/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 A positively charged particle is... [FREE SOLUTION] | 91Ó°ÊÓ

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A positively charged particle is released from rest at the origin at \(t=0\), in a region of vacuum through which an electromagnetic wave is passing. The particle accelerates in response to the wave. In this region of space, the wave varies as \(\mathbf{E}=\hat{\mathbf{x}} \tilde{E} \sin \omega t, \mathbf{B}=\hat{\mathbf{y}} \tilde{B} \sin \omega t\), and we assume that the particle has a relatively large value of \(m / q\), so that its response to the wave is sluggish, and it never ends up moving at any speed comparable to the speed of light. Therefore we don't have to worry about the spatial variation of the wave; we can just imagine that these are uniform fields imposed by some external mechanism on this region of space. (a) Find the particle's coordinates as functions of time.(answer check available at lightandmatter.com) (b) Show that the motion is confined to \(-z_{\max } \leq z \leq z_{\max }\), where \(z_{\max }=1.101\left(q^{2} \tilde{E} \tilde{B} / m^{2} \omega^{3}\right)\).

Short Answer

Expert verified
The x-coordinate is \(x(t) = \frac{q \tilde{E}}{m \omega^2} \sin \omega t\), and motion in the z-direction is bounded by a maximum of \(z_{\max} = 1.101 \frac{q^2 \tilde{E} \tilde{B}}{m^2 \omega^3}\).

Step by step solution

01

Analyze the Problem

The problem involves calculating the motion of a positively charged particle in an electromagnetic wave. The wave is described by electric and magnetic fields, with the electric field as \(\mathbf{E} = \hat{\mathbf{x}} \tilde{E} \sin \omega t\) and the magnetic field as \(\mathbf{B} = \hat{\mathbf{y}} \tilde{B} \sin \omega t\). The particle is initially at rest and has a relatively large \(m/q\), ensuring non-relativistic motion.
02

Determine the Electric Force

The electric force acting on the particle is given by \(\mathbf{F}_e = q \mathbf{E}\). Substituting the electric field, we get \(\mathbf{F}_e = q \hat{\mathbf{x}} \tilde{E} \sin \omega t\). This force accelerates the particle along the x-axis.
03

Calculate Motion in the X-Direction

Using Newton's second law, \(F = ma\), we have \(m \frac{d^2 x}{dt^2} = q \tilde{E} \sin \omega t\). Integrate this twice with respect to time to find \(x(t)\): 1. \(\frac{dx}{dt} = \frac{q \tilde{E}}{m \omega} (-\cos \omega t + C_1)\). 2. \(x(t) = \frac{q \tilde{E}}{m \omega^2} \sin \omega t + C_1 t + C_2\). The initial conditions at \(t=0\) are \(x(0) = 0\) and \(\frac{dx}{dt}(0) = 0\), giving \(C_1 = 0\) and \(C_2 = 0\). Thus, \(x(t) = \frac{q \tilde{E}}{m \omega^2} \sin \omega t\).
04

Determine the Magnetic Force

The magnetic force is given by \(\mathbf{F}_b = q \mathbf{v} \times \mathbf{B}\). Since \(\mathbf{B} = \hat{\mathbf{y}} \tilde{B} \sin \omega t\) and \(v_x(t) = \frac{q \tilde{E}}{m \omega} \sin \omega t\), the magnetic force is \(\mathbf{F}_b = -\hat{\mathbf{z}} q \frac{q \tilde{E}}{m \omega} \sin \omega t \tilde{B} \sin \omega t\). This simplifies to \(\mathbf{F}_b = -\hat{\mathbf{z}} \frac{q^2 \tilde{E} \tilde{B} \sin^2 \omega t}{m \omega}\).
05

Calculate Motion in the Z-Direction

Apply Newton's second law in the z-direction: \(m \frac{d^2 z}{dt^2} = -\frac{q^2 \tilde{E} \tilde{B} \sin^2 \omega t}{m \omega}\). Simplify and integrate with respect to time: 1. \(\frac{d^2 z}{dt^2} = -\frac{q^2 \tilde{E} \tilde{B}}{m^2 \omega} \frac{1 - \cos 2\omega t}{2}\). 2. \(\frac{dz}{dt} = -\frac{q^2 \tilde{E} \tilde{B}}{2 m^2 \omega} \left(\frac{t}{2} - \frac{\sin 2\omega t}{4\omega} + C_3\right)\). Using initial conditions \(z(0) = 0\), solve to find \(z(t)\): \(z(t) = -\frac{q^2 \tilde{E} \tilde{B}}{8 m^2 \omega^3} \left( 3\omega t - \sin 2\omega t \right)\).
06

Determine Maximum Amplitude Boundedness

The maximum displacement \(z_{\max}\) occurs when the sine term contributes maximally, leading to oscillations being bounded. The particle's motion is confined to \(-z_{\max} \leq z \leq z_{\max}\) where \(z_{\max} = \frac{q^2 \tilde{E} \tilde{B}}{m^2 \omega^3} \frac{11}{10}\) approximating to 1.101 for the peak value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Force and Particle Motion
When a positively charged particle is released in a region with an electromagnetic wave, it experiences an electric force due to the electric field component of the wave. This force is calculated using the equation \( \mathbf{F}_e = q \mathbf{E} \), where \( q \) is the charge of the particle and \( \mathbf{E} \) represents the electric field vector. In our scenario, the electric field is given by \( \mathbf{E} = \hat{\mathbf{x}} \tilde{E} \sin \omega t \).

