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In the electron gun of a TV picture tube the electrons (charge \(-e,\) mass \(m )\) are accelerated by a voltage \(V .\) After leaving the electron gun, the electron beam travels a distance \(D\) to the screen; in this region there is a transverse magnetic field of magnitude \(B\) and no electric field. (a) Sketch the path of the electron beam in the tube. (b) Show that the approximate deflection of the beam due to this magnetic field is $$ d=\frac{B D^{2}}{2} \sqrt{\frac{e}{2 m V}} $$ (Hint: Place the origin at the center of the electron beam's arc and compare an undefiected beam's path to the deflected beam's path.) (c) Evaluate this expression for \(V=750 \mathrm{V}, D=50 \mathrm{cm},\) and \(B=5.0 \times 10^{-5} \mathrm{T}\) (comparable to the earth's field). Is this deflection significant?

Short Answer

Expert verified
The deflection is approximately 2.9 mm, which may be noticeable on the TV screen.

Step by step solution

01

Sketch the Path

Draw a diagram showing the TV picture tube. Indicate the electron gun on the cathode side and the screen on the anode side. Show the path the electron beam takes in the absence of a magnetic field as a straight line. Illustrate the deflected path in the presence of the magnetic field as a curve, where the electron beam veers downwards or upwards, depending on the field direction.
02

Apply the Concept of Lorentz Force

Recognize that the magnetic field exerts a Lorentz force on the moving electrons, causing them to move in a circular path. This force is given by the equation:\[ F = e \cdot v \cdot B \]where \( v \) is the velocity of the electrons. Since this force also acts as a centripetal force, we can write:\[ e \cdot v \cdot B = \frac{m \cdot v^2}{R} \]From this, solve for the radius \( R \) of the circle.
03

Relate Voltage to Electron Velocity

Use the energy principle to relate the kinetic energy gained by the electrons to the accelerating voltage:\[ \frac{1}{2} m v^2 = e V \]From this equation, solve for the velocity \( v \):\[ v = \sqrt{\frac{2 e V}{m}} \].
04

Derive the Deflection Formula

Using the relationship derived in Step 2 for radius \( R \) and the expression for velocity \( v \) from Step 3, and assuming small angles of deflection where \( \theta \approx \frac{x}{R} \) and \( x \) is the deflection, approximate the deflection of the beam \( d \) as:\[ d \approx \frac{1}{2} \cdot \frac{D^2}{R} \]Substitute \( R = \frac{mv}{eB} \) from Step 2 into this formula and simplify to obtain:\[ d = \frac{B D^2}{2} \sqrt{\frac{e}{2 m V}} \].
05

Evaluate the Deflection

Substitute the given values \( V = 750 \text{ V}, D = 0.5 \text{ m}, \text{ and } B = 5.0 \times 10^{-5} \text{ T} \) into the derived formula:\[ d = \frac{5.0 \times 10^{-5} \times 0.5^2}{2} \sqrt{\frac{1.6 \times 10^{-19}}{2 \times 9.1 \times 10^{-31} \times 750}} \]Calculate to find:\[ d \approx 2.9 \times 10^{-3} \text{ m} \text{ or } 2.9 \text{ mm}. \]
06

Interpret the Results

Determine whether the calculated deflection of \( 2.9 \text{ mm} \) is significant. Given that the dimensions of a typical TV screen are much larger, this deflection could cause noticeable distortion, indicating that magnetic shielding may be necessary.

