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

Step-by-step 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 * (l)/(A)`

`R = 2.44* 10^(-8)* (0.14)/(3.14* 0.45* 0.45* 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

Answer 2

0.0360531138247 V/m

Step-by-step explanation:

`\rho` = Resistivity of gold =
`2.44* 10^(-8)\ \Omega .m` (General value)

I = Current = 940 mA

d = Diameter = 0.9 mm

A = Area =
`(\pi)/(4)d^2`

E = Electric field

Resistivity is given by

`\rho=(EA)/(I)\\\Rightarrow E=(\rho I)/(A)\\\Rightarrow E=(2.44* 10^(-8)* 940* 10^(-3))/((\pi)/(4)(0.9* 10^(-3))^2)\\\Rightarrow E=0.0360531138247\ V/m`

The electric field in the wire is 0.0360531138247 V/m