A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers per unit vol

Question

A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers per unit volume is 8.49 × 1028 m-3. Assuming the current is uniform, calculate the (a) current density and (b) electron drift speed.

in progress 0
Khải Quang 3 years 2021-08-19T09:13:39+00:00 1 Answers 24 views 0

Answers ( )

  1. kiet
    0
    2021-08-19T09:14:45+00:00
    Reply

    Answer:

    a) 4.9*10^-6

    b) 5.71*10^-15

    Explanation:

    Given

    current, I = 3.8*10^-10A

    Diameter, D = 2.5mm

    n = 8.49*10^28

    The equation for current density and speed drift is

    J = I/A = (ne) Vd

    A = πD²/4

    A = π*0.0025²/4

    A = π*6.25*10^-6/4

    A = 4.9*10^-6

    Now,

    J = I/A

    J = 3.8*10^-10/4.9*10^-6

    J = 7.76*10^-5

    Electron drift speed is

    J = (ne) Vd

    Vd = J/(ne)

    Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

    Vd = 7.76*10^-5/1.3584*10^10

    Vd = 5.71*10^-15

    Therefore, the current density and speed drift are 4.9*10^-6

    And 5.71*10^-15 respectively

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )