Current in a Wire

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[SOLVED] Current in a Wire

Homework Statement


A metallic wire has a diameter of 4.12 mm. When the current in the wire is 8 A, the drift velocity is 5.4*10^-5 m/s.

What is the density of free electrons in the metal?

I = 8
n = density??
A = (2.06)^2 * 3.14
v = 5.4 x 10^-5
q = 1.6x10^-19

Express your answer numerically in m^-3 to two significant figures.

Homework Equations



I = ne[tex]v_{d}[/tex]A

A = [tex]\pi r^2[/tex]

The Attempt at a Solution



A = [tex]\pi r^2[/tex] = [tex]\pi (2.06x10^-3)^2[/tex] = 1.3 x 10^-5

So manipulating the equation, I get n = I / evA

n = [tex]\frac{8}{(1.6x10^-19)(5.4x10^5)(1.3 x 10^-5)}[/tex] = 6.94 x 10^28

and since I need it in millimeters, I multiply it by 1000 (since I did the calculations in SI) and get 6.94 x 10 ^31, but it says it's wrong. Any ideas where I went wrong?
 
Last edited:

Answers and Replies

  • #2
Kurdt
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The question you've posted says express the answer in m-3. Does the answer need to be in meters cubed or millimeters cubed?
 
  • #3
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it's to the power of -3 so m^-3 is millimetres
 
  • #4
mukundpa
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it is not 10^-3 m it is 1/(m^3) where m^3 is the unit of volume (cubic meter)
 
  • #5
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so the answer would be 1 / 6.94 x 10^28 ?
 
  • #6
Kurdt
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The variable n is a number density or the number of electrons per unit volume so it is already expressed in terms of m-3
 
  • #7
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Thank You.
 

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