Answer:
Self inductance, [tex]L=127\ \mu H[/tex]
Explanation:
It is given that,
Length of the coil, l = 5 cm = 0.05 m
Area of cross section of the coil, [tex]A=3\ cm^2=0.0003\ m^2[/tex]
Number of turns in the coil, N = 130
The self inductance relates the magnetic flux linkage to the current through the coil and it is given by :
[tex]L=\dfrac{\mu_oN^2A}{l}[/tex]
[tex]L=\dfrac{4\pi \times 10^{-7}\times (130)^2\times 0.0003}{0.05}[/tex]
L = 0.000127 Henry
or
[tex]L=127\ \mu H[/tex]
So, the self inductance of the coil is 127 microhenry. Hence, this is the required solution.
