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NEET PHYSICSMedium

A copper wire of length 10 m10\text{ m} and radius 10−2π m\frac{10^{-2}}{\sqrt{\pi}}\text{ m} has electrical resistance of 10 Ω10\ \Omega. The current density in the wire for an electric field strength of 10 V/m10\text{ V/m} is

A

104 A/m210^4\text{ A/m}^2

B

106 A/m210^6\text{ A/m}^2

C

105 A/m210^5\text{ A/m}^2

D

105 A/m210^5\text{ A/m}^2

Step-by-Step Solution

Resistance R=ρLAR = \rho \frac{L}{A}. Current density j=σE=1ρEj = \sigma E = \frac{1}{\rho} E. Since R=ρLAR = \frac{\rho L}{A}, we have 1ρ=LRA\frac{1}{\rho} = \frac{L}{RA}. Thus, j=ELRAj = \frac{E L}{RA}. Substituting the values: j=10×1010×π(10−2π)2=10010×π×10−4π=10010−3=105 A/m2j = \frac{10 \times 10}{10 \times \pi (\frac{10^{-2}}{\sqrt{\pi}})^2} = \frac{100}{10 \times \pi \times \frac{10^{-4}}{\pi}} = \frac{100}{10^{-3}} = 10^5\text{ A/m}^2.

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