Question

Tarik winds a small paper tube uniformly with 163 turns of thin wire to form a solenoid. The tube's diameter is 6.13 mm and its length is 2.49 cm . What is the inductance, in microhenrys, of Tarik's solenoid?

Answer 1

Answer:

The  inductance is $L = 40\mu H$

Explanation:

From the question we are told that

    The number of turns is  $N = 163 \ turns$

    The  diameter is  $D = 6.13 \ mm = 6.13 *10^{-3} \ m$

    The  length is  $l = 2.49 \ cm = 0.0249 \ m$

     

The radius is evaluated as $r = \frac{d}{2}$

substituting values

        $r = \frac{6.13 *10^{-3}}{2}$

       $r = 3.065 *10^{-3} \ m$

The  inductance of the Tarik's solenoid is mathematically represented as

            $L = \frac{\mu_o * N^2 * A }{l }$

Here $\mu_o$ is the permeability of free space with value  

        $\mu_o = 4\pi *10^{-7} \ N/A^2$

A is the area which is mathematically evaluated as

         $A = \pi r^2$

substituting values

       $A = 3.142 * [ 3.065*10^{-3}]^2$

       $A = 2.952*10^{-5} \ m^2$

substituting values into formula for L  

      $L = \frac{ 4\pi *10^{-7} * [163]^2 * 2.952*10^{-5} }{0.0249 }$

     $L = 40\mu H$