The potential difference across a conducting wire of length $20 \ cm$ is $30 \ V$. If the electron mobility is $2 \times 10^{-6} \ m^2 \ V^{-1} \ s^{-1}$,then the drift velocity of the electrons is

$A$ $1\,m$ long copper wire carries a current of $1\,A$. If the cross-section of the wire is $2.0\,mm^{2}$ and the resistivity of copper is $1.7 \times 10^{-8}\,\Omega\,m$,the force experienced by a moving electron in the wire is $x \times 10^{-23}\,N$. Find the value of $x$. (Charge on electron $= 1.6 \times 10^{-19}\,C$)

$(a)$ The electron drift speed is estimated to be only a few $mm\; s^{-1}$ for currents in the range of a few amperes. How then is current established almost the instant a circuit is closed?
$(b)$ The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed?
$(c)$ If the electron drift speed is so small,and the electron's charge is small,how can we still obtain large amounts of current in a conductor?
$(d)$ When electrons drift in a metal from lower to higher potential,does it mean that all the 'free' electrons of the metal are moving in the same direction?
$(e)$ Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the $(i)$ absence of electric field,$(ii)$ presence of electric field?

In a wire with a cross-sectional area of $1 \, cm^2$ carrying a current of $24 \, mA$,the electron number density is $3 \times 10^{23} \, m^{-3}$. What is the drift velocity?

$A$ potential difference $V$ is applied across the ends of a copper wire of diameter $d$ and length $L$. If only the diameter $d$ is doubled,the drift velocity becomes:

The drift velocity of electrons in a conducting wire is of the order of $1 \, mm/s$. However,when the switch is turned on,the bulb glows almost instantaneously. This is because: