Question

A current-carrying gold wire has diameter 0.88 mm. The electric field in the wire is0.55 V/m. (Assume the resistivity ofgold is 2.44 10-8 Ω · m.)

(a) What is the current carried by thewire?
1 A
(b) What is the potential difference between two points in the wire6.3 m apart?
2 V
(c) What is the resistance of a 6.3 mlength of the same wire?
3 Ω

Answer 1

Hi there!

a)

We know that the resistance of the wire is equivalent to:

`R = (\rho L)/(A)`

R = Resistance (Ω)
ρ = Resistivity (Ωm)

L = Length (m)

A = Cross-sectional area (m²)

We can relate the voltage to an electric field by:

`V = Ed \\\\V = El`

V = Potential Difference (V)
E = Electric Field (V/m)

l = Length of wire (m)

And Ohm's Law:

`V =iR`

i = Current (A)

V = Potential Difference (V)

R = Resistance (Ω)

We do not know the length, so we can solve using the above relationships.

`V = iR\\\\V = i(\rho L)/(A)\\\\(V)/(L) = E = i(\rho)/(A)\\\\i = (EA)/(\rho) = (.55 * \pi (0.00044^2))/((2.44 * 10^(-8)))} = \boxed{13.71 A}`

b)

We know that V = Ed (Electric field × distance), so:

`V = 0.55 * 6.3 = \boxed{3.465 V}`

c)

Calculate the resistance using the above equation.

`R = (\rho L)/(A)\\\\R = ((2.44* 10^(-8))(6.3))/(\pi(0.044^2)) = \boxed{2.527 * 10^(-5) \Omega}`