Chapter 30 - Current and Resistance - Exercises and Problems - Page 889: 58

\begin{array}{|c|c|c|c|} \hline & 1-{\rm Top}& 2-{\rm Middle} & 3-{\rm Bottom} \\ \hline I& 10\;{\rm A} & 10\;{\rm A} & 10\;{\rm A} \\ \hline J & 3.18\times 10^6 \;{\rm A/m^2} & 1.27\times 10^7 \;{\rm A/m^2} & 3.18\times 10^6 \;{\rm A/m^2} \\ \hline E & 0.091 \;{\rm V/m}& 0.363 \;{\rm V/m}&0.091 \;{\rm V/m}\\ \hline v_{\rm d}& 0.331 \;{\rm mm/s}& 1.323\;{\rm mm/s}&0.331\;{\rm mm/s}\\ \hline i & 6.24\times 10^{19}\;{\rm s^{-1}}& 6.24\times 10^{19}\;{\rm s^{-1}}&6.24\times 10^{19}\;{\rm s^{-1}} \\ \hline \end{array}

We named the three parts from 1 to 3, where 1 for the top part, 2 for the middle part, and 3 for the bottom part. $$\color{blue}{\bf [a]}$$ The current is the same in both wires according to the current conservation principle. Hence, $$I_1=I_2=I_3=I=\color{red}{\bf 10}\;\rm A $$ $$\color{blue}{\bf [b]}$$ However, the current density is not the same. We can see that $J_1=J_3$ since they have the same diameter. $$J_1=J_3=\dfrac{I_1}{A_1}=\dfrac{10}{\pi r_1^2}=\dfrac{10}{\pi (1\times 10^{-3})^2}=\color{red}{\bf 3.18\times 10^6}\;\rm A/m^2$$ and hence, $$J_2=\dfrac{I_2}{A_2}=\dfrac{10}{\pi r_1^2}=\dfrac{10}{\pi (0.5\times 10^{-3})^2}=\color{red}{\bf 1.27\times 10^7}\;\rm A/m^2$$ $$\color{blue}{\bf [c]}$$ We know that the current density is given by $$J=\dfrac{I }{A }=\sigma E$$ So, $$E=\dfrac{J}{ \sigma}$$ Hence, $$E_1=E_3=\dfrac{J_1}{\sigma_{\rm aluminum}}=\dfrac{3.18\times 10^6}{3.5\times 10^7}=\color{red}{\bf 0.091}\;\rm V/m$$ $$E_2 =\dfrac{J_2}{\sigma_{\rm aluminum}}=\dfrac{1.27\times 10^7}{3.5\times 10^7}=\color{red}{\bf 0.363}\;\rm V/m$$ $$\color{blue}{\bf [d]}$$ We know that the drift speed is given by $$v_{\rm d}=\dfrac{J}{n_ee}$$ Hence, $$v_{\rm 1d}=v_{\rm 3d}=\dfrac{J_1}{n_ee}=\dfrac{(3.18\times 10^6)}{(6\times 10^{28})(1.6\times 10^{-19})}=\color{red}{\bf 0.331}\;\rm mm/s$$ $$v_{\rm 2d} =\dfrac{J_2}{n_ee}=\dfrac{(1.27\times 10^7)}{(6\times 10^{28})(1.6\times 10^{-19})}=\color{red}{\bf 1.323}\;\rm mm/s$$ $$\color{blue}{\bf [e]}$$ We know that the electron current is given by $$i=n_eAv_{\rm d}$$ So, $$i_1=i_3=n_eA_1v_{\rm 1d}=\pi r_1^2 n_e v_{\rm 1d}$$ $$i_1=i_3=\pi (1\times 10^{-3})^2( (6\times 10^{28})(0.331\times 10^{-3})$$ $$i_1=i_3=\color{red}{\bf 6.24\times 10^{19}}\;\rm s^{-1}$$ $$i_2=n_eA_2v_{\rm 2d}=\pi r_2^2 n_e v_{\rm 2d}$$ $$i_2=\pi (0.5\times 10^{-3})^2( (6\times 10^{28})( 1.323\times 10^{-3})$$ $$i_2=\color{red}{\bf 6.24\times 10^{19}}\;\rm s^{-1}$$ And as the author told us, here is the table of the results. \begin{array}{|c|c|c|c|} \hline & 1-{\rm Top}& 2-{\rm Middle} & 3-{\rm Bottom} \\ \hline I& 10\;{\rm A} & 10\;{\rm A} & 10\;{\rm A} \\ \hline J & 3.18\times 10^6 \;{\rm A/m^2} & 1.27\times 10^7 \;{\rm A/m^2} & 3.18\times 10^6 \;{\rm A/m^2} \\ \hline E & 0.091 \;{\rm V/m}& 0.363 \;{\rm V/m}&0.091 \;{\rm V/m}\\ \hline v_{\rm d}& 0.331 \;{\rm mm/s}& 1.323\;{\rm mm/s}&0.331\;{\rm mm/s}\\ \hline i & 6.24\times 10^{19}\;{\rm s^{-1}}& 6.24\times 10^{19}\;{\rm s^{-1}}&6.24\times 10^{19}\;{\rm s^{-1}} \\ \hline \end{array}