Current density in a cylindrical wire of radius R varies with radial distance as β r + r 0 2 . The…

Current density is current per unit area. If we take a circular element of radius $r$ and thickness $dr$, the current through it will be,

$$\begin{gathered} di=2\pi rdr\beta {\left(r+r_0\right)}^2 \\ \Rightarrow di=2\pi rdr\beta \left(r^2+2r_{0}r+{r_0}^2\right) \\ \Rightarrow di=2\pi \beta \left(r^{3}dr+2r_0{r}^{2}dr+{r_0}^{2}rdr\right) \end{gathered}$$

Integrating the above equation from $0$ to $R$ we will get current through the complete wire.

$$\begin{gathered} {\int }_0^{I}di={\int }_0^{R}2\pi \beta \left(r^{3}dr+2r_0{r}^{2}dr+{r_0}^{2}rdr\right) \\ \Rightarrow I=2\pi \beta \left[\frac{R^4}{4}+\frac{2r_0{R}^3}{3}+\frac{r_0{R}^2}{2}\right] \end{gathered}$$

For current through the shaded area,

$$\begin{gathered} i=\frac{\pi }{3\times \left(2\pi \right)}I=\frac{\pi }{3\times \left(2\pi \right)}\times 2\pi \beta \left[\frac{R^4}{4}+\frac{2r_0{R}^3}{3}+\frac{r_0{R}^2}{2}\right] \\ \Rightarrow i=\pi \beta \left[\frac{R^4}{12}+\frac{2r_0{R}^3}{9}+\frac{r_0{R}^2}{6}\right] \end{gathered}$$