Question

56. A hollow metal cylinder has inner radius $a$, outer radius $b$, length $L$, and conductivity$\sigma$. The current $I$is radially outward from the inner surface to the outer surface.
a. Find an expression for the electric field strength inside the metal as a function of the radius $r$ from the cylinder's axis.
b. Evaluate the electric field strength at the inner and outer surfaces of an iron cylinder if $a=1.0cm,b=2.5cm,L=10cm,$and $I=25A$.

Short answer

a. An expression for the electric field strength inside the metal as a function of the radius $r$from the cylinder's axis is $E=\frac{I}{2\pi \sigma r{L}^{\text{'}}}$
b. The electric field strength at the inner and outer surfaces of an iron cylinder is$0.40\mathrm{mV}/\mathrm{m},0.159\mathrm{mV}/\mathrm{m}\text{.}$

Step-by-step solution

Given information Part(a)

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 $r$.

Explanation Part (a)

Since,$J=\sigma E\Rightarrow E=\frac{J}{\sigma }$
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,$A=2\pi rL$
Now to find the current density as,
$J=\frac{I}{A}=\frac{I}{2\pi rL}$
Finally,
$E=\frac{I}{2\pi \sigma rL}$

Given information Part (b)

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 $b=2.5cm$.

Explanation Part (b)

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

$=\frac{25A}{2\times \pi \times 1.0\times {10}^7\times 0.01m\times 0.10m}$

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

The electric field strength at the outer boundary as,

$E=\frac{I}{2\pi \sigma aL}$

$=\frac{25A}{2\times \pi \times 1.0\times {10}^7\times 0.025m\times 0.10m}$

$=0.159mV/m$