/*! 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 4 An alpha particle (charge + 2e) ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 4

An alpha particle (charge + 2e) and an electron move in opposite directions from the same point, each with the speed of \(2.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) (Fig. \(28.32 ) .\) Find the magnitude and direction of the total magnetic field these charges produce at point \(P,\) which is 1.75 \(\mathrm{nm}\) from each of them.

Short Answer

Expert verified
The total magnetic field is \(1.81 \times 10^{-9} \text{ T}\) directed into the page.

Step by step solution

01

Identify Charge and Velocities

The problem involves two particles: an alpha particle with a charge of \(+2e\) and an electron with a charge of \(-1e\). Both particles move with a speed of \(2.50 \times 10^5 \text{ m/s}\) in opposite directions.
02

Compute the Magnetic Field from the Alpha Particle

Using the Biot-Savart Law, the magnetic field due to a moving charge is given by:\[B_{\alpha} = \frac{\mu_0}{4\pi} \frac{qv \sin \theta}{r^2}\]where \(q = +2e\), \(v = 2.50 \times 10^5 \text{ m/s}\), \(\theta = 90^\circ\) (since perpendicular to \(\vec{r}\) from point P), and \(r = 1.75 \times 10^{-9} \text{ m}\). Calculate \(B_{\alpha}\).- Plugging in the values, you find that:\[B_{\alpha} = \frac{10^{-7}}{(1.75 \times 10^{-9})^2} \times 2 \times 1.6 \times 10^{-19} \times 2.50 \times 10^5 \approx 3.63 \times 10^{-9} \text{ T}\]
03

Compute the Magnetic Field from the Electron

Similarly, compute the magnetic field from the electron using:\[B_{e} = \frac{\mu_0}{4\pi} \frac{qv \sin \theta}{r^2}\]where \(q = -1e\), and other values remain the same. The electron contributes:\[B_{e} = \frac{10^{-7}}{(1.75 \times 10^{-9})^2} \times (-1.6 \times 10^{-19}) \times 2.50 \times 10^5 \approx -1.82 \times 10^{-9} \text{ T}\]
04

Find the Total Magnetic Field at Point P

Since the alpha particle and electron move in opposite directions, their magnetic fields at point P add up algebraically due to vector directions, knowing that both are perpendicular:\[B_{\text{total}} = B_{\alpha} + B_{e}\]Substitute the calculated values:\[B_{\text{total}} = 3.63 \times 10^{-9} + (-1.82 \times 10^{-9}) = 1.81 \times 10^{-9} \text{ T}\]
05

Determine the Direction

Using the right-hand rule, the magnetic field from a positive charge moving in a given direction circulates counterclockwise, and for a negative charge, it moves clockwise. At point P, the alpha particle's contribution would point into the page, while the electron's contribution points out of the page. The net result, due to the larger field from the alpha particle, points into the page.

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 91影视!

Key Concepts

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

Biot-Savart Law
The Biot-Savart Law is a fundamental principle used to calculate the magnetic field produced by a moving charge. It describes how the magnetic field \( B \) is established at a point in space, due to a specific charge moving with velocity \( v \). The formula for the magnetic field \( B \) produced by a moving charge is:
  • \[B = \frac{\mu_0}{4\pi} \frac{qv \sin \theta}{r^2} \]
In this equation:
  • \( \mu_0 \) is the permeability of free space.
  • \( q \) represents the charge.
  • \( v \) is the speed of the charge.
  • \( \theta \) is the angle between the velocity of the charge and the vector pointing from the charge to the point where the field is calculated.
  • \( r \) is the distance from the charge to the point of measurement.
The Biot-Savart Law helps in calculating the magnitude of the magnetic field, and understanding its direction can make various physics problems easier to solve. For the given exercise, the law was employed to determine the magnetic fields contributed by both an alpha particle and an electron.
Alpha Particle
An alpha particle is a type of nuclear particle consisting of two protons and two neutrons, giving it a net positive charge of \(+2e\). It is significantly more massive compared to a single proton or neutron. In our exercise, the movement of an alpha particle contributes to the magnetic field at a specific point.
  • The charge of the alpha particle is \(+2e\) which doubles the effect it might have in comparison to a single charge of \(+1e\).
  • The velocity given is \(2.50 \times 10^{5} \, \text{m/s} \), a relatively high speed for subatomic particles.
  • The direction is perpendicular to the radius vector pointing to the point where the magnetic field is measured, which means \(\theta = 90^\circ\).
Due to its larger charge and the high speed, the alpha particle contributes substantially to the magnetic field observed at point P. Its direction was found using the right-hand rule, which indicated that if moving in a specific direction, it would cause a magnetic field oriented into the page at the point of observation.
Electron Motion
An electron is a subatomic particle with a negative charge, denoted as \(-1e\). Due to its relatively smaller mass compared to an alpha particle, electrons are highly responsive to electric and magnetic fields.
  • In the exercise, the electron moves at a speed of \(2.50 \times 10^{5} \, \text{m/s} \), initially from the same point as the alpha particle but in the opposite direction.
  • The contribution to the magnetic field due to the electron is calculated using similar formulas to that of the alpha particle but accounting for its single negative charge.
  • This charge means the direction of its magnetic field contribution differs from the alpha particle, being opposite when viewed from the same perspective.
The calculated magnetic field influence from the electron was found to be \(-1.82 \times 10^{-9} \, \text{T}\), suggesting its field points out of the page. While the alpha particle's larger charge results in a stronger field, the electron still plays a crucial part in the total field at point P.
Right-Hand Rule
The right-hand rule is a useful mnemonic for determining the direction of a magnetic field relative to the direction of current or, in this case, the velocity of moving charges. It is applied by following these simple steps:
  • Point your thumb in the direction of the conventional current (positive to negative) or velocity of the positive charge.
  • Curl your fingers around; the direction in which your fingers curl is the direction of the magnetic field lines.
In the context of this exercise:
  • For the alpha particle moving in one direction, the right-hand rule showed that its field at point P points into the page.
  • The electron, being negatively charged and moving in the opposite direction, produces a magnetic field pointing out of the page.
The right-hand rule helps to easily visualize and determine the directions of these magnetic fields, which is imperative when calculating the resultant magnetic field at any given point. Being able to apply this rule correctly allows students to predict the influence of various currents or moving charges in multiple situations.

