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particle of mass 0.195 g carries a charge of \(-2.50 \times\) \(10^{-8} \mathrm{C} .\) The particle is given an initial horizontal velocity that is due north and has magnitude \(4.00 \times 10^{4} \mathrm{m} / \mathrm{s}\) . What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

Short Answer

Expert verified
The magnetic field magnitude is 1.9 T, directed to the east.

Step by step solution

01

Understanding the Problem

We have a charged particle moving with an initial velocity under Earth's gravity. We want a magnetic field that maintains the particle's horizontal northward trajectory. Specifically, we need the magnetic field to counterbalance gravity.
02

Applying Magnetic Force

The magnetic force must counteract the gravitational force. Use the equation for magnetic force: \( F_B = qvB \), where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field. This force should equal the gravitational force \( F_g = mg \), where \( m \) is mass and \( g \) is acceleration due to gravity.
03

Equating the Forces

Set the magnetic force equal to the gravitational force: \[ qvB = mg \]. Solving for \( B \), we find that \[ B = \frac{mg}{qv} \].
04

Inserting Known Values

Substitute the known values into the equation: \( m = 0.195 \times 10^{-3} \) kg, \( g = 9.8 \) m/s\(^2\), \( q = -2.50 \times 10^{-8} \) C, and \( v = 4.00 \times 10^{4} \) m/s. That gives us \[ B = \frac{0.195 \times 10^{-3} \times 9.8}{-2.50 \times 10^{-8} \times 4.00 \times 10^{4}} \].
05

Calculating the Magnetic Field

Perform the calculation: \( B = \frac{0.195 \times 10^{-3} \times 9.8}{-2.50 \times 10^{-8} \times 4.00 \times 10^{4}} = 1.9 \) T. The negative sign indicates direction; however, magnitude is required as positive.
06

Determining Direction Using Right Hand Rule

The velocity is northward, and force acts upward, opposing gravity. Use the right-hand rule: Point thumb in the direction of velocity (north), and fingers in direction of magnetic force (up). The palm faces left, so the magnetic field is eastward.

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Key Concepts

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

Lorentz Force Equation
The Lorentz Force Equation is essential for understanding how charged particles move in electromagnetic fields. It is given by the equation: \( F_B = qvB \), where:
  • \( F_B \) is the magnetic force acting on the charge.
  • \( q \) represents the charge of the particle.
  • \( v \) is the velocity of the particle.
  • \( B \) denotes the magnetic field.

This equation reveals that the force on a charged particle is directly proportional to its charge, its velocity, and the magnetic field's strength. In solving problems like the one given, we often set the magnetic force equal to other forces, such as gravitational force, to find equilibrium conditions. This approach helps find the magnitude of the magnetic field needed to maintain a certain particle motion.
Understanding this equation fundamentally helps in applications such as designing devices like cyclotrons, where charged particles are accelerated in a magnetic field.
Right Hand Rule
The Right Hand Rule is a simple yet pivotal tool for determining the direction of the magnetic force on a moving charge. Here's how you can apply it:
  • Point your thumb in the direction of the velocity of the charged particle.
  • Extend your fingers in the direction of the magnetic field.
  • Your palm will then face in the direction of the force exerted on a positive charge. For a negative charge, the force direction is opposite.

In the context of our exercise, the thumb points north, representing the velocity, and the palm faces left, indicating the magnetic field's direction when accounting for a negative charge. This ultimately aligns the magnetic force with the upward direction needed to counteract gravity, keeping the trajectory horizontal.
Grasping the Right Hand Rule makes it easier to visualize and predict the movement of charged particles in a magnetic field, crucial for modern physics and engineering.
Gravitational Force
Gravitational Force is a well-known force that acts on any mass, drawing it towards the Earth's center. It is calculated using:\[ F_g = mg \]where:
  • \( F_g \) is the gravitational force.
  • \( m \) is the mass of the particle.
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \) on Earth.

In the scenario provided, the gravitational force pulls the charged particle downward. To maintain a northward horizontal motion, the magnetic force must counteract this downward pull exactly. By setting \( F_B = F_g \), we attain the balance required, illustrating how these forces interact dynamically in physics problems.
This understanding is vital in fields like aerospace, where understanding and manipulating gravitational forces are part of the everyday challenges.
Charged Particle Dynamics
Charged Particle Dynamics is the study of how electric and magnetic fields affect the motion of charged particles. It combines fundamental forces such as Lorentz, electric, and gravitational forces to determine particle trajectory, speed, and energy.
  • In magnetic fields, charged particles experience a perpendicular force altering their trajectory.
  • This can cause circular, spiral, or helical paths, depending on the initial velocity and the field's configuration.
  • By calculating forces applied, scientists can design precise paths for particles in devices like particle accelerators.

In the given exercise, dynamics are analyzed by equating gravitational and magnetic forces. This achievement illustrates how charged particle dynamics is applied to maintain specific paths or trajectories in various scientific and engineering scenarios. Understanding these interactions opens doors to advanced technologies in medical physics and energy solutions.
Mastery of these dynamics is crucial for students and professionals eager to excel in electromagnetism and its myriad applications.

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Most popular questions from this chapter

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?

You wish to hit a target from several meters away with a charged coin having a mass of 4.25\(\mathrm { g }\) and a charge of \(+ 2500 \mu \mathrm { C }\) . The coin is given an initial velocity of 12.8\(\mathrm { m } / \mathrm { s }\) , and a downward, uniform electric field with field strength 27.5\(\mathrm { N } / \mathrm { C }\) exists through-out the region. If you aim directly at the target and fire the coin horizontally, what magnitude and direction of uniform magnetic field are needed in the region for the coin to hit the target?

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?

A singly charged ion of \(^{7} \mathrm{Li}\) (an isotope of lithium) has a mass of \(1.16 \times 10^{-26} \mathrm{kg}\) . It is accelerated through a potential dif- ference of 220 \(\mathrm{V}\) and then enters a magnetic field with magnitude 0.723 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

An electron moves at \(2.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) through a region in which there is a magnetic field of unspecified direction and magnitude \(7.40 \times 10^{-2} \mathrm{T}\) (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

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