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A copper wire of length \mathrm{10m} and radius \mathrm{\left ( 10^{-2}/\sqrt{\pi} \right )m} has an electrical resistance of \mathrm{10\Omega}. The current density in the wire for an electric field strength of \mathrm{10\left ( V/m \right )} is :

Option: 1

\mathrm{10^{6}A/m^{2}}


Option: 2

\mathrm{10^{-5}A/m^{2}}


Option: 3

\mathrm{10^{5}A/m^{2}}


Option: 4

\mathrm{10^{3}A/m^{2}}


Answers (1)

best_answer

We know that ,

\mathrm{J =\frac{E}{\rho} }\\

\mathrm{\text { Also, } R =\frac{\rho l}{A}} \\

\mathrm{\therefore J =\frac{E}{\left(\frac{R A}{l}\right)}}

\mathrm{J =\frac{10}{\frac{10 \times \pi\left(10^{-4} / \pi\right)}{10}}} \\

\mathrm{J =10^{5} \mathrm{~A} / \mathrm{m}^{2} }

Hence correct option is 3

Posted by

Ajit Kumar Dubey

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