Question

27. The current density in a wire is 10 A/cm2 and
the electric field in the wire is 5 V/cm. If p
resistivity of material, then (in S.L. units):
(a) p = 5 x 10-3 (b) p = 200
(c) p =2 x 10-3 (d) p =500​

Answer 1

Answer:

The resistivity of the material is $5\times10^{-3}\ \Omega-m$

(a) is correct.

Explanation:

Given that,

Current density = 10 A/cm²

Electric field = 5 V/cm

The resistance of the wire is the product of resistivity of the of the material and length divided by the area of cross section.

The formula is defined as:

$R=\dfrac{\rho l}{A}$

Using Ohm's law

$\dfrac{V}{I}=\dfrac{\rho l}{A}$

$\rho=\dfrac{V\times A}{I\times l}$

$\rho=\dfrac{5\times10^{2}}{10\times10^{4}}$

$\rho=5\times10^{-3}\ \Omega-m$

Hence, The resistivity of the material is $5\times10^{-3}\ \Omega-m$