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

Answer:

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

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 :

$L=\dfrac{\mu_oN^2A}{l}$

$L=\dfrac{4\pi \times 10^{-7}\times (130)^2\times 0.0003}{0.05}$

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.