Question

The self inductance relates the magnetic flux linkage to the current through the coil. Calculate the self inductance L in units of μH. The coil is d = 5 cm long and has a cross-sectional area of A = 3 cm2 and consists of N = 130 turns.

Answer 1

Self inductance,
`L=127\ \mu H`

Step-by-step explanation:

It is given that,

Length of the coil, l = 5 cm = 0.05 m

Area of cross section of the coil,
`A=3\ cm^2=0.0003\ m^2`

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 :

`L=(\mu_oN^2A)/(l)`

`L=(4\pi * 10^(-7)* (130)^2* 0.0003)/(0.05)`

L = 0.000127 Henry

or

`L=127\ \mu H`

So, the self inductance of the coil is 127 microhenry. Hence, this is the required solution.