# What are (a) the current density and (b) the electron drift speed?

## Homework Statement

"The current in a 1.70mm x 1.70mm square aluminum wire is 1.70 A. What are (a) the current density [in MA/m^2] and (b) the electron drift speed [in micrometers/second]?"

J = I/A

## The Attempt at a Solution

I plugged in the values for current (1.7 Amps) and the area (I got 2.89*10^-6 meters squared) into J = I/A, and I get an answer of 588235 A/m^2, which ended up being wrong. I rather saw this coming, since the question asked for an answer in MA/m^2...but I have no idea what they're talking about with that capital M! Does anyone have any idea what's going on with that?

(of course, also - could someone help me get started with the second part?)

rock.freak667
Homework Helper
M would stand for "mega" which is the prefix for 106

So 106 N (newtons) would be 1 MN.

Hah, I should've known; thanks, I got the first part. The second part, however, I got wrong:

I used J = n_e * e * v_d with the values:

588235 = 6*10^28 * 1.6*10^-19 * v_d and solved for v_d

I got 6.13*10^-5 m/s and then converted to 6.13*10^-11 micrometers / second, and got it wrong. I can't see my error; could someone help me out?

rock.freak667
Homework Helper
Where did you get the number of electrons from?

But you had 6.13*10^-5 m/s, "micro" is 10^-6 so, if you re-write it as 61.3*10^-6 m/s you will get it in micrometers/second.

Where did you get the number of electrons from?

But you had 6.13*10^-5 m/s, "micro" is 10^-6 so, if you re-write it as 61.3*10^-6 m/s you will get it in micrometers/second.

No, that wasn't it...

It's really quite frustrating. This looks to be essentially the only part on the assignment I won't understand/get. Does anyone know what's going on with it?

rock.freak667
Homework Helper
No, that wasn't it...

It's really quite frustrating. This looks to be essentially the only part on the assignment I won't understand/get. Does anyone know what's going on with it?

I still don't get where you got the number of electrons per unit volume from I still don't get where you got the number of electrons per unit volume from Oh man lol, sorry about that! It's the electron density for aluminum: 6.0*10^28 1/m^3.

But yea, like...I don't know. I've tried 6.13 * 10^-5, 6.13*10^-11, 6.13*10^-6, and 61.3*10^-6, with no luck. Should I try one of these again just in case, or what do you think it is?

rock.freak667
Homework Helper
Oh man lol, sorry about that! It's the electron density for aluminum: 6.0*10^28 1/m^3.

But yea, like...I don't know. I've tried 6.13 * 10^-5, 6.13*10^-11, 6.13*10^-6, and 61.3*10^-6, with no luck. Should I try one of these again just in case, or what do you think it is?

You should try plugging in the all the numbers into the formula instead of using the approximations, that may be causing an error.

Yea I....really just can't see another way out besides dong the following:

J = n_e * e * v_d

588235 = 6*10^28 * 1.6*10^-19 * v_d

v_d = 6.13*10^-5 m/s = 6.13*10^-11 um/s

Where is this wrong...? /o_o\ 1.6*10^-19 is the charge of an electron...6*10^28 1/m^3 is the electron density, and 588235 is *definitely* right (even plugging in what I'm told is 100% right by the computer, 588000 with limited sig figs) gives me the same answer. I just don't get it.

rock.freak667
Homework Helper
Yea I....really just can't see another way out besides dong the following:

J = n_e * e * v_d

588235 = 6*10^28 * 1.6*10^-19 * v_d

v_d = 6.13*10^-5 m/s = 6.13*10^-11 um/s

Where is this wrong...? /o_o\ 1.6*10^-19 is the charge of an electron...6*10^28 1/m^3 is the electron density, and 588235 is *definitely* right (even plugging in what I'm told is 100% right by the computer, 588000 with limited sig figs) gives me the same answer. I just don't get it.

vd = 6.13*10^-5 m/s =

1 um = 10-6 m so 1 m = 106 um

so vd = 6.13*10-5*106 um/s = 61.3 um/s

Oh god...lol, I've been really stressed and all over the place lately, I feel dumb for missing that XD Thanks a lot rock, got it!