The current density varies with radial distance $$r$$ as $$J = a{r^2},$$   in a cylindrical wire of radius $$R.$$ The current passing through the wire between radial distance $$\frac{R}{3}$$ and $$\frac{R}{2}$$ is

Solution :
Given ; $$J = a{r^2}.$$
$$\eqalign{ & i = \int_2^2 {J \times 2\pi \,r\,dr} = \int_{\frac{R}{3}}^{\frac{R}{2}} {a{r^2} \times 2\pi \,r\,dr} \cr & = 2\pi a\int_{\frac{R}{3}}^{\frac{R}{2}} {{r^3}dr} = 2\pi \,a\left| {\frac{{{r^4}}}{4}} \right|_{\frac{R}{3}}^{\frac{R}{2}} \cr & = \frac{{\pi a}}{2}\left[ {{{\left( {\frac{R}{2}} \right)}^4} - {{\left( {\frac{R}{3}} \right)}^4}} \right] \cr & = \frac{{\pi a{R^4}}}{2} \times \frac{{65}}{{81 \times 16}} = \frac{{65\pi \,a{R^4}}}{{2592}} \cr} $$