Deflection of beam due to magnetic field

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Homework Help Overview

The problem involves the deflection of an electron beam in a transverse magnetic field, specifically within the context of a TV picture tube's electron gun. The task is to find the deflection of the beam due to the magnetic field, using variables such as charge, mass, voltage, and magnetic field strength.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the Lorentz force and its implications on the motion of charged particles in a magnetic field. Questions arise regarding the specific direction of the force and the nature of the resulting motion, leading to considerations of circular motion and the derivation of relevant equations.

Discussion Status

The discussion is progressing with participants actively engaging in exploring the concepts of force and motion related to the problem. Some guidance has been offered regarding the derivation of formulas, and there is an acknowledgment of the need for visual aids to enhance understanding.

Contextual Notes

There is mention of a specific formula for deflection that may not be commonly remembered, and participants express a need for diagrams to assist in understanding the derivation process. One participant is new to the forum and seeks additional resources for clarification.

ttiger2k7
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[SOLVED] Deflection of beam due to magnetic field

Homework Statement



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.

Find the approximate deflection of the beam due to this magnetic field. (Hint: Place the origin at the center of the electron beam’s arc and compare an undeflected beam’s path to the deflected beam’s path.)
Express your answer in terms of the variables B, D, e, m, and V.

Homework Equations



E/B = ((2eV)/m)

e/m = 1.758E-11 C/kg

The Attempt at a Solution



I would show my work, but I really have no idea how to. I tried just typing in the equation for E/B, but it didn't work. Is there a specific formula for deflection?
 
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There is a specific formula, but it's not an equation that one usually remembers. In any case, I would imagine you are expected to derive such a formula in your working. I'll help you along the way if you like.

A good place to start would be the Lorentz force. What is the magnitude of the force? In what direction does the force act? If a force acts in this direction, what sort of motion results?
 
Thanks for your help.

Let's see, the magnitude of the Lorentz force is qvB. It acts perpendicular to the force and velocity..am I supposed to know which direction specifically from the information given?
 
ttiger2k7 said:
Thanks for your help.

Let's see, the magnitude of the Lorentz force is qvB. It acts perpendicular to the force and velocity.
Correct. So, do you know of a certain 'type' of motion where the force always acts perpendicular to the velocity?
 
Is the motion of the lorentz force circular?
 
ttiger2k7 said:
Is the motion of the lorentz force circular?
Indeed, so if the motion is circular can you write down an equation for the radius of the circular path the electron follows whilst in the magnetic field?
 
Hmm, I believe it's

R = mv/qB
 
Correct, here's a link to the rest of the derivation, http://www.drchaos.net/drchaos/Whit/Lab_Manual/node20.html. It's very difficult to lead someone through the derivation without a diagram (as I've only just found out).
 
Last edited:
Thanks, got it.
 
  • #10
Hootenanny said:
Correct, here's a link to the rest of the derivation, http://www.drchaos.net/drchaos/Whit/Lab_Manual/node20.html. It's very difficult to lead someone through the derivation without a diagram (as I've only just found out).

Hi.. I'm new on this forum.. I'm reading this tread and try to learn it. Can you give any other reference link that have a diagram on it. Maybe it was removed from the link you state here because I can't see any diagram.

Thanks
 

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