Question

When the tail light of a car is turned on, 0.5 A flows through the wires. The length of one wire from the battery to the tail light is 5.0 meters.Estimate how long you think it will take an electron to travel from the battery to the tail light and how many electrons flow through the tail in one minute. Thank you!

Given :

14AWG copper wire is used, which has a cross sectional area of 2.1e-6 m^2

2 The density of copper is 8.96 grams/cm^3

3 The atomic mass of copper is 63.5.

Each copper atom has one free electron.

(should be 3.3 days and 1.9e20 electrons per minute. How? ) Please show each steps. Thank you so much!

Answer 1

Relation between Current I and drift Velocity V_d is given by

where n = no of coduction electron/volume

e is electron charge = 1.6*10^-19 C

A is cross sectional area = 2.1*10^-6 m²

I is current = 0.5 A (Given)

length of wire = 5m

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Firstly Find n

A wire is actually a thin cylinder , Volume = cross sectional area * length

V = 2.1*10^-6 * 5

Volume= 1.05*10^-5 m³

Given density of copper = 8.96 g/cm³ = 8.96*10⁶ g/m³

Mass = Volume*density = 1.05*10^-5 * 8.96*10⁶

Mass =94.08 grams

Given atomic mass of Cu = 63.5 gram/mol

so No of moles in the wire = Mass/Atomic mass = 94.08 / 63.5 = 1.481574803 mol

So no of atoms = 6.022 *10²³ * no of moles

= 8.92204346*10²³

Since copper has 1 conducting eletron per atom

No of electron =8.92204346*10²³

Finally n =No of electron/Volume

n = 8.92204346*10²³/1.05*10^-5

n= 8.49718*10²⁸

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Find Drift Velocity

0.5 = 8.49718 *10²⁸ *1.6*10^-19 *2.1*10^-6 *

= 1.7512*10^-5 m/s

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Speed = distance /time

Time = distance/speed

T = 5 /1.7512*10^-5

T = 285505.248 sec  

(divide by 60*60*24)

T = 3.304 days

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No of Electrons passing through tail in 1 min

lets find no of electron required for1 C of charge

n*charge of elctron =1 C

n = 1 /e = 1/(1.6*10^-19)

n =6.25*10¹⁸ electrons

Here we use 0.5 A current = 0.5C/s

so no elctron passing in 1 sec = 6.25*10¹⁸ / 2 = 3.125*10¹⁸ electrons

in 1 min ,no of electon = 3.125*10¹⁸ *60

n=1.875 *10²⁰ electrons

ANSWER: n=1.875 *10²⁰ electrons

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