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Chapter 10: Problem 3

The general equation of motion of a (non-relativistic) particle of mass \(m\) and charge \(q\) when it is placed in a region where there is a magnetic field \(\mathbf{B}\) and an electric field \(\mathbf{E}\) is $$ m \ddot{\mathbf{r}}=q(\mathbf{E}+\dot{\mathbf{r}} \times \mathbf{B}) $$ here \(\mathbf{r}\) is the position of the particle at time \(t\) and \(\dot{\mathbf{r}}=d \mathbf{r} / d t\) etc. Write this as three separate equations in terms of the Cartesian components of the vectors involved. For the simple case of crossed uniform fields \(\mathbf{E}=E \mathbf{i}, \mathbf{B}=B \mathbf{j}\) in which the particle starts from the origin at \(t=0\) with \(\dot{\mathbf{r}}=v_{0} \mathbf{k}\), find the equations of motion and show the following: (a) if \(v_{0}=E / B\) then the particle continues its initial motion; (b) if \(v_{0}=0\) then the particle follows the space curve given in terms of the parameter \(\xi\) by $$ x=\frac{m E}{B^{2} q}(1-\cos \xi), \quad y=0, \quad z=\frac{m E}{B^{2} q}(\xi-\sin \xi) $$ Interpret this curve geometrically and relate \(\xi\) to \(t\). Show that the total distance travelled by the particle after time \(t\) is $$ \frac{2 E}{B} \int_{0}^{t}\left|\sin \frac{B q t^{\prime}}{2 m}\right| d t^{\prime} $$

Short Answer

Expert verified
Components are solved into cycloid motion described parametrically. For the initial conditions provided, particle motion can be equated to described values.

Step by step solution

01

Break Down the General Equation into Cartesian Components

Start from the equation of motion: \[ m \frac{d^2 \textbf{r}}{dt^2} = q (\textbf{E} + \frac{d \textbf{r}}{dt} \times \textbf{B}) \]Expand the equation in Cartesian coordinates:For the x-component: \[ m \frac{d^2 x}{dt^2} = q (E_x + (\frac{d y}{dt} B - \frac{d z}{dt} B)) \]For the y-component: \[ m \frac{d^2 y}{dt^2} = q (E_y + (\frac{d z}{dt} A - \frac{d x}{dt} B)) \]For the z-component: \[ m \frac{d^2 z}{dt^2} = q (E_z + (\frac{d x}{dt} B - \frac{d y}{dt} A)) \]
02

Apply Given Fields and Initial Conditions

Use the given fields \( \textbf{E} = E \textbf{i} \) and \( \textbf{B} = B \textbf{j} \):This simplifies the Cartesian equations to:\[ m \frac{d^2 x}{dt^2} = q E \]\[ m \frac{d^2 y}{dt^2} = q (\frac{d z}{dt} B) \]\[ m \frac{d^2 z}{dt^2} = - q (\frac{d y}{dt} B) \]At \( t = 0 \), \( \textbf{r} = 0 \) and \( \frac{d \textbf{r}}{dt} = v_0 \textbf{k} \) or \( \frac{d z}{dt} = v_0 \) and \( \frac{d x}{dt} = \frac{d y}{dt} = 0 \)
03

Solve for Cartesian Components with \(v_{0} = E/B\)

For the x-component:\[ m \frac{d^2 x}{dt^2} = q E \]This is simply:\[ \frac{d^2 x}{dt^2} = \frac{q E}{m} \]Integrating twice gives:\[ x = \frac{q E}{2 m} t^2 \]For the y-component, since \(y = 0 \), no force acts in this direction:\[ y = 0 \]For the z-component:Since \( v_{0} = \frac{E}{B} \), the particle continues its motion.
04

Solve for Cartesian Components with \(v_{0} = 0\)

Apply initial conditions for x-component:\[ x = \frac{m E}{B^2 q} (1 - \text{cos}\theta) \]For y-component:\[ y = 0 \]For z-component:\[ z = \frac{m E}{B^2 q} (\theta - \text{sin}\theta) \]
05

Integrate and Geometric Interpretation

Given \( \theta = \frac{B q t}{m} \) thus \( \theta \rightarrow t \).The curve described is a cycloid. Distance travels as:\[ \frac{2 E}{B} \times \text{integrate} (0 to t) (|\text{sin} \frac{B q t}{2 m}| d t) \]

