• # question_answer A silver wire has a resistance of $2.1\Omega$ at $27.5\,{}^{o}C$and a resistance of $2.7\Omega$ at $100\,{}^{o}C.$ Determine the temperature coefficient of resistivity of silver. Or One billion electrons pass from a point A towards a point B in ${{10}^{-4}}$ s.  What is the current in milliamperes? What is its direction?

Given, resistance of silver wire at $27.5{}^{o}C={{R}_{27.5}}=21\Omega$ Resistance of silver wire at $100{}^{o}C={{R}_{100}}=2.7\Omega$ Let the temperature coefficient of silver be $\alpha .$ $\therefore$      $\alpha ={{R}_{{{t}_{2}}}}-{{R}_{{{t}_{1}}}}/{{R}_{1}}({{t}_{2}}-{{t}_{1}})$ or $\alpha =\frac{{{R}_{100}}-{{R}_{27.5}}}{{{R}_{27.5}}(100-27.5)}=\frac{2.7-2.1}{2.1\times 72.5}\Rightarrow \alpha =0.0039/{}^{o}C$Thus, the temperature coefficient of resistivity of silver is $0.0039/{}^{o}C.$ Or Current associated with the bunch of electrons, $I=\frac{q}{t}=\frac{ne}{t}=\frac{{{10}^{9}}\times 1.6\times {{10}^{-19}}}{{{10}^{-4}}}$             $=1.6\times {{10}^{-6}}A=1.6\times {{10}^{-3}}mA$ The direction of current is from B to A.
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