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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

User Nlowe
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1 Answer

7 votes

Answer:

The resistivity is
\rho = 2.5 *10^(-8) \ \Omega \cdot m

Step-by-step explanation:

From the question we are told that

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

The current density is
J = 2.4 *10^(8) \ A/m^2

Generally the resistivity is mathematically represented as


\rho = (E)/(J)

substituting values


\rho = (6.2)/(2.4 *10^(8))


\rho = 2.5 *10^(-8) \ \Omega \cdot m

User Andy Heard
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