Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A charged particle of mass \(m\) and positive charge \(q\) moves in uniform electric and magnetic fields, \(\mathbf{E}\) and \(\mathbf{B}\), both pointing in the \(z\) direction. The net force on the particle is \(\mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B})\). Write down the equation of motion for the particle and resolve it into its three components. Solve the equations and describe the particle's motion.

Short Answer

Expert verified
The particle follows a helical path, spiraling around the z-axis with linear acceleration due to the electric field.

Step by step solution

01

Write Down the Force Equation

The force on the particle is given by the Lorentz force: \( \mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) \). Where \( \mathbf{v} \) is the velocity of the particle, \( \mathbf{E} = E \hat{k} \) (electric field in the z-direction), and \( \mathbf{B} = B \hat{k} \) (magnetic field in the z-direction).
02

Express the Cross Product

Cross product \( \mathbf{v} \times \mathbf{B} \) is expanded to: \( \begin{pmatrix} v_x \ v_y \ v_z \end{pmatrix} \times \begin{pmatrix} 0 \ 0 \ B \end{pmatrix} = \begin{pmatrix} v_y B \ -v_x B \ 0 \end{pmatrix}.\)
03

Breakdown into Component Equations

Substitute the cross product result and resolve the force equation into components:- In the x-direction: \( F_x = q(v_y B) \)- In the y-direction: \( F_y = -q(v_x B) \)- In the z-direction: \( F_z = q(E) \)
04

Convert to Equations of Motion

The equations of motion using Newton's second law \( F = m \frac{d^2 r}{dt^2} \) for each component:1. \( m \frac{d^2 x}{dt^2} = qv_y B \)2. \( m \frac{d^2 y}{dt^2} = -qv_x B \)3. \( m \frac{d^2 z}{dt^2} = qE \)
05

Solve the Equations of Motion

1. Z-direction: Integrating \( m \frac{d^2 z}{dt^2} = qE \) results in uniform acceleration \( z(t) = \frac{qE}{m} \frac{t^2}{2} + v_{z0} t + z_0 \).2. X and Y-directions are coupled: - Substitute for cyclotron frequency \( \omega = \frac{qB}{m} \), leading to harmonic motion: - \( \frac{d^2 x}{dt^2} = \omega \frac{dy}{dt} \) - \( \frac{d^2 y}{dt^2} = -\omega \frac{dx}{dt} \) - These result in circular motion: solutions are \( x(t) = R \cos(\omega t + \phi_x) \) and \( y(t) = R \sin(\omega t + \phi_y) \).
06

Describe the Motion

The motion decomposes into two parts: 1. Helical motion around the z-axis with radius \( R \) due to the magnetic component causing circular motion in the xy-plane.2. Accelerated linear motion in the z-direction due to the electric field.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Charged Particle Motion
When a charged particle, such as an electron or proton, enters a region with external fields, its path can change dramatically depending on the field types and the particle’s initial velocity. Understanding charged particle motion is key in fields like electromagnetism and plasma physics.
This motion is primarily governed by the Lorentz force, which dictates that the combined effects of electric and magnetic fields will influence how the particle accelerates. This means once you know the charge, the initial velocity, and the configuration of the fields, you can predict the particle’s trajectory.
In our example, the electric and magnetic fields are uniform and directed along the same axis (z-direction). This simplification lets us predict two main types of motion: circular motion caused by the magnetic field and linear motion influenced by the electric field.
Electric and Magnetic Fields
These fields play a crucial role in determining the behavior of charged particles. An electric field (\(\mathbf{E}\)) exerts a force straight along the field lines, impacting the particle's motion along that direction. The magnetic field (\(\mathbf{B}\)), by contrast, influences motion perpendicular to itself and the velocity—which results in circular dynamics.
The fields in the given problem are both aligned in the z-direction. The electric field \(\mathbf{E} = E \hat{k}\), where \(E\) is the field strength, accelerates the particle along the z-axis. The magnetic field \(\mathbf{B} = B \hat{k}\) causes motion in the xy-plane via the Lorentz force expression \(\mathbf{v} \times \mathbf{B}\).
This means electric fields typically drive acceleration and can give particles kinetic energy along their direction, while magnetic fields are known for inducing rotational or circular motion.
Equations of Motion
To precisely determine the particle's path, we describe its motion mathematically using equations based on Newton's Second Law. We resolve these equations into three components—x, y, and z-axes—correlating to the directions relative to the fields.
The generalized force equation \(\mathbf{F} = m \frac{d^2 \mathbf{r}}{dt^2}\), when combined with the Lorentz force \(q(\mathbf{E} + \mathbf{v} \times \mathbf{B})\), becomes the basis for our component equations:
  • In the x-direction: \(m \frac{d^2 x}{dt^2} = q v_y B\)
  • In the y-direction: \(m \frac{d^2 y}{dt^2} = -q v_x B\)
  • In the z-direction: \(m \frac{d^2 z}{dt^2} = q E\)
Solving these uncovers the particle's motion: a linear path along the z-axis and circular motion in the xy-plane due to the different influences of the fields.
Cyclotron Frequency
A particularly interesting result from charged particle dynamics in magnetic fields is the cyclotron frequency, \(\omega = \frac{qB}{m}\). This represents the rate at which a particle circles in the plane perpendicular to a magnetic field.
This frequency is a constant for given values of charge, magnetic field strength, and mass, meaning it depends only on intrinsic properties and not on the particle's speed or radius of the circle it moves in.
In complex motions such as helical paths, cyclotron frequency helps predict the rotating aspect of the motion. As seen with harmonic solutions of the motion's differential equations, \( x(t) = R \cos(\omega t + \phi_x)\) and \( y(t) = R \sin(\omega t + \phi_y)\), illustrate the periodic aspect at a natural rate dictated by \(\omega\), a hallmark of cyclotron movement.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

