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Chapter 27: Problem 6

An electron moves at \(1.40 \times 10^{6} \mathrm{~m} / \mathrm{s}\) through a region in which there is a magnetic field of unspecified direction and magnitude \(7.40 \times 10^{-2} \mathrm{~T}\). (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

Short Answer

Expert verified
The largest possible acceleration of the electron due to the magnetic field is \( 1.76 \times 10^{11} m/s^2 \) and smallest possible acceleration is 0. The angle between the electron velocity and the magnetic field when the acceleration is one-fourth of the maximum is approximately \( 14.48 \, degrees \).

Step by step solution

01

Calculate the maximum acceleration

We first need to calculate maximum acceleration of the electron. We know that \( F = ma \), where m is the mass of electron, \( 9.11 \times 10^{-31} kg \), and \( F = q(vB \sin \theta) \). At maximum acceleration, \( \sin\theta = 1 \). Therefore we can calculate \( a_{max} = qvB/m \). Plug in the values for q, v, B, and m, this gives us: \( a_{max} = 1.76 \times 10^{11} m/s^2 \).
02

Identify the minimum acceleration

The smallest possible magnitude for the acceleration is 0, when the velocity and the magnetic field are parallel, \( \sin\theta = 0 \). This is because there will be no force acting on the electron due to magnetic field if the electron moves parallel to the field.
03

Calculate the angle

In part b, we are given that the actual acceleration is \( a = a_{max}/4 \). We need to calculate the angle between the velocity and magnetic field. As we know \( F = q(vB \sin \theta) \), and \( F = ma \) therefore \( ma = qvB\sin \theta \). We can rearrange this to \( sin \theta = ma/qvB\) . Substituting given values, we will get \( \sin\theta = 0.25 \). The angle \( \theta = \arcsin(0.25) = 14.48 \, degrees \).

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Key Concepts

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

Electron Motion
When discussing an electron's motion, it involves understanding how electrons travel through space. Electrons are tiny particles with a negative charge and very small mass鈥攁pproximately \(9.11 \times 10^{-31} \mathrm{~kg}\). In our scenario, the electron is moving at a velocity of \(1.40 \times 10^{6} \mathrm{~m/s}\). One important thing to remember is that when electrons move, they create an electric current, which can interact with magnetic fields. This interaction affects the path that the electron takes.
  • An electron's velocity is a vector quantity, meaning it has both magnitude and direction.
  • Changes in these vectors can result from interactions with forces like magnetic fields, causing the electron to accelerate or decelerate.
It's crucial to know how this motion will change when under the influence of other forces, especially when discussing concepts such as magnetic induction or acceleration.
Magnetic Field
A magnetic field is an invisible field that exerts a force on certain particles like charged particles. The magnetic field here is specified as \(7.40 \times 10^{-2} \mathrm{~T}\), where \(\mathrm{T}\) stands for Tesla, the unit of magnetic field strength. This field can either pull or push on charged particles, like electrons, creating a motion that's perpendicular to both the field and the direction of the particle's velocity.
  • A magnetic field is described by its direction and magnitude.
  • Unlike electric fields, magnetic fields do not do work on the charged particle, hence they do not change the kinetic energy of the particle but only change its direction.
This characteristic leads to the fascinating phenomena of circular or spiral paths taken by charged particles under the influence of a uniform magnetic field.
Acceleration
Acceleration refers to the change in velocity of a particle, which can be due to speed or direction change. When an electron enters a magnetic field, it experiences a magnetic force that changes its velocity's direction, thus causing an acceleration. The force acting on the electron is given by Lorentz force, expressed as \( F = q(vB \sin \theta) \), where \( q \) is the charge of the electron, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field.
  • Maximum acceleration occurs when \( \sin \theta = 1 \) (or \( \theta = 90掳 \)), meaning the force is at its strongest as the electron moves perpendicular to the magnetic field.
  • Minimum acceleration, often zero, happens when the velocity and magnetic field are parallel (\( \sin \theta = 0 \)), as no force acts on the electron.
Understanding this helps us analyze the paths of charged particles in fields, which is foundational in technologies like cyclotrons and other particle accelerators.

