If the magnitude of the drift velocity of free electrons in a copper wire is 8.5
ID: 1538394 • Letter: I
Question
If the magnitude of the drift velocity of free electrons in a copper wire is 8.50x10^-4 m/s, what is the field in the conductor? [Electron mass, m = 9.11 x 10^-31 kg: Resistivity of Copper = = 1.70 times 10^-8 Ohm m; umber of electrons in copper = n = 8.46 x 10^28 electrons/m^3; Electron charge q = -1.60 x 10^-19 C/electron] (b) Suppose you wish to fabricate a uniform wire from 5.00 g of copper. If the wire is to have a resistance of R = 10.0 Ohm and all copper is to be used, what must be used, what must be (i) the length and (ii) the diameter of the is wire? (iii) If this wire is used in a circuit with source voltage of 110 V, how much heat is produced in it in one minute? [Resistivity of copper = = 1.7 times 10^-8 Ohm m; Mass density of copper d = 8.92 times 10^3 kg/m^3]
Explanation / Answer
Drift speed of the elctrons is Vd = 8.5*10^-4 m/sec
current density is j = i/A = n*e*Vd = 8.46*10^28*1.6*10^-19*8.5*10^-4 = 11.5*10^6 A/m^2
But current denisy j = E/rho
Electric field is E = j*rho = 1.7*10^-8*11.5*10^6 = 0.1955 V/m
relation between Resistance and resistivity is R = rho*l/A
10 = 1.7*10^-8*l/A
(l/A) = 5.88*10^8 m^-1............(1)
mass density is d = M/V
d = m/(l*A)
l*A = m/d = (5*10^-3)/(8.92*10^3) = 5.6*10^-7 m^3...(2)
(1)*(2)= (l/A)*l*A = 5.88*5.67*10
l^2 = 333.396 m^2
l = 18.25 m is the length of the wire
A = 5.6*10^-7/18.25 = 3.06*10^-8 m^2
pi*(D/2)^2 = 3.06*10^-8
Diameter D = 0.0001973 m = 0.01973 cm
heat produced in one second is P = V^2/R = 110^2/10 = 1210 W
in one minute heat produced is P = 1210*60 = 72600 J