Question

If resistivity of copper conductor is $1.7 \times 10^{-8} \Omega m$ and electric field is $100\, Vm ^{-1}$, then current density will be

• A. $6 \times 10^{9} Am ^{-2}$
• B. $1.7 \times 10^{-6} Am ^{-2}$
• C. $1.7 \times 10^{-10} Am ^{-2}$
• D. $6 \times 10^{7} Am ^{-2}$

Answer 1

Answer: (A)

Current density, $J=\sigma E$
where, $\sigma=$ conductivity $=\frac{1}{\text { resistivity }}=\frac{1}{\rho}=\frac{1}{1.7 \times 10^{-8} \Omega m }$
$\Rightarrow J=\frac{1}{1.7 \times 10^{-8}} \times 100=\frac{100}{17} \times 10^{9}=6 \times 10^{9} Am ^{-2}$