Physics - Current Electricity Question with Solution | TestHub

Given,

The diameter of the cylindrical conductor$\left(\mathrm{D}\right)=0.01 \mathrm{m}\mathrm{m}$

Radius$\left(\mathrm{r}\right)=\frac{\mathrm{D}}{2}=\frac{0.1}{2 } \mathrm{m}\mathrm{m}$

Current$\left(\mathrm{I}\right)=90 \mathrm{m}\mathrm{A}=90\times {10}^{\mathrm{-}3}A$

We know that,

Current density$\left(\mathrm{J}\right)=\frac{\mathrm{I}}{\mathrm{A}}$

$=\frac{\mathrm{I}}{{\mathrm{\pi r}}^2}$

$=\frac{90\times {10}^{-3}}{\frac{22}{7}\times {\left(\frac{0.1\times {10}^{-3} }{2}\right)}^2}$

$=\frac{90\times {10}^{-3} }{3.14\times {\left(\frac{0.1\times {10}^{-3} }{2}\right)}^2}$

$=12000\times {10}^3$

$=1.2\times {10}^7 \mathrm{A} {\mathrm{m}}^{-2}$