Question

The current density in a cylindrical wire of radius $4 \,mm$ is $4 \times 10^{6} Am ^{-2}$. The current through the outer portion of the wire between radial distance $\frac{R}{2}$ and $R$ is _______$\pi A .$

Answer 1

$J=\frac{I}{A}$
$I=J A$
$=4 \times 10^{6} \times\left[\pi R^{2}-\pi\left(\frac{R}{2}\right)^{2}\right]$
$=4 \times 10^{6} \times \pi R^{2} \times \frac{3}{4}$
$=4 \times 10^{6} \times \pi \times\left(4 \times 10^{-3}\right)^{2} \times \frac{3}{4}=48 \,\pi A .$