30: Current in a Wire

INTRO:
A metallic wire has a diameter of 4.12 mm. When the current in the wire is 8.00 A, the drift velocity is 5.40×10⁻⁵ m/s.
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PART A:
What is the density of free electrons in the metal?
Express your answer numerically in m⁻³ to two significant figures.

SOLUTION:

Givens/ Conversions:​

wire diameter ≡ d = 4.12 mm = 4.12×10^-3 m
Current ≡ I = 8 A
drift velocity ≡ v_d = 5.40×10⁻⁵ m/s.

Using eq. (30.13) ... J = I/A = n⋅e⋅v_d

→ n = I/(A⋅e⋅v_d)​

e = 1.6×10^-19 C
& A = π/4⋅d²

∴ n = I/(π/4⋅d²⋅e⋅v_d)

= (8 A)/(π/4⋅(4.12×10^-3 m)²⋅(1.6×10^-19 C)⋅(5.40×10⁻⁵ m/s))
⇒ n = 6.9×10²⁸ m^-3
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