Q17P - Fundamentals Of Physics

Question

A wire of Nichrome (a nickel–chromium–iron alloy commonly used in heating elements) is 1.0m long and $\mathbf{1}\mathbf{.}\mathbf{0}{\mathbf{mm}}^{\mathbf{2}}$ in cross-sectional area. It carries a current of 4.0 A when a 2.0 V potential difference is applied between its ends. Calculate the conductivity  of Nichrome.

Step-by-Step Solution

Answer

The conductivity of Nichrome is $2\times {10}^6/\mathrm{\Omega m}$.

Step 1: The given data

1. Length of the wire, L = 1.0 m
2. Cross-sectional area of the wire, $A=1.0{\mathrm{mm}}^2\mathrm{or} 1.0\times {10}^{-6}{\mathrm{m}}^2$
3. Current, l = 4.0 A
4. Potential difference, V = 2.0 V

Step 2: Understanding the concept of conductivity

Conductivity is the property of the material. It is the ability of a material to carry the current.

We can write the equation of resistivity in terms of voltage and current by using Ohm’s law. By taking the inverse of resistivity, we can find the conductivity of Nichrome.

 

Formulae:

The voltage equation from Ohm’s law, V = lR                                                             …(i)

Here, is the voltage, is current, and is resistance.

The resistivity of a material, $p=\frac{RA}{L}$                                                                            …(ii)

Here, p is the resistivity of the material, R is the resistance, A is the area of cross-section, L is the length of the conductor.

The conductivity of a material is related to its resistivity, $\sigma =\frac{1}{p}$                                  …(iii)

Step 3: Calculation of the conductivity of Nichrome

Substituting the given values in equation (ii) with resistance value from equation (i), we can get the resistivity of Nichrome as follows:

$$\begin{gathered} p=\frac{VA}{lL} \\ =\frac{92.0\mathrm{V}\left)\right(1.0\times {10}^{-6}{\mathrm{m}}^2)}{(4.0\mathrm{A})(1.0\mathrm{m})} \\ =5\times {10}^{-7}\mathrm{\Omega m} \end{gathered}$$

Now using this resistivity value in equation (iii), we can get the required value of the conductivity of Nichrome as follows:

$$\begin{gathered} \mathrm{\sigma }=\frac{1}{5\times {10}^{-7}\mathrm{\Omega }.\mathrm{m}} \\ =2\times {10}^6/\mathrm{\Omega m} \end{gathered}$$

Therefore, the conductivity of Nichrome is $2\times {10}^6/\mathrm{\Omega m}$.