Question

The current density in a cylindrical wire of radius $r =4.0\, mm$ is $1.0 \times 10^{6} A / m ^{2}$. The current through the outer portion of the wire between radial distances $r / 2$ and $r$ is $x \pi A$; where $x$ is ______

Answer 1

$I=\int JdA$
$=\int 10^{6} \times 2 \pi xdx$
$\left.=10^{6} \times 2 \pi \cdot x \frac{ x ^{2}}{2}\right]_{\frac{ r }{2}}^{ r }$
$=\pi \times 10^{6}\left[ r ^{2}-\frac{ r ^{2}}{4}\right]=12 \pi$
$x =12$