Mobility of free electrons in a conductor is

$A$ steady current $I$ is set up in a wire whose cross-sectional area decreases in the direction of the flow of the current. Then,as we examine the narrowing region,

The relationship between the electric field $E$ and the current density $J$ is given by:

The drift velocity of electrons in a conducting wire connected to a cell is $V_{d}$. If the length of the wire is doubled and the area of cross-section is halved,then the drift velocity of electrons becomes:

If a current of $80 \ A$ is passing through a straight conductor of length $10 \ m$,then the total momentum of electrons in the conductor is (mass of electron $= 9.1 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$).

$A$ conducting wire of cross-sectional area $1 \,cm^2$ has $3 \times 10^{23}$ charge carriers per $m^3$. If the wire carries a current of $24 \,mA$, then the drift velocity of the carriers is: