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Chapter 21: Problem 87

The figure shows a particle that carries a charge of \(q_{0}=-2.80 \times 10^{-6} \mathrm{C} .\) It is moving along the \(+y\) axis at a speed of \(v=4.8 \times 10^{6} \mathrm{m} / \mathrm{s} .\) A magnetic field \(\overrightarrow{\mathbf{B}}\) of magnitude \(3.35 \times 10^{-5} \mathrm{T}\) is directed along the \(+z\) axis, and an electric field \(\overrightarrow{\mathbf{E}}\) of magnitude \(123 \mathrm{N} / \mathrm{C}\) points along the \(-x\) axis. Concepts: (i) What forces make up the net force acting on the particle? (ii) How do you determine the direction of the magnetic force acting on the negative charge? (iii) How do you determine the direction of the electric force acting on the negative charge? (iv) Does the fact that the charge is moving affect the values of the magnetic and electric forces? Calculations: Determine the magnitude and direction of the net force that acts on the particle.

Short Answer

Expert verified
Net force is -0.7956 N along the -x axis.

Step by step solution

01

Identify the Forces

The particle experiences two types of forces: the magnetic force and the electric force. The magnetic force acting on a charged particle is given by the Lorentz force formula: \( \overrightarrow{F}_{B} = q \cdot (\overrightarrow{v} \times \overrightarrow{B}) \). The electric force is given by: \( \overrightarrow{F}_{E} = q \cdot \overrightarrow{E} \).
02

Calculate the Magnetic Force

Using the formula \( F_{B} = q \cdot v \cdot B \cdot \sin(\theta) \), where \( \theta = 90^\circ \) since the velocity \( \overrightarrow{v} \) is perpendicular to \( \overrightarrow{B} \), \( F_{B} = (-2.80 \times 10^{-6} \, C) \cdot (4.8 \times 10^{6} \, \text{m/s}) \cdot (3.35 \times 10^{-5} \, T) \cdot 1 \). Calculate to find \( F_{B} \approx -0.4512 \, N \).
03

Determine Direction of Magnetic Force

Use the right-hand rule. Point your fingers in the direction of velocity \( \overrightarrow{v} \), curl them towards the direction of \( \overrightarrow{B} \), but note the charge is negative. Thus, the direction of \( \overrightarrow{F}_{B} \) is along \(-x\) (reversing the positive direction from right-hand rule due to negative charge).
04

Calculate the Electric Force

Using the formula \( F_{E} = q \cdot E \), calculate \( F_{E} = (-2.80 \times 10^{-6} \, C) \cdot (123 \, N/C) = -0.3444 \, N \). The direction of \( \overrightarrow{F}_{E} \) is along the \(+x\) axis (opposite to \( \overrightarrow{E} \) because the charge is negative).
05

Determine Net Force and Direction

Calculate the net force: \( F_{net} = F_{B} + F_{E} = -0.4512 \, N + (-0.3444 \, N) = -0.7956 \, N \). This negative result along the \(x\)-axis indicates the net force is along the \(-x\) axis.
06

Moving Charge and Force Values

The motion of the charge affects the magnetic force because magnetic force is dependent on velocity. However, the electric force is independent of the charge's velocity.

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Key Concepts

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

Magnetic Field Effects
When a charged particle moves through a magnetic field, it experiences a force known as the magnetic force. This is a result of the magnetic field's influence on the moving charged particle. According to the Lorentz force law, the magnetic force can be calculated using the formula:
  • \( \overrightarrow{F}_{B} = q \cdot (\overrightarrow{v} \times \overrightarrow{B}) \)
Here, \( q \) represents the charge of the particle, \( \overrightarrow{v} \) is the velocity vector, and \( \overrightarrow{B} \) is the magnetic field vector. One critical aspect of this force is that it acts perpendicular to both the velocity of the particle and the magnetic field. This perpendicular action means that the force doesn't do work on the particle but changes its direction.

