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How long does it take electrons to get from the car battery to the starting motor? Assume the current is 143 A and the electrons travel through copper wire with cross sectional area 40.2 mm2 and length 84.5 cm . The mass density of copper is 8960 kg/m3 and the molar mass is 63.5 g/mol . Avogadro’s number is 6.022 × 1023 . Assume that each copper atom contributes one conduction electron.

User Shantae
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2 Answers

3 votes

Final answer:

To calculate the time it takes for electrons to travel from the car battery to the starting motor, you can use the formula I = nqAvd to determine the drift velocity of the electrons in the copper wire. By calculating the number of free electrons per cubic meter and using the given values for current, charge density, and cross-sectional area, you can solve for the drift velocity. Once you have the drift velocity, you can calculate the time based on the distance traveled.

Step-by-step explanation:

To calculate the time it takes for electrons to travel from the car battery to the starting motor, we need to determine the drift velocity of the electrons in the copper wire. The drift velocity can be calculated using the formula I = nqAvd, where I is the current, n is the charge density, q is the charge of an electron, A is the cross-sectional area of the wire, and vd is the drift velocity.

First, let's calculate the number of free electrons per cubic meter (n). The mass density of copper is given as 8960 kg/m3 and the molar mass is 63.5 g/mol. Using Avogadro's number (6.022 × 1023), we can determine the number of copper atoms per cubic meter, which is equal to n.

Once we have n, we can solve for vd by rearranging the formula as vd = I / (nqA). Substitute the given values for I, n, q, and A to find the drift velocity. The time it takes for the electrons to travel from the car battery to the starting motor can then be calculated using the distance and drift velocity.

User Slybitz
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7 votes

Answer

the concentration of electron/ volume (n) = 8.5 × 10²⁸ /m³

the drift velocity of electron


v = (I)/(neA)


v = (143)/(8.5* 10^(28)* 1.6* 10^(-19)* 40.2 * 10^(-6))

v = 2.615 × 10⁻⁴ m/s

time require is


t=(L)/(v)


t=(0.845)/(2.615* 10^(-4))

t = 3231.35 s

User Andrej
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