A wire has an electric field of 6.2 V/m and carries a current density of 2.4 x 108 A/m2. What is its resistivity

Respuesta :

Answer:

The resistivity is  [tex]\rho = 2.5 *10^{-8} \ \Omega \cdot m[/tex]

Explanation:

From the question we are told that

    The magnitude of the electric field is  [tex]E = 6.2 V/m[/tex]

     The  current density is  [tex]J = 2.4 *10^{8} \ A/m^2[/tex]

Generally  the resistivity is mathematically represented as

         [tex]\rho = \frac{E}{J}[/tex]

substituting values

        [tex]\rho = \frac{6.2}{2.4 *10^{8}}[/tex]

        [tex]\rho = 2.5 *10^{-8} \ \Omega \cdot m[/tex]

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