Every atom makes one free electron in copper. If 1.1 ampere current is flowing in the wire of copper having 1 mm diameter, then the drift velocity (approx.) will be (Density of copper `=9 xx 10^(3) kg m^(-3)` and atomic weight = 63)A. `0.01 mm s^(-1)`B. `0.02mm s^(-1)`C. `0.2 mm s^(-1)`D. `0.1 mm s^(-1)`

Every atom makes one free electron in copper. If 1.1 ampere current is

1.	Every atom makes one free electron in copper. If 1.1 ampere current is flowing in the wire of copper having 1 mm diameter, then the drift velocity (approx.) will be (Density of copper `=9 xx 10^(3) kg m^(-3)` and atomic weight = 63)A. `0.01 mm s^(-1)`B. `0.02mm s^(-1)`C. `0.2 mm s^(-1)`D. `0.1 mm s^(-1)`
Answer» Correct Answer - 4 No. of atom in 1 gm `Cu=(N_(A))/(63)` No. of atom/`cm^(3)=(N_(A))/( 63)xx9` `because` each atom contributes one electron So no. of electron `(n)//m^(3)=(N_(A))/(63)xx9xx10^(6)` Now I=`"ne"Av_(d)` Put values I,n,e,A and calculate for `v_(d)`

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