Show the Joule's law can be expressed as `Q=rhol^(2) or sigmaE^(2)` where Q is the heat generated per unit volume of the conductor, j=current density, `rho`=respectively of the material of the conductor, `sigma`=conductivity and E= electric field step up in the conductor.

If q is the charge per unit area on the surface of a conductor, then the electric field intensity at a point on the surface is

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

If vec(j) and vec(E ) are current density and electric field respectively inside a current carrying conductor , then correct relation is

A long cylindrical conductor of cross-sectional area A and radius a is made of material whose resistivity depends only on a distance r from the axis of conductor, given by rho = c/r^(2) , where c is a constant. Find the resistance per unit length of the conductor and the electric field strength due to which a current I flows in it.

The current density in a conductor is 5 A //m^2 when an electric field of 50 V/m is applied across it. Calculate the conductivity of the material.

The current density in a wire is 10A//cm^(2) and the electric field in the wire is 5 V/cm. If p = resistivity of material, sigma = conductivity of the material then (in SI unit)

If A is the area of cross section of conductor, e be the charge on the electrons, n be the number of electrons per unit volume and J be the current density then drift velocity of electrons is