/*! 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 6 A proton moves through a uniform... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 6

A proton moves through a uniform magnetic field given by \(\vec{B}=(10 \hat{\mathrm{i}}-20 \hat{\mathrm{j}}+30 \hat{\mathrm{k}}) \mathrm{mT} .\) At time \(t_{1},\) the proton has a velocity given by \(\vec{v}_{\rightarrow}=v_{x} \hat{i}+v_{y} \hat{j}+(2.0 \mathrm{~km} / \mathrm{s}) \hat{\mathrm{k}}\) and the magnetic force on the proton is \(\vec{F}_{B}=\left(4.0 \times 10^{-17} \mathrm{~N}\right) \hat{\mathrm{i}}+\left(2.0 \times 10^{-17} \mathrm{~N}\right) \mathrm{j} .\) At that instant, what are (a) \(v_{x}\) and (b) \(v_{y} ?\)

Short Answer

Expert verified
(a) \(v_x = -4833.33\) m/s; (b) \(v_y = 7000\) m/s.

Step by step solution

01

Understand the Formula for Magnetic Force

The magnetic force \( \vec{F}_B \) on a charged particle moving in a magnetic field \( \vec{B} \) with velocity \( \vec{v} \) is given by the cross product \( \vec{F}_B = q (\vec{v} \times \vec{B}) \). Here, \( q \) is the charge of a proton, which is \( 1.6 \times 10^{-19} \) C.
02

Set Up the Cross Product Equation

Express the velocity vector as \( \vec{v} = v_x \hat{i} + v_y \hat{j} + 2000 \hat{k} \) m/s and the magnetic field as \( \vec{B} = 10 \hat{i} - 20 \hat{j} + 30 \hat{k} \) mT, which is equivalent to \( \vec{B} = 10^{-2} \hat{i} - 2 \times 10^{-2} \hat{j} + 3 \times 10^{-2} \hat{k} \) T. Calculate \( \vec{v} \times \vec{B} \).
03

Calculate the Components of Cross Product

Use the determinant method to find \( \vec{v} \times \vec{B} \): \[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ v_x & v_y & 2000 \ 0.01 & -0.02 & 0.03 \end{vmatrix} \]This simplifies to \[ \vec{v} \times \vec{B} = (v_y \cdot 0.03 - 2000 \cdot (-0.02))\hat{i} - (v_x \cdot 0.03 - 2000 \cdot 0.01)\hat{j} + (v_x \cdot (-0.02) - v_y \cdot 0.01)\hat{k} \].
04

Equate Force Components

For the \( \hat{i} \) component of \( \vec{F}_B \): \[ q(v_y \cdot 0.03 + 40) = 4.0 \times 10^{-17} \]Substitute \( q = 1.6 \times 10^{-19} \) C to solve for \( v_y \). For the \( \hat{j} \) component:\[ q(-v_x \cdot 0.03 - 20) = 2.0 \times 10^{-17} \]Solve for \( v_x \).
05

Solve for \( v_y \)

Start from the equation: \[ 1.6 \times 10^{-19} (v_y \cdot 0.03 + 40) = 4.0 \times 10^{-17} \]Simplify to get: \[ v_y \cdot 0.03 + 40 = 250 \ v_y \cdot 0.03 = 210 \ v_y = 7000 \text{ m/s} \].
06

Solve for \( v_x \)

Use the equation: \[ 1.6 \times 10^{-19} (-v_x \cdot 0.03 - 20) = 2.0 \times 10^{-17} \]Reorganize to obtain: \[ -v_x \cdot 0.03 - 20 = 125 \ -v_x \cdot 0.03 = 145 \ v_x = -\frac{145}{0.03} \approx -4833.33 \text{ m/s} \].

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.

Cross Product in Physics
In physics, one of the critical mathematical operations used to analyze forces acting on particles in magnetic fields is the cross product. The cross product between two vectors results in a new vector, which is perpendicular to the plane containing the original vectors.

When it comes to magnetic forces, the cross product helps determine the direction and magnitude of the force acting on a charged particle moving through a magnetic field. The formula for magnetic force \(/\(\vec{F}_B = q (\vec{v} \times \vec{B})\)\\) captures this concept. Here, \( q \) is the charge of the particle, \( \vec{v} \) is its velocity vector, and \( \vec{B} \) represents the magnetic field.
  • The right-hand rule is often used to quickly find the direction of the resultant vector from a cross product.
  • In the context of our exercise, the cross product\( \(\vec{v} \times \vec{B}\) \\) determines the magnetic force direction on a proton, providing insight into how velocity components \( v_x \) and \( v_y \) influence this force.
By understanding the cross product, students can better follow the calculations that lead to describing forces such as those in this exercise.
Proton Velocity in Magnetic Fields
The velocity of a proton moving through a magnetic field is essential in determining the magnetic force experienced by the proton.

