/*! 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 66 A particle of charge \(q > 0\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A particle of charge \(q > 0\) is moving at speed \(v\) in the \(+ z\) -direction through a region of uniform magnetic field \(\vec { \boldsymbol { B } }\) . The magnetic force on the particle is \(\vec { \boldsymbol { F } } = F _ { 0 } ( 3 \hat { \boldsymbol { \imath } } + 4 \hat { \boldsymbol { J } } ) ,\) where \(F _ { 0 }\) is a positive constant. (a) Determine the components \(B _ { x } , B _ { y } ,\) and \(B _ { z }\) , or at least as many of the three components as is possible from the information given. (b) If it is given in addition that the magnetic field has magnitude \(6 F _ { 0 } / q v ,\) determine as much as you can about the remaining components of \(\vec { B } .\)

Short Answer

Expert verified
\(B_x = -\frac{4F_0}{qv}\), \(B_y = \frac{3F_0}{qv}\), \(B_z = \pm \frac{\sqrt{11}F_0}{qv}\)."

Step by step solution

01

Understand the Magnetic Force Equation

The magnetic force on a moving charge is given by the equation \( \vec{F} = q (\vec{v} \times \vec{B}) \), where \( \vec{v} \) is the velocity vector in the \( +z \)-direction and \( \vec{B} \) represents the magnetic field vector. Since the particle is moving in the \( z \)-direction, we write: \( \vec{v} = v \hat{k} \). The given force is \( \vec { oldsymbol { F } } = F _ { 0 } ( 3 \hat { oldsymbol { \i } } + 4 \hat { oldsymbol { J } } ) \).
02

Apply the Cross Product

Calculate \( \vec{v} \times \vec{B} \). With \( \vec{v} = v \hat{k} \) and \( \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \), the cross product is: \( \vec{v} \times \vec{B} = v (B_x \hat{k} \times \hat{i} + B_y \hat{k} \times \hat{j} + B_z \hat{k} \times \hat{k}) \). Using the right-hand rule, \( \hat{k} \times \hat{i} = -\hat{j} \), \( \hat{k} \times \hat{j} = \hat{i} \), and \( \hat{k} \times \hat{k} = \vec{0} \). Thus, \( \vec{v} \times \vec{B} = v (-B_x \hat{j} + B_y \hat{i}) \).
03

Equate to the Given Force

Equating the calculated cross product to the given force, we have: \( qv(-B_x \hat{j} + B_y \hat{i}) = F_0 (3 \hat{i} + 4 \hat{j}) \). Comparing components, we get \(-qvB_x = 4F_0 \) and \( qvB_y = 3F_0 \).
04

Solve for \(B_x \) and \(B_y \)

From the equations derived:1. \( B_x = -\frac{4F_0}{qv} \)2. \( B_y = \frac{3F_0}{qv} \)
05

Analyzing the Magnitude of \( \vec{B} \)

Given the magnitude of \( \vec{B} = \frac{6F_0}{qv} \), we use the formula for the magnitude of a vector: \( |\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2} \). Substitute the known components:\[\sqrt{\left(-\frac{4F_0}{qv}\right)^2 + \left(\frac{3F_0}{qv}\right)^2 + B_z^2} = \frac{6F_0}{qv}.\]
06

Solve for \( B_z \)

Squaring both sides of the magnitude equation: \[\left(\frac{4F_0}{qv}\right)^2 + \left(\frac{3F_0}{qv}\right)^2 + B_z^2 = \left(\frac{6F_0}{qv}\right)^2.\] Calculate and solve: \( \frac{16F_0^2}{q^2v^2} + \frac{9F_0^2}{q^2v^2} + B_z^2 = \frac{36F_0^2}{q^2v^2} \) \( B_z^2 = \frac{36F_0^2}{q^2v^2} - \frac{25F_0^2}{q^2v^2} \) \( B_z^2 = \frac{11F_0^2}{q^2v^2} \) \( B_z = \pm \frac{\sqrt{11}F_0}{qv} \)

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.

Particle Charge
In the world of electromagnetism, the concept of a particle charge is crucial. A particle, such as an electron or proton, carries a charge, often denoted as \( q \). The charge can be either positive or negative, determining how the particle interacts with electric and magnetic fields. In the exercise, we consider a particle with a positive charge, i.e., \( q > 0 \).

Understanding the charge of a particle helps in predicting the direction and magnitude of forces that act on it when subjected to magnetic or electric fields.
- Positive charges are repelled by other positive charges and attracted to negative ones.- The magnetic force on a charged particle also depends on its velocity, as it experiences a force perpendicular to both its motion and the magnetic field lines.

The relationship between the charge, velocity, and magnetic field is vital in producing a magnetic force that follows the equation: \[ \vec{F} = q(\vec{v} \times \vec{B}) \]This equation underlines how the charge interacts with the magnetic field through the velocity vector, making understanding particle charge foundational in electromagnetism.
Magnetic Field Components
Magnetic fields can be broken down into components along the coordinate axes: \( B_x \), \( B_y \), and \( B_z \). These components describe the magnetic field's strength and direction in space.

