Question

A current of I=8.0A is flowing in a typical extension cord of length L=3.00m. The cord is made of copper wire with diameter d=1.5mm. The charge of the electron is e=1.6×10−19C. The mass of the electron is me=9.1×10−31kg. The resisitivity of copper is rho=1.7×10−8Ω⋅m. The concentration of free electrons in copper is n=8.5×1028m−3.
Find the drift velocity vd of the electrons in the wire

Answer 1

Answer: $v_d=3.33 \times 10^{-4} m.s^-1$

Explanation: Drift velocity is the velocity attained by the chaged particles in a material due to an electric field.

Most electrical signals carried by currents travel at speeds on the order of $10^8 m.s_-1$, a significant fraction of the speed of light.

Mathematically, we have:

$I=n.A.q.v_d$....................................(1)

where:

• I= current in the conductor (A)
• A= cross-sectional area normal to the flow of current (m²)

• q= charge on each charge carrier (coulombs, C)
• n= no. of charge carriers per unit volume i.e., charge density
• $v_d$= drift velocity $m.s^-1$

Given:

• I= 8 A
• Diameter of wire, d= 1.5 mm =$1.5\times 10^{-3} m$
• q= $1.6\times 10^{-19} C$
• n= $8.5\times 10^{28} m^{-3}$

Asked:

• $v_d$

Firstly, we find area:

$A=\pi \frac{d^{2} }{4}$

$A=\pi \frac{1.5\times 10^{-3}}{4}$

$A= 1.767\times 10^{-6} m^{2}$

Now, putting the required values in eq. (1)

$8=(8.5\times 10^{28}) \times (1.767\times 10^{-6}) \times (1.6\times 10^{-19})\times v_d$

$v_d=\frac{8}{8.5\times 10^{28}\times 1.6\times 10^{-19}\times 1.767\times 10^{-6} }$

$v_d=3.33 \times 10^{-4} m.s^-1$