Question

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?

Answer 1

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, $v_{d}$ , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

$v_{d}$ = $\frac{I}{qnA}$             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = $\frac{I}{A}$ = $\frac{E}{p}$

Equation (i) can then be written as follows;

$v_{d}$ = $\frac{J}{qn}$ = $\frac{E}{qnp}$

$v_{d}$ = $\frac{E}{qnp}$      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

$v_{d}$ = $\frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}$

$v_{d}$ = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V