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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 \( (\pi \simeq 3) \)
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\( \left.\mathrm{Am}^{-2}\right) \) is \( (\pi \simeq 3) \)
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Verified Answer
The correct answer is:
\( 1.2 \times 10^{7} \)
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 \( J=\frac{I}{A} \)
\( \therefore J=\frac{90 \times 10^{-3}}{0.75 \times 10^{-8}}=1.2 \times 10^{7} \mathrm{Am}^{-2} \)
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