Answered

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 :

okpalawalter8

Answer:

The value is   [tex]\rho  =  4.02 *10^{-8} \  \Omega \cdot m[/tex]

Explanation:

From the question we are told that

   The radius is  [tex]r =   0.800 mm = 0.0008 \ m[/tex]

   The voltage of the battery is  [tex]emf  =  12.0 V[/tex]

    The slope is  [tex]s =  600 \ A \cdot m[/tex]

Generally the resistance is mathematically represented as

     [tex]R  =  \frac{\rho *  d }{A }[/tex]

Generally the current is mathematically represented as

      [tex]I = \frac{V}{R}[/tex]

=>    [tex]I = \frac{V}{\frac{\rho *  d }{A }}[/tex]

=>   [tex]I  =  \frac{V *  A }{\rho}  *  \frac{1}{d}[/tex]

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

      [tex]s =  \frac{V *  A }{\rho}[/tex]

So    [tex] 600  =  \frac{V *  A }{\rho}[/tex]

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

        [tex]A =  \pi r^2[/tex]

=>       [tex]A =  3.142 *   (0.0008 )^2[/tex]

=>       [tex]A =  2.011*10^{-6} \ m^2[/tex]    

So

     [tex] 600  =  \frac{12.0 *  (2.011*10^{-6}) }{\rho}[/tex]

=>   [tex]\rho  =  \frac{12 * 2.011*10^{-6} }{600}[/tex]

=>  [tex]\rho  =  4.02 *10^{-8} \  \Omega \cdot m[/tex]