AIIMS 2019 Physics Current Density MCQ Question

To find the total current $I$ in the radial region from $r = 0$ to $r = R/4$, we integrate the current density $J(r) = J_0 \left(1 - \frac{r}{R}\right)$ over the cross-sectional area. The differential current $dI$ through a ring of radius $r$ and thickness $dr$ is given by $dI = J(r) \cdot 2\pi r \, dr$. Integrating from $0$ to $R/4$ yields $I = \int_0^{R/4} J_0 \left(1 - \frac{r}{R}\right) 2\pi r \, dr = \frac{5J_0 \pi R^2}{96}$, confirming option B as correct. Other options are incorrect as they do not match the result of this integration, which accurately accounts for the varying current density across the specified region.