The self-inductance of an inductance coil having 100 turns is 20 mH.…

The correct answer is Option (1) → \( 8 \times 10^{-5}\,\mathrm{Wb} \)

The self-inductance is the ratio of the total flux linked with the coil to the current in it, so the total flux (flux linkage) is

    \( \displaystyle N\phi = L\,I \)

    \( L = 20\,\mathrm{mH} = 20\times10^{-3}\,\mathrm{H} \)
    \( I = 4\,\mathrm{mA} = 4\times10^{-3}\,\mathrm{A} \)

    \( \displaystyle N\phi = 20\times10^{-3} \times 4\times10^{-3} = 8\times10^{-5}\,\mathrm{Wb} \)

The number of turns is not needed for the total flux; it would be needed only for the flux through one turn, \( \phi = 8\times10^{-5}/100 = 8\times10^{-7}\,\mathrm{Wb} \).

Note: NTA dropped this question and gave full marks to every candidate, because none of the printed options was \( 8\times10^{-5}\,\mathrm{Wb} \): they were \( 8\times10^{-3} \), \( 4\times10^{-3} \), \( 8\times10^{-4} \) and \( 4\times10^{-4}\,\mathrm{Wb} \). Option (1), printed as \( 8\times10^{-3}\,\mathrm{Wb} \), has been changed to \( 8\times10^{-5}\,\mathrm{Wb} \) so that the question has one correct answer, for the benefit of students.