In wire A of electrical resistivity $\rho_A = 1.5\times 10^{-8} \Omega -m$ , the current density is $ J_A = 10^6 A/m^2 $ in wire B of electrical resistivity $\rho_B = 5\times 10^{-7} \Omega -m$ , the current density $ J_B = 10^6 A/m^2 $ .If the electric field in wire B is E_B and that in wire A is E_A, then $\frac{E_B}{E_A}$ is:

100

$ J = \frac{E}{\rho}$

$ E = \rho J$

$ \frac{E_B}{E_A} = \frac{\rho_B}{\rho_A} \times {J_B}{J_A} = \frac{50}{1.5}\times 3 =100$