In the current scenario, we have a magnetic field expressed in the format: \( \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \)The task is to find these components using the information given.

When the particle moves solely in the \(+z\) direction (with velocity \(v\hat{k}\)), and a magnetic force expressed as \( \vec{F} = F_0(3 \hat{i} + 4 \hat{j}) \) acts on it:
  • The component \(B_x\) impacts the force in the \(\hat{j}\) direction.
  • Similarly, \(B_y\) affects the force in the \(\hat{i}\) direction.
  • Interestingly, \(B_z\) does not contribute since the cross product with the same direction \(\hat{k}\) results in zero.
Understanding magnetic field components is key to determining how fields interact with charged particles and can help explain phenomena like the Lorentz force, crucial in many technological applications ranging from MRI machines to particle accelerators.
Cross Product Calculation
One of the cornerstones in dealing with forces in electromagnetic fields is the cross product, which involves the interaction between vectors. Here, the force that a magnetic field exerts on a moving charged particle is dictated by this operation.

The cross product \( \vec{v} \times \vec{B} \), involves the velocity vector \( \vec{v} = v \hat{k} \) and the magnetic field vector \( \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \). Let's break this down step-by-step:

- The term \( \hat{k} \times \hat{i} \) simplifies to \(-\hat{j} \).- The term \( \hat{k} \times \hat{j} \) results in \(\hat{i}\).- Finally, \( \hat{k} \times \hat{k} \) yields \(\vec{0}\).
Thus, the cross product is:\[ \vec{v} \times \vec{B} = v(-B_x \hat{j} + B_y \hat{i}) \]Evaluating this product allows us to solve for the unknown components of the magnetic field, as by equating it to the given magnetic force \( \vec{F} = F_0(3 \hat{i} + 4 \hat{j}) \), we derive the necessary expressions to find \(B_x\) and \(B_y\).

The cross product ultimately gives a vector perpendicular to the plane formed by the interacting vectors, providing insight into the direction and magnitude of forces applied during the interaction. This concept is broadly applicable in physics and engineering, making it an essential tool in understanding complex systems.

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 cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconducting electromagnet of the cyclotron produces a 2.9 -T magnetic field perpendicular to the proton orbits. (a) When the protons have achieved a kinetic energy of 2.7\(\mathrm { MeV }\) what is the radius of their circular orbit and what is their angular speed? (b) Repeat part (a) when the protons have achieved their final kinetic energy of 5.4\(\mathrm { MeV }\) .

A coil with magnetic moment 1.45\(\mathrm { A } \cdot \mathrm { m } ^ { 2 }\) is oriented initially with its magnetic moment antiparallel to a uniform \(0.835 - \mathrm { T }\) magnetic field. What is the change in potential energy of the coil when it is rotated \(180 ^ { \circ }\) so that its magnetic moment is parallel to the field?

The magnetic poles of a small cyclotron produce a magnetic field with magnitude 0.85\(\mathrm { T }\) . The poles have a radius of \(0.40 \mathrm { m } ,\) which is the maximum radius of the orbits of the accelerated particles. (a) What is the maximum energy to which protons \(\left( q = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m = 1.67 \times 10 ^ { - 27 } \mathrm { kg } \right)\) can be accelerated by this cyclotron? Give your answer in electron volts and in joules. (b) What is the time for one revolution of a proton orbiting at this maximum radius? (c) What would the magnetic-field magnitude have to be for the maximum energy to which a proton can be accelerated to be twice that calculated in part (a)? For \(B = 0.85 \mathrm { T } ,\) what is the maximum energy to which alpha particles \(\left( q = 3.20 \times 10 ^ { - 19 } \mathrm { C } , m = 6.65 \times 10 ^ { - 27 } \mathrm { kg } \right)\) can be accelerated by this cyclotron? How does this compare to the maximum energy for protons?

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section \(38.5 ) ,\) in the lowest energy state the electron orbits the proton at a speed of \(2.2 \times\) \(10 ^ { 6 } \mathrm { m } / \mathrm { s }\) in a circular orbit of radius \(5.3 \times 10 ^ { - 11 } \mathrm { m } .\) (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I\) (c) What is the magnetic moment of the atom due to the motion of the electron?

A particle with charge \(9.45 \times 10 ^ { - 8 } \mathrm { C }\) is moving in a region where there is a uniform magnetic field of 0.650 T in the \(+ x\) -direction. At a particular instant of time the velocity of the particle has components \(v _ { x } = - 1.68 \times 10 ^ { 4 } \mathrm { m } / \mathrm { s } , v _ { y } = - 3.11 \times\) \(10 ^ { 4 } \mathrm { m } / \mathrm { s } ,\) and \(v _ { z } = 5.85 \times 10 ^ { 4 } \mathrm { m } / \mathrm { s } .\) What are the components of the force on the particle at this time?

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.