Question

A metal wire has a circular cross section with radius 0.800 mm. You measure the resistivity of the wire in the following way: you connect one end of the wire to one terminal of a battery that has emf 12.0 V and negligible internal resistance. To the other terminal of the battery you connect a point along the wire so that the length of wire between the battery terminals is d. You measure the current in the wire as a function of d. The currents are small, so the temperature change of the wire is very small. You plot your results as I versus 1/d and find that the data lie close to a straight line that has slope 600 A⋅m.
Required:
What is the resistivity of the material of which the wire is made?

Answer 1

Answer:

The value is   $\rho = 4.02 *10^{-8} \ \Omega \cdot m$

Explanation:

From the question we are told that

   The radius is  $r = 0.800 mm = 0.0008 \ m$

   The voltage of the battery is  $emf = 12.0 V$

    The slope is  $s = 600 \ A \cdot m$

Generally the resistance is mathematically represented as

     $R = \frac{\rho * d }{A }$

Generally the current is mathematically represented as

      $I = \frac{V}{R}$

=>    $I = \frac{V}{\frac{\rho * d }{A }}$

=>   $I = \frac{V * A }{\rho} * \frac{1}{d}$

Comparing this equation to that of a straight line we see that the slope is  

      $s = \frac{V * A }{\rho}$

So    $600 = \frac{V * A }{\rho}$

Here A is the cross-sectional  area of the wire which is mathematically represented as

        $A = \pi r^2$

=>       $A = 3.142 * (0.0008 )^2$

=>       $A = 2.011*10^{-6} \ m^2$    

So

     $600 = \frac{12.0 * (2.011*10^{-6}) }{\rho}$

=>   $\rho = \frac{12 * 2.011*10^{-6} }{600}$

=>  $\rho = 4.02 *10^{-8} \ \Omega \cdot m$