Q 12-03-177JEE MainJEE Main 2022 (27 Jun, Shift 2)Easy
The current density in a cylindrical wire of radius $r = 4.0$ mm is $1.0\times10^6\ \text{A m}^{-2}$. The current through the outer portion of the wire between radial distances $\dfrac r2$ and $r$ is $x\pi$ A; where $x$ is ______ .
Numerical value type. Enter your answer.
Answer: 12
$$I = J\pi\left(r^2 - \frac{r^2}{4}\right) = \frac34 J\pi r^2 = \frac34\times10^6\times\pi\times16\times10^{-6} = 12\pi\ \text{A}$$
$x = 12$.
Solution by Sreeraj P, M.Sc Physics