Question

To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 900 loops of wire wound on a rod 6 cm long with radius 1 cm?

Answer 1

Answer:

The  self-inductance is  $L = 0.0053 \ H$

Explanation:

From the question we are told that  

      The number of loops is  $N = 900$

      The  length of the rod is  $l =6 \ cm = 0.06 \ m$

      The radius of the rod is  $r = 1 \ cm = 0.01 \ m$

The  self-inductance for the solenoid is mathematically represented as

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

Now the cross-sectional of the solenoid is mathematically evaluated as

        $A = \pi r^2$

substituting values  

         $A =3.142 * 0.01 ^2$

        $A = 3.142 *10^{-4} \ m^2$

and  $\mu_o$ is the permeability of free space with a value  $\mu_o = 4\pi * 10^{-7} N/A^2$

    substituting values into above equation

          $L = \frac{ 4\pi * 10^{-7} ^2* 3.142*10^{-4} * 900^2 }{0.06}$

          $L = 0.0053 \ H$