Charge passing through a conductor of cross-section $0.3 \, m^2$ is given by $q = (3t^2 + 5t + 2) \, C$ where '$t$' is in seconds. The drift velocity at $t = 2 \, s$ is (Concentration of electrons in the conductor $= 2 \times 10^{25} \, m^{-3}$)

$(a)$ 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?

What is current density? Derive Ohm's law in the form of current density.

Drift speed $(v_d)$ varies with the intensity of the electric field $(E)$ as per the relation:

$A$ conductor of length $\ell$ with a circular cross-section is shown in the figure,carrying a current $i$. The radius of the cross-section varies linearly from $a$ to $b$. Assuming $(b - a) << \ell$,calculate the current density at a distance $x$ from the left end.

$A$ current flows in a wire of circular cross-section with the free electrons travelling with a mean drift velocity $\vec v$. If an equal current flows in a wire of twice the radius,the new mean drift velocity is: