Question

A 170-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,010 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable

Answer 1

Answer:

The number of years is $t_y = 22.8 \ years$

Explanation:

From the question we are told that

    The length of the transmission line is  $L = 170 \ km = 170000 \ m$

    The diameter of the transmission line is  $d = 2.0 \ cm = 0.02 \ m$

     The current which the transmission line carry is  $I = 1,010 \ A$

      The charge density of the transmission line is  $j = 8.50 *10^{28 } \ electron/m^3$

Now  the cross-sectional  area of the transmission line is mathematically represented  as

           $A = \pi r^2$

Here r is the radius which is mathematically evaluated as

         $r = \frac{d}{2}$

substituting values

         $r = \frac{0.02}{2}$

         $r = 0.01 \ m$

Hence

       $A = 3.142 * (0.01)^2$

=>      $A = 3.142 *10^{-4} \ m^2$

Now the drift velocity of electron is mathematically evaluated as

        $v = \frac{I}{j* e * A }$

Where e is the charge on one electron and the values is  $e = 1.60 *10^{-19} \ C$

       So

              $v = \frac{ 1010}{8.50 *10^{28}* (1.60 *10^{-19}) * 3.142*10^{-4} }$

              $v = 2.363 *10^{-4} \ m/s$

Now the time taken is  mathematically evaluated as

                $t = \frac{L}{v }$

substituting values

                $t = \frac{170000}{2.363 *10^{-4} }$

                $t = 7.194*10^{8}\ s$

Converting to years

       $t_y = \frac{t}{365\ days * 24 \ hours * 3600\ seconds}$

substituting values

        $t_y =\frac{7.194 *10^{8}}{365 *24 * 3600}$

        $t_y = 22.8 \ years$