Find the drift velocity

  • #1
334
44

Homework Statement


A 5.00-A current runs through a 12 gauge copper wire (diameter 2.05mm) and through a light bulb. Copper has 8.5 x 1028 free electrons per cubic meter.

c) At what see does a typical electron pass by any given point in the wire?

Homework Equations


J = nqvd

The Attempt at a Solution


The previous parts of the question I solved for current density.

J = 3.105 x 106 A / m2

n = 8.5 x 1028 electrons / m3

vd = J / n|q|

I substituted all my values in:

vd = (3.105 x 106 ) / (8.5 x 1028)(1.602 x 10-28)

vd = (3.105 x 106) / 13.167 m/s

vd = 2.28 x 105 m/s

the correct answer is vd = 0.111 mm/s

I'm sorry this is probably an arithmetic error but i can't see what I did wrong
 

Answers and Replies

  • #2
gneill
Mentor
20,925
2,867
Can you show us how you arrived at your current density?
 
  • #3
334
44
Can you show us how you arrived at your current density?

b) What is the current density?

cross section area = A = πr2 = πd2 / 4 = π(2.05x106) / 4 = 1.61 x 10-6

I = 5 A

Then current density = J = I / A = 5 / 1.61 x 10-6 = 3.105 x 106

..
Also the very first part:
a) How many electrons pass through the lightbulb each second?

I = 5A = 5 C/s

5 C/s * 1s = 5C

5C / (-1.602 x 10-28) = -3.121 x 1028 electrons
 
  • #4
gneill
Mentor
20,925
2,867
b) What is the current density?

cross section area = A = πr2 = πd2 / 4 = π(2.05x106) / 4 = 1.61 x 10-6
You didn't square your diameter when you calculated the cross sectional area.

a) How many electrons pass through the lightbulb each second?

I = 5A = 5 C/s

5 C/s * 1s = 5C

5C / (-1.602 x 10-28) = -3.121 x 1028 electrons
Use the magnitude of the charge on the electron. The sign of the charge is unimportant if you're counting quantity, which should be a positive number.
 
  • #5
334
44
You didn't square your diameter when you calculated the cross sectional area.


Use the magnitude of the charge on the electron. The sign of the charge is unimportant if you're counting quantity, which should be a positive number.
Ok I see what you mean .. so the correct cross sectional area is:

A = π((2.05)2 x 10-3*2) / 4 = 3.30 x 10-6
J = 5 / 3.30 x 10-6 = 1.51 x 106 A / m2

So I know my J and |q| are correct..I also forgot to mention that the problem gave "Copper has 8.5 x 1028 free electrons per cubic meter" which I use as my n value.

So vd = J / n|q|.
J = 1.51 x 106
n = 8.5 x 1028 free electron / m3
|q| = 1.602 x 10-28 C


vd = (1.51 x 106) / (8.5)(1.602) = (1.51 x 106) / 13.617 = 110,890 m/s

never mind I just realized |q| = 1.602 x 10-19.. so 110,890 * 10-9 = .111 x 10-3 which is the answer in the book.

Thanks for helping me on this.
 

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