contestada

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

Respuesta :

Answer: [tex]v_d=3.33 \times 10^{-4} m.s^-1[/tex]

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 [tex]10^8 m.s_-1[/tex], a significant fraction of the speed of light.

Mathematically, we have:

[tex]I=n.A.q.v_d[/tex]....................................(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
  • [tex]v_d[/tex]= drift velocity [tex]m.s^-1[/tex]

Given:

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

Asked:

  • [tex]v_d[/tex]

Firstly, we find area:

[tex]A=\pi \frac{d^{2} }{4}[/tex]

[tex]A=\pi \frac{1.5\times 10^{-3}}{4}[/tex]

[tex]A= 1.767\times 10^{-6} m^{2}[/tex]

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

[tex]8=(8.5\times 10^{28}) \times (1.767\times 10^{-6}) \times (1.6\times 10^{-19})\times v_d[/tex]

[tex]v_d=\frac{8}{8.5\times 10^{28}\times 1.6\times 10^{-19}\times 1.767\times 10^{-6} }[/tex]

[tex]v_d=3.33 \times 10^{-4} m.s^-1[/tex]

nuhulawal20

The drift velocity [tex]V_D[/tex] of the electrons in the wire is 3.329 × 10^⁻⁴ meter per seconds.

Given the data in the question;

Current; [tex]I = 8.0A[/tex]

Length of cord; [tex]L=3.00m[/tex]

Diameter of copper wire; [tex]d = 1.5mm[/tex]

Charge of the electron; [tex]e = 1.6*10^{-19}C[/tex]

Mass of electron; [tex]m_e=9.1*10^{-31}kg[/tex]

Resistivity of copper; [tex]R=1.7*10^{-8}ohm[/tex]

Concentration of free electrons in copper; [tex]n= 8.5 * 10^{28}m^{-3}[/tex]

Drift velocity of the electrons in the wire = ?

Drift velocity is the average velocity with which an electrons drift in the opposite direction of its field.

[tex]V_D = \frac{I}{n*e*A}[/tex]

Where I is current flow,  n is the free electron density, e is the charge of an electron  and A is the cross sectional area.

First we cross-sectional area of the wire:

Area; [tex]A =\frac{\pi }{4} d^2[/tex]

We substitute in our given values

[tex]A =\frac{\pi }{4} (1.5*10^{-3}m)^2\\\\A = 1.767*10^{-6}m^2[/tex]

So,  Cross-sectional area of the wire is [tex]1.767*10^{-6}m^2[/tex]

Now we substitute all our values into equation for Drift velocity

[tex]V_D = \frac{8}{(8.5*10^{28})(1.6*10^{-19})(1.767*10^{-6})}[/tex]

[tex]V_D = 3.329 * 10^{-4}m/s[/tex]

Therefore, the drift velocity [tex]V_D[/tex] of the electrons in the wire is 3.329 × 10⁻⁴ m/s

Learn more; https://brainly.com/question/14997979

Otras preguntas