There is a current of $20 \, A$ in a copper wire of $10^{-6} \, m^2$ area of cross-section. If the number of free electrons per cubic meter is $10^{29}$,then the drift velocity is:

The lengths of two wires made of the same material are in the ratio $2:3$ and their radii are in the ratio $1:2$. If the two wires are connected in parallel to a battery,then the ratio of the drift velocities of free electrons in the two wires is

$(a)$ 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?

Drift speed of electrons,when $1.5 \, A$ of current flows in a copper wire of cross-section $5 \, mm^2$,is $v$. If the electron density in copper is $9 \times 10^{28} \, m^{-3}$,the value of $v$ in $mm/s$ is close to (Take charge of electron to be $1.6 \times 10^{-19} \, C$).

$A$ $1\,m$ 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 $= 1.6 \times 10^{-19}\,C$)

Assertion: The current density $\vec J$ at any point in an ohmic resistor is in the direction of the electric field $\vec E$ at that point.
Reason: $A$ point charge when released from rest in a region having only an electrostatic field always moves along electric lines of force.