Question

A solenoid 30 cm long is wound with 250 turns of wire. The cross-sectional area of the coil is 3.5 cm^2. What is the self-inductance of the solenoid?

Answer 1

Answer:

The value is  $L = 9.164 *10^{-5} \ H$

Explanation:

From the question we are told that

    The length of the solenoid is $l = 30 \ cm = 0.3 \ m$

     The number of turns is  $N = 250 \ turns$

    The cross-sectional  area is  $A = 3.5 \ cm^2 = 3.5 *10^{-4} \ m^2$

Generally the self  inductance of the solenoid is  

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

Here  $\mu_o$ is the permeability of free space with value  $\mu_o = 4\pi *10^{-7} \ T \cdot m/A$

So

       $L = \frac{ 4 \pi*0^{-7} * 250^2 * 3.5 *10^{-4} }{ 0.3 }$

=>     $L = 9.164 *10^{-5} \ H$