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

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SUMMARY

The discussion centers on the E cross B drift velocity in charged particle motion, specifically addressing the equation v = (E X B)/B². Participants clarify that this equation represents a condition where the acceleration (dv/dt) is zero, indicating a constant drift velocity when a charged particle is subjected to both electric (E) and magnetic (B) fields. The conversation highlights the importance of vector quantities in understanding the motion and the special condition under which this drift velocity remains constant, particularly when the velocity vector is perpendicular to the magnetic field vector.

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  • Understanding of vector calculus and cross products
  • Familiarity with the Lorentz force law
  • Knowledge of charged particle dynamics in electromagnetic fields
  • Basic principles of circular motion in magnetic fields
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alannakamura
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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?
 
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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?
 
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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.
 

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