Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A particle of mass 0.195 g carries a charge of -2.50 \(\times\) 10\(^{-8}\) C. The particle is given an initial horizontal velocity that is due north and has magnitude 4.00 \(\times\) 10\(^4\) m/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 has a magnitude of 0.191 T and is directed west.

Step by step solution

01

Understand the Forces

The particle experiences a gravitational force downwards due to gravity and a magnetic force acting perpendicular to both its velocity and the magnetic field. For the particle to move horizontally, the magnetic force must counteract the gravitational force.
02

Express the Gravitational Force

The gravitational force can be expressed as \( F_g = m imes g \), where \( m = 0.195 \times 10^{-3} \) kg and \( g = 9.81 \) m/s extsuperscript{2}. Compute \( F_g \) as follows:\[ F_g = 0.195 \times 10^{-3} \times 9.81 = 1.91195 \times 10^{-3} \text{ N} \].
03

Determine the Necessary Magnetic Force

Since the magnetic force \( F_m \) must balance the gravitational force to maintain horizontal movement, we have \( F_m = F_g = 1.91195 \times 10^{-3} \) N.
04

Use the Lorentz Force Equation

The magnetic force on a charged particle moving in a magnetic field is given by \( F_m = qvB \sin\theta \). For a perpendicular magnetic field, \( \theta = 90^{\circ} \), hence \( \sin\theta = 1 \).We have \( q = -2.50 \times 10^{-8} \) C and \( v = 4.00 \times 10^4 \) m/s. Thus, \( F_m = |q|vB = 1.91195 \times 10^{-3} \) N.
05

Solve for the Magnetic Field Magnitude

Rearranging the equation \( |q|vB = 1.91195 \times 10^{-3} \), solve for \( B \):\[ B = \frac{F_m}{|q|v} = \frac{1.91195 \times 10^{-3}}{2.50 \times 10^{-8} \times 4.00 \times 10^4} \]Calculate \( B \):\[ B = 1.91195 \times 10^{-3} / (2.50 \times 10^{-8} \times 4.00 \times 10^4) = 0.191195 \text{ T} \].
06

Determine the Direction of the Magnetic Field

Using the right-hand rule, since the particle's velocity is northward and the force must act upward to counteract gravity, the magnetic field should point towards the west.

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 Vaia!

Key Concepts

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

Lorentz Force
The Lorentz force is the force exerted on a charged particle moving through both electric and magnetic fields. In many practical scenarios, like the one examined in our exercise, we consider only the magnetic component. This force can be calculated using the equation:
  • \( F_m = qvB\sin\theta \)
Here, \( q \) is the charge of the particle, \( v \) is its velocity, \( B \) is the magnetic field, and \( \theta \) is the angle between the velocity and the magnetic field vectors.In situations where the magnetic field is perpendicular to the velocity (\( \theta = 90^\circ \)), the \( \sin\theta \) becomes 1, simplifying the equation to \( F_m = qvB \). This scenario is common in physics problems, making the Lorentz force dependent directly on the magnitude of \( q \), \( v \), and \( B \). This ability to influence the motion of charged particles, especially with adjustable field directions and magnitudes, is the principle underlying devices like cyclotrons and mass spectrometers.
Gravitational Force
The gravitational force is a fundamental force exerted by the Earth on objects, pulling them towards its center. It depends on the mass of the object and the acceleration due to gravity. The formula to calculate the gravitational force is:
  • \( F_g = mg \)
Where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \) on the Earth's surface.In this problem, the gravitational force acts downward, which conflicts with the magnetic force we want to carefully balance to keep the particle on a horizontal path. The gravitational force is a constant intrinsic to the mass and position of the particle relative to Earth, and it plays a crucial role in determining the necessity of a specific opposing force such as the magnetic force to achieve horizontal, stable motion.
Magnetic Field Direction
The direction of a magnetic field is significant when determining the net force acting on a charged particle. The exercise required maintaining a charged particle in horizontal motion while countering the downward gravitational pull. To understand the direction of the magnetic force, we use the right-hand rule:
  • Point your fingers in the direction of the particle's velocity (northward).
  • Align your palm to "push" in the direction of the force required to balance gravity (upwards).
  • Your thumb, extending perpendicular to both, indicates the magnetic field direction (west).
By directing the magnetic field to the west, the resulting magnetic force acts upwards to counterbalance the gravitational force. The right-hand rule is a helpful technique that physical scientists use to rapidly determine field directions without complex computations.
Charged Particle Motion
The motion of a charged particle in a magnetic field is distinctive because the magnetic force acts perpendicular to the motion of the particle. This force does not work on the particle (as work involves displacement in the direction of force), but it does change the direction of the velocity. For a particle like ours with an initial horizontal velocity:
  • The velocity vector points north.
  • The magnetic force, exerting force perpendicular to its direction, keeps the particle moving in the plane by offsetting other forces, such as gravity.
  • This keeps the particle in "circular" motion.
While the core aim in this context is to maintain a constant horizontal path, the principles apply broadly to understanding phenomena like loops of charged particles in familiar settings such as motors or the aurora borealis caused by charged particles from solar winds being steered by the Earth's magnetic field.

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 particle of charge \(q\) > 0 is moving at speed v in the \(+z\)-direction through a region of uniform magnetic field \(\overrightarrow{B}\). The magnetic force on the particle is \(\overrightarrow{F} =\) \(F_0\)(3\(\hat{\imath}\) + 4 \(\hat{\jmath}\)), 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\(F_0/qv\), determine as much as you can about the remaining components of \(\overrightarrow{B}\).

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an angle of 30.0\(^\circ\) above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10 \(\times\) 10\(^{-4}\) Wb through the surface?

The magnetic poles of a small cyclotron produce a magnetic field with magnitude 0.85 T. The poles have a radius of 0.40 m, which is the maximum radius of the orbits of the accelerated particles. (a) What is the maximum energy to which protons (\(q =\) 1.60 \(\times\) 10\(^{-19}\)C, \(m =\) 1.67 \(\times\) 10\(^{-27}\) kg) 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)? (d) For \(B =\) 0.85 T, what is the maximum energy to which alpha particles (\(q =\) 3.20 \(\times\) 10\(^{-19}\) C, \(m =\) 6.64 \(\times\) 10\(^{-27}\) kg) can be accelerated by this cyclotron? How does this compare to the maximum energy for protons?

In the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed of 2.2 \(\times\) 10\(^6\) m/s in a circular orbit of radius 5.3 \(\times\) 10\(^{-11}\) 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 flat, square surface with side length 3.40 cm is in the xy-plane at \(z =\) 0. Calculate the magnitude of the flux through this surface produced by a magnetic field \(\overrightarrow{B} =\) (0.200 T)\(\hat{\imath}\) + (0.300 T)\(\hat{\jmath}\) - (0.500 T)\(\hat{k}\).

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.

Sign-up for free