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Chapter 21: Problem 1

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.

Short Answer

Expert verified
The magnitude of the acceleration is \( 5.9 \times 10^{12} \mathrm{m/s^2} \).

Step by step solution

01

Identify the Known Values

We are given the horizontal component of the Earth's magnetic field, which is \( B = 1.6 \times 10^{-5} \) T. The speed of the electron is \( v = 2.1 \times 10^{6}\, \mathrm{m/s} \). The charge of an electron is \( e = 1.6 \times 10^{-19} \) C, and the electron's mass is \( m = 9.11 \times 10^{-31} \) kg.
02

Determine the Magnetic Force

The magnetic force \( F \) on a charged particle moving in a magnetic field is given by the formula:\[ F = qvB \sin \theta \]Since the electron is shot vertically and the magnetic field is horizontal, \( \theta = 90^\circ \) and thus \( \sin 90^\circ = 1 \). So, the force becomes:\[ F = qvB \]
03

Substitute the Values Into the Force Equation

Substitute the known values into the magnetic force equation:\[ F = (1.6 \times 10^{-19}\, \mathrm{C})(2.1 \times 10^{6} \, \mathrm{m/s})(1.6 \times 10^{-5} \, \mathrm{T}) \]Calculate \( F \).
04

Calculate the Force

Calculate the value:\[ F = 1.6 \times 2.1 \times 1.6 \times 10^{-19} \times 10^{6} \times 10^{-5} = 5.376 \times 10^{-18} \, \mathrm{N} \]
05

Calculate the Acceleration

Using Newton's Second Law \( F = ma \), solve for acceleration \( a \):\[ a = \frac{F}{m} \]Substitute the force and mass of the electron:\[ a = \frac{5.376 \times 10^{-18} \, \mathrm{N}}{9.11 \times 10^{-31} \, \mathrm{kg}} \]Calculate \( a \).
06

Calculate the Acceleration Value

Calculate the final acceleration:\[ a = 5.9 \times 10^{12} \, \mathrm{m/s^2} \]

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Key Concepts

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

Earth's Magnetic Field
The Earth's magnetic field is like a vast, invisible shield surrounding our planet. It's generated by movements of molten iron within Earth's outer core, creating a magnetic field that protects us from solar winds and cosmic radiation.
But it's not just a single uniform field. It has different components, including a horizontal component which affects how charged particles like electrons move across the Earth's surface. In New England, this horizontal component has a strength of about \(1.6 \times 10^{-5}\) Tesla.
This might seem small, but even such a weak field can exert a noticeable force on small charged particles moving at high speeds. Understanding this is crucial for applications involving navigation and technologies relying on Earth's natural magnetic field, such as compasses.
Electron Motion
Electrons are negatively charged particles that can be found in atoms, orbiting the nucleus. However, in many contexts, electrons also move freely, such as in electrical currents or while being fired in experiments.
When electrons move, such as being shot upwards at \(2.1 \times 10^{6} \mathrm{m/s}\), they interact with magnetic fields. The force of interaction depends on their speed, the strength of the magnetic field, and the angle at which they enter the field.
In our case, with a horizontal magnetic field and upward electron motion, the angle is \(90^\circ\), maximizing the force exerted on the electron. This principle supports technologies such as particle accelerators and contributes to studies in electromagnetism.
Newton's Second Law
Newton's Second Law is a cornerstone of classical physics, dictating that the force exerted on an object is equal to the product of its mass and the acceleration it experiences. This is mathematically represented as \( F = ma \).
When we consider an electron moving in a magnetic field, the force due to the magnetic field causes it to accelerate. Given the force has been calculated as \(5.376 \times 10^{-18} \, \mathrm{N}\) and knowing the mass of an electron is \(9.11 \times 10^{-31}\, \mathrm{kg}\), Newton's Second Law allows us to determine the acceleration.
  • First, by rearranging the formula to find acceleration \(a\), we have \(a = \frac{F}{m}\).
  • This results in an acceleration \(a = 5.9 \times 10^{12} \, \mathrm{m/s^2}\).
Understanding this law is essential, not just in electromagnetism, but in all areas of physics where forces and motion are involved.

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Most popular questions from this chapter

Suppose that a uniform magnetic field is everywhere perpendicular to this page. The field points directly upward toward you. A circular path is drawn on the page. Use Amp猫re's law to show that there can be no net current passing through the circular surface.

The drawing shows a parallel plate capacitor that is moving with a speed of \(32 \mathrm{m} / \mathrm{s}\) through a \(3.6-\mathrm{T}\) magnetic field. The velocity \(\overrightarrow{\mathbf{v}}\) is perpendicular to the magnetic field. The electric field within the capacitor has a value of \(170 \mathrm{N} / \mathrm{C},\) and each plate has an area of \(7.5 \times 10^{-4} \mathrm{m}^{2} .\) What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?

A wire has a length of \(7.00 \times 10^{-2} \mathrm{m}\) and is used to make a circular coil of one turn. There is a current of \(4.30 \mathrm{A}\) in the wire. In the presence of a \(2.50-\mathrm{T}\) magnetic field, what is the maximum torque that this coil can experience?

An ionized helium atom has a mass of \(6.6 \times 10^{-27} \mathrm{kg}\) and a speed of \(4.4 \times 10^{5} \mathrm{m} / \mathrm{s} .\) It moves perpendicular to a \(0.75-\mathrm{T}\) magnetic field on a circular path that has a 0.012-m radius. Determine whether the charge of the ionized atom is \(+e\) or \(+2 e\).

In the operating room, anesthesiologists use mass spectrometers to monitor the respiratory gases of patients undergoing surgery. One gas that is often monitored is the anesthetic isoflurane (molecular mass \(=\) \(3.06 \times 10^{-25} \mathrm{kg}\) ). In a spectrometer, a singly ionized molecule of isoflurane (charge \(=+e\) ) moves at a speed of \(7.2 \times 10^{3} \mathrm{m} / \mathrm{s}\) on a circular path that has a radius of \(0.10 \mathrm{m} .\) What is the magnitude of the magnetic field that the spectrometer uses?
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