A long solenoid with 15 turns per cm has a small loop of area \( \left(\displaystyle\frac{6}{\pi}\right) \) cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2 A to 4 A in 0.1 sec, the induced emf in the loop is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 7.2 μV
Inside a long solenoid the field is uniform, \( B = \mu_0 nI \), so the flux through the small loop is \( \phi = BA = \mu_0 nIA \). When the current changes, the flux changes and, by Faraday’s law, an emf is induced in the loop (this is mutual induction).
\( \displaystyle \varepsilon = \frac{d\phi}{dt} = \mu_0 nA\,\frac{dI}{dt} \)
Convert to SI units:
\( n = 15\,\mathrm{cm^{-1}} = 1500\,\mathrm{m^{-1}} \)
\( \displaystyle A = \frac{6}{\pi}\,\mathrm{cm^{2}} = \frac{6}{\pi}\times10^{-4}\,\mathrm{m^{2}} \)
\( \displaystyle \frac{dI}{dt} = \frac{4\,\mathrm{A} - 2\,\mathrm{A}}{0.1\,\mathrm{s}} = 20\,\mathrm{A\,s^{-1}} \)
\( \displaystyle \varepsilon = 4\pi\times10^{-7} \times 1500 \times \frac{6}{\pi}\times10^{-4} \times 20\,\mathrm{V} \)
\( \varepsilon = 4 \times 6 \times 1500 \times 20 \times 10^{-11}\,\mathrm{V} \)
\( \varepsilon = 7.2\times10^{-6}\,\mathrm{V} = 7.2\,\mu\mathrm{V} \)
The factor \( \pi \) cancels, which is why the area was given as \( 6/\pi \) cm2.
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 – …
