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The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temperature of the wire is 20.0∘C. Use the resistivity at room temperature for copper is rho = 1.72×10^−8 Ω⋅m.

Required:
a. What is the drift speed vdvd of the electrons in the wire?
b. What is the potential difference between two points in the wire that are separated by 25.0 cm?

Respuesta :

onyebuchinnaji

(a) The drift speed of electrons in the wire is 2.22 x 10⁻⁴ m/s.

(b) The potential difference between two points in the wire is 0.013 V.

The given parameters;

  • electron density of the copper wire, n = 8.5 x 10²⁸ electrons/m³.
  • electric field, E = 0.0520 N/C
  • temperature of the wire, t = 20 ⁰C
  • resistivity of the copper wire, ρ = 1.72 x 10⁻⁸ Ω⋅m

The drift speed of electrons in the wire is calculated as follows;

[tex]v_d = \frac{I}{qn A} \\\\but , \ \frac{I}{A} = \frac{E}{\rho} \\\\v_d = \frac{E}{qn \rho}[/tex]

where;

  • E is the electric field
  • q is charge of electron = 1.602 x 10⁻¹⁹ C

[tex]v_d = \frac{0.052}{1.602 \times 10^{-19} \times 8.5\times 10^{28} \times 1.72 \times 10^{-8}} \\\\v_d = 2.22 \times 10^{-4} \ m/s[/tex]

The potential difference between two points in the wire, separated by 25 cm;

V = Ed

where;

  • d is the distance of separation = 25 cm = 0.25 m

V = 0.052 x 0.25

V = 0.013 V

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