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 $\frac{r}{2}$ and $r$ is $xπ$ $A$; where $x$ is _________.

Answer 1

where $x$ is $\underline{12\pi}$

$i = A × j$

=$π\bigg(R^2–\frac{R^2}{4}\bigg)j$

=$\frac{3πR^2}{4×j}$

=$\frac{3π×(4×10^{−3})^2}{4}×1.0×10 ^6$

= $12π$