.65 \times 10^{-4} \text{ m} \).

Step 6: Calculate the Radius (r)

The radius of the helical path is given by \( r = \frac{mv_{\perp}}{qB} \).Substitute \( m = 9.11 \times 10^{-31} \, \text{kg}, v_{\perp} = 2.65 \times 10^7 \, \text{m/s} \), \( q = 1.6 \times 10^{-19} \, \text{C}, B = 0.100 \, \text{T} \):\( r = \frac{9.11 \times 10^{-31} \times 2.65 \times 10^7}{1.6 \times 10^{-19} \times 0.100} \approx 1.51 \times 10^{-3} \text{ m} \).

Key Concepts

Kinetic Energy

Kinetic energy plays a crucial role when analyzing the motion of a positron, especially in magnetic fields. This energy, symbolized by \( KE \), is the energy that a particle possesses due to its motion. For charged particles like positrons, kinetic energy is not only a measure of motion but also helps calculate velocity. To find the kinetic energy in joules from electron volts (eV), a common practice is to use the conversion factor of \( 1 ext{ eV} = 1.6 \times 10^{-19} ext{ Joules} \). In our example, the positron has \( 2.00 \text{ keV} \), which we convert to joules for further calculations. Using the formula for kinetic energy \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the positron and \( v \) is its velocity, we solve for \( v \). This step is essential, as it provides the total velocity required for calculating various path characteristics when the positron interacts with a magnetic field.

Magnetic Field

A magnetic field is a vital component in the study of charged particle motion. It represents the magnetic influence on moving charges, affecting their trajectories significantly. The magnetic field strength is represented by \( B \) and measured in Tesla (T).In the given exercise, a uniform magnetic field of magnitude \( 0.100 \text{ T} \) is present. This uniformity implies that both the direction and magnitude of the field are consistent through the space in which the positron moves. The magnetic force acting on the positron results from its charges moving within this field. This force is perpendicular to both the velocity of the particle and the magnetic field, ensuring that the path followed by the particle remains circular or helical, rather than straight. Hence, understanding both the nature and the influence of the magnetic field is crucial in predicting and explaining the motion of charged particles like the positron.

Helical Path

When a charged particle, such as a positron, moves through a magnetic field at an angle, it follows a helical path. This occurs due to the interaction between the particle's velocity and the magnetic force acting perpendicularly on it. The helix can be thought of as similar to a spring coil. The radius of this coil is determined by the component of the positron's velocity perpendicular to the magnetic field, known as \( v_{\perp} \). Meanwhile, the axial distance covered in one complete turn of the helix, called the pitch \( p \), depends on the component of velocity parallel to the magnetic field \( v_{\parallel} \). The formula \( r = \frac{mv_{\perp}}{qB} \) is used to calculate the radius, showing how it depends on the mass \( m \), the charge \( q \), and the magnetic field \( B \). The pitch is calculated using \( p = v_{\parallel}T \), where \( T \) is the period of one complete rotation. These calculations highlight the dance-like motion of the positron within the magnetic field, making it a fascinating aspect of electromagnetism.

Velocity Components

Understanding the components of velocity is fundamental in analyzing the motion of the positron in the magnetic field. The total velocity vector \( v \) can be broken down into two key parts: \( v_{\parallel} \) (parallel to the magnetic field) and \( v_{\perp} \) (perpendicular to the magnetic field). These components are calculated using the angle \( 89.0^{\circ} \), which the velocity vector makes with the magnetic field. The formulas used are:

• \( v_{\parallel} = v \cos(\theta) \)
• \( v_{\perp} = v \sin(\theta) \)

The parallel component \( v_{\parallel} \) determines how far along the field line the positron travels in one rotation, thus affecting the pitch of the helical path. Meanwhile, \( v_{\perp} \) impacts the radius of the circular component of the path.Calculating these components gives keen insight into how the positron's path shapes and evolves over time in response to the magnetic field. They help in predicting not just where, but how the positron will travel in the magnetic landscape.