Question

When you step on the brake pedal in your car, charge carriers flow from the battery to the rear brake lights. Suppose the wire connecting the switch at the pedal to the brake lights is made of copper and has a diameter of 1.2 mm . If the current through the wire is 2.5 A , how long on average does it take an electron to travel from the switch at the pedal to one of the brake lights? The number density of free electrons in copper is n = 8.4 ×1028 electrons/m3. Suppose that the length of the wire is 3 m. t = nothing h

Answer 1

Answer:

18240.3382742 s

Explanation:

I = Current = 2.5 A

d = Diameter = 1.2 mm

n = Number of electrons = $8.4\times 10^{28}\ electrons/m^3$

L = Length of the wire = 3 m

Drift speed is given by

$v_d=\dfrac{I}{neA}\\\Rightarrow v_d=\dfrac{2.5}{8.4\times 10^{28}\times 1.6\times 10^{-19}\times \dfrac{\pi}{4}\times (1.2\times 10^{-3})^2}\\\Rightarrow v_d=0.000164470633982\ m/s$

Time taken is given by

$t=\dfrac{L}{v_d}\\\Rightarrow t=\dfrac{3}{0.000164470633982}\\\Rightarrow t=18240.3382742\ s$

The time taken for an electron to travel from the switch at the pedal to one of the brake lights is 18240.3382742 s

Answer 2

Answer:

Explanation:

diameter of wire, d = 1.2 mm

radius of wire, r = 0.6 mm

current, i = 2.5 A

number density of electrons, n = 8.4 x 10^28 /m³

Length of wire, l = 3 m

resistivity of copper, ρ = 1.72 x 10^-8 ohm metre

Let R be the resistance of wire

$R = \rho \frac{l}{A}$

$R = 1.72\times 10^{-8} \frac{3}{3.14\times 0.6\times 0.6\times 10^{-6}}$

R = 0.046 ohm

i = n e A v

where, e is the charge of electron and v is the drift velocity of electrons

2.5 = 8.4 x 10^28 x 1.6 x 10^-19 x 3.14 x 0.6 x 0.6 x 10^-6 x v

v = 1.65 x 10^-4 m/s

Let t be the time taken

length, l = drift velocity x time

3 = 1.65 x 10^-4 x t

t = 18231.1 second

t = 5.1 hour