Question

A cylindrical conductor of diameter $0.1mm$ carries a current of $90mA$ . The current density (in $A{m}^{-2}$ ) is $\left(\pi \sim eq3\right)$

• A. $1.2\times 10^7$
• B. $3\times 10^6$
• C. $6\times 10^6$
• D. $2.4\times 10^7$

Answer 1

Answer: (A)

Given,
The diameter of the cylindrical conductor $\left(D\right)=0.01mm$
Radius $\left(r\right)=\frac{D}{2}=\frac{0.1}{2}mm$
Current $\left(I\right)=90mA=90\times {\left(10\right)}^{-3}A$
We know that,
Current density $\left(J\right)=\frac{I}{A}$
$=\frac{I}{\pi r^2}$
$=\frac{90\times {\left(10\right)}^{-3}}{\frac{22}{7}\times {\left(\frac{0.1\times {\left(10\right)}^{-3}}{2}\right)}^2}$
$=\frac{90\times {\left(10\right)}^{-3}}{3.14\times {\left(\frac{0.1\times {\left(10\right)}^{-3}}{2}\right)}^2}$
$=12000\times 10^3$
$=1.2\times 10^{7}A{m}^{-2}$