The current density is a solid cylindrical wire of radius $R ,$ as a function of radial distance $r$ is given by $J ( r )= J _{0}\left(1-\frac{ r }{ R }\right) .$ The total current in the radial region $r =0$ to $r =\frac{ R }{4}$ will be
AIIMS 2019, Diffcult
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Consider the expression,
$d i=J d A$
$=J_{0}\left(1-\frac{r}{R}\right) 2 \pi r d r$
Integrate on both the sides
$i=\int_{r=0}^{r=\frac{R}{4}} J_{0}\left(1-\frac{r}{R}\right) 2 \pi r d r$
$=\frac{J_{0} 5 \pi R^{2}}{96}$
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