Answer:
The conductivity of Nichrome is
.
Explanation:
Given:
Potential difference (V) = 2.0 V
Current flowing (I) = 4.0 A
Length of wire (L) = 1.0 m
Area of cross section of wire (A) = 1.0 mm² = 1 × 10⁻⁶ m² [1 mm² = 10⁻⁶ m²]
We know, from Ohm's law, that the ratio of voltage and current is always a constant and equal to the resistance of the resistor. Therefore, the resistance of the nichrome wire is given as:
![R=\frac{V}{A}=\frac{2.0}{4.0}=0.5\ \Omega](https://tex.z-dn.net/?f=R%3D%5Cfrac%7BV%7D%7BA%7D%3D%5Cfrac%7B2.0%7D%7B4.0%7D%3D0.5%5C%20%5COmega)
Now, resistance of the nichrome wire in terms of its resistivity, length and area of cross section is given as:
![R=\rho\frac{L}{A}](https://tex.z-dn.net/?f=R%3D%5Crho%5Cfrac%7BL%7D%7BA%7D)
Where, ![\rho\to resistivity\ of\ Nichrome](https://tex.z-dn.net/?f=%5Crho%5Cto%20resistivity%5C%20of%5C%20Nichrome)
Now, plug in all the values given and solve for
. This gives,
![0.5\ \Omega=\rho\frac{1.0\ m}{1\times 10^{-6}\ m^2}\\\\\rho=\frac{0.5\times 1\times 10^{-6}}{1.0}=0.5\times 10^{-6}\ \Omega-m](https://tex.z-dn.net/?f=0.5%5C%20%5COmega%3D%5Crho%5Cfrac%7B1.0%5C%20m%7D%7B1%5Ctimes%2010%5E%7B-6%7D%5C%20m%5E2%7D%5C%5C%5C%5C%5Crho%3D%5Cfrac%7B0.5%5Ctimes%201%5Ctimes%2010%5E%7B-6%7D%7D%7B1.0%7D%3D0.5%5Ctimes%2010%5E%7B-6%7D%5C%20%5COmega-m)
Now, conductivity of a material is the reciprocal of its resistivity. Therefore, the conductivity of Nichrome is given as:
![\sigma=\frac{1}{\rho}=\frac{1}{0.5\times 10^{-6}}=2\times 10^6\ S/m](https://tex.z-dn.net/?f=%5Csigma%3D%5Cfrac%7B1%7D%7B%5Crho%7D%3D%5Cfrac%7B1%7D%7B0.5%5Ctimes%2010%5E%7B-6%7D%7D%3D2%5Ctimes%2010%5E6%5C%20S%2Fm)
Conductivity is measured in Siemens per meter (S/m)
Therefore, the conductivity of Nichrome is
.