Question

20. What electric field strength is needed to create a $5.0A$current in a $2.0$-mm-diameter iron wire?

Short answer

The electric field strength is needed to create a $5.0A$current in a

$2.0$-mm-diameter iron wire is $0.159\mathrm{V}/\mathrm{m}$.

Step-by-step solution

Given information

A $5.0$ current in a $2.0$-mm-diameter iron wire with the electric field strength.

Explanation

The current density is given by the electric field and the conductance as, $J=\sigma E$

The current density $J=\frac{I}{A}$as,

$\Rightarrow \frac{I}{\pi r^2}=\frac{4I}{\pi d^2}$

Substituting the two expressions for the current density,

$\sigma E=\frac{4I}{\pi d^2}$

The electric field $E$to be,

$E=\frac{4I}{\pi \sigma d^2}$

$E=\frac{4\times 5}{\pi \times 1\times {10}^7\times 0.{002}^2}$

$=0.159\mathrm{V}/\mathrm{m}$

Final answer

The electric field strength is needed to create a $5.0A$current in a

$2.0$-mm-diameter iron wire is $0.159\mathrm{V}/\mathrm{m}$.