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

A magnetic field of 37.2 T has been achieved at the MTT Francis Bitter National Magnetic Laboratory. Find the current needed to achieve such a field (a) 2.00 \(\mathrm{cm}\) from a long, straight wire; \((b)\) at the center of a circular coil of radius 42.0 \(\mathrm{cm}\) that has 100 turns; \((\mathrm{c})\) near the center of a solenoid with radius \(2.40 \mathrm{cm},\) length \(32.0 \mathrm{cm},\) and \(40,000\) turns.

A closed curve encircles several conductors. The line integral \(\phi \overrightarrow{\boldsymbol{B}} \cdot d \vec{l}\) around this curve is \(3.83 \times 10^{-4} \mathrm{T} \cdot \mathrm{m} .\) (a) What is the net current in the conductors? (b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? Explain.

A long, straight, solid cylinder, oriented with its axis in the z-direction, carries a current whose current density is \(\overrightarrow{\boldsymbol{J}}\) . The current density, although symmetrical about the cylinder axis, is not constant and varies according to the relationship $$ \begin{array}{rlrl}{\overrightarrow{\boldsymbol{J}}} & {=\left(\frac{b}{r}\right) e^{(r-a) / \delta} \hat{\boldsymbol{k}}} & {} & {\text { for } r \leq a} \\ {} & {=\mathbf{0}} & {} & {\text { for } \boldsymbol{r} \geq a}\end{array} $$ where the radius of the cylinder is \(a=5.00 \mathrm{cm}, r\) is the radial distance from the cylinder axis, \(b\) is a constant equal to \(600 \mathrm{A} / \mathrm{m},\) and \(\delta\) is a constant equal to \(2.50 \mathrm{cm} .\) (a) Let \(I_{0}\) be the total current passing through the entire cross section of the wire. Obtain an expression for \(I_{0}\) in terms of \(b, \delta,\) and \(a .\) Evaluate your expression to obtain a numerical value for \(I_{0}\) . \((b)\) Using Ampere's law, derive an expression for the magnetic field \(\vec{B}\) in the region \(r \geq a\) . Express your answer in terms of \(I_{0}\) rather than \(b\) . (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. Express your answer in terms of \(I_{0}\) rather than \(b\) . (d) Using Ampere's law, derive an expression for the magnetic field \(\vec{B}\) in the region \(r \leq a\) (e) Evaluate the magnitude of the magnetic field at \(r=\delta, r=a,\) and \(r=2 a\) .

A long, straight, cylindrical wire of radius \(R\) carries a current uniformly distributed over its cross section. At what location is the magnetic field produced by this current equal to half of its largest value? Consider points inside and outside the wire.

A very long, straight horizontal wire carries a current such that \(3.50 \times 10^{18}\) electrons per second pass any given point going from west to east. What are the magnitude and direction of the magnetic field this wire produces at a point 4.00 \(\mathrm{cm}\) directly above it?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Physics of Motion

Read Explanation

Classical Mechanics

Read Explanation

Force

Read Explanation

Fundamentals of Physics

Read Explanation
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.