/*! 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 38 A particle of charge \(q>0\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle of charge \(q>0\) is moving at speed \(v\) in the \(+z\) -direction through a region of uniform magnetic field \(\overrightarrow{\boldsymbol{B}}\). The magnetic force on the particle is \(\overrightarrow{\boldsymbol{F}}=F_{0}(3 \hat{\boldsymbol{\imath}}+4 \hat{\boldsymbol{j}})\), 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 \(\overrightarrow{\boldsymbol{B}}\).

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 force exerted on a charged particle moving in a magnetic field is given by the Lorentz force equation: \( \overrightarrow{\boldsymbol{F}} = q\left(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}\right) \). Since the particle is moving in the \( +z \)-direction, \( \overrightarrow{\boldsymbol{v}} = v \hat{\boldsymbol{k}} \). The force on the particle is \( \overrightarrow{\boldsymbol{F}} = F_{0}(3 \hat{\boldsymbol{\imath}} + 4 \hat{\boldsymbol{j}}) \).
02

Apply Cross Product for Force Components

Using the cross product \( \overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}} \), we have:\[qv(B_{x}, B_{y}, B_{z}) \times \hat{\boldsymbol{k}} = qv(-B_{y} \hat{\boldsymbol{\imath}} + B_{x} \hat{\boldsymbol{j}}) = F_{0}(3 \hat{\boldsymbol{\imath}} + 4 \hat{\boldsymbol{j}})\]This gives us the equations:\(-qvB_{y} = F_{0} \times 3 \) and \( qvB_{x} = F_{0} \times 4 \).
03

Solve for Magnetic Field Components Bx and By

From the force components, solve for \( B_{x} \) and \( B_{y} \):\[ B_{x} = \frac{4F_{0}}{qv} \]\[ B_{y} = -\frac{3F_{0}}{qv} \]
04

Address the Magnitude of Magnetic Field B

Given that the magnetic field has magnitude \( |\overrightarrow{\boldsymbol{B}}| = \frac{6F_{0}}{qv} \), we use the equation for magnitude:\[ \sqrt{B_{x}^2 + B_{y}^2 + B_{z}^2} = \frac{6F_{0}}{qv} \]Substitute the known values for \( B_{x} \) and \( B_{y} \):\[ \sqrt{\left(\frac{4F_{0}}{qv}\right)^2 + \left(-\frac{3F_{0}}{qv}\right)^2 + B_{z}^2} = \frac{6F_{0}}{qv} \]
05

Solve for the Third Component Bz

Simplify and solve for \( B_{z} \):\[ \sqrt{\frac{16F_{0}^2}{q^2v^2} + \frac{9F_{0}^2}{q^2v^2} + B_{z}^2} = \frac{6F_{0}}{qv} \]This reduces to:\[ \sqrt{\frac{25F_{0}^2}{q^2v^2} + B_{z}^2} = \frac{6F_{0}}{qv} \]\[ \frac{25F_{0}^2}{q^2v^2} + B_{z}^2 = \frac{36F_{0}^2}{q^2v^2} \]Solve for \( B_{z}^2 \):\[ B_{z}^2 = \frac{11F_{0}^2}{q^2v^2} \]So, \( B_{z} = \pm\frac{\sqrt{11}F_{0}}{qv} \).

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Key Concepts

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

Lorentz Force
The Lorentz Force is a fundamental concept in physics that explains the force exerted on a charged particle when it moves through both electric and magnetic fields. This force is crucial for understanding how charges interact with magnetic environments. The equation for the Lorentz force is given by:
\( \overrightarrow{\boldsymbol{F}} = q\left(\overrightarrow{\boldsymbol{E}} + \overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}\right) \), where:
  • \( q \) is the charge of the particle.
  • \( \overrightarrow{\boldsymbol{E}} \) is the electric field vector.
  • \( \overrightarrow{\boldsymbol{v}} \) is the velocity of the particle.
  • \( \overrightarrow{\boldsymbol{B}} \) is the magnetic field vector.
  • \( \times \) denotes the cross product.
In situations where only a magnetic field is present, as in this exercise, the electric field \( \overrightarrow{\boldsymbol{E}} \) is zero. Thus, the equation simplifies to \( \overrightarrow{\boldsymbol{F}} = q\left(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}\right) \). This force can change the direction of the velocity of the particle, but not its speed. Therefore, the particle generally follows a circular or helical path, depending on its initial velocity relative to the magnetic field.
Understanding Lorentz Force is key to problems involving magnetic and electric forces acting on charged particles.
Magnetic Field Components
The magnetic field is a vector field around a magnetic body, or a moving electric charge, within which the force of magnetism acts. In physics, it is important to determine its components in various scenarios. Knowing the components allows for calculation of how the field interacts with charged particles.
In the exercise, the particle moves in the \( +z \) direction, and the magnetic force is specified. To find the magnetic field components, one can equate this to the components of the Lorentz force vector.
  • The force components give the equations:
  • \( -qvB_{y} = F_{0} \times 3 \) (giving \( B_{y} = -\frac{3 F_{0}}{qv} \)).
  • \( qvB_{x} = F_{0} \times 4 \) (giving \( B_{x} = \frac{4 F_{0}}{qv} \)).
When the magnitude of the magnetic field is provided, as in this case with \( \frac{6F_{0}}{qv} \), it allows solving for the unknown component \( B_{z} \). The constraint \( \sqrt{B_{x}^2 + B_{y}^2 + B_{z}^2} = \frac{6F_{0}}{qv} \) leads to calculating \( B_{z} \) and understanding the field better.
Magnetic field components are crucial for accurately describing and predicting the behavior of particles in various magnetic field configurations.
Cross Product in Physics
Cross product is an important mathematical operation in vector algebra, often used in physics to determine the relationship between vectors in three-dimensional space. It is especially useful in concepts like magnetic force and angular momentum.
The cross product \( \overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}} \) involves two vectors: the velocity of a particle and the magnetic field, and yields a new vector that indicates the direction of force acting on a charged particle. The magnitude of this new vector can be calculated by:
\[ |\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}| = |\overrightarrow{\boldsymbol{v}}||\overrightarrow{\boldsymbol{B}}|\sin\theta \]where \( \theta \) is the angle between \( \overrightarrow{\boldsymbol{v}} \) and \( \overrightarrow{\boldsymbol{B}} \).
  • Only components of \( \overrightarrow{\boldsymbol{v}} \) and \( \overrightarrow{\boldsymbol{B}} \) perpendicular to each other contribute to the result.
  • The direction follows the right-hand rule: if you point your fingers in the direction of \( \overrightarrow{\boldsymbol{v}} \) and curl them towards \( \overrightarrow{\boldsymbol{B}} \), your thumb points in the direction of the resulting vector.
This method is applied in determining the components of the Lorentz force, as seen in the exercise. Understanding and using the cross product allows us to solve problems involving forces and motions in a magnetic field effectively.

