6.96 · 10²⁸ m⁻³
Further explanation
Given:
- A metallic wire has a diameter of 4.12 mm → r = 2.06 mm = 2.06 · 10⁻³ m
- The current in the wire is I = 8.00 A = 8.00 C/s
- The average drift velocity is v = 5.40 · 10⁻⁵ m/s.
- The elementary charge Q or e = 1.60 ∙ 10⁻¹⁹ C
Question:
What is the density of free electrons in the metal?
Expressed as n = number of charges carriers per unit volume.
The Process:
Drift velocity, or it can be also referred to as axial drift velocity, represents the average velocity with which the free electrons drift under the influence of an external field.
The relationship between I, n, q, A, and v can be written as a formula, i.e.,
[tex]\boxed{ \ I = nAQv \ }[/tex] and for electrons, [tex]\boxed{ \ I = nAev \ }[/tex]
We set n as the subject being asked, so [tex]\boxed{ \ n = \frac{I}{Aev} \ }[/tex]
Prepare the cross-sectional area.
[tex]\boxed{ \ A = \pi r^2 \rightarrow A = \pi (2.06 \cdot 10^{-3} )^2=1.33 \cdot 10^{-5} \ m^2 \ }[/tex]
And now, let us find out the number of n. Substitute all the data above into the formula.
[tex]\boxed{ \ n = \frac{8.00 \ Cs^{-1}}{(1.33 \cdot 10^{-5} \ m^2)(1.60 \cdot 10^{-19} \ C)(5.40 \cdot 10^{-5} \ ms^{-1})} \ }[/tex]
Thus, the density of free electrons in the metal is [tex]\boxed{ \ n=6.96 \cdot 10^{28} \ m^{-3} \ }[/tex]
Learn more
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