Question
A current carrying aluminum wire, with diameter of D=0.10mm and length L=2 m, has a uniform field of E=0.2V/m imposed along its entire length. The temperature of the wire is T=50 C. Assume one free electron per atom.
(T=20 C, r= 2.82x10^-8 ohm*m^2, a= 3.9 x 10^-3/C. Take the atomic mass of the Al as mA= 27 g/mole, density of the Al d= 2.7 g/cm^3 and the Avogadro's number as 6.02 x 10^23 atoms/mole.)
A. Determine resistivity of the wire.
B. What is the current density in the wire?
C. What is the total current in the wire?
D. what is the drift speed of the conduction electrons?
Explanation / Answer
length of the wire L = 2 m diameter of the wire D = 0.1 mm = 0.1*10-3 m radius of the wire r = D/2 = 0.05*10-3 m magnetic field E = 0.2 V/m temperature T = 500C when one free electron for atom , temperature T0 = 200C , 0 = 2.82*10-8 m , temperature coefficient = 3.9 x 10-3 /C a) the resistivity of the wire is = 0[1 + (T - T0)] = (2.82*10-8)[1 + (3.9*10-3)(50 - 20)] = 31.49*10-9 m b) current density J = E (since = 1/) J = E/ = 0.2/31.49*10-9 = 0.63*107 A/m2 c) total current in the wire is I = JA = J(r2) = (0.63*107)(3.14)(0.05*10-3)2 = 0.0494 A d) the drift velocity vd = J/ne where , free electron density n = [(2.7*103 kg/m3)(6.023*1023 /mol)] / (27*10-3 kg/mol) = 6.023*1028 /m3 the drift velocity vd = J/ne = (0.63*107 ) / (1.6*10-19)(6.023*1028) = 0.653*10-3 m/s the drift velocity vd = J/ne = (0.63*107 ) / (1.6*10-19)(6.023*1028) = 0.653*10-3 m/s