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 $\dots \; \pi A .$
JEE MAIN 2022, Medium
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$J=\frac{I}{A}$
$I=J A$
$=4 \times 10^{6} \times\left[\pi R^{2}-\pi\left(\frac{R}{2}\right)^{2}\right]$
$=4 \times 10^{6} \times \pi R^{2} \times \frac{3}{4}$
$=4 \times 10^{6} \times \pi \times\left(4 \times 10^{-3}\right)^{2} \times \frac{3}{4}=48 \; \pi A .$
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