Electrons in a television's CRT are accelerated from rest by an electric field through a potential difference of \(2.5 \mathrm{kV} .\) In contrast to an oscilloscope, where the electron beam is deflected by an electric ficld, the beam is deflected by a magnetic field. (a) What is the specd of the electrons? (b) The beam is deflected by a perpendicular magnetic field of magnitude $0.80 \mathrm{T}$. What is the magnitude of the acceleration of the electrons while in the field? (c) What is the speed of the electrons after they travel 4.0 mm through the magnetic field? (d) What strength electric field would give the electrons the same magnitude acceleration as in (b)? (c) Why do we have to use an clectric ficld in the first place to get the electrons up to speed? Why not use the large acceleration due to a magnetic field for that purpose?

Short Answer

Expert verified
Answer: We use an electric field to accelerate the electrons because electric fields act parallel to the electron's motion, which allows them to change the speed (either increase or decrease) of the electrons. In contrast, magnetic fields act perpendicular to the electron's motion, so they can only change the direction of the electrons but not their magnitude. Therefore, electric fields are used to accelerate the electrons, while magnetic fields are used to deflect (change the direction) of the electrons.

Step by step solution

01

Find the speed of the electrons after acceleration

To find the speed of the electrons after acceleration through the electric field, first, find their kinetic energy using the formula \(KE = eV\), where \(e\) is the charge of the electron, and \(V\) is the potential difference. Then, use the kinetic energy and the mass of the electron to find their speed using the formula \(KE = \frac{1}{2}mv_a^2\).
02

Calculate the magnetic force and acceleration

To find the magnetic force acting on the electrons, use the formula \(F_m = ev_aB\), where \(F_m\) is the magnetic force, \(v_a\) is the speed of the electrons calculated in step 1, and \(B\) is the magnetic field strength. After finding the magnetic force, find the acceleration using the formula \(a_m = \frac{F_m}{m}\), where \(m\) is the mass of the electron.
03

Find the speed of the electrons after traveling through the magnetic field

Since the magnetic force acts perpendicular to the speed of the electrons, it does not change their magnitude. Thus, the speed of the electrons after traveling through the magnetic field remains the same as the speed calculated in step 1 \((v_a)\).
04

Calculate the strength of an equivalent electric field

To find the strength of an electric field that gives the same magnitude of acceleration as the magnetic field, use the formulas \(F_e = ea_e\) and \(F_e = eE\), where \(F_e\) is the electric force, \(E\) is the electric field strength, and \(a_e\) is the equivalent acceleration. With the calculated magnetic acceleration, \(a_m\), we can find the equivalent electric field strength \(E = \frac{a_m}{e}\).
05

Discuss the role of electric and magnetic fields

Electric fields are used to accelerate the electrons because they act parallel to the electric field, and thus, they can change the speed (either increase or decrease) of the electrons. In contrast, magnetic fields always act perpendicular to the electron's speed, so they can only change the direction of the electrons but not their magnitude. That is why electric fields are used to accelerate the electrons, and magnetic fields are used to deflect (change the direction) of the electrons.

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!

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

In a certain region of space, there is a uniform electric field \(\overrightarrow{\mathbf{E}}=3.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) directed due east and a uniform magnetic field \(\mathbf{B}=0.080\) T also directed due east. What is the electromagnetic force on an electron moving due south at \(5.0 \times 10^{6} \mathrm{m} / \mathrm{s} ?\)
An electron moves at speed \(8.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a plane perpendicular to a cyclotron's magnetic field. The magnitude of the magnetic force on the electron is \(1.0 \times 10^{-13} \mathrm{N}\) What is the magnitude of the magnetic field?
An electromagnetic flowmeter is to be used to measure blood speed. A magnetic field of \(0.115 \mathrm{T}\) is applied across an artery of inner diameter \(3.80 \mathrm{mm}\) The Hall voltage is measured to be \(88.0 \mu \mathrm{V} .\) What is the average speed of the blood flowing in the artery?
A strip of copper \(2.0 \mathrm{cm}\) wide carries a current \(I=\) $30.0 \mathrm{A}\( to the right. The strip is in a magnetic field \)B=5.0 \mathrm{T}$ into the page. (a) What is the direction of the average magnetic force on the conduction electrons? (b) The Hall voltage is \(20.0 \mu \mathrm{V}\). What is the drift velocity?
Point \(P\) is midway between two long, straight, parallel wires that run north- south in a horizontal plane. The distance between the wires is $1.0 \mathrm{cm} .\( Each wire carrics a current of \)1.0 \mathrm{A}$ toward the north. Find the magnitude and direction of the magnetic field at point \(P\)
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