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Key Concepts

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

Non-Relativistic Particle Motion
When we're talking about non-relativistic particle motion, we're focusing on scenarios where particles move at speeds significantly lower than the speed of light. This assumption simplifies our equations because we can ignore relativistic effects like time dilation. Such particles are governed by Newton's laws of motion. For a particle with mass \(\text{m}\) and charge \(q\) in an electromagnetic field, its equation of motion can be written as: \[ m \frac{d^2 \textbf{r}}{dt^2} = q (\textbf{E} + \frac{d \textbf{r}}{dt} \times \textbf{B}) \] Here:
  • \( \textbf{r} \text{represents the position vector of the particle.} \)
  • \( \textbf{E} \text{is the electric field vector.} \)
  • \( \textbf{B} \text{is the magnetic field vector.} \)

This equation tells us how the particle's motion evolves under the influence of electric and magnetic fields. Understanding this fundamental concept is crucial for solving problems in electromagnetism and understanding particle dynamics in various physics applications.
Lorentz Force
The Lorentz force is the force experienced by a charged particle in an electromagnetic field. It's given by: \[ \textbf{F} = q (\textbf{E} + \textbf{v} \times \textbf{B}) \] Here's what each term means:
  • \textbf{q} is the charge of the particle.

  • \textbf{E} is the electric field.
  • \textbf{v} is the velocity of the particle.
  • \textbf{B} is the magnetic field.

The equation shows that the total force on a charged particle is the sum of the forces due to the electric field and the magnetic field. The magnetic part of the force is always perpendicular to both the velocity and the magnetic field, which often results in circular or helical paths for the particle. In situations with only magnetic fields and no electric fields, the charged particle will follow a circular motion. This is core to understanding many phenomena in electromagnetism and helps us solve particle motion problems accurately.
Cartesian Components
To simplify solving equations of motion, we often break them down into their Cartesian components (x, y, and z directions). This makes it easier to handle the complex vectors involved. For a particle in an electric and magnetic field, we can expand the general equation in terms of Cartesian coordinates:
For the x-component:
\[ m \frac{d^2 x}{dt^2} = q (E_x + (\frac{d y}{dt} B - \frac{d z}{dt} B)) \]
For the y-component:
\[ m \frac{d^2 y}{dt^2} = q (E_y + (\frac{d z}{dt} B - \frac{d x}{dt} B)) \]
For the z-component:
\[ m \frac{d^2 z}{dt^2} = q (E_z + (\frac{d x}{dt} B - \frac{d y}{dt} B)) \]
Breaking down the motion into these components allows us to deal with each direction separately. This separation is particularly useful when handling uniform fields and initial conditions, as it allows us to solve each coordinate equation independently before combining them for the final motion path.
Crossed Electric and Magnetic Fields
Crossed electric and magnetic fields refer to a setup where the electric field \(\textbf{E}\) and magnetic field \(\textbf{B}\) are perpendicular to each other. Imagine an electric field pointing in the x-direction and a magnetic field pointing in the y-direction. This results in unique particle trajectories. Using the given fields \(\textbf{E} = E \textbf{i}\) and \(\textbf{B} = B \textbf{j}\), we can simplify the motion equations: \[ m \frac{d^2 x}{dt^2} = q E \] \[ m \frac{d^2 y}{dt^2} = q (\frac{d z}{dt} B) \] \[ m \frac{d^2 z}{dt^2} = -q (\frac{d y}{dt} B) \] When a particle starts from rest in such a field setup, it experiences forces due to both fields, leading to complex motion paths. Crossed fields are fundamental in many devices, such as velocity selectors, where they help select particles of specific velocities. This allows for practical applications like mass spectrometry and particle accelerators.
Cycloidal Motion
Cycloidal motion occurs when a particle in crossed electric and magnetic fields follows a path resembling a series of linked arcs. This happens when the initial velocity in one direction is zero, making the motion periodic. Given no initial velocity, we can express the motion:
  • For x-direction: \[ x = \frac{m E}{B^2 q}(1 - \text{cos}(\frac{B q t}{m})) \]

  • For y-direction: \[ y = 0 \]

  • For z-direction: \[ z = \frac{m E}{B^2 q}(\frac{B q t}{m} - \text{sin}(\frac{B q t}{m})) \]
The curve described is a cycloid. Over time, the parameter \(\frac{B q t}{m}\) represents the phase of the motion. The total path distance traveled by the particle can be calculated using an integral of the absolute velocity components. Cycloidal paths are crucial in certain types of accelerators and plasma confinement devices. Understanding this concept helps predict and control the behavior of charged particles in varied scientific and engineering applications.

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Most popular questions from this chapter

Prove that for a space curve \(\mathbf{r}=\mathbf{r}(s)\), where \(s\) is the arc length measured along the curve from a fixed point, the triple scalar product $$ \left(\frac{d \mathbf{r}}{d s} \times \frac{d^{2} \mathbf{r}}{d s^{2}}\right) \cdot \frac{d^{3} \mathbf{r}}{d s^{3}} $$ at any point on the curve has the value \(\kappa^{2} \tau\), where \(\kappa\) is the curvature and \(\tau\) the torsion at that point.

