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9 April, 16:43

Consider a platinum wire (σ = 1.0 * 107 Ω-1·m-1) with a cross-sectional area of 1 mm2 (similar to your connecting wires) and carrying 0.3 amperes of current, which is about what you get in a circuit with a round bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.

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  1. 9 April, 17:00
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    Answer: 0.03 N/C

    Explanation:

    We use the current density formula to solve this question.

    I/A = σ * E

    Where,

    I = current flowing in the circuit = 0.3 A

    A = cross sectional area of the wire = 1 mm²

    σ = resistivity of the wire = 1*10^7 Ω^-1·m^-1

    E = strength of the electric field required

    I/A = σ * E

    E = I / (A * σ)

    First we convert area from mm to m, so that, 1*10^-3 mm = 1*10^-6 m

    E = 0.3 A / (1*10^-6 m * 1*10^7 Ω^-1·m^-1)

    E = 0.3 A / 10 Ω^-1

    E = 0.03 N/C
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