Explanation:
Consider the inductor of inductance \(L\)

The current flowing through the inductor is \(i\).
Now, we can write,
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\phi = Li\)
where, \(\phi \) is magnetic flux linked with
the inductor.
\( \Rightarrow \,\,\,\,\,\left| {\frac{{d\phi }}{{dt}}} \right| = L\frac{{di}}{{dt}}\)
Given, \(L = 40\,mH\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dt = \) net time difference
\({\mkern 1mu} \,\,\,\,\,\,\,\, = {t_2} - {t_1} = 4\;ms = \Delta t{\mkern 1mu} \)
\({\mkern 1mu} di\,\,\, = {i_2} - {i_1}\)
\(\,\,\,\,\,\,\, = 11 - 1 = 10\;A = \Delta i\)
So, \({\mkern 1mu} {\text{|emf|}} = \left| {\frac{{d\phi }}{{dt}}} \right| = \frac{{\Delta \phi }}{{\Delta t}}{\mkern 1mu} {\mkern 1mu} \)
\(\,\,\,\,{\mkern 1mu} = L\frac{{\Delta i}}{{\Delta t}} = (40mH)\frac{{(10)A}}{{4\;ms}}{\mkern 1mu} \)
or induced emf \({\mkern 1mu} \,\,\,\,\, = 10 \times 10 = 100\;V\)