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Key Concepts

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

Lorentz Force
The Lorentz force is a fundamental concept in electromagnetism that describes the influence of a magnetic field on a moving charged particle. This force acts perpendicular to both the velocity of the particle and the magnetic field itself. It is mathematically represented by the equation:\[ F = e \cdot v \cdot B \]Where:- \( F \) is the force exerted on the particle.- \( e \) is the charge of the particle (in this case, an electron).- \( v \) is the velocity of the particle.- \( B \) is the magnetic field strength.In the scenario of an electron beam within a TV picture tube, this force causes the electrons to deviate from their initial straight path. This deviation is circular due to the centripetal nature of the Lorentz force. Understanding this concept helps clarify why electron beams follow curved paths in magnetic fields, a principle directly applied in the behavior of cathode ray tubes.
Electron Beam
An electron beam is essentially a stream of electrons moving through space or a vacuum, as in the electron gun of a TV screen. These beams are utilized in various devices, not just old TV tubes. Electrons emitted from an electron gun are accelerated by applying a voltage. This acceleration increases their velocity. Once emitted, these electrons travel towards a screen, often manipulated by electric or magnetic fields to create an image. In this context, the electron beam travels across a distance while experiencing magnetic fields that can alter its path. Engineers and scientists control these beams precisely to direct them along desired paths. This control is essential in traditional CRT displays, where beams need precise alignment to illuminate different parts of the screen accurately.
Electron Velocity
Electron velocity refers to the speed at which electrons travel in a given direction. In a TV picture tube, electrons are accelerated by a voltage, and their velocity can be calculated using the kinetic energy principle. The equation to find this velocity is: \[ v = \sqrt{\frac{2 e V}{m}} \]Where:- \( v \) is the velocity of the electron.- \( e \) is the electron charge.- \( V \) is the accelerating voltage.- \( m \) is the mass of the electron.This equation shows that electron velocity increases with higher accelerating voltages, providing them with more energy. Understanding electron velocity is crucial for predicting how they will behave in fields, as faster electrons will be less deflected by a magnetic field compared to slower ones. This principle is pivotal in devices using electron beam technology, ensuring the accurate placing of images and data on screens.
Kinetic Energy
Kinetic energy is the energy that a body possesses due to its motion. For electrons in an electron beam, this energy is acquired through acceleration by an electric field within a cathode ray tube. The connection between kinetic energy and velocity is outlined by the formula:\[ \frac{1}{2} m v^2 = e V \]Where:- \( m \) is the mass of the electron.- \( v \) is the velocity.- \( e \) is the electron charge.- \( V \) is the accelerating voltage.Here, the kinetic energy of the electrons is equated to the energy gained from the voltage applied across the electron gun. By rearranging this formula, it allows us to derive the velocity of the beam electrons. The significance of kinetic energy in this context lies in its ability to dictate the speed and subsequent path of electrons under the influence of external fields, a principle that comes into play in adjusting screens effectively in various electronic apparatuses.

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Most popular questions from this chapter

\(\mathbf{A}\) thin 50.0 -cm-long metal bar with mass 750 g rests on, but is not attached to, two metallic supports in a uniform 0.450 - T magnetic field, as shown in Fig. 27.51 . A battery and a \(25.0-\Omega\) resistor in series are connected to the supports. (a) What is the highest voltage the battery can have without breaking the circuit at the supports? (b) The battery voltage has the maximum value calculated in part (a). If the resistor suddenly gets partially short- circuited, decreasing its resistance to 2.0\(\Omega\) , find the initial acceleration of the bar.

Crussed \(\vec{E}\) and \(\vec{B}\) Fields. A particle with initial velocity \(\overrightarrow{\boldsymbol{v}}_{0}=\left(5.85 \times 10^{3} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{j}}\) enters a region of uniform electric and magnetic fields. The magnetic field in the region is \(\overrightarrow{\boldsymbol{B}}=\) \(-(1.35 \mathrm{T}) \hat{\boldsymbol{k}} .\) Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge \((\mathrm{a})+0.640 \mathrm{nC}\) and \((\mathrm{b})-0.320 \mathrm{nC}\) . You can ignore the weight of the particle.

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A magnetic field exerts a torque \(\tau\) on a round current-carrying loop of wire. What will be the torque on this loop (in terms of \(\tau )\) if its diameter is tripled?

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