Estimate the approximate maximum deflection of the electron...

Estimate the approximate maximum deflection of the electron beam near the center of a CRT television screen due to the Earth's $5.0 \times 10^{-5} \mathrm{~T}$ field. Assume the screen is $18 \mathrm{~cm}$ from the electron gun, where the electrons are accelerated (a) by $2.0 \mathrm{kV}$, or $(b)$ by $28 \mathrm{kV}$. Note that in color TV sets, the beam must be directed accurately to within less than $1 \mathrm{~mm}$ in order to strike the correct phosphor. Because the Earth's field is significant here, mu-metal shields are used to reduce the Earth's field in the CRT. (See Section $23-9 .)$

Estimate the approximate maximum deflection of the electron beam near the center of a CRT television screen due to the Earth's $5.0 \times 10^{-5} \mathrm{~T}$ field. Assume the screen is $18 \mathrm{~cm}$ from the electron gun, where the electrons are accelerated
(a) by $2.0 \mathrm{kV}$, or $(b)$ by $28 \mathrm{kV}$. Note that in color TV sets, the beam must be directed accurately to within less than $1 \mathrm{~mm}$ in order to strike the correct phosphor. Because the Earth's field is significant here, mu-metal shields are used to reduce the Earth's field in the CRT. (See Section $23-9 .)$

Key Concepts

Radius of Curvature

The radius of the circular path (r) of a charged particle moving in a magnetic field can be determined by equating the magnetic Lorentz force to the centripetal force, leading to the formula r = mv/(qB). This concept helps in quantifying how the strength of the magnetic field and the velocity of the particle affect its path curvature.

Circular Motion in a Magnetic Field

When the magnetic Lorentz force acts perpendicular to the electron’s velocity, the electron undergoes uniform circular motion. This concept is essential because it provides the basis for calculating the radius of the electron’s curved trajectory under the influence of a magnetic field.

Deflection Estimation

The deflection of the electron beam can be approximated by considering the portion of the circular arc subtended over the distance to the screen. By relating the geometric parameters of the circular trajectory to the linear displacement, one can estimate how much the electron beam deviates from its intended path due to the magnetic field.

Lorentz Force

This is the force experienced by a moving charged particle in a magnetic field, given by F = qvB sin(?). In many applications, including those involving electron beams, the force due to the magnetic field is responsible for altering the particle’s trajectory when the velocity vector is perpendicular to the magnetic field lines.

Electron Acceleration

Electrons are accelerated through an electric potential difference, which converts electrical potential energy into kinetic energy. The relationship ½ mv² = eV is used to determine the speed of the electron after acceleration, which is critical for understanding subsequent motion in a magnetic environment.

Transcript

00:01 Hi there, so we need to estimate the approximate maximum deflection of the electron beam near the center of a crt television screen.

00:11 Due to the earth's magnetic field, we are given the magnitude for that, that is 5 times 10 to the minus 5 tesla.

00:21 And assume that the screen is 18 centimeters from the electron gone.

00:26 So the value that we are given for this.

00:31 Is what we are going to call delta x that will be 18 centimeters that in meters is 0 .18 meters and now for part a when the electrons are accelerated using a potential difference or a voltage of two kilo bolts or which is the same to thousand volts.

01:10 Then with that set, we note that the speed of the electrons is found by assuming that the energy supplied by the accelerating voltage becomes kinetic energy of the electrons.

01:22 We assume that those electrons are initially directed horizontally, and the television set is orientated so that the electron velocity is perpendicular to the earth magnetic field, resulted in the largest possible force.

01:38 Finally, we assume that the magnetic force on these electrons is small enough that the electron velocity is essentially perpendicular to the earth field for the entire trajectory of these electrons.

01:52 Now, this results in a constant acceleration for those electrons.

01:56 So those are the considerations in here.

02:00 So, first of all, we consider that the initial potential energy due to the accelerating potential difference.

02:10 Is equal to the final kinetic energy.

02:14 This is by conservation of energy.

02:16 Now we define these two, so we know that the potential energy is just a product between the charge of an electron and the voltage, and then the kinetic energy is 1 divided by 2 times the mass of the electron times the speed, the horizontal speed to the square.

02:38 If we're solving here for the speed in the xx disruption we obtain 2 times a charge times a voltage this divided by the mass and then we take the score of this now for the deflection this was for the acceleration now for the deflection part of this we will have that the time in the field is delta which is the initial speed at the speed in that or is simply times the time in that field so if we solve for the time in the field we will have the that that is delta x divided by the speed in that direction.

03:19 Now, we start with the force in the vertical direction, so that will be the charge times the speed, times the magnetic field of the earth.

03:30 This is equal to the mass times the acceleration in the y direction.

03:35 So, so then for the acceleration in the y direction, we will have that that is the charge, the speed in the x eruption, the magnetic field, this divided by the mass.

03:46 Now we're substituting here the expression that we obtain for the speed horizontally.

03:54 So that will be then the charge the square root of two times the charge of an electron, the potential difference, this divided by the mass, this times the magnetic field, this divided by the mass.

04:08 Then in here, rewriting this, that will be two times the charge of an electrum to the 3 times the potential difference.

04:19 This divided by the mass to the 3, which is the square root of all of this, times the magnetic field.

04:35 With that said, then the vertical deflection of this is just 1 divided by 2 times the acceleration vertically times the time square.

04:52 Then in here we subsubsternation.

04:59 Distitute acceleration that we obtained from before, so that will be 1 divided by 2...

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