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(B) The current $I$ passing through a cross-section is given by the integral of current density $J$ over the area $A$: $I = \int J \cdot dA$.For a cyl

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$2\pi J_0 \ln 2$

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(B) The current $I$ passing through a cross-section is given by the integral of current density $J$ over the area $A$: $I = \int J \cdot dA$.For a cylindrical shell of radius $r$ and thickness $dr$,the elemental area is $dA = 2\pi r \cdot dr$.Substituting the given current density $J = \frac{J_0}{r^2}$ and the elemental area $dA$ into the integral,we get:$I = \int_{a}^{2a} \left( \frac{J_0}{r^2} \right) (2\pi r \cdot dr)$$I = 2\pi J_0 \int_{a}^{2a} \frac{1}{r} \cdot dr$$I = 2\pi J_0 [\ln r]_{a}^{2a}$$I = 2\pi J_0 (\ln(2a) - \ln(a))$$I = 2\pi J_0 \ln\left( \frac{2a}{a} \right)$$I = 2\pi J_0 \ln 2$.

In the given hollow cylindrical conductor,the current density is $J = \frac{J_0}{r^2}$,where $J_0$ is a constant and $r$ is the distance from the axis of the cylinder. If the radius of the inner surface is $a$ and the radius of the outer surface is $2a$,find the current passed through the conductor.

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