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Chapter 21: Problem 4

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 \(2 F ?\)

Short Answer

Expert verified
The angle is approximately \(57^{\circ}\).

Step by step solution

01

Understanding the Magnetic Force

The force exerted on a charged particle moving through a magnetic field is given by the equation \( F = qvB \sin(\theta) \), where \( q \) is the charge of the particle, \( v \) is its velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field.
02

Relating Force to the Given Angle

The initial force experienced is \( F = qvB \sin(25^{\circ}) \). We need to find another angle \( \theta \) such that the force becomes \( 2F = qvB \sin(\theta) \).
03

Set up the Equation for New Force

Since we want the new force to be double the original force, set the equation \( qvB \sin(\theta) = 2 \times qvB \sin(25^{\circ}) \). This simplifies to \( \sin(\theta) = 2 \sin(25^{\circ}) \).
04

Solving for the New Angle

To find \( \theta \), we need to take the inverse sine (arcsin) of both sides. Calculate \( \sin(25^{\circ}) \), which is approximately \( 0.4226 \). Thus, \( \sin(\theta) = 2 \times 0.4226 = 0.8452 \). Use a calculator to find \( \theta = \arcsin(0.8452) \), resulting in \( \theta \approx 57.0^{\circ} \).

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Key Concepts

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

Angle with Magnetic Field
When a charged particle moves in a magnetic field, the angle between its velocity and the magnetic field lines greatly affects the experienced magnetic force. This angle, denoted as \( \theta \), determines how much of the particle's motion actually interacts with the field.
Since magnetism acts perpendicular to motion, not all the movement contributes to the force. Only the component of velocity that is perpendicular to the magnetic field line interacts.
In the provided exercise, the particle initially moves at an angle of \( 25^{\circ} \) to the magnetic field. This means part of its velocity is contributing to the magnetic force, resulting in a force \( F \).
If the angle changes, the effect on force changes, which is why understanding this relationship is critical.
Trigonometric Functions in Physics
Trigonometric functions, especially the sine function, play a key role in physics to describe how different components of vectors influence outcomes.
The magnetic force on a particle is calculated using the sine of the angle \( \theta \) between its velocity and the magnetic field. The force formula is \( F = qvB\sin(\theta) \), where each symbol has a specific meaning:
  • \( F \) is the magnetic force.
  • \( q \) is the charge of the particle.
  • \( v \) is the velocity of the particle.
  • \( B \) is the magnetic field strength.
  • \( \theta \) is the angle with the magnetic field.

In our exercise, we started with \( \sin(25^{\circ}) \) and needed to find an angle \( \theta \) where \( \sin(\theta) = 2\sin(25^{\circ}) \). Using a calculator, we saw that \( \sin(25^{\circ}) \approx 0.4226 \), making \( \sin(\theta) \approx 0.8452 \). Thus, \( \theta \approx 57.0^{\circ} \).
This calculation demonstrates how trigonometry aids in translating angles into meaningful physical phenomena.
Magnitude of Magnetic Force
Understanding the magnitude of the magnetic force experienced by a particle requires an insight into both the physical properties involved and the geometric positioning, particularly the angle.
The magnetic force magnitude is directly proportional to the product of the charge \( q \), the particle鈥檚 velocity \( v \), the magnetic field strength \( B \), and the sine of the angle \( \theta \).
The equation \( F = qvB\sin(\theta) \) tells us that non-zero force only occurs when the charge is moving at an angle other than parallel to the magnetic field. At \( \theta = 0^{\circ} \), no force acts on the particle, as \( \sin(0^{\circ}) = 0 \).
With our aim of doubling the original force to \( 2F \), we used the angle such that the sine of that angle doubled, which directly increased the force magnitude. This emphasizes the power of understanding not just the quantities, but the trigonometric relationships that govern physical interactions.

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Most popular questions from this chapter

Two isotopes of carbon, carbon- 12 and carbon- \(13,\) have masses of \(19.93 \times 10^{-27} \mathrm{~kg}\) and \(21.59 \times 10^{-27} \mathrm{~kg},\) respectively. These two isotopes are singly ionized \((+e)\) and each is given a speed of \(6.667 \times 10^{5} \mathrm{~m} / \mathrm{s}\). The ions then enter the bending region of a mass spectrometer where the magnetic field is \(0.8500 \mathrm{~T}\). Determine the spatial separation between the two isotopes after they have traveled through a half-circle.

The ion source in a mass spectrometer produces both singly and doubly ionized species, \(\mathrm{X}^{+}\) and \(\mathrm{X}^{2+}\). The difference in mass between these species is too small to be detected. Both species are accelerated through the same electric potential difference, and both experience the same magnetic field, which causes them to move on circular paths. The radius of the path for the species \(\mathrm{X}^{+}\) is \(r_{1}\), while the radius for species \(\mathrm{X}^{2+}\) is \(r_{2} .\) Find the ratio \(r_{1} / r_{2}\) of the radii.

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.

In New England, the horizontal component of the earth's magnetic field has a magnitude of \(1.6 \times 10^{-5} \mathrm{~T}\). An electron is shot vertically straight up from the ground with a speed of \(2.1 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron.

A particle of mass \(6.0 \times 10^{-8} \mathrm{~kg}\) and charge \(+7.2 \mu \mathrm{C}\) is traveling due east. It enters perpendicularly a magnetic field whose magnitude is \(3.0 \mathrm{~T}\). After entering the field, the particle completes one-half of a circle and exits the field traveling due west. How much time does the particle spend traveling in the magnetic field?
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