- #1
Abdullah Almosalami
- 49
- 15
Homework Statement
A current of 1600A exist in a rectangular (0.4 x 16 cm) bus bar. The electrons move at an average velocity of v. If the concentration of electrons is 1029 per cubic meter, and they are uniformly distributed, what is v?
Knowns
- Current (i) = 1600A = 1600 x 1018 aA
- Charge per Electron (e) = 1.6022 x 10-19C = 0.16022 aC
- Area of Cross-Section of Bar (A) = 0.4 (w) x 16 cm (h) = 0.004 x 0.16 m --> A = w x h
- Velocity of Current (v)
Homework Equations
- Current = Charge Moving Through a Cross-sectional Area per Unit Time --> i = C/t
- Number of Electrons Passing Through Said Cross-Sectional Area Per Unit Time (N) = Current / (Charge / Electron) --> N = i / e
- Volume of a Rectangle (V) = Length (l) x Width (w) x Height (h) --> V = l x w x h --> l = V / (w x h)
- Speed = Distance / Time
The Attempt at a Solution
Ok, just to state this beforehand, my issue is that the velocity I end up getting seems way too low.
To help visualize this problem, I drew this:
(in case that doesn't work, here is the link.)
That's the rectangular bar; I'm calling the longest part of the bar the length and the other two sides, which are given (0.4cm by 16cm), width and height respectively. Since we're given the density of electrons relative to a cubic meter, that section from the left of the bar to the cross-section is 1 cubic meter of the bar. The length of this section is l. I am calling the length of the volume of electrons cleared per second as x. This is all obviously not drawn to scale, but just to help visualize.
My approach is:
- Find the number of electrons that go through that given cross-sectional area per second N given the current i at 1600A
- Find the length l of a cubic meter of the bar (<-- actually realized this step is not needed in my approach)
- Find how much of the volume of the bar is cleared per second V based on the results of step 1 and the density of electrons we're given (1029 per cubic meter).
- Find the length x of that volume using volume formula and two known lengths 0.004m and 0.16m.
- Since we've been setting the time at 1s, v = x / t = x / (1s) = x m/s.
- N = 9.9862 x 1021, so that is 9.9862 x 1021 electrons moving through any given cross-section per second.
- l = 1562.5m, so that is 1562.5 meters of length per cubic meter of volume (--> again, this was not needed in the end).
- V = N / (1029 / m3) = 9.9862 x 10-8 m3, so that is 9.9862 x 10-8 m3 cleared every second from the bar.
- x = V / (0.004m x 0.16m) = 1.56 x 10-4.
- v = 1.56 x 10-4 m/s = 156 µm / s.
Before you answer, the only solution I've seen after checking this forum, Chegg, and Yahoo Answers, is people using the fact that i = C/t, then solving for t and setting the charge to be the charge of an electron, then using that in v = x/t (using the x as I've defined above). That can work, and I'll go through that approach in this paragraph, but the way I've seen it done, people assume x = 1562.5m, or the likes, which seems to me a wrong assumption. The t you would get there is the amount of time it takes for the charge equivalent to an electron to pass through, which you can assume indicates 1 electron passing through, but the assumption that that the electron started out at x distance from the cross-sectional area is totally random. It could have been 1 ym away and that still wouldn't break any rules as far as we're given. The only way that approach works is if you knew how far the next electron from the cross-sectional area was. Using the density given in the problem, and the fact that we can assume uniform distribution, that would give us 1 electron per 10-29 m3, which would give us x = 1.56 x 10-26 m. Solving for time using i = C/t --> t = C/i gives us t = 1.0013 x 10-22 s. Using x and t, v = x / t = 1.56 x 10-4 m/s = 156 µm/s. Same deal here...
If the approach isn't wrong, then what's going on here? Shouldn't electrons be moving near the speed of light?