A current of `5` ampere is passing through a metallic wire of cross-section area `4 xx 10^-6 m^2`. If the density of the charge carries in the wire is `5 xx 10^26 m^3`, find the drift speed of the electrons.
(b) In copper there are `10^22` free electrons per `c.c.`, all of which contribute to a current of `1` ampere in a wire of copper of `0.01 cm^2` cross-sectional area :
(i) What is the average speed of electrons in the copper ?
(ii) What is the electric field in the wire ?
Given specific resistance of copper `1.6 xx 10^-8 ohm-meter`.
( c) The area of cross-section, length and density of a piece of a metal of atomic weight `60` are `10^-6 m^2, 1.0 m` and `5 xx 10^3 kg//m^3` respectively. Find the number of free electrons per unit volume if every atom contributes one free electron. Also, find the drift velocity of electrons in the metal when a current of `16 A` passes through it. (Given : Avogadro number `6 xx 10^26//kg-"mole"`).

Explanation

(a) `1 = 5A, A = 5 xx 10^-6 m^2, n = 5 xx 10^26//m^3`
`I = "ne"Av_d`
`v_d = (i)/(n e A) = (5)/(5 xx 10^26 xx 1.6 xx 10^-19 xx 5 xx 10^-6)`
=`(1)/(80) m//s = 0.0125 m//sec`
=`1.25 xx 10^-2 m//sec`
(b) (i) `I = n eA v_(d)`
`v_(d) = (i)/(n eA) = (1)/((10^22)/(10^(-6)) xx 1.6 xx 10^-19 xx 0.01 xx 10^(-4))`
=`(1)/(16) xx 10^(-2) = 0.0625 xx 10^(-2) m//sec`
=`6.25 xx 10^(-4) m//sec`
(ii) `j = sigma E("Ohm's law")`
`E = (j)/(sigma) = (i//A)/(1//rho) = (rho i)/(A) = (1.6 xx 10^-8 xx 1)/(0.01 xx 10^(-4))`
=`1.6 xx 10^-2 vol t//m`
( c) Volume `V = lA = 1 xx 10^-6 m^3 = 10^-6 m^3`
Mass = Density `xx` Volume
=`5 xx 10^3 xx 10^-6 = 5 xx 10^-3 kg = 5 g`
Number of atoms `= ("Mass")/("At.wt") xx` Avogadro's no.
=`(5)/(60) xx 6 xx 10^23 = 5 xx 10^22`
Number of free electrons `= 5 xx 10^22`
Number of free electrons/volume
`n = (5 xx 10^22)/(10^-6) = 5 xx 10^28`
`i = n eA v_d`
`v_d = (i)/(n eA) = (16)/(5 xx 10^28 xx 1.6 xx 10^-19 xx 10^-6)`
=`2 xx 10^-3 m//sec`
=`2 mm//sec`.