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  • The current density in a cylindrical wire of radius is <img alt="\mathrm{4\times10^{6}Am^{-2}}" s

The current density in a cylindrical wire of radius \mathrm{4mm} is \mathrm{4\times10^{6}Am^{-2}}. The current through the outer portion of the wire between radial distances \mathrm{\frac{R}{2}} and \mathrm{R} is ____________\mathrm{\pi A}.

Option: 1

48


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Current Density

\mathrm{J=\frac{I}{A}} \\

\mathrm{I=J A}

\mathrm{I =J \times\left[\pi R^{2}-\pi\left(\frac{R^{2}}{2}\right)^{2}\right] }\\

   \mathrm{=J \times\left(\pi R^{2}-\frac{\pi R^{2}}{4}\right)} \\

   \mathrm{=J \times \frac{3 \pi R^{2}}{4}} \\

   \mathrm{=4 \times 10^{6} \times \frac{3 \pi}{4} \times 16 \times 10^{-6}} \\

    \mathrm{=48 \pi \mathrm{A}}

Hence the answer is \mathrm{48 \pi \mathrm{A}}

Posted by

Rishabh

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