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Chapter 21: Problem 73

An electron is moving through a magnetic field whose magnitude is \(8.70 \times 10^{-4} \mathrm{T}\). The electron experiences only a magnetic force and has an acceleration of magnitude \(3.50 \times 10^{14} \mathrm{m} / \mathrm{s}^{2} .\) At a certain instant, it has a speed of \(6.80 \times 10^{6} \mathrm{m} / \mathrm{s}\). Determine the angle \(\theta\) (less than \(90^{\circ}\) ) between the electron's velocity and the magnetic field.

Short Answer

Expert verified
The angle \( \theta \) is approximately \( 15.29^\circ \).

Step by step solution

01

Understand the problem

We need to find the angle \( \theta \) between the velocity \( \vec{v} \) of the electron and the magnetic field \( \vec{B} \) using given information: magnetic field magnitude (\( B = 8.70 \times 10^{-4} \mathrm{T} \)), electron's acceleration (\( a = 3.50 \times 10^{14} \mathrm{m/s}^2 \)), and speed (\( v = 6.80 \times 10^{6} \mathrm{m/s} \)).
02

Recall the relationship of force, acceleration and magnetic field

The force on an electron due to the magnetic field can be expressed using the Lorentz force equation: \( F_B = qvB\sin\theta \). Also, recall from Newton's second law that \( F = ma \).
03

Equate Magnetic force and Newton's second law

Since the only force acting on the electron is the magnetic force, set \( ma = qvB\sin\theta \), where \( m \) is the mass of an electron (\( 9.11 \times 10^{-31} \mathrm{kg} \)) and \( q \) is the charge of an electron (\( 1.60 \times 10^{-19} \mathrm{C} \)).
04

Rearrange the equation to solve for \( \sin\theta \)

Rearrange the equation to get \( \sin\theta = \frac{ma}{qvB} \). Substitute the values: \( m = 9.11 \times 10^{-31} \mathrm{kg} \), \( a = 3.50 \times 10^{14} \mathrm{m/s}^2 \), \( q = 1.60 \times 10^{-19} \mathrm{C} \), \( v = 6.80 \times 10^{6} \mathrm{m/s} \), and \( B = 8.70 \times 10^{-4} \mathrm{T} \).
05

Calculate the value of \( \sin\theta \)

Plug in the numbers: \( \sin\theta = \frac{(9.11 \times 10^{-31} \mathrm{kg})(3.50 \times 10^{14} \mathrm{m/s}^2)}{(1.60 \times 10^{-19} \mathrm{C})(6.80 \times 10^{6} \mathrm{m/s})(8.70 \times 10^{-4} \mathrm{T})} \). This gives \( \sin\theta \approx 0.264 \).
06

Find \( \theta \) by taking the inverse sine

Compute \( \theta \) by taking the inverse sine of 0.264: \( \theta = \sin^{-1}(0.264) \approx 15.29^\circ \).

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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 key to understanding how charged particles like electrons move in a magnetic field. This equation tells us that the magnetic force (\( F_B \)) on a moving charged particle is given by:
\[ F_B = qvB\sin(\theta)\]where:
  • \( q \) is the charge of the particle
  • \( v \) is the velocity of the particle
  • \( B \) is the magnetic field strength
  • \( \theta \) is the angle between the particle's velocity and the magnetic field.
This equation tells us that the force depends on how fast the particle is moving, how strong the magnetic field is, and crucially, the angle \( \theta \). The sin(\theta) term shows us that the force is strongest when the velocity is perpendicular to the field, and weakest when they are parallel.
Angle between velocity and magnetic field
When dealing with magnetic forces, the angle \( \theta \) between the velocity \( \vec{v} \) of a charged particle and the magnetic field \( \vec{B} \) is crucial.
This angle affects how much force the particle feels as it moves through the magnetic field. To compute the force accurately, you need to know this angle because of the term \( \sin(\theta) \) in the Lorentz force equation.
Let's consider:
  • If \( \theta = 0掳 \) or \( 180掳 \) (meaning velocity is parallel to the field), \( \sin(\theta) = 0 \), and there is no force.
  • If \( \theta = 90掳 \), \( \sin(\theta) = 1 \), and the force reaches its maximum possible value.
Understanding this phenomenon is essential, especially in applications like magnetic confinement in fusion reactors, where maintaining precise angles ensures that particles remain controlled within their intended paths.
Newton's second law
Newton's second law provides a direct link between force, mass, and acceleration, stating that the force applied on an object is equal to the product of its mass and acceleration:
\[ F = ma \]In the context of magnetic forces, this fundamental principle helps us relate the magnetic force experienced by an electron moving in a magnetic field to its resultant acceleration.
  • In our exercise, this means equating the magnetic force given by the Lorentz equation to \( ma \), the expression from Newton's second law.
  • When the only force acting is magnetic, setting \( ma = qvB\sin(\theta) \) allows us to solve for the component of interest, which in this scenario, is \( \theta \).
Therefore, Newton's second law acts as the bridge that allows us to determine how changes in force affect the motion of charged particles within a magnetic field.

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Most popular questions from this chapter

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?

A \(45-\mathrm{m}\) length of wire is stretched horizontally between two vertical posts. The wire carries a current of \(75 \mathrm{A}\) and experiences a magnetic force of 0.15 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of \(60.0^{\circ}\) with respect to the wire.

In New England, the horizontal component of the earth's magnetic field has a magnitude of \(1.6 \times 10^{-5}\) 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 charged particle enters a uniform magnetic field and follows the circular path shown in the drawing. (a) Is the particle positively or negatively charged? Why? (b) The particle's speed is \(140 \mathrm{m} / \mathrm{s},\) the magnitude of the magnetic field is \(0.48 \mathrm{T}\), and the radius of the path is \(960 \mathrm{m}\). Determine the mass of the particle, given that its charge has a magnitude of \(8.2 \times 10^{-4} \mathrm{C}\).

Multiple-Concept Example 7 discusses how problems like this one can be solved. A \(+6.00 \mu \mathrm{C}\) charge is moving with a speed of \(7.50 \times 10^{4} \mathrm{m} / \mathrm{s}\) parallel to a very long, straight wire. The wire is \(5.00 \mathrm{cm}\) from the charge and carries a current of \(67.0 \mathrm{A}\) in a direction opposite to that of the moving charge. Find the magnitude and direction of the force on the charge.
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