Question

A cylindrical conductor of diameter $0.1 \, m m \, $ carries a current of $90 \, m A$ . The current density (in $A \, m^{- 2}$ ) is $\left(\pi \sim eq 3\right)$

• 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 $\left(D\right)=0.01 \, mm$
Radius $\left(r\right)=\frac{D}{2}=\frac{0 .1}{2 \, } \, mm$
Current $\left(I\right)=90 \, mA=90\times \left(10\right)^{- 3}A$
We know that,
Current density $\left(J\right)=\frac{I}{A}$
$=\frac{I}{\pi r^{2}}$
$=\frac{90 \times \left(10\right)^{- 3}}{\frac{22}{7} \times \left(\frac{0.1 \times \left(10\right)^{- 3} \, }{2}\right)^{2}}$
$=\frac{90 \times \left(10\right)^{- 3} \, }{3.14 \times \left(\frac{0.1 \times \left(10\right)^{- 3} \, }{2}\right)^{2}}$
$=12000\times 10^{3}$
$=1.2\times 10^{7} \, A \, m^{- 2}$