A metal wire of cross-sectional area 0.5 mm^2 and length 100 m is connected across a battery of e.m.f. 2 V and internal resistance 1\ . The density, atomic mass and electrical conductivity of the metal are 6.35 10^3 kg m^ -3 , 63.5 gm/mole and 2 10^8 mho m^ -1 , respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in mm s^ -1 ) of the electrons in the wire is: [Take Avogadro's number as 6 10^ 23 and charge of the electron as 1.6 10^ -19 C.]
C. 0.208
The number density of conduction electrons n is given by the number of atoms per unit volume, since there is one conduction electron per atom: n = d N_A M Substituting the given values: n = 6.35 10^3 6 10^ 23 63.5 10^ -3 = 6 10^ 28 m ^ -3 The resistance of the wire R is: R = L A R = 100 2 10^8 0.5 10^ -6 = 100 10^2 = 1\ The total resistance of the circuit is R_ total = R + r = 1 + 1 = 2\ . The current in the circuit is: I = E R_ total = 2 2 = 1 A The drift velocity v_d is given by the relation I = n e A v_d: v_d = I n e A v_d = 1 6 10^ 28 1.6 10^ -19 0.5 10^ -6 v_d = 1 4.8 10^3 m s ^ -1 Converting to mm s^ -1 : v_d = 1000 4800 mm s ^ -1 = 5 24 mm s ^ -1 0.208 mm s ^ -1 Answer: 0.208
Related: Physics — Current Electricity · All PYQ Banks