If the number density of electrons in a conductor is `3 xx10^(28)` and thickness of the conductor is 5 mm, then calculate the current in that conductor for a drift velocity of `2 xx 10^(-4) ms^(-1)`. If the voltage applied is `500 V`, then calculate relaxation time. Given specific charge of electron = `1.76xx 10^(11)` C `kg^(-1)` , length of the conductor = 10 m

Explanation

Given : `n=3xx10^(28)` electrons/`m^(3),d=5xx10^(-3)m`,
`v_(d)=2xx10^(-4)ms^(-1),V=500V,e/m=1.76xx10^(11)Ckg^(-1)`
`tau=?, L=10m`
`A=(pid^(2))/4=(3.142xx25xx10^(-6))/4=19.6375xx10^(-6)m^(2)`
We know that, `I=nAev_(d)`
i.e., `I=3xx10^(28)xx19.64xx10^(-6)xx1.6xx10^(-19)xx2xx10^(-4)A`
`thereforeI=1.88xx10.=18.8A.,`
Since, ma=aE where `E=V/L`
`(mv_(d))/(tau)=(eV)/L`
`tau=(v_(d)L)/((e/m)V)`
`tau=(2xx10^(-4)xx10)/(1.76xx10^(11)xx500)s`
i.e., `tau=0.00227xx10^(-14)`
i.e., `tau=2.27xx10^(-17)` s