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Most popular questions from this chapter

A circular loop of wire with area \(A\) lies in the \(x y\) -plane. As viewed along the \(z\) -axis looking in the \(-z\) -direction toward the origin, a current \(I\) is circulating clockwise around the loop. The torque produced by an external magnetic field \(\overrightarrow{\boldsymbol{B}}\) is given by \(\overrightarrow{\boldsymbol{\tau}}=D(4 \hat{\imath}-3 \hat{\jmath}),\) where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U=-\overrightarrow{\boldsymbol{\mu}} \cdot \overrightarrow{\boldsymbol{B}}\) is negative. The magnitude of the magnetic field is \(B_{0}=13 D / I A\). (a) Determine the vector magnetic moment of the current loop. (b) Determine the components \(B_{x}, B_{y}\) and \(B_{z}\) of \(\overrightarrow{\boldsymbol{B}}\)

A particle with a charge of \(-1.24 \times 10^{-8} \mathrm{C}\) is moving with instantaneous velocity \(\overrightarrow{\boldsymbol{v}}=\left(4.19 \times 10^{4} \mathrm{~m} / \mathrm{s}\right) \hat{\imath}+\left(-3.85 \times 10^{4} \mathrm{~m} / \mathrm{s}\right) \hat{\jmath}\) What is the force exerted on this particle by a magnetic field (a) \(\overrightarrow{\boldsymbol{B}}=(1.40 \mathrm{~T}) \hat{\imath}\) and (b) \(\overrightarrow{\boldsymbol{B}}=(1.40 \mathrm{~T}) \hat{\boldsymbol{k}} ?\)

Point \(a\) is on the \(+y\) -axis at \(y=+0.200 \mathrm{~m}\) and point \(b\) is on the \(+x\) -axis at \(x=+0.200 \mathrm{~m}\). A wire in the shape of a circular arc of radius \(0.200 \mathrm{~m}\) and centered on the origin goes from \(a\) to \(b\) and carries current \(I=5.00 \mathrm{~A}\) in the direction from \(a\) to \(b\). (a) If the wire is in a uniform magnetic field \(B=0.800 \mathrm{~T}\) in the \(+z\) -direction, what are the magnitude and direction of the net force that the magnetic field exerts on the wire segment? (b) What are the magnitude and direction of the net force on the wire if the field is \(B=0.800 \mathrm{~T}\) in the \(+x\) -direction?

A particle with charge \(2.15 \mu \mathrm{C}\) and mass \(3.20 \times 10^{-11} \mathrm{~kg}\) is initially traveling in the \(+y\) -direction with a speed \(v_{0}=1.45 \times 10^{5} \mathrm{~m} / \mathrm{s} .\) It then enters a region containing a uniform magnetic field that is directed into, and perpendicular to, the page in Fig. \(\mathrm{P} 27.81 .\) The magnitude of the field is \(0.420 \mathrm{~T}\). The region extends a distance of \(25.0 \mathrm{~cm}\) along the initial direction of travel; \(75.0 \mathrm{~cm}\) from the point of entry into the magnetic field region is a wall. The length of the field-free region is thus \(50.0 \mathrm{~cm} .\) When the charged particle enters the magnetic field, it follows a curved path whose radius of curvature is \(R .\) It then leaves the magnetic field after a time \(t_{1},\) having been deflected a distance \(\Delta x_{1} .\) The particle then travels in the field-free region and strikes the wall after undergoing a total deflection \(\Delta x\). (a) Determine the radius \(R\) of the curved part of the path. (b) Determine \(t_{1}\), the time the particle spends in the magnetic field. (c) Determine \(\Delta x_{1}\), the horizontal deflection at the point of exit from the field. (d) Determine \(\Delta x\), the total horizontal deflection.

A Cycloidal Path. A particle with mass \(m\) and positive charge \(q\) starts from rest at the origin shown in Fig. \(\mathbf{P 2 7 . 8 2 .}\). There is a uniform electric field \(\vec{E}\) in the \(+y\) -direction and a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}\) directed out of the page. It is shown in more advanced books that the path is a cycloid whose radius of curvature at the top points is twice the \(y\) -coordinate at that level. (a) Explain why the path has this general shape and why it is repetitive. (b) Prove that the speed at any point is equal to \(\sqrt{2 q E y / m}\). (Hint: Use energy conservation.) (c) Applying Newton's second law at the top point and taking as given that the radius of curvature here equals \(2 y,\) prove that the speed at this point is \(2 E / B\)
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