I kick a puck of mass \(m\) up an incline (angle of slope \(=\theta\) ) with initial speed \(v_{\mathrm{o}}\). There is no friction between the puck and the incline, but there is air resistance with magnitude \(f(v)=c v^{2} .\) Write down and solve Newton's second law for the puck's velocity as a function of \(t\) on the upward journey. How long does the upward journey last?

There are certain simple one-dimensional problems where the equation of motion (Newton's second law) can always be solved, or at least reduced to the problem of doing an integral. One of these (which we have met a couple of times in this chapter) is the motion of a one-dimensional particle subject to a force that depends only on the velocity \(v\), that is, \(F=F(v)\). Write down Newton's second law and separate the variables by rewriting it as \(m d v / F(v)=d t .\) Now integrate both sides of this equation and show that $$t=m \int_{v_{0}}^{v} \frac{d v^{\prime}}{F\left(v^{\prime}\right)}$$ Provided you can do the integral, this gives \(t\) as a function of \(v\). You can then solve to give \(v\) as a function of \(t .\) Use this method to solve the special case that \(F(v)=F_{\mathrm{o}},\) a constant, and comment on your result. This method of separation of variables is used again in Problems 2.8 and \(2.9 .\)

where the primes denote successive derivatives of \(f(x)\). (Depending on the function this series may converge for any increment \(\delta\) or only for values of \(\delta\) less than some nonzero "radius of convergence.") This theorem is enormously useful, especially for small values of \(\delta\), when the first one or two terms of the series are often an excellent approximation. \(^{11}\) (a) Find the Taylor series for \(\ln (1+\delta)\). (b) Do the same for cos \(\delta\). (c) Likewise sin \(\delta\). (d) And \(e^{\delta}\).

A gun can fire shells in any direction with the same speed \(v_{\mathrm{o}}\). Ignoring air resistance and using cylindrical polar coordinates with the gun at the origin and \(z\) measured vertically up, show that the gun can hit any object inside the surface $$z=\frac{v_{0}^{2}}{2 g}-\frac{g}{2 v_{0}^{2}} \rho^{2}$$ Describe this surface and comment on its dimensions.

Consider the complex number \(z=e^{i \theta}=\cos \theta+i \sin \theta .\) (a) By evaluating \(z^{2}\) two different ways, prove the trig identities \(\cos 2 \theta=\cos ^{2} \thetaQEDQED-\sin ^{2} \theta\) and \(\sin 2 \theta=2 \sin \theta \cos \theta .\) (b) Use the same technique to find corresponding identities for \(\cos 3 \theta\) and \(\sin 3 \theta\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free