91Ó°ÊÓ

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" $\left(\mathrm{Î\copyright }\right)$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" $R=\frac{V}{I}or\frac{\mathrm{ÒÏ}L}{A}$

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

If the area of cross-section islocalid="1655724005712" $\mathrm{Ï€}{r}^2$, 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,

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

The resistivity of the material is:

$$\begin{gathered} \mathrm{ÒÏ}=\frac{RA}{L} \\ =\frac{R\left(\mathrm{Ï€}{r}^2\right)}{L} \\ =\frac{\left(0.104\right)\left\{\mathrm{Ï€}{\left({\displaystyle \frac{2.5}{2}}×{\displaystyle \frac{1}{1000}}\right)}^2\right\}}{14.0} \\ =3.65×{10}^{-8}\mathrm{Î\copyright }·\mathrm{m} \end{gathered}$$

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

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 $q=1.6×{10}^{-19}C$.

The average drift speed is

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

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