Question

the magnetic flux linked with a coil changes by 2×10^-2 wb when the current changes by 0.01A. the self inductance of the coil is..............

Answer 1

It is given that,

Magnetic flux linked with the coil, \(\phi=2\times10^{-2}\,Wb\)

Change in current, \(\Delta I=0.01\,A\) 

The relation between the magnetic flux and the self inductance is given by :

\(\phi=L\times\Delta I\)

\(L=\frac{\phi}{\Delta I}\)

\(L=\frac{2\times10^{-2}}{0.01}\)

L = 2 Henry

So, the self inductance of the coil is 2 Henry.