Respuesta :
Answer:
0.03605 V/m is the electric field in the gold wire.
Explanation:
Resistivity of the gold = [tex]\rho = 2.44\times 10^{-8} \Omega.m[/tex]
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 = [tex]0.45\times 0.001 m[/tex]
( 1mm = 0.001 m)
Area of the cross-section = [tex]A=\pi r^2=\pi r^(0.45\times 0.001 m)^2[/tex]
Resistance of the wire = R
Current in the gold wire = 940 mA = 0.940 A ( 1 mA = 0.001 A)
[tex]R=\rho\times \frac{L}{A}[/tex]
[tex]V(voltage)=I(current)\times R(Resistance)[/tex] ( Ohm's law)
[tex]\frac{V}{I}=\rho\times \frac{L}{A}[/tex]
We know, Electric field is given by :
[tex]E=\frac{dV}{dr}[/tex]
[tex]E=\frac{V}{L}[/tex]
[tex]E=\frac{V}{L}=\rho\times \frac{I}{A}[/tex]
[tex]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[/tex]
0.03605 V/m is the electric field in the gold wire.
