Question

An aluminum wire with a diameter of 0.095 mm has a uniform electric field of 0.205 V/m imposed along its entire length. The temperature of the wire is 35.0°C. Assume one free electron per atom. (a) Use the information in this Table of Resistivities and Temperature Coefficients to determine the resistivity (in . m) of aluminum at this temperature. p= 2.m (b) What is the current density (in MA/m2) in the wire? = C MA/m2 (c) What is the total current in ma) in the wire? - IL MA (d) What is the drift speed of the conduction electrons ? um/s Va=0 (e) What potential difference must exist between the ends of a 2.10 m length of the wire to produce the stated electric field? AV =

Answer 1

at 7:20 the bi the Desistivity aluminium ej - 3.20 x10 here and the termal Coefficient a= 3.90 x103 so sesuadress of alluminium oc 35°C ?s etto [!+ (7-7)] = a8a%16% [1- 3.99x155 35-20)] = 8.98495x15dom 9. The current density is given by. 8,98X108 = 0.066677x108 = 5.25==x10 t/m = 5,86 oAlma bial Congrent in de were is :- I tre I =IÐ = 3,2652106, T210.954104)? = 4,2555 x1504 = 42.655 mof

e'] the roolar mass of aluminium is 86.98 g/mol. angel density of 2.7 x 100 g/m3 n = 9.0ax1083x2.73106 26.98 = 6.02 x 1038 electrons/mo The drifte velocity of electrons is then a ne 6.86X106 6.02x1028x160x10-19 = 7. 19x104 mis = 699712.a umis = 72.2 um/s The potential difference is AV - exe = 0,205% 2.10 = 0.4305 V