Question

The resistivity of gold is 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 1

Answer:

0.0360531138247 V/m

Explanation:

$\rho$ = Resistivity of gold = $2.44\times 10^{-8}\ \Omega .m$ (General value)

I = Current = 940 mA

d = Diameter = 0.9 mm

A = Area = $\dfrac{\pi}{4}d^2$

E = Electric field

Resistivity is given by

$\rho=\dfrac{EA}{I}\\\Rightarrow E=\dfrac{\rho I}{A}\\\Rightarrow E=\dfrac{2.44\times 10^{-8}\times 940\times 10^{-3}}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}\\\Rightarrow E=0.0360531138247\ V/m$

The  electric field in the wire is 0.0360531138247 V/m

Answer 2

Answer:

Explanation:

Current in the wire, i = 940 mA = 0.94 A

Length of the wire, l = 14 cm = 0.14 m

diameter of wire = 0.9 mm

radius of wire, r = 0.45 mm = 0.45 x 10^-3 m

resistivity of gold, ρ = 2.44 x 10^-8 ohm metre

Let R is the resistance of the wire.

$R = \rho \times \frac{l}{A}$

$R = 2.44\times 10^{-8}\times \frac{0.14}{3.14\times 0.45\times 0.45\times 10^{-6}}$

R = 5.37 x 10^-3 ohm

By the Ohm's law

V = i x R

V = 0.94 x 5.37 x 10^-3

V = 5.05 x 10^-3 V

E = V / l = (5.05 x 10^-3) / 0.14 = 0.036 V/m