/*! 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 A particle with mass \(1.81 \tim... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A particle with mass \(1.81 \times 10^{-3} \mathrm{kg}\) and a charge of \(1.22 \times\) \(10^{-8} \mathrm{C}\) has, at a given instant, a velocity \(\vec{v}=\left(3.00 \times 10^{4} \mathrm{m} / \mathrm{s}\right) \hat{j}\) What are the magnitude and direction of the particle's accoleration produced by a uniform magnetic field \(\vec{B}=(1.63 \mathrm{T}) \hat{\imath}+\) \((0.980 \mathrm{T}) \hat{\jmath} ?\)

Short Answer

Expert verified
The acceleration is 0.330 m/s² in the negative z-axis direction.

Step by step solution

01

Identify the Known Values

Start by listing all the known values we have from the problem:- Mass of the particle, \( m = 1.81 \times 10^{-3} \text{ kg} \)- Charge of the particle, \( q = 1.22 \times 10^{-8} \text{ C} \)- Velocity of the particle, \( \vec{v}=(3.00 \times 10^{4} \mathrm{m} / \mathrm{s}) \hat{j} \)- Magnetic field, \( \vec{B}=(1.63 \mathrm{T}) \hat{\imath}+(0.980 \mathrm{T}) \hat{\jmath} \)
02

Formula for Magnetic Force

We use the Lorentz force formula to find the magnetic force on a charged particle in a magnetic field: \[ \vec{F} = q(\vec{v} \times \vec{B}) \]This force will then be used to find the acceleration of the particle.
03

Cross Product Calculation

Calculate the cross product \( \vec{v} \times \vec{B} \) using the components:1. \( \vec{v} = (0)\hat{i} + (3.00 \times 10^{4})\hat{j} + (0)\hat{k} \)2. \( \vec{B} = (1.63)\hat{i} + (0.980)\hat{j} + (0)\hat{k} \)The cross product is:\[\vec{v} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \0 & 3.00 \times 10^{4} & 0 \1.63 & 0.980 & 0\end{vmatrix}= (0)\hat{i} + (0)\hat{j} + (-4.89 \times 10^{4})\hat{k} \]
04

Calculate the Magnetic Force

Substitute the cross product into the Lorentz force formula:\[ \vec{F} = (1.22 \times 10^{-8} \text{ C}) \times (0 \hat{i} + 0 \hat{j} - 4.89 \times 10^{4} \hat{k}) \]This gives \[ \vec{F} = (0) \hat{i} + (0) \hat{j} + (-5.96 \times 10^{-4} \text{ N}) \hat{k} \]
05

Calculate the Acceleration

To find the acceleration, use Newton's second law, \( \vec{F} = m\vec{a} \).\[ \vec{a} = \frac{\vec{F}}{m} = \frac{-5.96 \times 10^{-4} \hat{k}}{1.81 \times 10^{-3}} \]This results in:\[ \vec{a} = 0 \hat{i} + 0 \hat{j} - 0.330 \hat{k} \text{ m/s}^2\]
06

Determine the Magnitude and Direction

The magnitude of the acceleration vector is calculated as:\[ |\vec{a}| = \sqrt{(0)^2 + (0)^2 + (-0.330)^2} = 0.330 \, \text{m/s}^2 \]Since only the \(\hat{k}\) component is present, the direction of the acceleration is along the negative \(z\)-axis.

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 91Ó°ÊÓ!

Key Concepts

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

Lorentz Force
The concept of Lorentz force is central to understanding how charged particles behave in electromagnetic fields. It is the force exerted on a charged particle moving within these fields. The formula for Lorentz force is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), where \(q\) is the charge of the particle, \(\vec{v}\) is its velocity, and \(\vec{B}\) represents the magnetic field. This formula shows that the force depends on the velocity and magnetic field's orientation.
The cross product \(\vec{v} \times \vec{B}\) implies that the force is perpendicular to both the velocity of the particle and the magnetic field. This perpendicularity is a key point, meaning the particle's path will curve rather than align directly along any one of those vectors. The Lorentz force does not work in isolation; rather, it is a fundamental component when calculating acceleration in magnetic contexts. Understanding this concept is crucial for solving many physics problems involving charged particles.
Cross Product
The cross product in vector mathematics allows us to calculate a vector that is orthogonal to two given vectors. In the context of magnetic forces, this operation is used to find the direction of the force acting on a charged particle within a magnetic field.
Given vectors \(\vec{v} = (0)\hat{i} + (3.00 \times 10^{4})\hat{j} + (0)\hat{k}\) and \(\vec{B} = (1.63)\hat{i} + (0.980)\hat{j} + (0)\hat{k}\), the cross product \(\vec{v} \times \vec{B}\) is evaluated using a determinant:
  • Place vectors in a matrix along with unit vectors \(\hat{i}, \hat{j}, \hat{k}\)
  • Determinant calculation gives \((0)\hat{i} + (0)\hat{j} + (-4.89 \times 10^{4})\hat{k}\)
Here, the resulting vector from the cross product indicates the direction of the magnetic force, showing it acts along the negative \(z\)-axis. This vector operation is essential for interpreting the physical direction of forces in fields.
Acceleration Calculation
Once the force on the particle is determined using the Lorentz force law, the next step is to calculate the particle's acceleration. According to Newton's second law, acceleration \(\vec{a}\) is determined by dividing the force \(\vec{F}\) by the mass \(m\) of the particle: \(\vec{a} = \frac{\vec{F}}{m}\).
In this problem, \(\vec{F} = (0)\hat{i} + (0)\hat{j} + (-5.96 \times 10^{-4})\hat{k}\) newtons, and with \(m = 1.81 \times 10^{-3}\) kg, the acceleration is calculated as:
  • \(\vec{a} = \frac{-5.96 \times 10^{-4} \hat{k}}{1.81 \times 10^{-3}}\)
  • \(\vec{a} = 0 \hat{i} + 0 \hat{j} - 0.330 \hat{k}\, \text{m/s}^2\)
This provides the acceleration's magnitude and confirms its direction is aligned along the negative \(z\)-axis. Understanding this step shows how forces relate to changes in motion, a pivotal concept in mechanics.
Physics Problem Solving
Physics problem solving involves breaking down each part of a complex scenario into manageable steps. For this magnetic force problem, the process is as follows:
  • Identify and list known quantities like mass, charge, and velocity.
  • Apply key formulas such as the Lorentz force equation to find forces.
  • Carry out vector operations like cross products for force direction.
  • Use Newton’s laws to relate force and acceleration.
Each of these steps builds on the last, consolidating understanding and leading to a solution. This systematic approach aids in tackling similar physics problems, fostering deeper insights into the behavior of systems under different forces.

