Calculating Density of Free Electrons in a Metal Wire

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To calculate the density of free electrons in a metallic wire with a diameter of 4.12 mm and a current of 8 A, the relevant formula is I = nqva, where I is current, n is the density of free electrons, q is the charge of an electron, and va is the drift velocity. The cross-sectional area A can be calculated using the diameter, resulting in A = (2.06 mm)^2 * π. The drift velocity is given as 5.4 x 10^-5 m/s, and the charge of an electron is approximately 1.6 x 10^-19 C. The discussion emphasizes the need to determine n, with the SI unit for density being m^-3, and clarifies that q is necessary for solving the equation.
kyang002
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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 have considered a wire of cross-sectional area A, with current I. V is the drift velocity of the free charges in the conductor and n is the number of charged particles per unit volume. Then the current is given by I = nqva, where q is the magnitude of the charge on a charge carrier.

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

I am unsure of how to finish the problem. We are looking for n. But what is q? Are all my other numbers correct?
 
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All u need to know (and apparently u don't) is that
j=n_{el}e v

Daniel.
 
I am unsure of what the equation you gave was. In my book, I was given I = nqva.

Care to explain on how to do the problem?
 
Yes,the two formulas are obviously equivalent.It's better to use yours as it has already built in the numbers/variables u need to plug in.
What is "n" (or as i denoted it,n_{el.}) & what is its SI unit...?

Daniel.
 
The SI unit is m^-3.

How do I find q? Or do I even need q in this equation?
 
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