The drift velocity of the electrons in a copper wire of length $2\ m$ under the application of a potential difference of $200\ V$ is $0.5\ m/s$. Their mobility (in $m^2 V^{-1} s^{-1}$) is

Write Ohm's law in the form of current density (vector form).

As shown in the figure,a current is passing through a conducting wire. The radius of the cross-section of the wire at point $A$ is $3r$ and at point $B$ is $r$. Find the ratio of the drift velocity at point $A$ to that at point $B$.

If the ratio of the radii of two wires of the same material is $1 : 2$ and the ratio of the current flowing through them is $4 : 1$,what is the ratio of their drift velocities?

$A$ current of $10 \, A$ is passing through an aluminum wire of cross-sectional area $4 \times 10^{-6} \, m^{2}$. If the density of aluminum is $2.7 \, g/cm^{3}$ and it provides $1$ free electron per atom for conduction,find the drift speed of the electrons in $\times 10^{-4} \, m/s$. (Given: molecular weight of aluminum = $27 \, g/mol$,Avogadro's number $N_{A} = 6.022 \times 10^{23} \, mol^{-1}$)

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: