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(A) When a steady current $I$ flows through a junction between two different metals,charge accumulation occurs at the interface due to the difference

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(A) When a steady current $I$ flows through a junction between two different metals,charge accumulation occurs at the interface due to the difference in their conductivities.The current density $J$ is given by $J = \sigma E$,where $\sigma$ is the conductivity and $E$ is the electric field. Since the current $I$ and the cross-sectional area $A$ are the same for both rods,the current density $J = I/A$ is constant.At the junction,the continuity of current requires that the normal component of the current density be continuous. However,the electric field $E = J/\sigma$ must change because the conductivity $\sigma$ of copper is higher than that of iron $(\sigma_{Cu} > \sigma_{Fe})$.According to Gauss's law,the surface charge density $\rho_s$ at the interface is given by $\rho_s = \epsilon_0 (E_{Fe} - E_{Cu})$. Since $\sigma_{Cu} > \sigma_{Fe}$,it follows that $E_{Fe} > E_{Cu}$. Thus,$\rho_s$ is positive,meaning positive charge accumulates on the copper side and negative charge accumulates on the iron side of the junction to maintain the steady current flow.Therefore,the correct representation is shown in option $(A)$ and $(B)$ (as they are identical in the provided images). Based on the standard interpretation,the positive charge accumulates on the side of higher conductivity $(Cu)$ and negative charge on the side of lower conductivity $(Fe)$.

Two rods of copper $(Cu)$ and iron $(Fe)$ with the same cross-sectional area are joined at $S$ and a steady current $I$ flows through the rods as shown in the figure. Choose the most appropriate representation of charges accumulated near the junction $S$.

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