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(A) The current density $J$ is given as $4 	imes 10^{6} \; A \cdot m^{-2}$.The radius of the wire is $R = 4 \; mm = 4 	imes 10^{-3} \; m$.The curren

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$48$

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(A) The current density $J$ is given as $4 	imes 10^{6} \; A \cdot m^{-2}$.The radius of the wire is $R = 4 \; mm = 4 	imes 10^{-3} \; m$.The current $I$ through an area $A$ is given by $I = \int J \cdot dA$.For the outer portion between radial distance $\frac{R}{2}$ and $R$,the area is $A' = \pi R^{2} - \pi \left(\frac{R}{2}ight)^{2} = \pi R^{2} - \frac{\pi R^{2}}{4} = \frac{3}{4} \pi R^{2}$.Substituting the values:$I = J 	imes A' = 4 	imes 10^{6} 	imes \frac{3}{4} \pi R^{2}$.$I = 4 	imes 10^{6} 	imes \frac{3}{4} 	imes \pi 	imes (4 	imes 10^{-3})^{2}$.$I = 3 	imes 10^{6} 	imes \pi 	imes 16 	imes 10^{-6}$.$I = 48 \; \pi \; A$.

The current density in a cylindrical wire of radius $4 \; mm$ is $4 \times 10^{6} \; A \cdot m^{-2}$. The current through the outer portion of the wire between radial distance $\frac{R}{2}$ and $R$ is $\dots \; \pi \; A$.

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