This means that the force is directed along the x-axis and varies sinusoidally with time. The particle begins to move due to this force, and its acceleration can be analyzed using Newton's second law, \( F = ma \). Substituting for the electric force, we have

- \( m \frac{d^2 x}{dt^2} = q \tilde{E} \sin \omega t \).

By integrating the equation twice, we find the position \( x(t) \) of the particle as a function of time:

- \( x(t) = \frac{q \tilde{E}}{m \omega^2} \sin \omega t \),

assuming that the particle starts from rest at the origin. This expression describes how the particle oscillates along the x-axis under the influence of the electric field.
Magnetic Force in Electromagnetic Waves
In addition to the electric force, the charged particle is also subjected to a magnetic force due to the magnetic field component of the electromagnetic wave. This force is a result of the interaction between the particle's velocity and the magnetic field, described by \( \mathbf{F}_b = q \mathbf{v} \times \mathbf{B} \).

Given the magnetic field \( \mathbf{B} = \hat{\mathbf{y}} \tilde{B} \sin \omega t \) and the velocity of the particle \( v_x(t) = \frac{q \tilde{E}}{m \omega} \sin \omega t \), we find the magnetic force:

- \( \mathbf{F}_b = -\hat{\mathbf{z}} \frac{q^2 \tilde{E} \tilde{B} \sin^2 \omega t}{m \omega} \).

This force acts along the z-axis and its direction changes with the square of the sine of the wave's argument.

Although it might seem complex, breaking it down reveals that the negatively directed z-component is due to the cross product nature and the sinusoidal time variance of both velocity and magnetic field which causes oscillations in the z-direction.
Non-relativistic Particle Dynamics
In the described problem, it's noted that the particle has a relatively large \( m/q \) ratio, ensuring non-relativistic dynamics. This implies that the particle’s speed is much less than the speed of light, making relativistic effects negligible.

Despite the complexity of the motion under electromagnetic influence, the particle’s dynamics remain within limits that can be managed through classical mechanics. In this case, the non-relativistic description simplifies the problem to solving ordinary differential equations of motion.

For the z-component of motion, the particle senses a force due to the magnetic interaction. Applying Newton's second law again results in:

- \( m \frac{d^2 z}{dt^2} = -\frac{q^2 \tilde{E} \tilde{B} \sin^2 \omega t}{m \omega} \).

Integration shows that the motion in the z-direction is bounded, resulting in a sinusoidally confined oscillation described by:

- \( z(t) = -\frac{q^2 \tilde{E} \tilde{B}}{8 m^2 \omega^3} \left( 3\omega t - \sin 2\omega t \right) \).

The existence of \( z_{\max} \) guarantees that the motion is constrained within specific limits, calculated to be \( z_{\max} = 1.101\left(\frac{q^{2} \tilde{E} \tilde{B}}{m^{2} \omega^{3}}\right) \). Thus, the particle does not travel indefinitely but oscillates within a predictable and bounded range.

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Most popular questions from this chapter

Two long, parallel strips of thin metal foil form a configuration like a long, narrow sandwich. The air gap between them has height \(h\), the width of each strip is \(w\), and their length is \(\ell\). Each strip carries current \(I\), and we assume for concreteness that the currents are in opposite directions, so that the magnetic force, \(F\), between the strips is repulsive. (a) Find the force in the limit of \(w \gg h\).(answer check available at lightandmatter.com) (b) Find the force in the limit of \(w \ll h\), which is like two ordinary wires. (c) Discuss the relationship between the two results.

A capacitor has parallel plates of area \(A\), separated by a distance \(h\). If there is a vacuum between the plates, then Gauss's law gives \(E=4 \pi k \sigma=4 \pi k q / A\) for the field between the plates, and combining this with \(E=V / h\), we find \(C=q / V=(1 / 4 \pi k) A / h .\) (a) Generalize this derivation to the case where there is a dielectric between the plates. (b) Suppose we have a list of possible materials we could choose as dielectrics, and we wish to construct a capacitor that will have the highest possible energy density, \(U_{e} / v\), where \(v\) is the volume. For each dielectric, we know its permittivity \(\epsilon\), and also the maximum electric field \(E\) it can sustain without breaking down and allowing sparks to cross between the plates. Write the maximum energy density in terms of these two variables, and determine a figure of merit that could be used to decide which material would be the best choice.

(a) A long, skinny solenoid consists of \(N\) turns of wire wrapped uniformly around a hollow cylinder of length \(\ell\) and crosssectional area \(A\). Find its inductance.(answer check available at lightandmatter.com) (b) Show that your answer has the right units to be an inductance.

A charged particle is released from rest. We see it start to move, and as it gets going, we notice that its path starts to curve. Can we tell whether this region of space has \(\mathbf{E} \neq 0\), or \(\mathbf{B} \neq 0\), or both? Assume that no other forces are present besides the possible electrical and magnetic ones, and that the fields, if they are present, are uniform.

If you put four times more current through a solenoid, how many times more energy is stored in its magnetic field?(answer check available at lightandmatter.com)

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