A long solenoid with 30 turns/cm has a small loop of area \( 3\,\mathrm{cm^2} \) placed inside it normal to its axis. If the current carried by the solenoid changes steadily from 3.0 A to 6.0 A in 0.1 s, the induced emf in the loop is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 108\pi \times 10^{-7}\,\mathrm{V} \)
Inside a long solenoid the field is \( B = \mu_0 n I \). The changing current changes the field, so the flux through the loop changes, and Faraday's law gives the emf.
\( n = 30\,\mathrm{cm^{-1}} = 3000\,\mathrm{m^{-1}} \)
\( A = 3\,\mathrm{cm^2} = 3\times10^{-4}\,\mathrm{m^2} \)
\( \displaystyle \Delta B = \mu_0\,n\,\Delta I = 4\pi\times10^{-7} \times 3000 \times 3 = 3.6\pi\times10^{-3}\,\mathrm{T} \)
\( \displaystyle \Delta\Phi = \Delta B\,A = 3.6\pi\times10^{-3} \times 3\times10^{-4} = 1.08\pi\times10^{-6}\,\mathrm{Wb} \)
\( \displaystyle \varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{1.08\pi\times10^{-6}}{0.1} = 108\pi\times10^{-7}\,\mathrm{V} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
