A negligibly small current is passed through a wire of length \(15\text { m}\) and uniform cross-section \(6.0\times 10^{-7}\text{ m}^{2}\) , and its resistance is measured to be \(5.0\;\Omega \) . What is the resistivity of the material at the temperature of the experiment?
Answer
Here, we have length as \(15\text{ m}\) , area of \(6\times 10^{-17}\text{ m}^2\) and resistance as \(6\) ohms.
\(l=15\text{ m}\)
\(A=6\times 10^{-17}\text{ m}^2\)
\(R=5\;{\Omega}\)
To find the resistivity of the material, we use the following formula-
\(R=\rho \{\frac l A\}\)
\(5=\rho \left\{\frac{15} 6\right\}10^{17}\)
\(\rho =2\times 10^{-17}{\Omega-m}\) .
\(l=15\text{ m}\)
\(A=6\times 10^{-17}\text{ m}^2\)
\(R=5\;{\Omega}\)
To find the resistivity of the material, we use the following formula-
\(R=\rho \{\frac l A\}\)
\(5=\rho \left\{\frac{15} 6\right\}10^{17}\)
\(\rho =2\times 10^{-17}{\Omega-m}\) .
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