Current of 5A passes through a copper conductor (resistivity = $1.7 \times {10^{ - 8}}\Omega m$) of radius of cross-section 5mm. Find the mobility of the charges if their drift velocity is $1.1 \times {10^{ - 3}}\dfrac{m}{s}$.
A) $1.3\dfrac{{{m^2}}}{{Vs}}$.
B) $1.5\dfrac{{{m^2}}}{{Vs}}$.
C) $1.8\dfrac{{{m^2}}}{{Vs}}$.
D) $1.0\dfrac{{{m^2}}}{{Vs}}$.

Hint: Mobility of charges is defined as the ability of the charge to move freely .The drift velocity is the average velocity of the charged particle due to the influence of the electric field. The unit of the mobility of particles is $\dfrac{{{m^2}}}{{Vs}}$.

Formula used:
The formula of the mobility of particles is given by $\mu = \dfrac{{{{\text{V}}_d}}}{{\text{E}}}$, where $\mu $ is the mobility ${{\text{V}}_d}$ is the drift velocity and ${\text{E}}$ is the electric field. Also the formula of the electric field is ${\text{E}} = \rho J$ where $\rho $ is the resistivity and $J$ is current density.

Complete step by step answer:
As it given that the resistivity is $\rho = 1.7 \times {10^{ - 8}}\Omega m$ and the drift velocity is ${{\text{V}}_d} = 1.1 \times {10^{ - 3}}\dfrac{m}{s}$
Also the electric field is given by ${\text{E}} = \rho J$,
$ \Rightarrow {\text{E}} = \left( {1.7 \times {{10}^{ - 8}}} \right) \times \left( {\dfrac{5}{{\pi \times {{\left( {0.005} \right)}^2}}}} \right)$
$ \Rightarrow {\text{E}} = \dfrac{{5\left( {1.7 \times {{10}^{ - 8}}} \right)}}{{\pi. \left( {25 \times {{10}^{ - 6}}} \right)}}$
$ \Rightarrow E = \dfrac{{8.5 \times {{10}^{ - 2}}}}{{\pi \left( {25} \right)}}$
$ \Rightarrow E = \dfrac{{8.5 \times {{10}^{ - 2}}}}{{78. 54}}$
$ \Rightarrow E = 0.00108$.
The mobility is given by $\mu = \dfrac{{{{\text{V}}_d}}}{{\text{E}}}$.
$ \Rightarrow \mu = \dfrac{{1.1 \times {{10}^{ - 3}}}}{{0.00108}}$
$ \Rightarrow \mu = 1.0185\dfrac{{{m^2}V}}{s}$
The mobility of charge particles is given by $\mu = 1.0185\dfrac{{{m^2}V}}{s}$.

Therefore, the correct option for this problem is option D.

Note:
The students should remember the formula of the mobility of charges and also the formula of the electric field in different physical quantities. The electric density is the ratio of the current to the cross section of the wire.