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(C) The current $I$ through an area element $dA$ is given by $I = \int J \cdot dA$.For a cylindrical wire,the area element of a ring at radius $r'$ wi

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$12$

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(C) The current $I$ through an area element $dA$ is given by $I = \int J \cdot dA$.For a cylindrical wire,the area element of a ring at radius $r'$ with thickness $dr'$ is $dA = 2 \pi r' dr'$.Given $J = 1.0 	imes 10^{6} \, A/m^{2}$ and $r = 4.0 \, mm = 4.0 	imes 10^{-3} \, m$.The current $I$ through the region between $r/2$ and $r$ is:$I = \int_{r/2}^{r} J (2 \pi r') dr' = 2 \pi J \int_{r/2}^{r} r' dr'$$I = 2 \pi J \left[ \frac{(r')^{2}}{2} ight]_{r/2}^{r} = \pi J \left[ r^{2} - \left( \frac{r}{2} ight)^{2} ight]$$I = \pi J \left[ r^{2} - \frac{r^{2}}{4} ight] = \pi J \left( \frac{3r^{2}}{4} ight)$Substituting the values $J = 10^{6} \, A/m^{2}$ and $r = 4 	imes 10^{-3} \, m$:$I = \pi 	imes 10^{6} 	imes \frac{3}{4} 	imes (4 	imes 10^{-3})^{2}$$I = \pi 	imes 10^{6} 	imes \frac{3}{4} 	imes 16 	imes 10^{-6} = 12 \pi \, A$Comparing this with $x \pi \, A$,we get $x = 12$.

The current density in a cylindrical wire of radius $r = 4.0 \, mm$ is $J = 1.0 \times 10^{6} \, A/m^{2}$. The current through the outer portion of the wire between radial distances $r/2$ and $r$ is $x \pi \, A$,where $x$ is ..........

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