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(B) The mobility $\mu$ is defined as the ratio of drift velocity $V_d$ to the electric field $E$: $\mu = \frac{V_d}{E}$.From Ohm's law in vector form,

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(B) The mobility $\mu$ is defined as the ratio of drift velocity $V_d$ to the electric field $E$: $\mu = \frac{V_d}{E}$.From Ohm's law in vector form,$E = ho J$,where $ho$ is resistivity and $J$ is current density.Current density $J = \frac{I}{A} = \frac{I}{\pi r^2}$.Given: $I = 5\, A$,$ho = 1.7 	imes 10^{-8}\, \Omega \, m$,$r = 5\, mm = 5 	imes 10^{-3}\, m$,and $V_d = 1.1 	imes 10^{-3}\, m/s$.First,calculate the electric field $E$:$E = ho 	imes \frac{I}{\pi r^2} = 1.7 	imes 10^{-8} 	imes \frac{5}{\pi 	imes (5 	imes 10^{-3})^2} = 1.7 	imes 10^{-8} 	imes \frac{5}{\pi 	imes 25 	imes 10^{-6}} = \frac{1.7 	imes 10^{-8} 	imes 5}{78.54 	imes 10^{-6}} \approx 1.08 	imes 10^{-3}\, V/m$.Now,calculate mobility $\mu$:$\mu = \frac{1.1 	imes 10^{-3}}{1.08 	imes 10^{-3}} \approx 1.01\, m^2/Vs$.Rounding to the nearest given option,the correct value is $1.0\, m^2/Vs$.

$A$ current of $5\, A$ passes through a copper conductor (resistivity $= 1.7 \times 10^{-8}\, \Omega \, m$) of radius of cross-section $5\, mm$. Find the mobility of the charges if their drift velocity is $1.1 \times 10^{-3}\, m/s$.

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