Show the drift velocity is ExB/B^2

AI Thread Summary
A charged particle experiences drift velocity in uniform electric and magnetic fields, with the electric field perpendicular to the magnetic field. The drift velocity is derived as vd = (E×B)/B^2, starting from the equation F = e(E + v×B) = 0, indicating uniform motion. The discussion emphasizes the assumption of two-dimensional drift, clarifying that motion in the third dimension does not contribute to drift but may represent acceleration. Participants debate the implications of this two-dimensional assumption and its relevance to the concept of drift velocity. Ultimately, "drift" refers to the velocity component in the defined two-dimensional plane, distinguishing it from motion in other dimensions.
shedrick94
Messages
30
Reaction score
0
HW Template missing as it was moved from another forum
A charged particle drifts in uniform, constant magnetic and electric fields. The electric field, E, is perpendicular to the magnetic field, B.

Show that the drift velocity is given by vd = (E×B)/B2

Heres where I get to:
F=e(E+vxB)=0 as v is uniform.Therefore E+vxB=0.

Take vector product of B with both sides.

BxE +Bx(vxB)=0.

Using identity Ax(BxC) = B(A.C)-C(A.B)

I get BxE+v(B.B)-B(B.v)=0
Then I don't know where to go from here.
 
Physics news on Phys.org
You can assume that the drift happens in two dimensions, so you can say something about the relative direction of v and B which simplifies (B.v).
In three dimensions the total velocity does not have to follow the initial equation, so you need that assumption.
 
mfb said:
You can assume that the drift happens in two dimensions, so you can say something about the relative direction of v and B which simplifies (B.v).
In three dimensions the total velocity does not have to follow the initial equation, so you need that assumption.
Why are you allowed to assume the drift is in two dimensions though?
 
The component in the third dimension is not a drift.
 
mfb said:
The component in the third dimension is not a drift.
Would that be an acceleration then?
 
A motion in the third direction would stop quickly in matter. In vacuum, the charged particle could freely keep moving in that direction, without influencing the two-dimensional motion in the other directions.
 
mfb said:
A motion in the third direction would stop quickly in matter. In vacuum, the charged particle could freely keep moving in that direction, without influencing the two-dimensional motion in the other directions.
Can I ask what it means by a drift velocity then?
 
The velocity in the two-dimensional plane I mentioned.
 
mfb said:
The velocity in the two-dimensional plane I mentioned.
yes but what does it specifically mean by 'drift'. You said the velocity in the 3rd dimension is not a drift.
 
  • #10
I guess it is just convention to call that drift.
 
Back
Top