ETS extra, Q#29

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hk
Posts: 16
Joined: Wed Jun 24, 2009 3:43 am

ETS extra, Q#29

Postby hk » Thu Oct 22, 2009 5:10 am

A long, thin, vertical wire has a net positive charge L per unit length. In addition, there is a current I in the wire. A charged particle moves with speed u in a straight line trajectory, parallel to the wire and at a distance r from the wire. Assume 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.

Q:
The particle is later observed to move in a straight line trajectory, parallel 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 L, the speed of the particle is:
A) 4u
B) 2u
C) u
D) u/2
E) u/4

Correct answer is (C).

Actually this is related to previous problem (Q28) in which the current was reduced to I/2, then doubling the speed of the particle is necessary to keep it in the same trajectory at distance r. So in this problem the magnetic force is towards the wire direction and electric force is away from the wire -- if both forces are of the same strength then the particle will keep moving straight.
What I don't understand is why it is still moving at speed u at distance 2r, while the other parameters are still the same value. The factor 2 increase of distance to the wire will decrease magnetic force by 2 (B ~ 1/r), while it will decrease electric force by 4 (F ~ 1/r^2). So how come the speed is still the same?? I expect it to be u/2. What did I miss here? :shock:

sravanskarri
Posts: 58
Joined: Sat Jun 14, 2008 10:19 pm

Re: ETS extra, Q#29

Postby sravanskarri » Thu Oct 22, 2009 12:05 pm

Think, the catch here is the E field due to a charge distribution of infinite length at distance "r" is proportional to "1/r " (Gauss's law & cylindrcial symmetry) rather than "1/r^2" so "u" will be the same

kroner
Posts: 218
Joined: Fri Sep 25, 2009 1:58 am

Re: ETS extra, Q#29

Postby kroner » Fri Oct 23, 2009 4:01 pm

Or an alternate way to think about it is that in the rest frame of the charged particle, the charge density on the wire must be zero, so any other charged particle at rest in that frame will also have no force acting on it.

Incidentally, the condition that u << c is not necessary.

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hk
Posts: 16
Joined: Wed Jun 24, 2009 3:43 am

Re: ETS extra, Q#29

Postby hk » Sun Oct 25, 2009 10:18 pm

Thanks for the comments.

sravanskarri: That explains it very well! So both forces depend by 1/r then, and at any distance it will travel at speed u..

kroner: Sorry I don't understand. How come the charge density becomes zero? Furthermore, if it is zero then there would be no electric force to oppose the magnetic force and the particle would be deflected.

kroner
Posts: 218
Joined: Fri Sep 25, 2009 1:58 am

Re: ETS extra, Q#29

Postby kroner » Mon Oct 26, 2009 12:09 am

We know the net force on the original particle is zero. In the rest frame of the particle there is no magnetic force (there is a magnetic field but the particle has zero velocity), so the electric force must be zero as well. Therefore the charge density on the wire in that frame is zero.




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