/*! 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 26 Motion in \(E\) and B fields \(*... [FREE SOLUTION] | 91Ó°ÊÓ

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Motion in \(E\) and B fields \(* * *\) The task of Exercise \(6.29\) is to show that if a charged particle moves in the \(x y\) plane in the presence of a uniform magnetic field in the \(z\) direction, the path will be a circle. What does the path look like if we add on a uniform electric field in the \(y\) direction? Let the particle have mass \(m\) and charge \(q\). And let the magnitudes of the electric and magnetic fields be \(E\) and \(B\). Assume that the velocity is nonrelativistic, so that \(\gamma \approx 1\) (this assumption isn't necessary in Exercise 6.29, because \(v\) is constant there). Be careful, the answer is a bit counterintuitive.

Short Answer

Expert verified
The path of the particle is a cycloid, which can be counterintuitive. The particles tends to drift in the y direction, but circulates around an average position due to the magnetic field.

Step by step solution

01

Consider forces acting on the particle

The charged particle moving in both electric and magnetic field experiences two forces. The electric force \(F_E\) is \(F_E = qE\), and it acts in the y-direction. The magnetic force \(F_B\) is \(F_B = qvB\), and it acts perpendicular to the velocity vector and magnetic field according to the right-hand rule. Since the initial velocity is in the x-direction and the magnetic field is in the z-direction, the magnetic force at t=0 acts in the y-direction.
02

Write force equations

The equation of motion in the y-direction is \(m(dv_y/dt) = F_E + F_B = qE +qv_xB\). The motion in the x-direction is only influenced by the magnetic force, hence \(m(dv_x/dt) = -qv_yB \).
03

Solve differential equations

We need to solve these differential equations to find the path. First, take the time derivative of the first equation to eliminate \(v_x\) and get a second order differential equation, \( dd^2v_y/dt^2 + (qB/m)^2*v_y = qB(E/m)\). The solution of this equation gives \(v_y(t)\) as a function of time. The velocity in the x direction can then be obtained by integrating the second equation, \(dv_x/dt+ (qB/m)v_y = 0\), using \(v_y(t)\). Once \(v_x(t)\) and \(v_y(t)\) are known, the path of the particle can be obtained by integrating these functions.

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Key Concepts

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

Lorentz Force
The Lorentz Force is an essential concept in understanding how charged particles move through electromagnetic fields. It combines the effects of electric and magnetic fields on the particle and is the total force acting on it. This force is given by the formula:
  • \( \mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) \)
Here, \( q \) is the charge of the particle, \( \mathbf{E} \) represents the electric field, \( \mathbf{v} \) is the velocity of the particle, and \( \mathbf{B} \) is the magnetic field. The expression \( \mathbf{v} \times \mathbf{B} \) indicates the cross product, highlighting that the magnetic force component is always perpendicular to both \( \mathbf{v} \) and \( \mathbf{B} \).

Understanding the Lorentz Force helps predict how a charged particle will accelerate and change its path under the influence of these fields. In our exercise, we see that the Lorentz Force results from two components: the electric force \( F_E = qE \) acting in the direction of the electric field, and the magnetic force \( F_B = qvB \), acting perpendicular to the velocity.
Electric Field
An electric field \( \mathbf{E} \) is a vector field surrounding a charge that exerts a force on other charges within the field. The force due to an electric field is straightforward and linear, defined as:
  • \( \mathbf{F}_E = q\mathbf{E} \)
This means a particle with charge \( q \) in an electric field \( \mathbf{E} \) experiences a force proportional to the field strength and its charge.

In typical applications like our exercise, the electric field's direction determines the acceleration of the particle in that direction. For example, an electric field in the positive y-direction will push a positively charged particle upward in the same direction. This simple relationship makes predicting motion in a uniform electric field more intuitive compared to a magnetic field.
Magnetic Field
Magnetic fields, described by \( \mathbf{B} \), are vector fields that influence moving charges and magnetic materials. Different from electric fields, magnetic fields do not do work on particles; instead, they alter the trajectory. The force experienced by a charged particle in a magnetic field is specified by:
  • \( \mathbf{F}_B = q \mathbf{v} \times \mathbf{B} \)
This force is perpendicular to both the velocity \( \mathbf{v} \) and the magnetic field \( \mathbf{B} \), resulting in the charged particle undergoing circular motion when in a uniform \( \mathbf{B} \) field.

