3 years ago

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Physics

2 answers:

Allushta [10]3 years ago

5 0

Yes, jsfkgskjdfhkdsjfhksjdhfjkshkjdsjk.

lesya692 [45]3 years ago

4 0

Answer:

Explanation:

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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 cur

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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)

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( 1mm = 0.001 m)

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

$V(voltage)=I(current)\times R(Resistance)$ ( 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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