The number density of free electrons in a copper conductor is $8.5 \times 10^{28} \; m^{-3}$. How long does an electron take to drift from one end of a wire $3.0 \; m$ long to its other end? The area of cross-section of the wire is $2.0 \times 10^{-6} \; m^2$ and it is carrying a current of $3.0 \; A$.

The current density in a solid cylindrical wire of radius $R$,as a function of radial distance $r$,is given by $J(r) = J_{0}(1 - \frac{r}{R})$. The total current in the radial region $r = 0$ to $r = \frac{R}{4}$ will be:

$A$ 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 $\rho = 44 \times 10^{-8}\, \Omega m$).

$(a)$ Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0 \times 10^{-7} \; m^{2}$ carrying a current of $1.5 \; A$. Assume that each copper atom contributes roughly one conduction electron. The density of copper is $9.0 \times 10^{3} \; kg/m^{3}$ and its atomic mass is $63.5 \; u$.
$(b)$ Compare the drift speed obtained above with,$(i)$ thermal speeds of copper atoms at ordinary temperatures,$(ii)$ speed of propagation of electric field along the conductor which causes the drift motion.

Current density in a cylindrical wire of radius $R$ varies with radial distance as $J(r) = \beta(r + r_0)^2$. The current through the shaded section of the wire shown in the figure is:

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.