Explanation:
Given that,
Diameter of the gold wire, d = 0.84 mm
Radius, r = 0.42 mm
Electric field in the wire, E = 0.49 V/m
(a) Electric current density is given by :
[tex]J=\dfrac{I}{A}[/tex]
And electric field is :
[tex]E=\rho J[/tex]
[tex]\rho[/tex] is resistivity of Gold wire
So,
[tex]E=\dfrac{\rhi I}{A}\\\\I=\dfrac{EA}{\rho}\\\\I=\dfrac{0.49\times \pi (0.42\times 10^{-3})^2}{2.44\times 10^{-8}}\\\\I=11.12\ A[/tex]
(b) The potential difference between two points in the wire is given by :
[tex]V=E\times l\\\\V=0.49 \times 6.4\\\\V=3.136\ V[/tex]
(c) Resistance of a wire is given by :
[tex]R=\rho \dfrac{l}{A}\\\\R=2.44\times 10^{-8}\times \dfrac{6.4}{\pi (0.42\times 10^{-3})^2}\\\\R=0.281\ \Omega[/tex]
Hence, this is the required solution.
