Answer:
Electric field, E = 0.064 V/m
Explanation:
It is given that,
Resistivity of silver wire, [tex]\rho=1.59\times 10^{-8}\ \Omega-m[/tex]
Current density of the wire, [tex]J=4\ A/mm^2=4\times 10^6\ A/m^2[/tex]
We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :
[tex]E=J\times \rho[/tex]
[tex]E=4\times 10^6\times 1.59\times 10^{-8}[/tex]
E = 0.0636 V/m
or
E = 0.064 V/m
So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.