Verify by direct calculation that $$ \nabla \cdot(\mathbf{a} \times \mathbf{b})=\mathbf{b} \cdot(\nabla \times \mathbf{a})-\mathbf{a} \cdot(\nabla \times \mathbf{b}) $$

For a description in spherical polar coordinates with axial symmetry of the flow of a very viscous fluid, the components of the velocity field \(\mathbf{u}\) are given in terms of the stream function \(\psi\) by $$ u_{r}=\frac{1}{r^{2} \sin \theta} \frac{\partial \psi}{\partial \theta}, \quad u_{\theta}=\frac{-1}{r \sin \theta} \frac{\partial \psi}{\partial r} $$ Find an explicit expression for the differential operator \(E\) defined by $$ E \psi=-(r \sin \theta)(\nabla \times \mathbf{u})_{\phi} $$ The stream function satisfies the equation of motion \(E^{2} \psi=0\) and, for the flow of a fluid past a sphere, takes the form \(\psi(r, \theta)=f(r) \sin ^{2} \theta\). Show that \(f(r)\) satisfies the (ordinary) differential equation $$ r^{4} f^{(4)}-4 r^{2} f^{\prime \prime}+8 r f^{\prime}-8 f=0 $$

(a) Parameterising the hyperboloid $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1 $$ by \(x=a \cos \theta \sec \phi, y=b \sin \theta \sec \phi, z=c \tan \phi\), show that an area element on its surface is $$ d S=\sec ^{2} \phi\left[c^{2} \sec ^{2} \phi\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)+a^{2} b^{2} \tan ^{2} \phi\right]^{1 / 2} d \theta d \phi $$ (b) Use this formula to show that the area of the curved surface \(x^{2}+y^{2}-z^{2}=a^{2}\) between the planes \(z=0\) and \(z=2 a\) is $$ \pi a^{2}\left(6+\frac{1}{\sqrt{2}} \sinh ^{-1} 2 \sqrt{2}\right) $$

If two systems of coordinates with a common origin \(O\) are rotating with respect to each other, the measured accelerations differ in the two systems. Denoting by \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) position vectors in frames \(O X Y Z\) and \(O X^{\prime} Y^{\prime} Z^{\prime}\) respectively, the connection between the two is $$ \ddot{\mathbf{r}}^{\prime}=\ddot{\mathbf{r}}+\dot{\omega} \times \mathbf{r}+2 \omega \times \dot{\mathbf{r}}+\omega \times(\omega \times \mathbf{r}) $$ where \(\omega\) is the angular velocity vector of the rotation of \(O X Y Z\) with respect to \(O X^{\prime} Y^{\prime} Z^{\prime}\) (taken as fixed). The third term on the RHS is known as the Coriolis acceleration, whilst the final term gives rise to a centrifugal force. Consider the application of this result to the firing of a shell of mass \(m\) from a stationary ship on the steadily rotating earth, working to the first order in \(\omega\left(=7.3 \times 10^{-5} \mathrm{rad} \mathrm{s}^{-1}\right)\). If the shell is fired with velocity \(\mathbf{v}\) at time \(t=0\) and only reaches a height that is small compared to the radius of the earth, show that its acceleration, as recorded on the ship, is given approximately by $$ \ddot{\mathbf{r}}=\mathbf{g}-2 \omega \times(\mathbf{v}+\mathbf{g} t) $$ where \(m \mathbf{g}\) is the weight of the shell measured on the ship's deck. The shell is fired at another stationary ship (a distance \(\mathbf{s}\) away) and \(\mathbf{v}\) is such that the shell would have hit its target had there been no Coriolis effect. (a) Show that without the Coriolis effect the time of flight of the shell would have been \(\tau=-2 \mathbf{g} \cdot \mathbf{v} / g^{2}\) (b) Show further that when the shell actually hits the sea it is off target by approximately $$ \frac{2 \tau}{g^{2}}[(\mathbf{g} \times \omega) \cdot \mathbf{v}](\mathbf{g} \tau+\mathbf{v})-(\omega \times \mathbf{v}) \tau^{2}-\frac{1}{3}(\omega \times \mathbf{g}) \tau^{3} $$ (c) Estimate the order of magnitude \(\Delta\) of this miss for a shell for which \(v=300\) \(\mathrm{m} \mathrm{s}^{-1}\), firing close to its maximum range ( \(\mathbf{v}\) makes an angle of \(\pi / 4\) with the vertical) in a northerly direction, whilst the ship is stationed at latitude \(45^{\circ}\) North.
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