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Sum

Calculate the electric field in a copper wire of cross-sectional area 2.0 mm^{2} carrying a current of 1 A.

The resistivity of copper = 1.7 × 10^{–8} Ω m

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#### Solution

Given:-

Area of cross-section, A = 2 × 10^{–6 }m^{2}

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,

\[E = \frac{V}{l}\]

\[ \Rightarrow E = \frac{i\rho l}{Al} = \frac{i\rho}{A}\]

\[ \Rightarrow E = \frac{1 \times 1 . 7 \times {10}^{- 8}}{2 \times {10}^{- 6}}\]

\[ \Rightarrow E = 8 . 5 V/m\]

Concept: Ohm's Law

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