The current density in a cylindrical wire of radius r = 4.0 mm is $1.0 \times 10^6 \mathrm{~A} / \mathrm{m}^2$. The current through the outer portion of the wire between radial distances $r / 2$ and $r$ is $x \pi$ A; where x is _______ .
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Solution
$\begin{aligned} & I=\int J d A \\ & =\int 10^6 \times 2 \pi x d x \\ & \left.=10^6 \times 2 \pi \cdot x \frac{x^2}{2}\right]_{\frac{r}{2}}^r \\ & =\pi \times 10^6\left[r^2-\frac{r^2}{4}\right]=12 \pi \\ & x=12\end{aligned}$
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$\begin{aligned} & I=\int J d A \\ & =\int 10^6 \times 2 \pi x d x \\ & \left.=10^6 \times 2 \pi \cdot x \frac{x^2}{2}\right]_{\frac{r}{2}}^r \\ & =\pi \times 10^6\left[r^2-\frac{r^2}{4}\right]=12 \pi \\ & x=12\end{aligned}$