To determine the direction of the magnetic force on a negatively charged particle, we can use the right-hand rule. For a positive charge, you point your thumb in the velocity direction, fingers in the magnetic field direction, and the palm shows the force direction. For negative charges, the resulting force direction is opposite to the palm direction. In this scenario, the magnetic field is oriented along the \(+z\) axis, and the velocity directed along \(+y\) axis, making the magnetic force act along the \(-x\) axis due to the negative charge.
Electric Force Calculation
The electric force is a fundamental interaction between charged particles and electric fields. It can be calculated using the formula:
  • \( \overrightarrow{F}_{E} = q \cdot \overrightarrow{E} \)
This equation states that the force on a charge \( q \) within an electric field \( \overrightarrow{E} \) is proportional to the product of the charge magnitude and the electric field intensity. Notably, the direction of the electric force depends on the sign of the charge. Positive charges experience force in the direction of the electric field, while negative charges (like in this exercise) are pushed in the opposite direction.

In our specific case, the charge experiences an electric field pointing towards the \(-x\) axis. Therefore, since the charge is negative, the electric force will act along the \(+x\) axis. The magnitude is straightforward to calculate by multiplying the charge value with the electric field strength, resulting in \(-0.3444 \, N\). This highlights the electric force's independence from the velocity of the charge.
Lorentz Force
The Lorentz force combines both the electric and magnetic forces that act upon a charged particle. Together, these forces determine the net force acting on a moving charge in electromagnetic fields. The equation for Lorentz force is:
  • \( \overrightarrow{F}_{net} = \overrightarrow{F}_{E} + \overrightarrow{F}_{B} \)
It is essential to consider both components to fully understand how charged particles move in electromagnetic environments.

In our scenario, let's combine the calculated magnetic and electric forces. The magnetic force is \(-0.4512 \, N\) acting along \(-x\), while the electric force is \(-0.3444 \, N\) along \(+x\). Considering both directions and magnitudes gives us:
  • \( F_{net} = -0.4512 \, N + (-0.3444 \, N) = -0.7956 \, N \) along the \(-x\) axis
This negative result indicates that the net force culminates along the \(-x\) axis, primarily influenced by the magnetic component. Hence, the Lorentz force governs the motion and trajectory of the charged particle in a combined magnetic and electric field environment.

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Most popular questions from this chapter

When beryllium-7 ions \(\left(m=11.65 \times 10^{-27} \mathrm{kg}\right)\) pass through a mass spectrometer, a uniform magnetic field of \(0.283 \mathrm{T}\) curves their path directly to the center of the detector (see Figure 21.14 ). For the same accelerating potential difference, what magnetic field should be used to send beryllium-10 ions \(\left(m=16.63 \times 10^{-27} \mathrm{kg}\right)\) to the same location in the detector? Both types of ions are singly ionized \((q=+e)\).

A particle has a charge of \(q=+5.60 \mu \mathrm{C}\) and is located at the coordinate origin. As the drawing shows, an electric field of \(E_{x}=+245 \mathrm{N} / \mathrm{C}\) exists along the \(+x\) axis. A magnetic field also exists, and its \(x\) and \(y\) components are \(B_{x}=+1.80 \mathrm{T}\) and \(B_{y}=+1.40 \mathrm{T} .\) Calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is (a) stationary, (b) moving along the \(+x\) axis at a speed of \(375 \mathrm{m} / \mathrm{s},\) and \((\mathrm{c})\) moving along the \(+z\) axis at a speed of \(375 \mathrm{m} / \mathrm{s}\)

A square coil and a rectangular coil are each made from the same length of wire. Each contains a single turn. The long sides of the rectangle are twice as long as the short sides. Find the ratio \(\tau_{\text {xquar }} / \tau_{\text {rectangle }}\) of the maximum torques that these coils experience in the same magnetic field when they contain the same current.

A magnetic field has a magnitude of \(1.2 \times 10^{-3} \mathrm{T}\), and an electric field has a magnitude of \(4.6 \times 10^{3} \mathrm{N} / \mathrm{C}\). Both fields point in the same direction. A positive \(1.8 \mu \mathrm{C}\) charge moves at a speed of \(3.1 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.

When a charged particle moves at an angle of \(25^{\circ}\) with respect to a magnetic field, it experiences a magnetic force of magnitude \(F .\) At what angle (less than \(90^{\circ}\) ) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude \(2 F ?\)
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