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A conductor of length $100 \mathrm{~cm}$ and area of cross-section $1 \mathrm{~mm}^2$ carries a current of $5 \mathrm{~A}$. If the resistivity of the material of the conductor is $3.0 \times 10^{-8} \Omega-\mathrm{m}$, then the electric field across the conductor is
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Verified Answer
The correct answer is:
0.15 V/m
Current density is related to electric field as
$\mathbf{J}=\frac{\mathbf{E}}{\rho}$
where, $\rho=$ resistivity.
Here, $I=5 \mathrm{~A}, A=1 \mathrm{~mm}^2=1 \times 10^{-6} \mathrm{~m}^2$
and $\rho=3 \times 10^{-8} \Omega-\mathrm{m}$
So, electric field inside the conductor is
$\begin{aligned}
\mathbf{E} & =\mathbf{J} \cdot \rho=\frac{I}{A} \cdot \rho \\
& =\frac{5 \times 3 \times 10^{-8}}{1 \times 10^{-6}}=0.15 \mathrm{~V} / \mathrm{m}
\end{aligned}$
$\mathbf{J}=\frac{\mathbf{E}}{\rho}$
where, $\rho=$ resistivity.
Here, $I=5 \mathrm{~A}, A=1 \mathrm{~mm}^2=1 \times 10^{-6} \mathrm{~m}^2$
and $\rho=3 \times 10^{-8} \Omega-\mathrm{m}$
So, electric field inside the conductor is
$\begin{aligned}
\mathbf{E} & =\mathbf{J} \cdot \rho=\frac{I}{A} \cdot \rho \\
& =\frac{5 \times 3 \times 10^{-8}}{1 \times 10^{-6}}=0.15 \mathrm{~V} / \mathrm{m}
\end{aligned}$
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