/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Electrons in a television's CRT ... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 15

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.

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Most popular questions from this chapter

A positron \((q=+e)\) moves at \(5.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) in a magnetic field of magnitude 0.47 T. The magnetic force on the positron has magnitude \(2.3 \times 10^{-12} \mathrm{N} .\) (a) What is the component of the positron's velocity perpendicular to the magnetic field? (b) What is the component of the positron's velocity parallel to the magnetic field? (c) What is the angle between the velocity and the field?
The strength of Earth's magnetic field, as measured on the surface, is approximately \(6.0 \times 10^{-5} \mathrm{T}\) at the poles and $3.0 \times 10^{-5} \mathrm{T}$ at the equator. Suppose an alien from outer space were at the North Pole with a single loop of wire of the same circumference as his space helmet. The diameter of his helmet is \(20.0 \mathrm{cm} .\) The space invader wishes to cancel Earth's magnetic field at his location. (a) What is the current required to produce a magnetic ficld (due to the current alone) at the center of his loop of the same size as that of Earth's field at the North Pole? (b) In what direction does the current circulate in the loop, CW or CCW, as viewed from above, if it is to cancel Earth's field?
A straight wire segment of length \(0.60 \mathrm{m}\) carries a current of $18.0 \mathrm{A}$ and is immersed in a uniform external magnetic ficld of magnitude \(0.20 \mathrm{T}\). (a) What is the magnitude of the maximum possible magnetic force on the wire segment? (b) Explain why the given information enables you to calculate only the maximum possible force.
A magnet produces a \(0.30-\mathrm{T}\) field between its poles, directed to the east. A dust particle with charge \(q=-8.0 \times 10^{-18} \mathrm{C}\) is moving straight down at \(0.30 \mathrm{cm} / \mathrm{s}\) in this field. What is the magnitude and direction of the magnetic force on the dust particle?
An early cyclotron at Cornell University was used from the 1930 s to the 1950 s to accelerate protons, which would then bombard various nuclei. The cyclotron used a large electromagnet with an iron yoke to produce a uniform magnetic field of \(1.3 \mathrm{T}\) over a region in the shape of a flat cylinder. Two hollow copper dees of inside radius \(16 \mathrm{cm}\) were located in a vacuum chamber in this region. (a) What is the frequency of oscillation necessary for the alternating voltage difference between the dees? (b) What is the kinetic energy of a proton by the time it reaches the outside of the dees? (c) What would be the equivalent voltage necessary to accelerate protons to this energy from rest in one step (say between parallel plates)? (d) If the potential difference between the dees has a magnitude of $10.0 \mathrm{kV}$ each time the protons cross the gap, what is the minimum number of revolutions each proton has to make in the cyclotron?
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