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29 July, 21:00

How long does it take electrons to get from a car battery to the starting motor? Assume the current is 280 A and the electrons travel through a copper wire with cross-sectional area 0.44 cm2 and length 0.85 m. The number of charge carriers per unit volume is 8.49 ✕ 1028 m-3.

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Answers (2)
  1. 29 July, 23:29
    0
    Answer:1.86*10^ (-22) seconds

    Explanation:

    The copper wire has a cross sectional area (CSA) of 0.44cm² and a length of 0.85m ... we first need to convert the (CSA) to m². To do this 0.44cm² is divided by 10000 (since 100cm make 1m) ... 0.44/10000=0.000044m² ...

    Volume=CSA * length

    Volume=0.000044m²*0.85m

    Volume (m³) = 0.0000374m³

    Therefore, the total charge carriers is the product of the volume and the charge per unit volume

    Total charge carriers = 0.0000374m³ * 8.48*1028m^ (-3)

    Total charge carriers = 0.3260charge carriers

    Recall however, that the charge per unit charge carrier is 1.60*10^ (-19)

    Therefore, the total charge on the copper wire is=

    1.60*10^ (-19) coulombs*0.3260

    Total quantity of charge=5.216*10^ (-20) coulumbs

    And Quantity of charge (Q) = current (I) * time (t)

    Time={5.216*10^ (-20) }C/280A

    Time=1.86*10^ (-22) seconds
  2. 30 July, 00:31
    0
    Given:

    Current, i = 280 A

    Cross-sectional area, A = 0.44 cm^2

    Converting from cm^2 to m^2,

    0.44 cm^2 * (1 m) ^2 / (100 cm) ^2

    = 0.44 * 10^-4 m^2

    Length, L = 0.85 m

    Number of charge carriers per unit volume, nv = 8.49 ✕ 10^28 m^-3

    Current density, J = nv * q * v

    I/A = nv * q * v

    I = nv * A * q * (L/t)

    t = (nv * A * q * L) / I

    = (8.49 ✕ 10^28 * 0.44 * 10^-4 * 1.602 * 10^-19 * 0.85) / 280

    = 1816.7 s

    Since 60 seconds = 1 minute

    = 30.28 minutes
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