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

Question

asked Jan 26, 2021 235k views

1 Answer

Andy Heard · Answer 1 · 2021-01-31T06:47:37+0000

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

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

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