Question

A current is 1.5 ampere in a 800 turns coil produce 1.5 x 18–⁵ Weber magnetised flux in each turn. Find out the self induction of coil

Answer 1

Gives

Current=1.5A

Turn=500

Magnetic flux= $1.5\times10^{-5}Wb$

$\phi=nLi$

$L=\frac{\phi}{ni}$

Put value

$$L=\frac{1.5\times10^{-5}}{500\times1.5}=0.02\times10^{-6} H$$

$L=0.02\mu H$