. A current-carrying gold wire has diameter 0.87 mm. The electric field in the wire is 0.54 V>m. What are (a) the current carried by the wire; (b) the potential difference between two points in the wire 7.0 m apart; (c) the resistance of a 7.0-m length of this wire?
Question
Answer:
a) I = 13.156 A
b) V = 3.78 V
c) R = 0.2873 Ω
Explanation:
The diameter(d) of the wire is 0.87 mm = 0.87 × 10⁻³ m.
The electric field(E) = 0.54 V/m
The cross sectional area(A) = πd²/4 = π × (0.87 × 10⁻³)² /4 = 5.945 × 10⁺⁷ m²
Gold resistivity(ρ) = 2.44 × 10⁻⁸
(a) the current carried by the wire
E = Jρ
Where J is the current density = I/A and I is the current
E = Iρ/A
I = EA/ρ
Substituting values:
I = 0.54 × 5.945 × 10⁺⁷/ 2.44 × 10⁻⁸ = 13.156 A
(b) the potential difference between two points in the wire 7.0 m apart
The potential difference(V) = EL
Where L is the length = 7 m
V = EL
Substituting values:
V = 0.54 × 7 = 3.78 V
(c) the resistance of a 7.0-m length of this wire
The resistance(R) = ρL/A
R = ρL/A
Substituting values:
R = 2.44 × 10⁻⁸ × 7 / 5.945 × 10⁺⁷ = 0.2873 Ω