Question

For what electric field strength would the current in a $2.0mm$diameter nichrome wire be the same as the current in a $1.0mm$diameter aluminum wire in which the electric field strength is$0.0080\mathrm{V}/\mathrm{m}$?

Short answer

The electric field strength is$0.104\mathrm{V}/\mathrm{m}$.

Step-by-step solution

Given Information

Nichrome wire diameter$=2.0mm$

Aluminum wire diameter$=1.0mm$

Electric field strength$=0.0080V/m$

Explanation

Express the electric field strength as,

$J=\sigma E\Rightarrow E=\frac{J}{\sigma }$

Electric current density is,

$J=\frac{I}{A}$

Here, the cross-sectional area is a circle,

$J=\frac{4I}{\pi D^2}.$

Combining the expression,

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

So, the product $E\pi \sigma D^2$will be constant for both wires.

Simplify the $\pi ,$

$E_1{\sigma }_1{D}_1^2=E_2{\sigma }_2{D}_2^2$

The aluminum is represented by $1$and the nichrome is represented by $2$.

In the case of the nichrome, the electric field is calculated as follows:

$E_2=\frac{{\sigma }_1}{{\sigma }_2}{\left(\frac{D_1}{D_2}\right)}^2{E}_1$

Substitute the expression,

localid="1649072796279" $E_2=\left(\frac{350s}{6.7s}\times 0.5^{2}mm\times 0.008V/m\right)$

localid="1649072915118" $=0.104\mathrm{V}/\mathrm{m}$

Final Answer

Hence, the electric field strength will be$104mV/m.$