The inductance of a solenoid is given by:
[tex]L= \frac{\mu_0 N^2 A}{l} [/tex]
where
[tex]\mu_0[/tex] is the vacuum permeability
N is the number of turns of the solenoid
A is the area of one loop of the solenoid
l is its length
For the solenoid in our problem, N=500, and the radius of one loop is r=0.025 m, so the area of one loop is
[tex]A=\pi r^2 = \pi (0.025 m)^2 = 1.96 \cdot 10^{-3} m^2[/tex]
and its length is l=0.20 m
So the inductance of the solenoid is
[tex]L= \frac{(4 \pi \cdot 10^{-7} NA^{-2})(500)^2(1.96 \cdot 10^{-3} m^2)}{ 0.20 m}=3.08 \cdot 10^{-3} H[/tex]
