Question

A conductor of uniform radius 1.20 cm carries a current of 3.00 A produced by an electric field of 120 V/m. What is the resistivity of the material?

Answer 1

Answer:

The value is  $\rho = 0.0180 97 9 \ \Omega m^{-1}$

Explanation:

From the question we are told that

   The radius of the conductor is $r = 1.20 \ cm = 0.012 \ m$

    The current it is carrying is  $I = 3.00 \ A$

    The electric field is  $E = 120 \ V/m$

Generally the resistivity of the material is mathematically represented as

        $\rho = \frac{EA}{ I }$

Here A is the area of the conductor which is mathematically represented as

      $A = \pi r^2$

=>  $A =3.142 * (0.012)^2$

=>  $A = 0.000452 \ m^2$

So

    $\rho = \frac{120 * 0.000452}{ 3 }$

=> $\rho = 0.0180 97 9 \ \Omega m^{-1}$

Answer 2

The resistivity of the material will be "0.0180979 Ωm⁻¹".

Given values,

• Radius, r = 1.20 m or 0.012 m
• Current, I = 3.00 A
• Electric field, E = 120 V/m

Now,

The area of conductor,

→ $A = \pi r^2$

      $= 3.142\times (0.012)^2$

      $= 0.000452 \ m^2$

hence,

The resistivity will be:

→ $\rho = \frac{EA}{I}$

By substituting the values, we get

     $= \frac{120\times 0.000452}{3}$

     $= 0.0180979 \ \Omega m^{-1}$

Thus the above response is right.  

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