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 wire 25.0 \(\mathrm{cm}\) long lies along the \(z\) -axis and carries a current of 9.00 \(\mathrm{A}\) in the \(+z\) -direction. The magnetic field is uniform and has components \(B_{x}=-0.242 \mathrm{T}, B_{y}=-0.985 \mathrm{T}\) , and \(B_{z}=\) \(-0.336 \mathrm{T} .\) (a) Find the components of the magnetic force on the wire. (b) What is the magnitude of the net magnetic force on the wire?

In the electron gun of a TV picture tube the electrons (charge \(-e,\) mass \(m )\) are accelerated by a voltage \(V .\) After leaving the electron gun, the electron beam travels a distance \(D\) to the screen; in this region there is a transverse magnetic field of magnitude \(B\) and no electric field. (a) Sketch the path of the electron beam in the tube. (b) Show that the approximate deflection of the beam due to this magnetic field is $$ d=\frac{B D^{2}}{2} \sqrt{\frac{e}{2 m V}} $$ (Hint: Place the origin at the center of the electron beam's arc and compare an undefiected beam's path to the deflected beam's path.) (c) Evaluate this expression for \(V=750 \mathrm{V}, D=50 \mathrm{cm},\) and \(B=5.0 \times 10^{-5} \mathrm{T}\) (comparable to the earth's field). Is this deflection significant?

An electromagnet produces a magnetic field of 0.550 T in a cylindrical region of radius 2.50 \(\mathrm{cm}\) between its poles. A straight wire carrying a current of 10.8 \(\mathrm{A}\) passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force is exerted on the wire?

A circular ring with area \(4.45 \mathrm{~cm}^{2}\) is carrying a current of 12.5 A. The ring is free to rotate about a diameter. The ring, initially at rest, is immersed in a region of uniform magnetic field given by \(\vec{B}=\left(1.15 \times 10^{-2} \mathrm{~T}\right)(12 \hat{\imath}+3 \hat{\jmath}-4 \hat{k}) .\) The ring is positioned initially such that its magnetic moment is given by \(\vec{\mu}_{1}=\mu(-0.800 \hat{\imath}+0.600 \hat{\jmath}),\) where \(\mu\) is the (positive) magnitude of the magnetic moment. The ring is released and turns through an angle of \(90.0^{\circ},\) at which point its magnetic moment is given by \(\vec{\mu}_{f}=-\mu \hat{k} .\) (a) Determine the decrease in potential energy. (b) If the moment of inertia of the ring about a diameter is \(8.50 \times 10^{-7} \mathrm{~kg} \cdot \mathrm{m}^{2},\) determine the angular speed of the ring as it passes through the second position.

Force on a Current Loop in a Nonuniform Magnetic Field. It was shown in Section 27.7 that the net force on a current loop in a uniform magnetic field is zero. But what if \(\overrightarrow{\boldsymbol{B}}\) is not uniform? Figure 27.70 shows a square loop of wire that ties in the \(x y\) -plane. The loop has corners at \((0,0),(0, L),(L, 0)\) , and \((L, L)\) and carries a constant current \(I\) in the clockwise direction. The magnetic field has no \(x\) -component but has both \(y-\) and \(z\) -components: \(\vec{B}=\) \(\left(B_{0} z / L\right) \hat{j}+\left(B_{0} y / L\right) \hat{k},\) where \(B_{0}\) is a positive constant. (a) Sketch the magnetic field lines in the \(y z\) -plane. (b) Find the magnitude and direction of the magnetic force exerted on each of the sides of the loop by integrating \(\mathrm{Eq} .(27.20) .\) (c) Find the magnitude and direction of the net magnetic force on the loop.

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.