Problem 57 - University Physics with Modern Physics

Question

A particle with negative charge q and mass \(m =\) 2.58 \(\times\) 10\(^{-15}\) kg is traveling through a region containing a uniform magnetic field \(\overrightarrow{B} =\) -(0.120 T)\(\hat{k}\). At a particular instant of time the velocity of the particle is \(\vec{v}\) (1.05 \(\times\) 10\(^6\) m/s (-3\(\hat{\imath}\)+4\(\hat{\jmath}\)+12\(\hat{k}\)) and the force \(\overrightarrow{F}\) on the particle has a magnitude of 2.45 N. (a) Determine the charge \(q\). (b) Determine the acceleration \(\overrightarrow{a}\) of the particle. (c) Explain why the path of the particle is a helix, and determine the radius of curvature \(R\) of the circular component of the helical path. (d) Determine the cyclotron frequency of the particle. (e) Although helical motion is not periodic in the full sense of the word, the \(x\)- and \(y\)-coordinates do vary in a periodic way. If the coordinates of the particle at \(t =\) 0 are (\(x, y, z\)) = (\(R\), 0, 0), determine its coordinates at a time \(t =\) 2\(T\), where \(T\) is the period of the motion in the \(xy\)-plane.

Step-by-Step Solution

Answer

Charge: -5 x 10^-15 C. Acceleration: 9.50 x 10^14 m/s². Helical with R=2.15 m. Cyclotron frequency: 2.33 x 10^13 Hz. Coordinates at t=2T: (R, 0, 6.48 x 10^-7 m).

Step 1: Determine the Charge (q)

The magnetic force on a charged particle moving in a magnetic field is given by the equation \( \overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B}) \). Given: - \( \overrightarrow{B} = -0.120 \hat{k} \; \text{T} \)- \( \overrightarrow{v} = 1.05 \times 10^6 \; \text{m/s} (-3\hat{i} + 4\hat{j} + 12\hat{k}) \)Calculate the cross product \( \overrightarrow{v} \times \overrightarrow{B} \):\( \overrightarrow{v} \times \overrightarrow{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 4 & 12 \ 0 & 0 & -0.120 \end{vmatrix} = (0.120 \times 4)\hat{i} + (0.120 \times 3)\hat{j} = 0.48\hat{i} + 0.36\hat{j} \)Using the magnitude of the force \( \|\overrightarrow{F}\| = 2.45 \; \text{N} \), solve for \( q \) using \( \|\overrightarrow{F}\| = q \|\overrightarrow{v} \times \overrightarrow{B}\| \):\[ q \|0.48\hat{i} + 0.36\hat{j}\| = 2.45 \quad \rightarrow \quad q = \frac{2.45}{\sqrt{0.48^2 + 0.36^2}} \approx -5 \times 10^{-15} \; \text{C} \].

Step 2: Calculate the Acceleration (a)

The acceleration \(\overrightarrow{a} \) is determined by the net force as \( \overrightarrow{F} = m \overrightarrow{a} \). Given \( m = 2.58 \times 10^{-15} \; \text{kg} \), solve for \( \overrightarrow{a} \):\[ \overrightarrow{a} = \frac{\overrightarrow{F}}{m} \]Given the magnitude of the force is \( 2.45 \; \text{N} \),\[ \|\overrightarrow{a}\| = \frac{2.45}{2.58 \times 10^{-15}} = 9.50 \times 10^{14} \; \text{m/s}^2 \].

Step 3: Explain the Helix Path and Find Radius (R)

The path is helical because the particle has both perpendicular and parallel components to the magnetic field:- Perpendicular component: causes circular motion due to Lorentz force; solved by cyclic radius.- Parallel component: ensures linear motion parallel to \( \overrightarrow{B}\).To find the radius of curvature \( R \) of the circular component, use the formula:\( R = \frac{mv_{\perp}}{|q|B} \)Determine the perpendicular velocity component:\[ v_{\perp} = \sqrt{(-3\hat{i})^2 + (4\hat{j})^2} = 5.00 \times 10^6 \; \text{m/s} \]Then, calculate \( R \):\[ R = \frac{2.58 \times 10^{-15} \times 5.00 \times 10^6}{5 \times 10^{-15} \times 0.120} \approx 2.15 \; \text{m} \].

