. 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

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?

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Thái Dương 3 years 2021-08-26T08:49:27+00:00 1 Answers 5 views 0

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  1. Sigrid
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    2021-08-26T08:51:02+00:00
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    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

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