An electrical conductor designed to carry large currents has a circular cross section 2.50 mm in diameter and is 14.0 m long. The resistance between its ends is $\mathbf{0}\mathbf{.}\mathbf{104}\mathbf{}\mathbf{\Omega }$. (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is 1.28 V/m, what is the total current? (c) If the material has $\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}$free electrons per cubic meter, find the average drift speed under the conditions of part (b).

According to Ohm’s law the current flowing through conductor is directly proportional to the voltage across the two points.

$V=IR$

Where, $I$is current in ampere localid="1655724078434" $\left(A\right)$, localid="1655724083468" $R$is resistance in ohms localid="1655724088023" $\left(\mathrm{\Omega }\right)$and localid="1655724093311" $V$is the potential difference volt localid="1655724097962" $\left(V\right)$.

Also, localid="1655724101438" $V=EL$

Where, localid="1655724105667" $E$is the electric field across conductor and localid="1655724111023" $L$is length of the conductor.

The ratio of V to I for a particular conductor is called its resistance R:

localid="1655724116244" $R=\frac{V}{I}or\frac{\rho L}{A}$

Where, localid="1655724026382" $\rho$is resistivity localid="1655724030493" $\mathrm{\Omega }·\mathrm{m},L$is length in m and localid="1655724036681" $A$is area in localid="1655724042282" ${\mathrm{m}}^2$.

If the area of cross-section islocalid="1655724005712" $\pi r^2$, concentration islocalid="1655724011647" $n$and localid="1655724017124" $I$is current then drift velocitylocalid="1655724021912" $V_d$is given by equation:

localid="1655723999554" $v_d=\frac{l}{\mathrm{mn}\left|q\right|r^2}$

Given that,

$\begin{array}{l}R=0104\mathrm{\Omega }\\ d=250\mathrm{mm}\\ L=14\mathrm{m}\end{array}$

The resistivity of the material is:

$$\begin{gathered} \rho =\frac{RA}{L} \\ =\frac{R\left(\pi r^2\right)}{L} \\ =\frac{\left(0.104\right)\left\{\pi {\left({\displaystyle \frac{2.5}{2}}\times {\displaystyle \frac{1}{1000}}\right)}^2\right\}}{14.0} \\ =3.65\times {10}^{-8}\mathrm{\Omega }·\mathrm{m} \end{gathered}$$

Hence, the resistivity of the material is $3.65\times {10}^{-8}\mathrm{\Omega }·\mathrm{m}$.

Given that,

$E=1.28\mathrm{V}/\mathrm{m}$

The current through conductor is

$$\begin{gathered} I=\frac{V}{R} \\ =\frac{EL}{R} \\ =\frac{\left(1.25\right)\left(14\right)}{0.104} \\ =172.0\mathrm{A} \end{gathered}$$

Hence, if the electric-field magnitude in the conductor is 1.28 V/m , then the total current is 172.0 A .

The charge of electron $q=1.6\times {10}^{-19}C$.

The average drift speed is

role="math" localid="1655723824487" $$\begin{gathered} V_d=\frac{1}{\pi n\left|q\right|r^2} \\ =\frac{172.0}{\mathrm{\pi }\left(8.5\times {10}^{28}\right)\left(1.6\times {10}^{-19}\right){\left(0.00125\right)}^2} \\ =2.58\times {10}^{-3}\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, if the material has $8.5\times {10}^{28}$free electrons per cubic meters, the average drift speed is$2.58\times {10}^{-3}\mathrm{m}/\mathrm{s}$