Question

. 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?

Answer 1

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