Q 12-03-298NEETNEET 2022Top questionEasy
A copper wire of length $10$ m and radius $\left(\dfrac{10^{-2}}{\sqrt{\pi}}\right)$ m has electrical resistance of $10\ \Omega$. The current density in the wire for an electric field strength of $10$ (V/m) is
Answer: (A) $10^5\ \text{A/m}^2$
Area: $A = \pi r^2 = \pi \times \dfrac{10^{-4}}{\pi} = 10^{-4}\ \text{m}^2$.
Resistivity: $\rho = \dfrac{RA}{l} = \dfrac{10 \times 10^{-4}}{10} = 10^{-4}\ \Omega\,\text{m}$.
$$J = \frac{E}{\rho} = \frac{10}{10^{-4}} = 10^5\ \text{A/m}^2$$
Solution by Sreeraj P, M.Sc Physics