Question

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

• A. $10^{5} A / m ^{2}$
• B. $10^{4} A / m ^{2}$
• C. $10^{6} A / m ^{2}$
• D. $10^{-5} A / m ^{2}$

Answer 1

Answer: (A)

Radius of wire $=\frac{10^{-2}}{\sqrt{\pi}}$
Cross sectional area $A=\pi r ^{2}=10^{-4} m ^{2}$
$j =\frac{ i }{ A }=\left(\frac{ V }{ R }\right) \cdot \frac{1}{ A }=\frac{ E \ell}{ RA } R =\frac{\rho \ell}{ A }$
$j =\frac{10 \times 10}{10 \times 10^{-4}}=10^{5} A / m ^{2}$
or
$ J =\sigma E \Rightarrow \frac{E}{\rho}=\frac{ E \ell}{ RA }=\frac{10 \times 10 \times \pi}{10 \times 10^{-4} \times \pi} $
$\Rightarrow 10^{5} A / m ^{2}$