Question

The resistivity of gold is 2.44 x 10-8 ohms.m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 mA. What is the electric field in the wire?
Answer

0.090 V/m

0.028 V/m

0.046 V/m

0.0090 V/m

0.036 V/m

Answer 1

Answer:

E=0.036 V/m

Explanation:

Given that

Resistivity ,ρ=2.44 x 10⁻⁸ ohms.m

d= 0.9 mm

L= 14 cm

I = 940 m A = 0.94 A

We know that electric field E

E= V/L

V= I R

R=ρL/A

So we can say that

E= ρI/A

Now by putting the values

$E=\dfrac{ 2.44\times 10^{-8}\times 0.94}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}$

E=0.036 V/m