Q142 / 180 · KCET 2017 · Physics
A cylindrical conductor of diameter \( 0.1 \mathrm{~mm} \) carries a current of \( 90 \mathrm{~mA} \). The current density (in
\( \left.\mathrm{Am}^{-2}\right) \) is \( (\pi \simeq 3) \)
\( \left.\mathrm{Am}^{-2}\right) \) is \( (\pi \simeq 3) \)
- Correct answerA\( 1.2 \times 10^{7} \)
- B\( 2.4 \times 10^{7} \)
- C\( 3 \times 10^{6} \)
- D\( 6 \times 10^{6} \)
Solution
Diameter of the wire \( d=0.1 \mathrm{~mm}=10^{-4} \mathrm{~m}^{2} \)
Cross-sectional area of the wire \( A=\frac{\Pi d^{2}}{4} \)
\( \therefore A=\frac{3 \times\left(10^{-4}\right)^{2}}{4}=0.75 \times 10^{-8} \mathrm{~m}^{2} \)
Current flowing through the wire \( I=90 \mathrm{~mA}=90 \times 10^{-3} \mathrm{~A} \)
Thus current density \( J=\frac{I}{A} \)
\( \therefore J=\frac{90 \times 10^{-3}}{0.75 \times 10^{-8}}=1.2 \times 10^{7} \mathrm{Am}^{-2} \)
Cross-sectional area of the wire \( A=\frac{\Pi d^{2}}{4} \)
\( \therefore A=\frac{3 \times\left(10^{-4}\right)^{2}}{4}=0.75 \times 10^{-8} \mathrm{~m}^{2} \)
Current flowing through the wire \( I=90 \mathrm{~mA}=90 \times 10^{-3} \mathrm{~A} \)
Thus current density \( J=\frac{I}{A} \)
\( \therefore J=\frac{90 \times 10^{-3}}{0.75 \times 10^{-8}}=1.2 \times 10^{7} \mathrm{Am}^{-2} \)
Hence, the correct answer is option: 'A'.