In wire A of electrical resistivity \( \rho_A=1.5\times10^{-8}\,\Omega\,\mathrm{m} \), the current density is \( J_A=10^{6}\,\mathrm{A/m^{2}} \). In wire B of electrical resistivity \( \rho_B=5\times10^{-7}\,\Omega\,\mathrm{m} \), the current density is \( J_B=10^{6}\,\mathrm{A/m^{2}} \). If the electric field in wire B is \( E_B \) and that in wire A is \( E_A \), then \( E_B/E_A \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 33.33
The current density and the electric field in a conductor are related by \( j=\sigma E \). Since the resistivity is \( \rho=1/\sigma \), this gives \( E=\rho j \). With the same current density in both wires, the field is proportional to the resistivity.
\( \begin{aligned} \frac{E_B}{E_A} &= \frac{\rho_B\,J_B}{\rho_A\,J_A} \\ &= \frac{5\times10^{-7}\,\Omega\,\mathrm{m}\times10^{6}\,\mathrm{A/m^{2}}}{1.5\times10^{-8}\,\Omega\,\mathrm{m}\times10^{6}\,\mathrm{A/m^{2}}} \\ &= \frac{50}{1.5} \\ &\approx 33.33 \end{aligned} \)