In our exercise, the magnetic field is in the \( z \)-direction, causing a charged particle moving in the \( xy \)-plane to experience a force perpendicular to its velocity. This causes the motion to bend in a circular path initially with no electric field. The inclusion of a magnetic field is crucial for understanding particle dynamics in devices like cyclotrons and mass spectrometers.
Differential Equations
Differential equations are mathematical tools used to describe the rate of change of variables involved in physical phenomena. In our exercise, we use them to model the motion of a charged particle in electromagnetic fields. These equations are derived from Newton's second law applied to the forces involved.

By setting up equations for forces in both x- and y-directions, we express the acceleration of the particle in terms of its velocity \((dv/dt)\) as a function of time. For example, the equation in the y-direction includes forces from both the electric and magnetic fields:
  • \( m\frac{dv_y}{dt} = qE + qv_xB \)
By solving these differential equations, we determine how the particle's velocity and position change over time. This process involves finding solutions that satisfy the equations, which, in turn, describe the unique path, or trajectory, of the particle.
Circular Motion
Circular motion occurs when a particle moves along a circular path under the influence of a force perpendicular to its velocity. This is common in scenarios involving magnetic fields, as these fields exert a perpendicular force that continuously alters the direction of velocity but not its magnitude.

The centripetal force needed for circular motion in a magnetic field is provided by the magnetic component of the Lorentz Force. The force requirement matches the expression for centripetal force \( F = \frac{mv^2}{r} \), which determines the radius \( r \) of the path, showing it depends on particle velocity, charge, and magnetic field strength. The uniform circular path observed when only a magnetic field acts is disrupted by an electric field, changing the resultant path to a spiral or helical trajectory depending on the fields' relative directions and magnitudes.

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Most popular questions from this chapter

Magnetic scalar "potential" \(* *\) (a) Consider an infinite straight wire carrying current \(I\). We know that the magnetic field outside the wire is \(\mathbf{B}=\left(\mu_{0} I / 2 \pi r\right) \hat{\boldsymbol{\theta}}\) There are no currents outside the wire, so \(\nabla \times \mathbf{B}=0 ;\) verify this by explicitly calculating the curl. (b) Since \(\nabla \times \mathbf{B}=0\), we should be able to write \(\mathbf{B}\) as the gradient of a function, \(\mathbf{B}=\nabla \psi .\) Find \(\psi\), but then explain why the usefulness of \(\psi\) as a potential function is limited.

E and B for a point charge ** (a) Use the Lorentz transformations to show that the \(\mathbf{E}\) and \(\mathbf{B}\) fields due to a point charge moving with constant velocity \(\mathbf{v}\) are related by \(\mathbf{B}=\left(\mathbf{v} / c^{2}\right) \times \mathbf{E}\) (b) If \(v \ll c\), then \(\mathbf{E}\) is essentially obtained from Coulomb's law, and \(\mathbf{B}\) can be calculated from the Biot-Savart law. Calculate \(\mathbf{B}\) this way, and then verify that it satisfies \(\mathbf{B}=\left(\mathbf{v} / c^{2}\right) \times\) E. (It may be helpful to think of the point charge as a tiny rod of charge, in order to get a handle on the \(d l\) in the BiotSavart law.)

Far field from a square loop ** Consider a square loop with current \(I\) and side length \(a\). The goal of this problem is to determine the magnetic field at a point a large, distance \(r\) (with \(r \gg a\) ) from the loop. (a) At the distant point \(P\) in Fig. 6.36, the two vertical sides give essentially zero Biot-Savart contributions to the field, because they are essentially parallel to the radius vector to \(P\). What are the Biot-Savart contributions from the two horizontal sides? These are easy to calculate because every little interval in these sides is essentially perpendicular to the radius vector to \(P\). Show that the sum (or difference) of these contributions equals \(\mu_{0} I a^{2} / 2 \pi r^{3}\), to leading order in \(a\). (b) This result of \(\mu_{0} I a^{2} / 2 \pi r^{3}\) is not the correct field from the loop at point \(P\). The correct field is half of this, or \(\mu_{0} I a^{2} / 4 \pi r^{3} .\) We will eventually derive this in Chapter 11, where we will show that the general result is \(\mu_{0} I A / 4 \pi r^{3}\), where \(A\) is the area of a loop with arbitrary shape. But we should be able to calculate it via the Biot-Savart law. Where is the error in the reasoning in part (a), and how do you go about fixing it? This is a nice one - don't peek at the answer too soon!

Constant magnitude of \(B\) ** How should the current density inside a thick cylindrical wire depend on \(r\) so that the magnetic field has constant magnitude inside the wire?

Field at the center of a disk * A disk with radius \(R\) and surface charge density \(\sigma\) spins with angular frequency \(\omega\). What is the magnetic field at the center?

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