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

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 . . . . . . .

The current density in a cylindrical wire of radius $r = 4.0 \, mm$ is $J = 1.0 \times 10^{6} \, A/m^{2}$. The current through the outer portion of the wire between radial distances $r/2$ and $r$ is $x \pi \, A$,where $x$ is ..........

Two wires $A$ and $B$ of the same material,having radii in the ratio $1: 2$,carry currents in the ratio $4: 1$. The ratio of drift speed of electrons in $A$ and $B$ is .......

$A$ current of $5\; A$ is passing through a non-linear magnesium wire of cross-section $0.04\; m^2$. At every point,the direction of current density is at an angle of $60^{\circ}$ with the unit vector of the area of cross-section. The magnitude of the electric field at every point of the conductor is .... $V/m$ (Resistivity of magnesium is $\rho = 44 \times 10^{-8}\, \Omega m$).

Explain the drift of electrons and drift velocity. Derive the equation for electric current in terms of the cross-sectional area of a conductor.