the resistivity of gold is 2.44×10−8Ω⋅m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940

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the resistivity of gold is 2.44×10−8Ω⋅m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 mA. What is the electric field in the wire

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Adela 4 years 2021-08-29T13:53:41+00:00 1 Answers 204 views 0

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  1. Xavia
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    2021-08-29T13:55:00+00:00
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    Answer:

    0.03605 V/m is the electric field in the gold wire.

    Explanation:

    Resistivity of the gold = \rho = 2.44\times 10^{-8} \Omega.m

    Length of the gold wire = L = 14 cm = 0.14 m ( 1 cm = 0.01 m)

    Diameter of the wire = d = 0.9 mm

    Radius of the wire = r = 0.5 d = 0.5 × 0.9 mm = 0.45 mm = 0.45\times 0.001 m

    ( 1mm = 0.001 m)

    Area of the cross-section = A=\pi r^2=\pi r^(0.45\times 0.001 m)^2

    Resistance of the wire = R

    Current in the gold wire = 940 mA = 0.940 A ( 1 mA = 0.001 A)

    R=\rho\times \frac{L}{A}

    V(voltage)=I(current)\times R(Resistance) ( Ohm’s law)

    \frac{V}{I}=\rho\times \frac{L}{A}

    We know, Electric field is given by :

    E=\frac{dV}{dr}

    E=\frac{V}{L}

    E=\frac{V}{L}=\rho\times \frac{I}{A}

    E=2.44\times 10^{-8} \Omega.m\times \frac{0.940 A}{\pi r^(0.45\times 0.001 m)^2}=0.03605 V/m

    0.03605 V/m is the electric field in the gold wire.

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