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Most popular questions from this chapter

(a) An \({ }^{16} \mathrm{O}\) nucleus (charge \(+8 e\) ) moving horizontally from west to east with a speed of \(500 \mathrm{~km} / \mathrm{s}\) experiences a magnetic force of \(0.00320 \mathrm{nN}\) vertically downward. Find the magnitude and direction of the weakest magnetic field required to produce this force. Explain how this same force could be caused by a larger magnetic field. (b) An electron moves in a uniform, horizontal, \(2.10-\mathrm{T}\) magnetic field that is toward the west. What must the magnitude and direction of the minimum velocity of the electron be so that the magnetic force on it will be \(4.60 \mathrm{pN}\), vertically upward? Explain how the velocity could be greater than this minimum value and the force still have this same magnitude and direction.

A thin, uniform rod Figure \(P 7.45\) with negligible mass and length \(0.200\) \(\mathrm{m}\) is attached to the floor by less hinge at point \(P\) (Fig. horizontal spring with force \(k=4.80 \mathrm{~N} / \mathrm{m}\) connects the of the rod to a vertical wall. in a uniform magnetic field \(0.340 \mathrm{~T}\) directed into the plane figure. There is current \(I=6.50 \mathrm{~A}\) in the rod, in the direction shown. (a) Calculate the torque due to the magnetic force on the rod, for an axis at \(P\). Is it correct to take the total magnetic force to act at the center of gravity of the rod when calculating the torque? Explain. (b) When the rod is in equilibrium and makes an angle of \(53.0^{\circ}\) with the floor, is the spring stretched or compressed? (c) How much energy is stored in the spring when the rod is in equilibrium?

Force on a Current Loop in a Nonuniform Magnetic Field. It was shown in Section \(7.7\) that the net force on a current loop in a uniform magnetic field is zero. But what if \(\vec{B}\) is not uniform? Figure P7.85 shows a square loop of wire that lies 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 *-component: \(\overrightarrow{\boldsymbol{B}}=\left(B_{0^{2}} / L\right) \hat{\jmath}+\left(B_{0} y / L\right) \hat{\boldsymbol{k}}\), 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 Eq. (7.20). (c) Find the magnitude and direction of the net magnetic force on the loop.

A particle with negative charge \(q\) and mass \(m=2.58 \times 1\) \(10^{-15} \mathrm{~kg}\) is traveling through a region containing a uniform magnetic field \(\vec{B}=-(0.120 \mathrm{~T}) \hat{k}\). At a particular instant of time the velocity of the particle is \(\vec{V}=\left(1.05 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)(-3 \hat{\boldsymbol{i}}+4 \hat{\boldsymbol{j}}+12 \hat{\boldsymbol{k}})\) and the force \(\vec{F}\) on the particle has a magnitude of \(2.45 \mathrm{~N}\). (a) Determine the charge \(q .\) (b) Determine the acceleration \(\vec{a}\) of the particle. (c) Explain why the path of the particle is a helix, and determine the radius of curvature \(R\) of the circular component of the helical path. (d) Determine the cyclotron frequency of the particle. (e) Although helical motion is not periodic in the full sense of the word, the \(x\) - and \(y\)-coordinates do vary in a periodic way. If the coordinates of the particle at \(t=0\) are \((x, y, z)=(R, 0,0)\), determine its coordinates at a time \(t=2 T\), where \(T\) is the period of the motion in the \(x y\)-plane.

A particle with charge \(q\) is moving with speed \(v\) in the \(-y\) -direction. It is moving in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=\) \(B_{x} \hat{\imath}+B_{y} \hat{J}+B_{z} \hat{k} .\) (a) What are the components of the force \(\vec{F}\) exerted on the particle by the magnetic field? (b) If \(q>0\), what must the signs of the components of \(\overrightarrow{\boldsymbol{B}}\) be if the components of \(\overrightarrow{\boldsymbol{F}}\) are all nonnegative? (c) If \(q<0\) and \(B_{x}=B_{y}=B_{z}>0\), find the direction of \(\overrightarrow{\boldsymbol{F}}\) and find the magnitude of \(\overrightarrow{\boldsymbol{F}}\) in terms of \(|q|, v\), and \(B_{x}\)

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