The electric field required is 0.0604 N/C
It is given that the diameter of the wire is d = 0.326cm.
So radius r = [tex]0.163\times10^{-2}m[/tex].
Therefore the cross-sectional area:
[tex]A=\pi r^2=\pi(0.163\times10^{-2})^2=8.35\times10^{-6}m^2[/tex]
Now the current in the wire is [tex]I=30A[/tex]
So the electric field E needed is:
[tex]E=\rho J[/tex]
here [tex]\rho=1.68\times10^{-8}\Omega m[/tex] is the resistivity of copper.
and J is the current density which is given by:
[tex]J=\frac{I}{A}[/tex]
thus:
[tex]E=1.68\times10^{-8}\times\frac{30}{8.35\times10^{-6}} N/C\\\\E=0.0604N/C[/tex]
Learn more about current density:
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