Question

A copper wire of cross-sectional area $ =2.0\text{ }m{{m}^{2}}, $ resistivity $ =1.7\times {{10}^{-8}}\Omega m, $ carries a current of 1 A. The electric field in the copper wire is

• A. $ 8.5\times {{10}^{-5}}V{{m}^{-1}} $
• B. $ 8.5\times {{10}^{-4}}V{{m}^{-1}} $
• C. $ 8.5\times {{10}^{-3}}V{{m}^{-1}} $
• D. $ 8.5\times {{10}^{-2}}V{{m}^{-1}} $

Answer 1

Answer: (C)

From symmetry considerations and also from theory $ \frac{{{V}_{a}}}{{{V}_{d}}}=\frac{{{V}_{b}}}{{{V}_{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}}V{{m}^{-1}} $