The current density in a solid cylindrical wire of radius $R$, as a function of radial distance $r$ is given…
The current density in 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
$\frac{5 J_0 \pi R^2}{32}$
$\frac{5 J_0 \pi R^2}{96}$
$\frac{3 J_0 \pi R^2}{94}$
$\frac{J_0 \pi R^2}{128}$
Solution
$\begin{aligned} \text { } & : d i=J d A=J_0\left(1-\frac{r}{R}\right) 2 \pi r d r \\ i & =\int_{r=0}^{r=\frac{R}{4}} J_0\left(1-\frac{r}{R}\right) 2 \pi r d r \\ & =\frac{J_0 5 \pi R^2}{96}\end{aligned}$
