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(A) The current $dI$ through a thin cylindrical shell of radius $r$ and thickness $dr$ is given by $dI = J \cdot dA$,where $dA = 2 \pi r dr$.Substitut

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$\frac{2J_0A}{3}$

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(A) The current $dI$ through a thin cylindrical shell of radius $r$ and thickness $dr$ is given by $dI = J \cdot dA$,where $dA = 2 \pi r dr$.Substituting the given current density $J = J_0 \frac{r}{R}$:$dI = (J_0 \frac{r}{R}) \cdot (2 \pi r dr) = \frac{2 \pi J_0}{R} r^2 dr$.To find the total current $I$,we integrate from $r = 0$ to $r = R$:$I = \int_{0}^{R} \frac{2 \pi J_0}{R} r^2 dr = \frac{2 \pi J_0}{R} \int_{0}^{R} r^2 dr$.$I = \frac{2 \pi J_0}{R} \left[ \frac{r^3}{3} \right]_{0}^{R} = \frac{2 \pi J_0}{R} \cdot \frac{R^3}{3} = \frac{2}{3} J_0 (\pi R^2)$.Since the cross-sectional area $A = \pi R^2$,the total current is $I = \frac{2}{3} J_0 A$.

For a cylinder of radius $R$,the current density is given by $J = J_0 \frac{r}{R}$,where $J_0$ is a constant and $r$ is the distance from the axis. Calculate the total current.

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