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:
$R = \frac{\rho L}{A}$

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.

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

b)

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

$V = 0.55 \times 6.3 = \boxed{3.465 V}$

c)

Calculate the resistance using the above equation.

$R = \frac{\rho L}{A}\\\\R = \frac{(2.44\times 10^{-8})(6.3)}{\pi(0.044^2)} = \boxed{2.527 \times 10^{-5} \Omega}$