91Ó°ÊÓ

In a hollow metal cylinder has inner radius is $a$, outer radius is $b$, and the electric field strength inside the metal as a function of the radius is $r$.

Since,$J=\mathrm{σ}E⇒E=\frac{J}{\mathrm{σ}}$
The current will flow around to uncover the current density in the area. This area will be a rectangle, with one side the length of the height of the cylinder and the other the perimeter of the circle with a radius$r$as,$A=2\mathrm{Ï€}rL$
Now to find the current density as,
$J=\frac{I}{A}=\frac{I}{2\mathrm{Ï€}rL}$
Finally,
$E=\frac{I}{2\mathrm{π}\mathrm{σ}rL}$

The electric field strength at the inner surfaces of an iron cylinder is $a=1.0cm$ and the outer surfaces of an iron cylinder is $b=2.5cm$.

The electric field strength at the inner boundary as,
$E=\frac{I}{2\mathrm{π}\mathrm{σ}aL}$

$=\frac{25A}{2×\mathrm{π}×1.0×{10}^7×0.01m×0.10m}$

$=0.40\mathrm{mV}/\mathrm{m}$

The electric field strength at the outer boundary as,

$E=\frac{I}{2\mathrm{π}\mathrm{σ}aL}$

$=\frac{25A}{2×\mathrm{π}×1.0×{10}^7×0.025m×0.10m}$

$=0.159mV/m$