Question

An eighteen gauge copper wire has a nominal diameter of 1.02mm. This wire carries a constant current of 1.67A to a 200w lamp. The density of free electrons is 8.5 x 1028 electrons per cubic metre. Find the magnitude of:
i. The current density ii. The drift velocity

Answer 1

Answer:

The current density is  $J = 2.04 * 10^{6} A /m^2$

The drift velocity is  $v_d = 1.5 * 10^{-4} m/s$

Explanation:

From the question we are told that

  The nominal diameter of the wire is $d = 1.02 mm= \frac{1.02}{1000} = 0.00102 \ m$

   The current carried by the wire is $I = 1.67 A$

    The power rating of the lamp is $P = 200 W$

    The density of electron is $n = 8.5 * 10^{28} \ e/m^3$

   

The current density is mathematically represented as

       $J = \frac{I}{A}$

Where A is the area which is mathematically evaluated as

          $A = \pi \frac{d^2}{4}$

Substituting values

         $A = 3.142 * \frac{(1.02 * 10^{-3})^2 }{4}$

       $A = 8.0*10^{-4}m^2$

So

         $J = \frac{1.67}{8.0*10^{-4}}$

       $J = 2.04 * 10^{6} A /m^2$

The drift velocity is mathematically represented as

       $v_d = \frac{J}{ne}$

Where e is the charge on one electron which has a value  $e = 1.602 *10^{-19} C$

So

         $v_d =\frac{2.04 * 10^6 }{8.5 *10^{28} * 1.6 * 10^{-19}}$

        $v_d = 1.5 * 10^{-4} m/s$