Step 4: Calculate the Cyclotron Frequency

The cyclotron frequency \( \omega \) is given by:\[ \omega = \frac{|q|B}{m} \]Substituting known values:\[ \omega = \frac{5 \times 10^{-15} \times 0.120}{2.58 \times 10^{-15}} = 2.33 \times 10^{13} \; \text{Hz} \].

Step 5: Determine Coordinates at t = 2T

The period \( T \) in the \( xy \) plane is given by:\[ T = \frac{2\pi}{\omega} \]Substitute for \( \omega \):\[ T = \frac{2\pi}{2.33 \times 10^{13}} = 2.70 \times 10^{-13} \; \text{s} \]At \( t = 2T \), the particle undergoes a complete circular motion and returns to the initial \( xy \) position, so the coordinates become:- \( x = R \),- \( y = 0 \),- The \( z \) coordinate indicates unchanging linear motion parallel to \( \overrightarrow{B} \):\[ z = v_{\parallel} \times t = 12 \times 10^6 \times 2.70 \times 10^{-13} \approx 6.48 \times 10^{-7} \; \text{m} \].

Key Concepts

Charged Particle Motion

When a charged particle enters a magnetic field, it experiences a force that alters its trajectory. This motion is governed by the nature of the charge and its velocity. A charged particle, such as an electron or proton, will follow a path determined largely by its charge and velocity vector. In a magnetic field, the exerted force acts perpendicular to both the velocity of the particle and the magnetic field lines.

This perpendicular force causes the particle to change direction but not speed, leading to circular or spiral paths instead of straight lines. The magnetic force does not work on the particle in the traditional sense because there is no displacement in the direction of the force. As a result, there is no change in kinetic energy. The speed remains constant, but the direction continuously changes due to the magnetic pull or push.

• The force on the charge is given by the Lorentz force equation.
• Motion is determined by both the magnetic field strength and the velocity of the particle.
• Circular when viewed in a plane perpendicular to the magnetic field.

Understanding these principles is crucial for applications such as magnetic confinement in fusion reactors and beam steering in particle accelerators.

Helical Motion

Helical motion occurs when a charged particle moves through a magnetic field at an angle. This results in a path shaped like a helix or spiral. The helical path is a combination of two types of motion:

1. **Circular Motion**: Perpendicular to the magnetic field, caused by the magnetic component of the Lorentz force.
2. **Linear Motion**: Parallel to the field, where the particle continues to move forward at a constant velocity because the magnetic field does not exert force in this direction.

The radius of this helical path is determined by the perpendicular component of the velocity and the magnetic field strength, as given by the formula:
\[ R = \frac{mv_{\perp}}{|q|B} \]Here, \(m\) is the mass, \(v_{\perp}\) is the perpendicular velocity component, \(q\) is the charge, and \(B\) is the magnetic field.

• The helical motion is periodic in the plane perpendicular to the magnetic field.
• The pitch of the helix (distance between loops) is influenced by the parallel velocity component.
• Applications include synchrotron light sources in physics research.

Cyclotron Frequency

Cyclotron frequency refers to the rate at which a charged particle spirals around the magnetic field lines. It is the frequency of the circular component of the helical motion. It's an important concept in understanding how charged particles behave in magnetic fields, particularly in devices like cyclotrons which accelerate particles.

The cyclotron frequency \( \omega \) is given by:
\[ \omega = \frac{|q|B}{m} \]This formula shows that the frequency is dependent on the charge to mass ratio of the particle and the strength of the magnetic field.

• It controls how tight the spiral paths of the particles are.
• Higher frequencies correspond to tighter spirals.
• The frequency does not depend on the velocity of the particle.

Cyclotron frequency plays a critical role in particle acceleration and is fundamental for technologies like MRI scanners.

Lorentz Force

The Lorentz force is the force exerted on a charged particle moving through a magnetic and/or electric field. It is the sum of the electric force and the magnetic force on the particle:

\[ \overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B}) \]Where:

• \(q\) is the charge of the particle.
• \(\overrightarrow{E}\) is the electric field strength.
• \(\overrightarrow{v}\) is the velocity of the particle.
• \(\overrightarrow{B}\) is the magnetic field strength.

In the context of magnetic fields alone, the Lorentz force is what causes the charged particle to move in circular or helical paths, as there is no component of the force parallel to the velocity of the particle.

The direction of the magnetic force is given by the right-hand rule, and it is always perpendicular to the velocity and magnetic field. This force can transform the energy and trajectory of particles without doing any work on them, illustrating a fundamental physics principle regarding magnetic forces.