The current density in a cylindrical wire of radius r = 4.0 mm is 1.0×106 A/m2. The current through the outer portion of the wire between radial distances r / 2 and r is xπA; where x is ________.

J=1×106 A/m2I =∫JdA =∫106×2πxdx =106×2π·xx22r2r =π×106r2-r24=12π x=12

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

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answer is 12.

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Detailed Solution

$J = 1 \times 10^{6} A / m^{2}$

$I = \int JdA = \int 10^{6} \times 2 π xdx {= 10^{6} \times 2 π \cdot x \frac{x^{2}}{2}]}_{\frac{r}{2}}^{r} = π \times 10^{6} [r^{2} - \frac{r^{2}}{4}] = 12 π x = 12$

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