Question

all of 8 please.

8. An aluminum wire carrying a current of 5 A has a cross sectional area of 4 x 10'6 m2. Find the drift speed of the electrons in the wire. The density of aluminum is 2.7 g/cm3. (Assume three electrons are supplied by each atom). a) Recall the mass flow rate equation from semester 1. Rms Ro v b) Write the equivalent current relation for charges in a wire, I- c) What is the electrical analogy with fluid density, p? d) Given that the molar mass of aluminum is M 27 g/mol and the density of aluminum is drift mol/cm3 ρ-27 g/cm3, determine the number of mol cm3 for aluminum. ρ/M Convert into atoms/m3, N/V-p M (mol/cm3) x(10"cm3/m3) (6.02E23 atoms mol) = e) atoms/m esnr edoom criutes 3 conducting electrons. n = 3(electrons/atom) ×N/V (atoms/m3) g) Get the charge density, nq h) Solve the equation in b) for the drift velocity, Vdrit i) Plug in the numbers to get electrons/m3 C/m Vdriftー m/s j) How long would it take an electron to drift 1 meter through this wire? t- sec

Answer 1

a)Mass flow rate is  $R_{mass\, flow}=\rho *vA$  

Where $\rho$ is density

b) Current $I=nq\, v_{drift}A$

c) Electrical analogy with fluid density $\rho$ is charge density  $nq$

d) $\rho/M=\frac{2.7}{27}=0.1\,mol/cm^3$

e) $N/V=\rho/M*10^6*6.02*10^{23}=0.1*10^6*6.02*10^{23}=6.02*10^{28}\,atom/m^3$

f) $n=3*N/V=3*6.02*10^{28}=1.806*10^{29}\,electrons/m^3$

g)Charge density $nq=1.806*10^{29}*1.6*10^{-19}=2.9*10^{10}\,C/m^3$

h)$I=nq\, v_{drift}A$

$v_{drift}=\frac{I}{nq A}$

i)

$v_{drift}=\frac{I}{nq A}=\frac{5}{2.9*10^{10}*4*10^{-6}}=4.33*10^{-5}\,m/s$

j) Time taken $t=\frac{d}{v_{drift}}=\frac{1}{4.33*10^{-5}}=2.31*10^4\,s$