Hi, I was working my way through the practice test booklet that ETS has available for d/l on their website (30 questions with answers and info about the test).
Anyway, I was wondering about question 29 which reads:
A long thin vertical wire has a net positive charge lambda per unit length and current I in the wire. A charged particle moves with speed u in a straight line trajectory parrallel to the wire and at a distance r from the wire. Auume that the only forces on the particle are those that result from the charge on and the current in the wire and that u is much less than c, the speed of light.
29. he particle is later observed to move in a straight line trajectory parrallel to the wire but at a distance 2r from the wire. If the wire carries a current I and the charge per unit length is still lambda, what is the speed of the particle?
I don't really have a good idea of how to approach the problem or why the answer is what it is (u).
In general, I was wondering if there are complete solutions to the GRE sample tests on the web anywhere?
Thanks!
(ETS booklet #29) Question from ETS GRE Physics Booklet.
For question 29:
The particle is moving parallel to the wire, so the Lorentz force (F=qvB) equals the Coulomb force (F=Eq).
The magnetic field due to a current carrying wire is B=u_0 I / 2 pi r
The Lorentz force then becomes F=q v u_0 I / 2 pi r
The electric field due to a long, charged wire is E=(1 / 2 pi e_0)(lam/r).
The Coulomb force becomes F= (1 / 2 pi e_0)(lam q/r).
Setting the two forces equal and ignoring constants, we get
q v I / r (is proportional to) lam q / r
Notice the r terms would cancel, implying that the speed would stay constant irrespective of r.
The particle is moving parallel to the wire, so the Lorentz force (F=qvB) equals the Coulomb force (F=Eq).
The magnetic field due to a current carrying wire is B=u_0 I / 2 pi r
The Lorentz force then becomes F=q v u_0 I / 2 pi r
The electric field due to a long, charged wire is E=(1 / 2 pi e_0)(lam/r).
The Coulomb force becomes F= (1 / 2 pi e_0)(lam q/r).
Setting the two forces equal and ignoring constants, we get
q v I / r (is proportional to) lam q / r
Notice the r terms would cancel, implying that the speed would stay constant irrespective of r.