Question

A hollow copper wire with an inner diameter of 1.1 mm and an outer diameter of 1.8 mm carries a current of 15 A. What is the current density in the wire? 

Answer 1

Current density in the wire is given by, \(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\)

Here,

J- Current Density

I- Current A- Cross section of the wire

The cross section of the wire is given by, \(\mathrm{A}=\pi\left(\left(\frac{1.8 \times 10^{-3}}{2}\right)^{2}-\left(\frac{1.1 \times 10^{-3}}{2}\right)^{2}\right)\)

\(=\pi\left(0.81 \times 10^{-6}\right)-\left(0.302 \times 10^{-6}\right)\)

\(=1.594 \times 10^{-6} \mathrm{~m}^{2}\)

Thus the current density is given by, \(\mathrm{J}=\frac{15}{1.594 \times 10^{-6}}\)

\(=9.41 \times 10^{6} \mathrm{~A} / \mathrm{m}^{2}\)

The current density of the wire is \(9.41 \times 10^{6} \mathrm{~A} / \mathrm{m}^{2}\)