In our scenario, the proton has a velocity vector described with components \( v_x \), \( v_y \), and a known \( z \)-component of 2000 m/s. The cross product of the velocity vector \( \vec{v} \) and the magnetic field vector \( \vec{B} \) results in the magnetic force. However, it's the interaction of the vector components of \( \vec{v} \) with \( \vec{B} \) that intricately influences the force.
  • The relationship between velocity and force is pivotal: variations in \( v_x \) and \( v_y \) will change the cross product with \( \vec{B} \), thus affecting \( \vec{F}_B \).
  • The steps involved in solving for \( v_x \) and \( v_y \) include simplifying the cross products and equating them to the known force product.
By isolating \( v_x \) and \( v_y \) in separate equations, you can solve for these unknowns with known magnetic force components, describing the proton鈥檚 motion accurately within the magnetic field.
Magnetic Field Vector Calculation
Calculating the magnetic field vector is a fundamental step in solving problems where forces on moving charges are considered.

The vector \( \vec{B} \) indicates the direction and magnitude of the field, typically broken down into three-dimensional components. In this exercise, \(\vec{B} \) was given in millitesla (mT) and converted to tesla (T), which included multiplying each component by \( 10^{-2} \).
  • Understanding and converting units correctly is critical for calculating other forces in physics, such as the magnetic force.
  • Magnetic field vectors determine how the paths of moving particles like protons are altered. The stronger the perpendicular component of the field to the velocity, the stronger the magnetic force experienced.
Precise calculation and manipulation of \( \vec{B} \) in vector form is crucial for advancing through the problem, leading to correct and comprehensible solutions of the forces and velocity components involved.

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

What uniform magnetic field, applied perpendicular to a beam of electrons moving at \(1.30 \times 10^{6} \mathrm{~m} / \mathrm{s},\) is required to make the electrons travel in a circular arc of radius \(0.350 \mathrm{~m} ?\)

An electron with kinetic energy \(2.5 \mathrm{keV}\) moving along the positive direction of an \(x\) axis enters a region in which a uniform electric field of magnitude \(10 \mathrm{kV} / \mathrm{m}\) is in the negative direction of the \(y\) axis. A uniform magnetic field \(\vec{B}\) is to be set up to keep the electron moving along the \(x\) axis, and the direction of \(\vec{B}\) is to be chosen to minimize the required magnitude of \(\vec{B}\). In unit-vector notation, what \(\vec{B}\) should be set up?

A proton, a deuteron \((q=+e, m=2.0 \mathrm{u}),\) and an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) are accelerated through the same potential difference and then enter the same region of uniform magnetic field \(\vec{B},\) moving perpendicular to \(\vec{B}\). What is the ratio of (a) the proton's kinetic energy \(K_{p}\) to the alpha particle's kinetic energy \(K_{a}\) and \((\mathrm{b})\) the deuteron's kinetic energy \(K_{d}\) to \(K_{a} ?\) If the radius of the proton's circular path is \(10 \mathrm{~cm},\) what is the radius of (c) the deuteron's path and (d) the alpha particle's path?

A cyclotron with dee radius \(53.0 \mathrm{~cm}\) is operated at an oscillator frequency of \(12.0 \mathrm{MHz}\) to accelerate protons. (a) What magnitude \(B\) of magnetic field is required to achieve resonance? (b) At that field magnitude, what is the kinetic energy of a proton emerging from the cyclotron? Suppose, instead, that \(B=1.57 \mathrm{~T}\). (c) What oscillator frequency is required to achieve resonance now? (d) At that frequency, what is the kinetic energy of an emerging proton?

A particle of mass \(6.0 \mathrm{~g}\) moves at \(4.0 \mathrm{~km} / \mathrm{s}\) in an \(x y\) plane, in a region with a uniform magnetic field given by \(5.0 \hat{\mathrm{i}} \mathrm{mT}\). At one instant, when the particle's velocity is directed \(37^{\circ}\) counterclockwise from the positive direction of the \(x\) axis, the magnetic force on the particle is \(0.48 \hat{\mathrm{k}} \mathrm{N}\). What is the particle's charge?
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Circular Motion and Gravitation

Read Explanation

Radiation

Read Explanation

Particle Model of Matter

Read Explanation

Scientific Method 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.