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13 July, 09:19

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.

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  1. 13 July, 12:02
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    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
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