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:

33.33

The correct answer is Option : 33.33

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

$E = J\rho$

$E_A = J_A \rho_A = 10^6 \times 1.5 \times 10^{-8} = 1.5 \times 10^{-2}$

$E_B = J_B \rho_B = 10^6 \times 5 \times 10^{-7} = 5 \times 10^{-1}$

$\frac{E_B}{E_A} = \frac{5 \times 10^{-1}}{1.5 \times 10^{-2}}$

$= \frac{5}{1.5} \times 10^{1}$

$= \frac{50}{1.5}$

$= \frac{100}{3}$

The value of $\frac{E_B}{E_A}$ is $\frac{100}{3}$.