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

In summary, the current density in the 1.70mm x 1.70mm square aluminum wire is 588235 A/m^2 and the electron drift speed is 61.3 um/s. The electron density of aluminum is 6.0*10^28 1/m^3 and the charge of an electron is 1.6*10^-19 C.
  • #1
Netsurfer733
17
0

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]?"

Homework Equations



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?)
 
Physics news on Phys.org
  • #2
M would stand for "mega" which is the prefix for 106

So 106 N (Newtons) would be 1 MN.
 
  • #3
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?
 
  • #4
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.
 
  • #5
rock.freak667 said:
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?
 
  • #6
Netsurfer733 said:
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 :confused:

But your numbers are correct.
 
  • #7
rock.freak667 said:
I still don't get where you got the number of electrons per unit volume from :confused:

But your numbers are correct.

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?
 
  • #8
Netsurfer733 said:
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.
 
  • #9
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.
 
  • #10
Netsurfer733 said:
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
 
  • #11
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!
 

What are the current density and electron drift speed?

The current density refers to the amount of electric current per unit area, while the electron drift speed refers to the average speed at which electrons move through a material due to an applied electric field.

How are the current density and electron drift speed related?

The current density is directly proportional to the electron drift speed, meaning that as the electron drift speed increases, so does the current density.

What factors affect the current density and electron drift speed?

The current density is affected by the amount of current flowing through a material and the cross-sectional area of the material. Meanwhile, the electron drift speed is influenced by the strength of the electric field, the type of material, and the temperature.

How can the current density and electron drift speed be measured?

The current density can be measured using a device called an ammeter, which measures the amount of current passing through a given area. The electron drift speed can be indirectly measured by using the material's conductivity and the applied electric field.

Why are the current density and electron drift speed important in electrical circuits?

The current density and electron drift speed are important because they help us understand how electricity moves through a material, which is essential in designing efficient and safe electrical circuits. They also help us analyze and troubleshoot any issues that may arise in a circuit.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
10K
  • Introductory Physics Homework Help
Replies
7
Views
6K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top