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

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)

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Comments

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

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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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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