(a) Esttimate the average drift speed of conduction electrons in a copper wite of cross-secttonal area `1.0 xx 10 ^(-7)m^(2)` carrying a current of `1.5A.` Assume the each copper atom contrbutes roughly one conduction electron. The density of copper is `9.0 xx 10 ^(3)kg//m^(3),` and its atomic mass is `63.5 u.` (b) Compare the drift speed obyained above with, (1) thermal speeds of copper atoms at ordinary temperaturtes. (ii) speed of propagation of electric field along the conductor which causes the drift motion.

Explanation

(a) If valency of given element is p then its one atom will have p no. of free electrons. Hence, its total N no. of atoms, in total volume V will have pN no. of free electrons. Hence, free electron number density of this element would be :
`n = (pN)/(V)`
`therefore (N)/(V) = (n)/(p) " " `.... (1)
Now if mass of one atom of given element is m (which is known as atomic mass) then its total N no. of atoms will have total mass mN. Hence, volume density of mass for given element would be,
`rho = (M)/(V) = (mN)/(V) = (mn)/(p) " " `[ From equation (1)]
`therefore n = (p rho )/(m) " "` .... (2)
Now, current passing through given wire is,
I = `Av_(d)`ne
`therefore I = Av_(d) ((p rho )/(m) )e " " ` [ From equation (2) ]
`therefore v_(d) = (Im )/(Ap rho e)`
For copper, m = 63.5 u
`therefore m = 63.5 xx 1.66 xx 10^(-27) ` kg
(`because ` 1 u = one atomic mass unit = 1.66 `xx 10^(27)` kg )
p = 1 (valency of copper)
`rho ` =9000 kg /`m^(3)`
For given wire, I = 1.5 A
A = `10^(-7) m^(2)`
Substituting these value in above equation,
`v_(d) = ((1.5) (63.5 xx 1.66 xx 10^(-27)))/((10^(-7)) (1) (9000) (1.6 xx 10^(-19)))`
`therefore v_(d) = 1.098 xx 10^(-3) `m/s
Above is the drift speed of free electron in copper under given conditions.
(b) (i) At ordinary temperature (room temperature), T = `27^(@) `C
= (27 + 273) K = 300 K, if rms value of vibrational speed of copper atom (known as thermal speed because vibrations of atoms are due to absorption of thermal energy from surrounding) is `v_(rms)` then we have the formula,
`(1)/(2) m lt v^(2) gt = (3)/(2) k_(B) ` T
`("Where" K_(B)` = Boltzmann constant
= 1.38 `xx 10^(-23) (J)/(("molecule") (K))`
T = absolute temperature )
`therefore lt v^(2) gt = (3k_(B)T)/(m)`
`therefore sqrt(lt v^(2) gt ) = sqrt((3K_(B)T)/(m))`
`therefore v_(rms) = sqrt((3K_(B) T)/(m)) " " (because sqrt(lt v^(2) gt ) = v_(rms) )`
`therefore v_(rms) = sqrt((3 xx 1.38 xx 10^(-23) xx 300)/(63.5 xx 1.66 xx 10^(-27))`
`therefore v_(rms) = 343.2 ` m/s
Now,
`(v_(d))/(v_(rms)) = (1.098 xx 10^(-3))/(343.2) = 3.2 xx 10^(-6)`
`therefore v_(d) = 3.2 xx 10^(-6) v_(rms)`
(ii) When a metallic wire is connected across a battery, electric field is found to be established inside the wire with a speed almost equal to velocity of light in vacuum, c = 3`xx 10^(8)` m/s.
Now,
`(v_(d))/(c) = (1.098 xx 10^(3))/(3xx 10^(8)) = 3.66 xx 10^(-12)`
`therefore v_(d) = 3.66 xx 10^(-12) `c