The drift velocity of an electron is $v_d$ in a conductor of area of cross-section $A$ and carries a current $I$. Now, the area of cross-section and current flowing through the conductor are doubled, then the new drift velocity of the electron is . . . . . . .

$A$ copper wire of length $1\, m$ and radius $1\, mm$ is joined in series with an iron wire of length $2\, m$ and radius $3\, mm$ and a current is passed through the wires. The ratio of the current density in the copper and iron wires is

There is a current of $1.344 \, A$ in a copper wire whose area of cross-section normal to the length of the wire is $1 \, mm^2$. If the number of free electrons per $cm^3$ is $8.4 \times 10^{22}$, then the drift velocity would be

$A$ copper wire of cross-sectional area $1.0 \ mm^2$ carries a current of $1.34 \ A$. Assuming that each copper atom contributes one free electron,calculate the drift velocity of the free electrons in the wire in $mm/s$. (Given: density of copper = $8990 \ kg/m^3$,atomic mass = $63.50 \ g/mol$)

We are able to obtain fairly large currents in a conductor because

In a copper wire,each atom releases one free electron. If the diameter is $1\, mm$ and the current is $1.1\, A$,find the drift velocity. (Given: Density $\rho = 9 \times 10^{3}\, kg/m^3$,Molar mass $M = 63\, g/mol$,Avogadro number $N_A = 6.023 \times 10^{23}\, mol^{-1}$,Elementary charge $e = 1.6 \times 10^{-19}\, C$)