Why is the E cross B drift velocity constant in charged particle motion?

Click For Summary

Homework Help Overview

The discussion revolves around the concept of E cross B drift velocity in the context of charged particle motion within electric and magnetic fields. The original poster presents a scenario involving an electron in a uniform magnetic field, which subsequently experiences a uniform electric field, leading to questions about the conditions under which the drift velocity remains constant.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the forces acting on a charged particle in magnetic and electric fields, questioning the condition under which the acceleration is zero. They discuss the implications of vector notation and the significance of the velocity being perpendicular to the magnetic field.

Discussion Status

Participants are actively questioning the assumption that the acceleration is zero (dv/dt = 0) and whether this is a general condition or merely a hypothesis. Some suggest that the scenario described is specific and may not apply universally, while others reference related concepts such as velocity selectors to provide context.

Contextual Notes

There is an indication that the discussion may be constrained by the original poster's presentation of the problem, particularly regarding the completeness of the information provided about the particle's motion and the fields involved.

alannakamura
Messages
4
Reaction score
0

Homework Statement



good afternoon,

I am a problem with the E cross B drift velocity calculus from a charged particle. At first, an eletron is submitted a uniform magnetic field and have a circular trajetory. After, appear a uniform eletric field that create a drift velocity. The equations are:

m dv/dt = qE + qv X B.

Taking v constant or the mean aceleration zero, arrive the equation:

v = (E X B)/B^2

But I don't understand why the v is constant or mean aceleration is zero. Anyone can help me?
 
Physics news on Phys.org
I think you first have to realize that you are dealing with vector quantities.

With just a magnetic field of strength B you have the force acting on a charge q moving with velocity v is mdv/dt = q vXB (where vectors are bold face).

In this situation if you assume that the charge moves perpendicular to the magnetic field (what does that mean for the term vXB?) then, as you have stated, the charge moves in a circular path.

Question: What if the charge of the particle has a component of its velocity in the same direction as the magnetic field?

When you apply an electric field, E, the total force acting on the charge is mdv/dt = qE + q vXB.

The velocity you determine (making use of vector notation) v = EXB/(B.B) is the condition for dv/dt = 0 and its not clear to me whether you have presented all the information?
 
  • Like
Likes   Reactions: alannakamura
sorry for I don't put in vector notation, but I write thinking in vectors. The problem is why dv/dt=0?
 
No problem about the vector notation.

The reason I wondered if there was more to the question because, in general, dv/dt wouldn't be zero.

Look up "velocity selector", for example https://en.wikipedia.org/wiki/Wien_filter (must admit I learned something as well, I never realized that velocity selectors in this case were referred to as Wien filters).

If you look at the attached figure (taken from Wikipedia) you can see that the choice v = EXB/(B.B) is quite special.

velocity selector.png
 
I think nobody erason for dv/dt=0 too. But, in all places that I read they use this condition. Is it only a hipothesis? I want to know if it was a hypothesis or something general of the problem.
 
Something else you need to consider: To derive the expression for v you used the result
vXB = - E
then I guess you took the cross product with B giving
BX(vXB) = -BXE = EXB
Using the rules for vector cross product gives the left hand side as
v B2 + B v.B
so that you can only determine v for the case v.B = 0, namely that v and B are perpendicular
 
v is the perpendicular component of velocity related tho magnetic field.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
4K
Replies
19
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
1K
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K