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(A) The current $I$ flowing through the wire is constant at all cross-sections.Current density is defined as $J = \frac{I}{A}$,where $A$ is the cross-

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$j_A > j_B; E_A > E_B$

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(A) The current $I$ flowing through the wire is constant at all cross-sections.Current density is defined as $J = \frac{I}{A}$,where $A$ is the cross-sectional area.Since $I$ is constant,$J \propto \frac{1}{A}$.From the figure,the area at $A$ is smaller than the area at $B$ $(A_A  j_B$.According to Ohm's law in microscopic form,$\vec{J} = \sigma \vec{E}$,where $\sigma$ is the conductivity of the material.Since the material is homogeneous and isotropic,$\sigma$ is constant throughout the wire.Thus,$E = \frac{J}{\sigma}$,which implies $E \propto J$.Since $j_A > j_B$,it follows that $E_A > E_B$.

An electric current passes through a wire of non-uniform cross-section,made of a homogeneous and isotropic material. If $j_A$ and $j_B$ are the current densities and $E_A$ and $E_B$ are the electric field intensities at $A$ and $B$ respectively,then $-$

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