The electric field $E$,current density $J$ and conductivity $\sigma$ of a conductor are related as

In the given hollow cylindrical conductor,the current density is $J = \frac{J_0}{r^2}$,where $J_0$ is a constant and $r$ is the distance from the axis of the cylinder. If the radius of the inner surface is $a$ and the radius of the outer surface is $2a$,find the current passed through the conductor.

When $5\ V$ potential difference is applied across a wire of length $0.1\ m$,the drift speed of electrons is $2.5 \times 10^{-4} \ m/s$. If the electron density in the wire is $8 \times 10^{28} \ m^{-3}$,the resistivity of the material is close to:

An electric current passes through a wire of non-uniform cross-section,made of a homogeneous and isotropic material. If $j_A$ and $j_B$ are the current densities and $E_A$ and $E_B$ are the electric field intensities at $A$ and $B$ respectively,then $-$

The area of cross-section of a copper wire is $4 \times 10^{-7} \,m^2$ and the number of free electrons per cubic metre in copper is $8 \times 10^{28}$. If the wire carries a current of $6.4 \,A$,then the drift velocity of the electrons (in $10^{-3} \,m \,s^{-1}$) is:

An electric cell of emf $E$ is connected across a copper wire of diameter $d$ and length $l$. The drift velocity of electrons in the wire is $v_{d}$. If the length of the wire is changed to $2l$, the new drift velocity of electrons in the copper wire will be