What is the self - inductance of an air solenoid 50cm long and 2cm radius if it has 500 turns?
Solution:
The self-inductance is given as,
L = (µo) N2 A / l
l = 50 cm = 0.5 m
r = 2cm = 0.02 m
A = pi * r^2 = pi * (0.02)^2
N = 500
Hence, L = (4 pi * 10^-7) * (500)^2 * (pi * (0.02)^2) / (0.5)
= 7.896 * 10^-4 Henry