Q. A copper wire of cross-sectional area 2.0 $mm^{2}$, resistivity =1.7$\times 10^{- 8}\Omega m$, carries a current of 1A. The electric field in the copper wire is
A) $8.5 \times 10^{- 5}Vm^{- 1}$
B) $8.5 \times 10^{- 4}Vm^{- 1}$
C) $8.5 \times 10^{- 3}Vm^{- 1}$
D) $8.5 \times 10^{- 2}Vm^{- 1}$
Explanation
(c) $E = \frac{V}{l} = \frac{iR}{l} = \frac{i\rho l}{Al} = \frac{i\rho}{A} = \frac{1 \times 1.7 \times 10^{- 8}}{2 \times 10^{- 6}}$
$= 8.5 \times 10^{- 3}Vm^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.