GREPhysics.NET: GR8677 #9

GR8677 #9

Problem

GREPhysics.NET Official Solution

Electromagnetism $\Rightarrow$ }Current Density

There are (at least) two ways to solve this problem. If one has a decent memory or might make for a decent experimenter, one should remember that the electron drift velocity is on the order of $10^{-4}m/s$ . Choice D is the only thing that fits this order. If one is a theorist who can't remember jack, then one would have to do a bit more work:

Recall the basic equation:

$j=\frac{I}{A}=nev,<br />$

where $n=\frac{N}{V}=1E28$ , i.e., the number of electron per volume, and $v$ is the drift velocity. The area is $A=\pi(\frac{d}{2})^2$ .

Solving for $v$ , and making the appropriate approximate plug-in's:

$\frac{4I}{\pi d^{2}}\frac{1}{ne}\approx\frac{4I}{\pi (\frac{2}{100})^{2}}\frac{1}{(1E28)(1.5E-19)}\approx 3E-4, <br />$

which is closest to choice D.

Alternate Solutions

thinkexist 2012-10-12 14:37:10	You can solve this by common sense and a little physics. Firstly, it doesn't make much sense for an electron to be traveling speeds comparable to the size of a proton per second. Exclude A, B, C. You also know that current works by the propagation of the coulomb force so when you put one electron into a filled tube of electrons a different electron on the other side of the tube pops out because of the coulomb force. Having said that, it doesn't really make much sense for the electron to be traveling 10^3 m/s when there are a ton of electrons it can bounce into including the atoms. Exclude E. What we have left is D. Reply to this comment
flyboy621 2010-11-09 20:51:08	$I=\frac{dq}{dt}=\frac{dq}{dx}\frac{dx}{dt}=(enA)(v_d)$ , where $e$ is the electron charge, $n$ is the electron density, $A$ is the cross-sectional area of the wire, and $v_d$ is the drift velocity. Solve for $v_d$ , plug in rough numbers, and you get: $v_d=\frac{I}{enA}=\frac{100}{\pi(.01^2)(10^{28})(1.6\times10^{-19})}\approx\frac{10^{-3}}{5}=2\times10^{-4}$ , which matches (D). Reply to this comment
Ben 2005-12-01 13:23:38	I did it using unit analysis sort of. Boils down to the same thing but I get much closer to 2E-4. 100A = 100 C/s (100 C/s)(1/1.6E-19 e-/C) = 6.25E20 e-/s ${pi}$ r^2 = ${pi}$ E-4 ( ${pi}$ E-4 m^2)(1E28 e-/m^3) = ${pi}$ E24 e-/m (of wire) Finally: (6.25E20 e-/s)*(1/ ${pi}$ E24 m/e-) which gives pretty close to 2E-4 (hopefully my latex pi's worked) Reply to this comment

Comments

ernest21
2019-08-10 03:09:47

I think m was mass not meter, otherwise it made no sense at all. strategy gamer

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joshuaprice153
2019-08-08 06:10:59

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thinkexist
2012-10-12 14:37:10

You can solve this by common sense and a little physics.
Firstly, it doesn't make much sense for an electron to be traveling speeds comparable to the size of a proton per second. Exclude A, B, C.

You also know that current works by the propagation of the coulomb force so when you put one electron into a filled tube of electrons a different electron on the other side of the tube pops out because of the coulomb force. Having said that, it doesn't really make much sense for the electron to be traveling 10^3 m/s when there are a ton of electrons it can bounce into including the atoms. Exclude E.

What we have left is D.

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natestree
2011-03-22 14:47:19

So, I figured the correct answer w/o any knownledge on drift velocity and some simple unit analysis.
Note the units for quantities given.
current is charge per time.
diameter is meters
density is # of particle per cubic meter

we want something in units of meters per second. Therefor, we can use the quantities given to come up with the relation:
$v_{drift}$ = $\frac{I}{\rho d^2 e}$
Where I is current, $\rho$ is the density, e is charge of an electron, and d is the diameter. This gives an answer ~ $10^{-3}$ which is closest to option D.

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flyboy621
2010-11-09 20:51:08

$I=\frac{dq}{dt}=\frac{dq}{dx}\frac{dx}{dt}=(enA)(v_d)$ ,

where $e$ is the electron charge, $n$ is the electron density, $A$ is the cross-sectional area of the wire, and $v_d$ is the drift velocity.

Solve for $v_d$ , plug in rough numbers, and you get:

$v_d=\frac{I}{enA}=\frac{100}{\pi(.01^2)(10^{28})(1.6\times10^{-19})}\approx\frac{10^{-3}}{5}=2\times10^{-4}$ , which matches (D).

walczyk
2011-04-06 16:27:46

this is by far the best way to do this. its perfect when you can derive it so quickly and easily like that, i just wish i recognized it too. forget all the equations. follow this. intuition is our friend

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kicksp
2007-10-29 09:42:47

Inspection shows that the choices are separated by several orders of magnitude. Hence, we need only an order of magnitude estimate:

$v \sim \frac{10^2}{10^{-4}10^{28}10^{-19}} =10^{-3}$

The answer is clearly (D). Throw in the factor $1/\pi$ and you obtain $v\sim10^{-4}$ .

Go maroons!

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Andresito
2006-03-14 22:16:53

Your solution stills showing the area proportional to d^(2*ne). Although, it is not to worry.

thanks Yosun.

Andresito
2006-03-14 22:18:21

"Nevermind"

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wishIwasaphysicist
2006-01-24 11:22:51

oops...you were right. I took the diamter squared and not radius squared. It comes out to 2 E -4 m/s and not 3 E - 4 m/s though.

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wishIwasaphysicist
2006-01-24 11:08:34

I don't see where the 4 in front of current (I) came from. If you remove it, you get exactly the answer choice: 2 E -4 m/s.

grae313
2007-10-07 15:11:30

It comes from squaring the radius (d/2) to get the cross sectional area

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Ben
2005-12-01 13:23:38

I did it using unit analysis sort of. Boils down to the same thing but I get much closer to 2E-4.

100A = 100 C/s

(100 C/s)*(1/1.6E-19 e-/C) = 6.25E20 e-/s

${pi}$ r^2 = ${pi}$ E-4

( ${pi}$ E-4 m^2)*(1E28 e-/m^3) = ${pi}$ E24 e-/m (of wire)

Finally:

(6.25E20 e-/s)*(1/ ${pi}$ E24 m/e-) which gives pretty close to 2E-4

(hopefully my latex pi's worked)

yosun
2005-12-02 01:10:35

ben: your LaTeX came out iffy. you should have used a \ (back-slash) sign before your pi's (and other latex commands) so that you type-set $\pi$ instead of $pi$ . it's been manually corrected.

(for everyone else with typos i didn't get to catch: over winter break, i'd code options for edit-posts, among other features.)

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