NEET-XII-Physics
32: Electric Current in Conductors
- #12Calculate the electric field in a copper wire of cross-sectional area 2.0 mm2 carrying a current of 1 A.
The resistivity of copper = 1.7 × 10-8 Ω mAns : Given:
Area of cross-section, A = 2 × 10-6 m2
Current through the wire, i = 1 A
Resistivity of copper, ρ = 1.7 × 10-8 Ωm
Resistance of a wire,
`` R=\rho \frac{l}{A}``
Also from Ohm's Law, voltage across a wire,
`` V=iR=\frac{i\rho l}{A}``
The electric field of the wire,
`` E=\frac{V}{l}``
`` \Rightarrow E=\frac{i\rho l}{Al}=\frac{i\rho }{A}``
`` \Rightarrow E=\frac{1\times 1.7\times {10}^{-8}}{2\times {10}^{-6}}``
`` \Rightarrow E=8.5\,\mathrm{\,V\,}/\,\mathrm{\,m\,}``
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