Question

A hollow copper wire with an inner diameter of 0.60mm and an outer diameter of 2.0mm carries a current of 14 A. What is the current density in the wire?

Answer 1

The expression for current density is,

$$ I=J A $$

The inner radius of the wire is,

$$ \begin{aligned} r_{1} &=\frac{0.6 \mathrm{~mm}}{2}\left(\frac{1.0 \mathrm{~m}}{1000 \mathrm{~mm}}\right) \\ &=0.3 \times 10^{-3} \mathrm{~m} \end{aligned} $$

The outer radius of the wire is,

$$ \begin{aligned} r_{2} &=\frac{2.0 \mathrm{~mm}}{2}\left(\frac{1.0 \mathrm{~m}}{1000 \mathrm{~mm}}\right) \\ &=1.0 \times 10^{-3} \mathrm{~m} \end{aligned} $$

The area of cross-section of conductor having inner radius and outer radius is,

$$ \begin{aligned} A &=\pi\left(\left(1.0 \times 10^{-3} \mathrm{~m}\right)^{2}-\left(0.3 \times 10^{-3} \mathrm{~m}\right)^{2}\right) \\ &=2.85 \times 10^{-6} \mathrm{~m}^{2} \end{aligned} $$

The current density is,

$$ \begin{aligned} J &=\frac{I}{A} \\ &=\frac{(14 \mathrm{~A})}{\left(2.85 \times 10^{-6} \mathrm{~m}^{2}\right)} \\ &=4.89 \times 10^{6} \mathrm{~A} / \mathrm{m}^{2} \end{aligned} $$